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Is there a way to make the values in the list as columns in pandas? I am not looking to rename the columns, but add the list values as columns [duplicate]

I want to change the column labels of a Pandas DataFrame from
['$a', '$b', '$c', '$d', '$e']
to
['a', 'b', 'c', 'd', 'e']

Rename Specific Columns
Use the df.rename() function and refer the columns to be renamed. Not all the columns have to be renamed:
df = df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'})
# Or rename the existing DataFrame (rather than creating a copy)
df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'}, inplace=True)
Minimal Code Example
df = pd.DataFrame('x', index=range(3), columns=list('abcde'))
df
a b c d e
0 x x x x x
1 x x x x x
2 x x x x x
The following methods all work and produce the same output:
df2 = df.rename({'a': 'X', 'b': 'Y'}, axis=1) # new method
df2 = df.rename({'a': 'X', 'b': 'Y'}, axis='columns')
df2 = df.rename(columns={'a': 'X', 'b': 'Y'}) # old method
df2
X Y c d e
0 x x x x x
1 x x x x x
2 x x x x x
Remember to assign the result back, as the modification is not-inplace. Alternatively, specify inplace=True:
df.rename({'a': 'X', 'b': 'Y'}, axis=1, inplace=True)
df
X Y c d e
0 x x x x x
1 x x x x x
2 x x x x x
From v0.25, you can also specify errors='raise' to raise errors if an invalid column-to-rename is specified. See v0.25 rename() docs.
Reassign Column Headers
Use df.set_axis() with axis=1 and inplace=False (to return a copy).
df2 = df.set_axis(['V', 'W', 'X', 'Y', 'Z'], axis=1, inplace=False)
df2
V W X Y Z
0 x x x x x
1 x x x x x
2 x x x x x
This returns a copy, but you can modify the DataFrame in-place by setting inplace=True (this is the default behaviour for versions <=0.24 but is likely to change in the future).
You can also assign headers directly:
df.columns = ['V', 'W', 'X', 'Y', 'Z']
df
V W X Y Z
0 x x x x x
1 x x x x x
2 x x x x x

Just assign it to the .columns attribute:
>>> df = pd.DataFrame({'$a':[1,2], '$b': [10,20]})
>>> df
$a $b
0 1 10
1 2 20
>>> df.columns = ['a', 'b']
>>> df
a b
0 1 10
1 2 20

The rename method can take a function, for example:
In [11]: df.columns
Out[11]: Index([u'$a', u'$b', u'$c', u'$d', u'$e'], dtype=object)
In [12]: df.rename(columns=lambda x: x[1:], inplace=True)
In [13]: df.columns
Out[13]: Index([u'a', u'b', u'c', u'd', u'e'], dtype=object)

As documented in Working with text data:
df.columns = df.columns.str.replace('$', '')

Pandas 0.21+ Answer
There have been some significant updates to column renaming in version 0.21.
The rename method has added the axis parameter which may be set to columns or 1. This update makes this method match the rest of the pandas API. It still has the index and columns parameters but you are no longer forced to use them.
The set_axis method with the inplace set to False enables you to rename all the index or column labels with a list.
Examples for Pandas 0.21+
Construct sample DataFrame:
df = pd.DataFrame({'$a':[1,2], '$b': [3,4],
'$c':[5,6], '$d':[7,8],
'$e':[9,10]})
$a $b $c $d $e
0 1 3 5 7 9
1 2 4 6 8 10
Using rename with axis='columns' or axis=1
df.rename({'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'}, axis='columns')
or
df.rename({'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'}, axis=1)
Both result in the following:
a b c d e
0 1 3 5 7 9
1 2 4 6 8 10
It is still possible to use the old method signature:
df.rename(columns={'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'})
The rename function also accepts functions that will be applied to each column name.
df.rename(lambda x: x[1:], axis='columns')
or
df.rename(lambda x: x[1:], axis=1)
Using set_axis with a list and inplace=False
You can supply a list to the set_axis method that is equal in length to the number of columns (or index). Currently, inplace defaults to True, but inplace will be defaulted to False in future releases.
df.set_axis(['a', 'b', 'c', 'd', 'e'], axis='columns', inplace=False)
or
df.set_axis(['a', 'b', 'c', 'd', 'e'], axis=1, inplace=False)
Why not use df.columns = ['a', 'b', 'c', 'd', 'e']?
There is nothing wrong with assigning columns directly like this. It is a perfectly good solution.
The advantage of using set_axis is that it can be used as part of a method chain and that it returns a new copy of the DataFrame. Without it, you would have to store your intermediate steps of the chain to another variable before reassigning the columns.
# new for pandas 0.21+
df.some_method1()
.some_method2()
.set_axis()
.some_method3()
# old way
df1 = df.some_method1()
.some_method2()
df1.columns = columns
df1.some_method3()

Since you only want to remove the $ sign in all column names, you could just do:
df = df.rename(columns=lambda x: x.replace('$', ''))
OR
df.rename(columns=lambda x: x.replace('$', ''), inplace=True)

Renaming columns in Pandas is an easy task.
df.rename(columns={'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}, inplace=True)

df.columns = ['a', 'b', 'c', 'd', 'e']
It will replace the existing names with the names you provide, in the order you provide.

Use:
old_names = ['$a', '$b', '$c', '$d', '$e']
new_names = ['a', 'b', 'c', 'd', 'e']
df.rename(columns=dict(zip(old_names, new_names)), inplace=True)
This way you can manually edit the new_names as you wish. It works great when you need to rename only a few columns to correct misspellings, accents, remove special characters, etc.

One line or Pipeline solutions
I'll focus on two things:
OP clearly states
I have the edited column names stored it in a list, but I don't know how to replace the column names.
I do not want to solve the problem of how to replace '$' or strip the first character off of each column header. OP has already done this step. Instead I want to focus on replacing the existing columns object with a new one given a list of replacement column names.
df.columns = new where new is the list of new columns names is as simple as it gets. The drawback of this approach is that it requires editing the existing dataframe's columns attribute and it isn't done inline. I'll show a few ways to perform this via pipelining without editing the existing dataframe.
Setup 1
To focus on the need to rename of replace column names with a pre-existing list, I'll create a new sample dataframe df with initial column names and unrelated new column names.
df = pd.DataFrame({'Jack': [1, 2], 'Mahesh': [3, 4], 'Xin': [5, 6]})
new = ['x098', 'y765', 'z432']
df
Jack Mahesh Xin
0 1 3 5
1 2 4 6
Solution 1
pd.DataFrame.rename
It has been said already that if you had a dictionary mapping the old column names to new column names, you could use pd.DataFrame.rename.
d = {'Jack': 'x098', 'Mahesh': 'y765', 'Xin': 'z432'}
df.rename(columns=d)
x098 y765 z432
0 1 3 5
1 2 4 6
However, you can easily create that dictionary and include it in the call to rename. The following takes advantage of the fact that when iterating over df, we iterate over each column name.
# Given just a list of new column names
df.rename(columns=dict(zip(df, new)))
x098 y765 z432
0 1 3 5
1 2 4 6
This works great if your original column names are unique. But if they are not, then this breaks down.
Setup 2
Non-unique columns
df = pd.DataFrame(
[[1, 3, 5], [2, 4, 6]],
columns=['Mahesh', 'Mahesh', 'Xin']
)
new = ['x098', 'y765', 'z432']
df
Mahesh Mahesh Xin
0 1 3 5
1 2 4 6
Solution 2
pd.concat using the keys argument
First, notice what happens when we attempt to use solution 1:
df.rename(columns=dict(zip(df, new)))
y765 y765 z432
0 1 3 5
1 2 4 6
We didn't map the new list as the column names. We ended up repeating y765. Instead, we can use the keys argument of the pd.concat function while iterating through the columns of df.
pd.concat([c for _, c in df.items()], axis=1, keys=new)
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 3
Reconstruct. This should only be used if you have a single dtype for all columns. Otherwise, you'll end up with dtype object for all columns and converting them back requires more dictionary work.
Single dtype
pd.DataFrame(df.values, df.index, new)
x098 y765 z432
0 1 3 5
1 2 4 6
Mixed dtype
pd.DataFrame(df.values, df.index, new).astype(dict(zip(new, df.dtypes)))
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 4
This is a gimmicky trick with transpose and set_index. pd.DataFrame.set_index allows us to set an index inline, but there is no corresponding set_columns. So we can transpose, then set_index, and transpose back. However, the same single dtype versus mixed dtype caveat from solution 3 applies here.
Single dtype
df.T.set_index(np.asarray(new)).T
x098 y765 z432
0 1 3 5
1 2 4 6
Mixed dtype
df.T.set_index(np.asarray(new)).T.astype(dict(zip(new, df.dtypes)))
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 5
Use a lambda in pd.DataFrame.rename that cycles through each element of new.
In this solution, we pass a lambda that takes x but then ignores it. It also takes a y but doesn't expect it. Instead, an iterator is given as a default value and I can then use that to cycle through one at a time without regard to what the value of x is.
df.rename(columns=lambda x, y=iter(new): next(y))
x098 y765 z432
0 1 3 5
1 2 4 6
And as pointed out to me by the folks in sopython chat, if I add a * in between x and y, I can protect my y variable. Though, in this context I don't believe it needs protecting. It is still worth mentioning.
df.rename(columns=lambda x, *, y=iter(new): next(y))
x098 y765 z432
0 1 3 5
1 2 4 6

Column names vs Names of Series
I would like to explain a bit what happens behind the scenes.
Dataframes are a set of Series.
Series in turn are an extension of a numpy.array.
numpy.arrays have a property .name.
This is the name of the series. It is seldom that Pandas respects this attribute, but it lingers in places and can be used to hack some Pandas behaviors.
Naming the list of columns
A lot of answers here talks about the df.columns attribute being a list when in fact it is a Series. This means it has a .name attribute.
This is what happens if you decide to fill in the name of the columns Series:
df.columns = ['column_one', 'column_two']
df.columns.names = ['name of the list of columns']
df.index.names = ['name of the index']
name of the list of columns column_one column_two
name of the index
0 4 1
1 5 2
2 6 3
Note that the name of the index always comes one column lower.
Artefacts that linger
The .name attribute lingers on sometimes. If you set df.columns = ['one', 'two'] then the df.one.name will be 'one'.
If you set df.one.name = 'three' then df.columns will still give you ['one', 'two'], and df.one.name will give you 'three'.
BUT
pd.DataFrame(df.one) will return
three
0 1
1 2
2 3
Because Pandas reuses the .name of the already defined Series.
Multi-level column names
Pandas has ways of doing multi-layered column names. There is not so much magic involved, but I wanted to cover this in my answer too since I don't see anyone picking up on this here.
|one |
|one |two |
0 | 4 | 1 |
1 | 5 | 2 |
2 | 6 | 3 |
This is easily achievable by setting columns to lists, like this:
df.columns = [['one', 'one'], ['one', 'two']]

Many of pandas functions have an inplace parameter. When setting it True, the transformation applies directly to the dataframe that you are calling it on. For example:
df = pd.DataFrame({'$a':[1,2], '$b': [3,4]})
df.rename(columns={'$a': 'a'}, inplace=True)
df.columns
>>> Index(['a', '$b'], dtype='object')
Alternatively, there are cases where you want to preserve the original dataframe. I have often seen people fall into this case if creating the dataframe is an expensive task. For example, if creating the dataframe required querying a snowflake database. In this case, just make sure the the inplace parameter is set to False.
df = pd.DataFrame({'$a':[1,2], '$b': [3,4]})
df2 = df.rename(columns={'$a': 'a'}, inplace=False)
df.columns
>>> Index(['$a', '$b'], dtype='object')
df2.columns
>>> Index(['a', '$b'], dtype='object')
If these types of transformations are something that you do often, you could also look into a number of different pandas GUI tools. I'm the creator of one called Mito. It’s a spreadsheet that automatically converts your edits to python code.

Let's understand renaming by a small example...
Renaming columns using mapping:
df = pd.DataFrame({"A": [1, 2, 3], "B": [4, 5, 6]}) # Creating a df with column name A and B
df.rename({"A": "new_a", "B": "new_b"}, axis='columns', inplace =True) # Renaming column A with 'new_a' and B with 'new_b'
Output:
new_a new_b
0 1 4
1 2 5
2 3 6
Renaming index/Row_Name using mapping:
df.rename({0: "x", 1: "y", 2: "z"}, axis='index', inplace =True) # Row name are getting replaced by 'x', 'y', and 'z'.
Output:
new_a new_b
x 1 4
y 2 5
z 3 6

Suppose your dataset name is df, and df has.
df = ['$a', '$b', '$c', '$d', '$e']`
So, to rename these, we would simply do.
df.columns = ['a','b','c','d','e']

Let's say this is your dataframe.
You can rename the columns using two methods.
Using dataframe.columns=[#list]
df.columns=['a','b','c','d','e']
The limitation of this method is that if one column has to be changed, full column list has to be passed. Also, this method is not applicable on index labels.
For example, if you passed this:
df.columns = ['a','b','c','d']
This will throw an error. Length mismatch: Expected axis has 5 elements, new values have 4 elements.
Another method is the Pandas rename() method which is used to rename any index, column or row
df = df.rename(columns={'$a':'a'})
Similarly, you can change any rows or columns.

If you've got the dataframe, df.columns dumps everything into a list you can manipulate and then reassign into your dataframe as the names of columns...
columns = df.columns
columns = [row.replace("$", "") for row in columns]
df.rename(columns=dict(zip(columns, things)), inplace=True)
df.head() # To validate the output
Best way? I don't know. A way - yes.
A better way of evaluating all the main techniques put forward in the answers to the question is below using cProfile to gage memory and execution time. #kadee, #kaitlyn, and #eumiro had the functions with the fastest execution times - though these functions are so fast we're comparing the rounding of 0.000 and 0.001 seconds for all the answers. Moral: my answer above likely isn't the 'best' way.
import pandas as pd
import cProfile, pstats, re
old_names = ['$a', '$b', '$c', '$d', '$e']
new_names = ['a', 'b', 'c', 'd', 'e']
col_dict = {'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}
df = pd.DataFrame({'$a':[1, 2], '$b': [10, 20], '$c': ['bleep', 'blorp'], '$d': [1, 2], '$e': ['texa$', '']})
df.head()
def eumiro(df, nn):
df.columns = nn
# This direct renaming approach is duplicated in methodology in several other answers:
return df
def lexual1(df):
return df.rename(columns=col_dict)
def lexual2(df, col_dict):
return df.rename(columns=col_dict, inplace=True)
def Panda_Master_Hayden(df):
return df.rename(columns=lambda x: x[1:], inplace=True)
def paulo1(df):
return df.rename(columns=lambda x: x.replace('$', ''))
def paulo2(df):
return df.rename(columns=lambda x: x.replace('$', ''), inplace=True)
def migloo(df, on, nn):
return df.rename(columns=dict(zip(on, nn)), inplace=True)
def kadee(df):
return df.columns.str.replace('$', '')
def awo(df):
columns = df.columns
columns = [row.replace("$", "") for row in columns]
return df.rename(columns=dict(zip(columns, '')), inplace=True)
def kaitlyn(df):
df.columns = [col.strip('$') for col in df.columns]
return df
print 'eumiro'
cProfile.run('eumiro(df, new_names)')
print 'lexual1'
cProfile.run('lexual1(df)')
print 'lexual2'
cProfile.run('lexual2(df, col_dict)')
print 'andy hayden'
cProfile.run('Panda_Master_Hayden(df)')
print 'paulo1'
cProfile.run('paulo1(df)')
print 'paulo2'
cProfile.run('paulo2(df)')
print 'migloo'
cProfile.run('migloo(df, old_names, new_names)')
print 'kadee'
cProfile.run('kadee(df)')
print 'awo'
cProfile.run('awo(df)')
print 'kaitlyn'
cProfile.run('kaitlyn(df)')

df = pd.DataFrame({'$a': [1], '$b': [1], '$c': [1], '$d': [1], '$e': [1]})
If your new list of columns is in the same order as the existing columns, the assignment is simple:
new_cols = ['a', 'b', 'c', 'd', 'e']
df.columns = new_cols
>>> df
a b c d e
0 1 1 1 1 1
If you had a dictionary keyed on old column names to new column names, you could do the following:
d = {'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}
df.columns = df.columns.map(lambda col: d[col]) # Or `.map(d.get)` as pointed out by #PiRSquared.
>>> df
a b c d e
0 1 1 1 1 1
If you don't have a list or dictionary mapping, you could strip the leading $ symbol via a list comprehension:
df.columns = [col[1:] if col[0] == '$' else col for col in df]

df.rename(index=str, columns={'A':'a', 'B':'b'})
pandas.DataFrame.rename

If you already have a list for the new column names, you can try this:
new_cols = ['a', 'b', 'c', 'd', 'e']
new_names_map = {df.columns[i]:new_cols[i] for i in range(len(new_cols))}
df.rename(new_names_map, axis=1, inplace=True)

Another way we could replace the original column labels is by stripping the unwanted characters (here '$') from the original column labels.
This could have been done by running a for loop over df.columns and appending the stripped columns to df.columns.
Instead, we can do this neatly in a single statement by using list comprehension like below:
df.columns = [col.strip('$') for col in df.columns]
(strip method in Python strips the given character from beginning and end of the string.)

It is real simple. Just use:
df.columns = ['Name1', 'Name2', 'Name3'...]
And it will assign the column names by the order you put them in.

# This way it will work
import pandas as pd
# Define a dictionary
rankings = {'test': ['a'],
'odi': ['E'],
't20': ['P']}
# Convert the dictionary into DataFrame
rankings_pd = pd.DataFrame(rankings)
# Before renaming the columns
print(rankings_pd)
rankings_pd.rename(columns = {'test':'TEST'}, inplace = True)

You could use str.slice for that:
df.columns = df.columns.str.slice(1)

Another option is to rename using a regular expression:
import pandas as pd
import re
df = pd.DataFrame({'$a':[1,2], '$b':[3,4], '$c':[5,6]})
df = df.rename(columns=lambda x: re.sub('\$','',x))
>>> df
a b c
0 1 3 5
1 2 4 6

My method is generic wherein you can add additional delimiters by comma separating delimiters= variable and future-proof it.
Working Code:
import pandas as pd
import re
df = pd.DataFrame({'$a':[1,2], '$b': [3,4],'$c':[5,6], '$d': [7,8], '$e': [9,10]})
delimiters = '$'
matchPattern = '|'.join(map(re.escape, delimiters))
df.columns = [re.split(matchPattern, i)[1] for i in df.columns ]
Output:
>>> df
$a $b $c $d $e
0 1 3 5 7 9
1 2 4 6 8 10
>>> df
a b c d e
0 1 3 5 7 9
1 2 4 6 8 10

Note that the approaches in previous answers do not work for a MultiIndex. For a MultiIndex, you need to do something like the following:
>>> df = pd.DataFrame({('$a','$x'):[1,2], ('$b','$y'): [3,4], ('e','f'):[5,6]})
>>> df
$a $b e
$x $y f
0 1 3 5
1 2 4 6
>>> rename = {('$a','$x'):('a','x'), ('$b','$y'):('b','y')}
>>> df.columns = pandas.MultiIndex.from_tuples([
rename.get(item, item) for item in df.columns.tolist()])
>>> df
a b e
x y f
0 1 3 5
1 2 4 6

If you have to deal with loads of columns named by the providing system out of your control, I came up with the following approach that is a combination of a general approach and specific replacements in one go.
First create a dictionary from the dataframe column names using regular expressions in order to throw away certain appendixes of column names and then add specific replacements to the dictionary to name core columns as expected later in the receiving database.
This is then applied to the dataframe in one go.
dict = dict(zip(df.columns, df.columns.str.replace('(:S$|:C1$|:L$|:D$|\.Serial:L$)', '')))
dict['brand_timeseries:C1'] = 'BTS'
dict['respid:L'] = 'RespID'
dict['country:C1'] = 'CountryID'
dict['pim1:D'] = 'pim_actual'
df.rename(columns=dict, inplace=True)

If you just want to remove the '$' sign then use the below code
df.columns = pd.Series(df.columns.str.replace("$", ""))

In addition to the solution already provided, you can replace all the columns while you are reading the file. We can use names and header=0 to do that.
First, we create a list of the names that we like to use as our column names:
import pandas as pd
ufo_cols = ['city', 'color reported', 'shape reported', 'state', 'time']
ufo.columns = ufo_cols
ufo = pd.read_csv('link to the file you are using', names = ufo_cols, header = 0)
In this case, all the column names will be replaced with the names you have in your list.

Here's a nifty little function I like to use to cut down on typing:
def rename(data, oldnames, newname):
if type(oldnames) == str: # Input can be a string or list of strings
oldnames = [oldnames] # When renaming multiple columns
newname = [newname] # Make sure you pass the corresponding list of new names
i = 0
for name in oldnames:
oldvar = [c for c in data.columns if name in c]
if len(oldvar) == 0:
raise ValueError("Sorry, couldn't find that column in the dataset")
if len(oldvar) > 1: # Doesn't have to be an exact match
print("Found multiple columns that matched " + str(name) + ": ")
for c in oldvar:
print(str(oldvar.index(c)) + ": " + str(c))
ind = input('Please enter the index of the column you would like to rename: ')
oldvar = oldvar[int(ind)]
if len(oldvar) == 1:
oldvar = oldvar[0]
data = data.rename(columns = {oldvar : newname[i]})
i += 1
return data
Here is an example of how it works:
In [2]: df = pd.DataFrame(np.random.randint(0, 10, size=(10, 4)), columns = ['col1', 'col2', 'omg', 'idk'])
# First list = existing variables
# Second list = new names for those variables
In [3]: df = rename(df, ['col', 'omg'],['first', 'ohmy'])
Found multiple columns that matched col:
0: col1
1: col2
Please enter the index of the column you would like to rename: 0
In [4]: df.columns
Out[5]: Index(['first', 'col2', 'ohmy', 'idk'], dtype='object')

Rename the dataframe columns [duplicate]

I want to change the column labels of a Pandas DataFrame from
['$a', '$b', '$c', '$d', '$e']
to
['a', 'b', 'c', 'd', 'e']

Rename Specific Columns
Use the df.rename() function and refer the columns to be renamed. Not all the columns have to be renamed:
df = df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'})
# Or rename the existing DataFrame (rather than creating a copy)
df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'}, inplace=True)
Minimal Code Example
df = pd.DataFrame('x', index=range(3), columns=list('abcde'))
df
a b c d e
0 x x x x x
1 x x x x x
2 x x x x x
The following methods all work and produce the same output:
df2 = df.rename({'a': 'X', 'b': 'Y'}, axis=1) # new method
df2 = df.rename({'a': 'X', 'b': 'Y'}, axis='columns')
df2 = df.rename(columns={'a': 'X', 'b': 'Y'}) # old method
df2
X Y c d e
0 x x x x x
1 x x x x x
2 x x x x x
Remember to assign the result back, as the modification is not-inplace. Alternatively, specify inplace=True:
df.rename({'a': 'X', 'b': 'Y'}, axis=1, inplace=True)
df
X Y c d e
0 x x x x x
1 x x x x x
2 x x x x x
From v0.25, you can also specify errors='raise' to raise errors if an invalid column-to-rename is specified. See v0.25 rename() docs.
Reassign Column Headers
Use df.set_axis() with axis=1 and inplace=False (to return a copy).
df2 = df.set_axis(['V', 'W', 'X', 'Y', 'Z'], axis=1, inplace=False)
df2
V W X Y Z
0 x x x x x
1 x x x x x
2 x x x x x
This returns a copy, but you can modify the DataFrame in-place by setting inplace=True (this is the default behaviour for versions <=0.24 but is likely to change in the future).
You can also assign headers directly:
df.columns = ['V', 'W', 'X', 'Y', 'Z']
df
V W X Y Z
0 x x x x x
1 x x x x x
2 x x x x x

Just assign it to the .columns attribute:
>>> df = pd.DataFrame({'$a':[1,2], '$b': [10,20]})
>>> df
$a $b
0 1 10
1 2 20
>>> df.columns = ['a', 'b']
>>> df
a b
0 1 10
1 2 20

The rename method can take a function, for example:
In [11]: df.columns
Out[11]: Index([u'$a', u'$b', u'$c', u'$d', u'$e'], dtype=object)
In [12]: df.rename(columns=lambda x: x[1:], inplace=True)
In [13]: df.columns
Out[13]: Index([u'a', u'b', u'c', u'd', u'e'], dtype=object)

As documented in Working with text data:
df.columns = df.columns.str.replace('$', '')

Pandas 0.21+ Answer
There have been some significant updates to column renaming in version 0.21.
The rename method has added the axis parameter which may be set to columns or 1. This update makes this method match the rest of the pandas API. It still has the index and columns parameters but you are no longer forced to use them.
The set_axis method with the inplace set to False enables you to rename all the index or column labels with a list.
Examples for Pandas 0.21+
Construct sample DataFrame:
df = pd.DataFrame({'$a':[1,2], '$b': [3,4],
'$c':[5,6], '$d':[7,8],
'$e':[9,10]})
$a $b $c $d $e
0 1 3 5 7 9
1 2 4 6 8 10
Using rename with axis='columns' or axis=1
df.rename({'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'}, axis='columns')
or
df.rename({'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'}, axis=1)
Both result in the following:
a b c d e
0 1 3 5 7 9
1 2 4 6 8 10
It is still possible to use the old method signature:
df.rename(columns={'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'})
The rename function also accepts functions that will be applied to each column name.
df.rename(lambda x: x[1:], axis='columns')
or
df.rename(lambda x: x[1:], axis=1)
Using set_axis with a list and inplace=False
You can supply a list to the set_axis method that is equal in length to the number of columns (or index). Currently, inplace defaults to True, but inplace will be defaulted to False in future releases.
df.set_axis(['a', 'b', 'c', 'd', 'e'], axis='columns', inplace=False)
or
df.set_axis(['a', 'b', 'c', 'd', 'e'], axis=1, inplace=False)
Why not use df.columns = ['a', 'b', 'c', 'd', 'e']?
There is nothing wrong with assigning columns directly like this. It is a perfectly good solution.
The advantage of using set_axis is that it can be used as part of a method chain and that it returns a new copy of the DataFrame. Without it, you would have to store your intermediate steps of the chain to another variable before reassigning the columns.
# new for pandas 0.21+
df.some_method1()
.some_method2()
.set_axis()
.some_method3()
# old way
df1 = df.some_method1()
.some_method2()
df1.columns = columns
df1.some_method3()

Since you only want to remove the $ sign in all column names, you could just do:
df = df.rename(columns=lambda x: x.replace('$', ''))
OR
df.rename(columns=lambda x: x.replace('$', ''), inplace=True)

Renaming columns in Pandas is an easy task.
df.rename(columns={'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}, inplace=True)

df.columns = ['a', 'b', 'c', 'd', 'e']
It will replace the existing names with the names you provide, in the order you provide.

Use:
old_names = ['$a', '$b', '$c', '$d', '$e']
new_names = ['a', 'b', 'c', 'd', 'e']
df.rename(columns=dict(zip(old_names, new_names)), inplace=True)
This way you can manually edit the new_names as you wish. It works great when you need to rename only a few columns to correct misspellings, accents, remove special characters, etc.

One line or Pipeline solutions
I'll focus on two things:
OP clearly states
I have the edited column names stored it in a list, but I don't know how to replace the column names.
I do not want to solve the problem of how to replace '$' or strip the first character off of each column header. OP has already done this step. Instead I want to focus on replacing the existing columns object with a new one given a list of replacement column names.
df.columns = new where new is the list of new columns names is as simple as it gets. The drawback of this approach is that it requires editing the existing dataframe's columns attribute and it isn't done inline. I'll show a few ways to perform this via pipelining without editing the existing dataframe.
Setup 1
To focus on the need to rename of replace column names with a pre-existing list, I'll create a new sample dataframe df with initial column names and unrelated new column names.
df = pd.DataFrame({'Jack': [1, 2], 'Mahesh': [3, 4], 'Xin': [5, 6]})
new = ['x098', 'y765', 'z432']
df
Jack Mahesh Xin
0 1 3 5
1 2 4 6
Solution 1
pd.DataFrame.rename
It has been said already that if you had a dictionary mapping the old column names to new column names, you could use pd.DataFrame.rename.
d = {'Jack': 'x098', 'Mahesh': 'y765', 'Xin': 'z432'}
df.rename(columns=d)
x098 y765 z432
0 1 3 5
1 2 4 6
However, you can easily create that dictionary and include it in the call to rename. The following takes advantage of the fact that when iterating over df, we iterate over each column name.
# Given just a list of new column names
df.rename(columns=dict(zip(df, new)))
x098 y765 z432
0 1 3 5
1 2 4 6
This works great if your original column names are unique. But if they are not, then this breaks down.
Setup 2
Non-unique columns
df = pd.DataFrame(
[[1, 3, 5], [2, 4, 6]],
columns=['Mahesh', 'Mahesh', 'Xin']
)
new = ['x098', 'y765', 'z432']
df
Mahesh Mahesh Xin
0 1 3 5
1 2 4 6
Solution 2
pd.concat using the keys argument
First, notice what happens when we attempt to use solution 1:
df.rename(columns=dict(zip(df, new)))
y765 y765 z432
0 1 3 5
1 2 4 6
We didn't map the new list as the column names. We ended up repeating y765. Instead, we can use the keys argument of the pd.concat function while iterating through the columns of df.
pd.concat([c for _, c in df.items()], axis=1, keys=new)
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 3
Reconstruct. This should only be used if you have a single dtype for all columns. Otherwise, you'll end up with dtype object for all columns and converting them back requires more dictionary work.
Single dtype
pd.DataFrame(df.values, df.index, new)
x098 y765 z432
0 1 3 5
1 2 4 6
Mixed dtype
pd.DataFrame(df.values, df.index, new).astype(dict(zip(new, df.dtypes)))
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 4
This is a gimmicky trick with transpose and set_index. pd.DataFrame.set_index allows us to set an index inline, but there is no corresponding set_columns. So we can transpose, then set_index, and transpose back. However, the same single dtype versus mixed dtype caveat from solution 3 applies here.
Single dtype
df.T.set_index(np.asarray(new)).T
x098 y765 z432
0 1 3 5
1 2 4 6
Mixed dtype
df.T.set_index(np.asarray(new)).T.astype(dict(zip(new, df.dtypes)))
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 5
Use a lambda in pd.DataFrame.rename that cycles through each element of new.
In this solution, we pass a lambda that takes x but then ignores it. It also takes a y but doesn't expect it. Instead, an iterator is given as a default value and I can then use that to cycle through one at a time without regard to what the value of x is.
df.rename(columns=lambda x, y=iter(new): next(y))
x098 y765 z432
0 1 3 5
1 2 4 6
And as pointed out to me by the folks in sopython chat, if I add a * in between x and y, I can protect my y variable. Though, in this context I don't believe it needs protecting. It is still worth mentioning.
df.rename(columns=lambda x, *, y=iter(new): next(y))
x098 y765 z432
0 1 3 5
1 2 4 6

Column names vs Names of Series
I would like to explain a bit what happens behind the scenes.
Dataframes are a set of Series.
Series in turn are an extension of a numpy.array.
numpy.arrays have a property .name.
This is the name of the series. It is seldom that Pandas respects this attribute, but it lingers in places and can be used to hack some Pandas behaviors.
Naming the list of columns
A lot of answers here talks about the df.columns attribute being a list when in fact it is a Series. This means it has a .name attribute.
This is what happens if you decide to fill in the name of the columns Series:
df.columns = ['column_one', 'column_two']
df.columns.names = ['name of the list of columns']
df.index.names = ['name of the index']
name of the list of columns column_one column_two
name of the index
0 4 1
1 5 2
2 6 3
Note that the name of the index always comes one column lower.
Artefacts that linger
The .name attribute lingers on sometimes. If you set df.columns = ['one', 'two'] then the df.one.name will be 'one'.
If you set df.one.name = 'three' then df.columns will still give you ['one', 'two'], and df.one.name will give you 'three'.
BUT
pd.DataFrame(df.one) will return
three
0 1
1 2
2 3
Because Pandas reuses the .name of the already defined Series.
Multi-level column names
Pandas has ways of doing multi-layered column names. There is not so much magic involved, but I wanted to cover this in my answer too since I don't see anyone picking up on this here.
|one |
|one |two |
0 | 4 | 1 |
1 | 5 | 2 |
2 | 6 | 3 |
This is easily achievable by setting columns to lists, like this:
df.columns = [['one', 'one'], ['one', 'two']]

Many of pandas functions have an inplace parameter. When setting it True, the transformation applies directly to the dataframe that you are calling it on. For example:
df = pd.DataFrame({'$a':[1,2], '$b': [3,4]})
df.rename(columns={'$a': 'a'}, inplace=True)
df.columns
>>> Index(['a', '$b'], dtype='object')
Alternatively, there are cases where you want to preserve the original dataframe. I have often seen people fall into this case if creating the dataframe is an expensive task. For example, if creating the dataframe required querying a snowflake database. In this case, just make sure the the inplace parameter is set to False.
df = pd.DataFrame({'$a':[1,2], '$b': [3,4]})
df2 = df.rename(columns={'$a': 'a'}, inplace=False)
df.columns
>>> Index(['$a', '$b'], dtype='object')
df2.columns
>>> Index(['a', '$b'], dtype='object')
If these types of transformations are something that you do often, you could also look into a number of different pandas GUI tools. I'm the creator of one called Mito. It’s a spreadsheet that automatically converts your edits to python code.

Let's understand renaming by a small example...
Renaming columns using mapping:
df = pd.DataFrame({"A": [1, 2, 3], "B": [4, 5, 6]}) # Creating a df with column name A and B
df.rename({"A": "new_a", "B": "new_b"}, axis='columns', inplace =True) # Renaming column A with 'new_a' and B with 'new_b'
Output:
new_a new_b
0 1 4
1 2 5
2 3 6
Renaming index/Row_Name using mapping:
df.rename({0: "x", 1: "y", 2: "z"}, axis='index', inplace =True) # Row name are getting replaced by 'x', 'y', and 'z'.
Output:
new_a new_b
x 1 4
y 2 5
z 3 6

Suppose your dataset name is df, and df has.
df = ['$a', '$b', '$c', '$d', '$e']`
So, to rename these, we would simply do.
df.columns = ['a','b','c','d','e']

Let's say this is your dataframe.
You can rename the columns using two methods.
Using dataframe.columns=[#list]
df.columns=['a','b','c','d','e']
The limitation of this method is that if one column has to be changed, full column list has to be passed. Also, this method is not applicable on index labels.
For example, if you passed this:
df.columns = ['a','b','c','d']
This will throw an error. Length mismatch: Expected axis has 5 elements, new values have 4 elements.
Another method is the Pandas rename() method which is used to rename any index, column or row
df = df.rename(columns={'$a':'a'})
Similarly, you can change any rows or columns.

If you've got the dataframe, df.columns dumps everything into a list you can manipulate and then reassign into your dataframe as the names of columns...
columns = df.columns
columns = [row.replace("$", "") for row in columns]
df.rename(columns=dict(zip(columns, things)), inplace=True)
df.head() # To validate the output
Best way? I don't know. A way - yes.
A better way of evaluating all the main techniques put forward in the answers to the question is below using cProfile to gage memory and execution time. #kadee, #kaitlyn, and #eumiro had the functions with the fastest execution times - though these functions are so fast we're comparing the rounding of 0.000 and 0.001 seconds for all the answers. Moral: my answer above likely isn't the 'best' way.
import pandas as pd
import cProfile, pstats, re
old_names = ['$a', '$b', '$c', '$d', '$e']
new_names = ['a', 'b', 'c', 'd', 'e']
col_dict = {'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}
df = pd.DataFrame({'$a':[1, 2], '$b': [10, 20], '$c': ['bleep', 'blorp'], '$d': [1, 2], '$e': ['texa$', '']})
df.head()
def eumiro(df, nn):
df.columns = nn
# This direct renaming approach is duplicated in methodology in several other answers:
return df
def lexual1(df):
return df.rename(columns=col_dict)
def lexual2(df, col_dict):
return df.rename(columns=col_dict, inplace=True)
def Panda_Master_Hayden(df):
return df.rename(columns=lambda x: x[1:], inplace=True)
def paulo1(df):
return df.rename(columns=lambda x: x.replace('$', ''))
def paulo2(df):
return df.rename(columns=lambda x: x.replace('$', ''), inplace=True)
def migloo(df, on, nn):
return df.rename(columns=dict(zip(on, nn)), inplace=True)
def kadee(df):
return df.columns.str.replace('$', '')
def awo(df):
columns = df.columns
columns = [row.replace("$", "") for row in columns]
return df.rename(columns=dict(zip(columns, '')), inplace=True)
def kaitlyn(df):
df.columns = [col.strip('$') for col in df.columns]
return df
print 'eumiro'
cProfile.run('eumiro(df, new_names)')
print 'lexual1'
cProfile.run('lexual1(df)')
print 'lexual2'
cProfile.run('lexual2(df, col_dict)')
print 'andy hayden'
cProfile.run('Panda_Master_Hayden(df)')
print 'paulo1'
cProfile.run('paulo1(df)')
print 'paulo2'
cProfile.run('paulo2(df)')
print 'migloo'
cProfile.run('migloo(df, old_names, new_names)')
print 'kadee'
cProfile.run('kadee(df)')
print 'awo'
cProfile.run('awo(df)')
print 'kaitlyn'
cProfile.run('kaitlyn(df)')

df = pd.DataFrame({'$a': [1], '$b': [1], '$c': [1], '$d': [1], '$e': [1]})
If your new list of columns is in the same order as the existing columns, the assignment is simple:
new_cols = ['a', 'b', 'c', 'd', 'e']
df.columns = new_cols
>>> df
a b c d e
0 1 1 1 1 1
If you had a dictionary keyed on old column names to new column names, you could do the following:
d = {'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}
df.columns = df.columns.map(lambda col: d[col]) # Or `.map(d.get)` as pointed out by #PiRSquared.
>>> df
a b c d e
0 1 1 1 1 1
If you don't have a list or dictionary mapping, you could strip the leading $ symbol via a list comprehension:
df.columns = [col[1:] if col[0] == '$' else col for col in df]

df.rename(index=str, columns={'A':'a', 'B':'b'})
pandas.DataFrame.rename

If you already have a list for the new column names, you can try this:
new_cols = ['a', 'b', 'c', 'd', 'e']
new_names_map = {df.columns[i]:new_cols[i] for i in range(len(new_cols))}
df.rename(new_names_map, axis=1, inplace=True)

Another way we could replace the original column labels is by stripping the unwanted characters (here '$') from the original column labels.
This could have been done by running a for loop over df.columns and appending the stripped columns to df.columns.
Instead, we can do this neatly in a single statement by using list comprehension like below:
df.columns = [col.strip('$') for col in df.columns]
(strip method in Python strips the given character from beginning and end of the string.)

It is real simple. Just use:
df.columns = ['Name1', 'Name2', 'Name3'...]
And it will assign the column names by the order you put them in.

# This way it will work
import pandas as pd
# Define a dictionary
rankings = {'test': ['a'],
'odi': ['E'],
't20': ['P']}
# Convert the dictionary into DataFrame
rankings_pd = pd.DataFrame(rankings)
# Before renaming the columns
print(rankings_pd)
rankings_pd.rename(columns = {'test':'TEST'}, inplace = True)

You could use str.slice for that:
df.columns = df.columns.str.slice(1)

Another option is to rename using a regular expression:
import pandas as pd
import re
df = pd.DataFrame({'$a':[1,2], '$b':[3,4], '$c':[5,6]})
df = df.rename(columns=lambda x: re.sub('\$','',x))
>>> df
a b c
0 1 3 5
1 2 4 6

My method is generic wherein you can add additional delimiters by comma separating delimiters= variable and future-proof it.
Working Code:
import pandas as pd
import re
df = pd.DataFrame({'$a':[1,2], '$b': [3,4],'$c':[5,6], '$d': [7,8], '$e': [9,10]})
delimiters = '$'
matchPattern = '|'.join(map(re.escape, delimiters))
df.columns = [re.split(matchPattern, i)[1] for i in df.columns ]
Output:
>>> df
$a $b $c $d $e
0 1 3 5 7 9
1 2 4 6 8 10
>>> df
a b c d e
0 1 3 5 7 9
1 2 4 6 8 10

Note that the approaches in previous answers do not work for a MultiIndex. For a MultiIndex, you need to do something like the following:
>>> df = pd.DataFrame({('$a','$x'):[1,2], ('$b','$y'): [3,4], ('e','f'):[5,6]})
>>> df
$a $b e
$x $y f
0 1 3 5
1 2 4 6
>>> rename = {('$a','$x'):('a','x'), ('$b','$y'):('b','y')}
>>> df.columns = pandas.MultiIndex.from_tuples([
rename.get(item, item) for item in df.columns.tolist()])
>>> df
a b e
x y f
0 1 3 5
1 2 4 6

If you have to deal with loads of columns named by the providing system out of your control, I came up with the following approach that is a combination of a general approach and specific replacements in one go.
First create a dictionary from the dataframe column names using regular expressions in order to throw away certain appendixes of column names and then add specific replacements to the dictionary to name core columns as expected later in the receiving database.
This is then applied to the dataframe in one go.
dict = dict(zip(df.columns, df.columns.str.replace('(:S$|:C1$|:L$|:D$|\.Serial:L$)', '')))
dict['brand_timeseries:C1'] = 'BTS'
dict['respid:L'] = 'RespID'
dict['country:C1'] = 'CountryID'
dict['pim1:D'] = 'pim_actual'
df.rename(columns=dict, inplace=True)

If you just want to remove the '$' sign then use the below code
df.columns = pd.Series(df.columns.str.replace("$", ""))

In addition to the solution already provided, you can replace all the columns while you are reading the file. We can use names and header=0 to do that.
First, we create a list of the names that we like to use as our column names:
import pandas as pd
ufo_cols = ['city', 'color reported', 'shape reported', 'state', 'time']
ufo.columns = ufo_cols
ufo = pd.read_csv('link to the file you are using', names = ufo_cols, header = 0)
In this case, all the column names will be replaced with the names you have in your list.

Here's a nifty little function I like to use to cut down on typing:
def rename(data, oldnames, newname):
if type(oldnames) == str: # Input can be a string or list of strings
oldnames = [oldnames] # When renaming multiple columns
newname = [newname] # Make sure you pass the corresponding list of new names
i = 0
for name in oldnames:
oldvar = [c for c in data.columns if name in c]
if len(oldvar) == 0:
raise ValueError("Sorry, couldn't find that column in the dataset")
if len(oldvar) > 1: # Doesn't have to be an exact match
print("Found multiple columns that matched " + str(name) + ": ")
for c in oldvar:
print(str(oldvar.index(c)) + ": " + str(c))
ind = input('Please enter the index of the column you would like to rename: ')
oldvar = oldvar[int(ind)]
if len(oldvar) == 1:
oldvar = oldvar[0]
data = data.rename(columns = {oldvar : newname[i]})
i += 1
return data
Here is an example of how it works:
In [2]: df = pd.DataFrame(np.random.randint(0, 10, size=(10, 4)), columns = ['col1', 'col2', 'omg', 'idk'])
# First list = existing variables
# Second list = new names for those variables
In [3]: df = rename(df, ['col', 'omg'],['first', 'ohmy'])
Found multiple columns that matched col:
0: col1
1: col2
Please enter the index of the column you would like to rename: 0
In [4]: df.columns
Out[5]: Index(['first', 'col2', 'ohmy', 'idk'], dtype='object')

asking about pandas dataframe regarding the column name changes in it [duplicate]

I want to change the column labels of a Pandas DataFrame from
['$a', '$b', '$c', '$d', '$e']
to
['a', 'b', 'c', 'd', 'e']

Rename Specific Columns
Use the df.rename() function and refer the columns to be renamed. Not all the columns have to be renamed:
df = df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'})
# Or rename the existing DataFrame (rather than creating a copy)
df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'}, inplace=True)
Minimal Code Example
df = pd.DataFrame('x', index=range(3), columns=list('abcde'))
df
a b c d e
0 x x x x x
1 x x x x x
2 x x x x x
The following methods all work and produce the same output:
df2 = df.rename({'a': 'X', 'b': 'Y'}, axis=1) # new method
df2 = df.rename({'a': 'X', 'b': 'Y'}, axis='columns')
df2 = df.rename(columns={'a': 'X', 'b': 'Y'}) # old method
df2
X Y c d e
0 x x x x x
1 x x x x x
2 x x x x x
Remember to assign the result back, as the modification is not-inplace. Alternatively, specify inplace=True:
df.rename({'a': 'X', 'b': 'Y'}, axis=1, inplace=True)
df
X Y c d e
0 x x x x x
1 x x x x x
2 x x x x x
From v0.25, you can also specify errors='raise' to raise errors if an invalid column-to-rename is specified. See v0.25 rename() docs.
Reassign Column Headers
Use df.set_axis() with axis=1 and inplace=False (to return a copy).
df2 = df.set_axis(['V', 'W', 'X', 'Y', 'Z'], axis=1, inplace=False)
df2
V W X Y Z
0 x x x x x
1 x x x x x
2 x x x x x
This returns a copy, but you can modify the DataFrame in-place by setting inplace=True (this is the default behaviour for versions <=0.24 but is likely to change in the future).
You can also assign headers directly:
df.columns = ['V', 'W', 'X', 'Y', 'Z']
df
V W X Y Z
0 x x x x x
1 x x x x x
2 x x x x x

Just assign it to the .columns attribute:
>>> df = pd.DataFrame({'$a':[1,2], '$b': [10,20]})
>>> df
$a $b
0 1 10
1 2 20
>>> df.columns = ['a', 'b']
>>> df
a b
0 1 10
1 2 20

The rename method can take a function, for example:
In [11]: df.columns
Out[11]: Index([u'$a', u'$b', u'$c', u'$d', u'$e'], dtype=object)
In [12]: df.rename(columns=lambda x: x[1:], inplace=True)
In [13]: df.columns
Out[13]: Index([u'a', u'b', u'c', u'd', u'e'], dtype=object)

As documented in Working with text data:
df.columns = df.columns.str.replace('$', '')

Pandas 0.21+ Answer
There have been some significant updates to column renaming in version 0.21.
The rename method has added the axis parameter which may be set to columns or 1. This update makes this method match the rest of the pandas API. It still has the index and columns parameters but you are no longer forced to use them.
The set_axis method with the inplace set to False enables you to rename all the index or column labels with a list.
Examples for Pandas 0.21+
Construct sample DataFrame:
df = pd.DataFrame({'$a':[1,2], '$b': [3,4],
'$c':[5,6], '$d':[7,8],
'$e':[9,10]})
$a $b $c $d $e
0 1 3 5 7 9
1 2 4 6 8 10
Using rename with axis='columns' or axis=1
df.rename({'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'}, axis='columns')
or
df.rename({'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'}, axis=1)
Both result in the following:
a b c d e
0 1 3 5 7 9
1 2 4 6 8 10
It is still possible to use the old method signature:
df.rename(columns={'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'})
The rename function also accepts functions that will be applied to each column name.
df.rename(lambda x: x[1:], axis='columns')
or
df.rename(lambda x: x[1:], axis=1)
Using set_axis with a list and inplace=False
You can supply a list to the set_axis method that is equal in length to the number of columns (or index). Currently, inplace defaults to True, but inplace will be defaulted to False in future releases.
df.set_axis(['a', 'b', 'c', 'd', 'e'], axis='columns', inplace=False)
or
df.set_axis(['a', 'b', 'c', 'd', 'e'], axis=1, inplace=False)
Why not use df.columns = ['a', 'b', 'c', 'd', 'e']?
There is nothing wrong with assigning columns directly like this. It is a perfectly good solution.
The advantage of using set_axis is that it can be used as part of a method chain and that it returns a new copy of the DataFrame. Without it, you would have to store your intermediate steps of the chain to another variable before reassigning the columns.
# new for pandas 0.21+
df.some_method1()
.some_method2()
.set_axis()
.some_method3()
# old way
df1 = df.some_method1()
.some_method2()
df1.columns = columns
df1.some_method3()

Since you only want to remove the $ sign in all column names, you could just do:
df = df.rename(columns=lambda x: x.replace('$', ''))
OR
df.rename(columns=lambda x: x.replace('$', ''), inplace=True)

Renaming columns in Pandas is an easy task.
df.rename(columns={'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}, inplace=True)

df.columns = ['a', 'b', 'c', 'd', 'e']
It will replace the existing names with the names you provide, in the order you provide.

Use:
old_names = ['$a', '$b', '$c', '$d', '$e']
new_names = ['a', 'b', 'c', 'd', 'e']
df.rename(columns=dict(zip(old_names, new_names)), inplace=True)
This way you can manually edit the new_names as you wish. It works great when you need to rename only a few columns to correct misspellings, accents, remove special characters, etc.

One line or Pipeline solutions
I'll focus on two things:
OP clearly states
I have the edited column names stored it in a list, but I don't know how to replace the column names.
I do not want to solve the problem of how to replace '$' or strip the first character off of each column header. OP has already done this step. Instead I want to focus on replacing the existing columns object with a new one given a list of replacement column names.
df.columns = new where new is the list of new columns names is as simple as it gets. The drawback of this approach is that it requires editing the existing dataframe's columns attribute and it isn't done inline. I'll show a few ways to perform this via pipelining without editing the existing dataframe.
Setup 1
To focus on the need to rename of replace column names with a pre-existing list, I'll create a new sample dataframe df with initial column names and unrelated new column names.
df = pd.DataFrame({'Jack': [1, 2], 'Mahesh': [3, 4], 'Xin': [5, 6]})
new = ['x098', 'y765', 'z432']
df
Jack Mahesh Xin
0 1 3 5
1 2 4 6
Solution 1
pd.DataFrame.rename
It has been said already that if you had a dictionary mapping the old column names to new column names, you could use pd.DataFrame.rename.
d = {'Jack': 'x098', 'Mahesh': 'y765', 'Xin': 'z432'}
df.rename(columns=d)
x098 y765 z432
0 1 3 5
1 2 4 6
However, you can easily create that dictionary and include it in the call to rename. The following takes advantage of the fact that when iterating over df, we iterate over each column name.
# Given just a list of new column names
df.rename(columns=dict(zip(df, new)))
x098 y765 z432
0 1 3 5
1 2 4 6
This works great if your original column names are unique. But if they are not, then this breaks down.
Setup 2
Non-unique columns
df = pd.DataFrame(
[[1, 3, 5], [2, 4, 6]],
columns=['Mahesh', 'Mahesh', 'Xin']
)
new = ['x098', 'y765', 'z432']
df
Mahesh Mahesh Xin
0 1 3 5
1 2 4 6
Solution 2
pd.concat using the keys argument
First, notice what happens when we attempt to use solution 1:
df.rename(columns=dict(zip(df, new)))
y765 y765 z432
0 1 3 5
1 2 4 6
We didn't map the new list as the column names. We ended up repeating y765. Instead, we can use the keys argument of the pd.concat function while iterating through the columns of df.
pd.concat([c for _, c in df.items()], axis=1, keys=new)
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 3
Reconstruct. This should only be used if you have a single dtype for all columns. Otherwise, you'll end up with dtype object for all columns and converting them back requires more dictionary work.
Single dtype
pd.DataFrame(df.values, df.index, new)
x098 y765 z432
0 1 3 5
1 2 4 6
Mixed dtype
pd.DataFrame(df.values, df.index, new).astype(dict(zip(new, df.dtypes)))
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 4
This is a gimmicky trick with transpose and set_index. pd.DataFrame.set_index allows us to set an index inline, but there is no corresponding set_columns. So we can transpose, then set_index, and transpose back. However, the same single dtype versus mixed dtype caveat from solution 3 applies here.
Single dtype
df.T.set_index(np.asarray(new)).T
x098 y765 z432
0 1 3 5
1 2 4 6
Mixed dtype
df.T.set_index(np.asarray(new)).T.astype(dict(zip(new, df.dtypes)))
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 5
Use a lambda in pd.DataFrame.rename that cycles through each element of new.
In this solution, we pass a lambda that takes x but then ignores it. It also takes a y but doesn't expect it. Instead, an iterator is given as a default value and I can then use that to cycle through one at a time without regard to what the value of x is.
df.rename(columns=lambda x, y=iter(new): next(y))
x098 y765 z432
0 1 3 5
1 2 4 6
And as pointed out to me by the folks in sopython chat, if I add a * in between x and y, I can protect my y variable. Though, in this context I don't believe it needs protecting. It is still worth mentioning.
df.rename(columns=lambda x, *, y=iter(new): next(y))
x098 y765 z432
0 1 3 5
1 2 4 6

Column names vs Names of Series
I would like to explain a bit what happens behind the scenes.
Dataframes are a set of Series.
Series in turn are an extension of a numpy.array.
numpy.arrays have a property .name.
This is the name of the series. It is seldom that Pandas respects this attribute, but it lingers in places and can be used to hack some Pandas behaviors.
Naming the list of columns
A lot of answers here talks about the df.columns attribute being a list when in fact it is a Series. This means it has a .name attribute.
This is what happens if you decide to fill in the name of the columns Series:
df.columns = ['column_one', 'column_two']
df.columns.names = ['name of the list of columns']
df.index.names = ['name of the index']
name of the list of columns column_one column_two
name of the index
0 4 1
1 5 2
2 6 3
Note that the name of the index always comes one column lower.
Artefacts that linger
The .name attribute lingers on sometimes. If you set df.columns = ['one', 'two'] then the df.one.name will be 'one'.
If you set df.one.name = 'three' then df.columns will still give you ['one', 'two'], and df.one.name will give you 'three'.
BUT
pd.DataFrame(df.one) will return
three
0 1
1 2
2 3
Because Pandas reuses the .name of the already defined Series.
Multi-level column names
Pandas has ways of doing multi-layered column names. There is not so much magic involved, but I wanted to cover this in my answer too since I don't see anyone picking up on this here.
|one |
|one |two |
0 | 4 | 1 |
1 | 5 | 2 |
2 | 6 | 3 |
This is easily achievable by setting columns to lists, like this:
df.columns = [['one', 'one'], ['one', 'two']]

Many of pandas functions have an inplace parameter. When setting it True, the transformation applies directly to the dataframe that you are calling it on. For example:
df = pd.DataFrame({'$a':[1,2], '$b': [3,4]})
df.rename(columns={'$a': 'a'}, inplace=True)
df.columns
>>> Index(['a', '$b'], dtype='object')
Alternatively, there are cases where you want to preserve the original dataframe. I have often seen people fall into this case if creating the dataframe is an expensive task. For example, if creating the dataframe required querying a snowflake database. In this case, just make sure the the inplace parameter is set to False.
df = pd.DataFrame({'$a':[1,2], '$b': [3,4]})
df2 = df.rename(columns={'$a': 'a'}, inplace=False)
df.columns
>>> Index(['$a', '$b'], dtype='object')
df2.columns
>>> Index(['a', '$b'], dtype='object')
If these types of transformations are something that you do often, you could also look into a number of different pandas GUI tools. I'm the creator of one called Mito. It’s a spreadsheet that automatically converts your edits to python code.

Let's understand renaming by a small example...
Renaming columns using mapping:
df = pd.DataFrame({"A": [1, 2, 3], "B": [4, 5, 6]}) # Creating a df with column name A and B
df.rename({"A": "new_a", "B": "new_b"}, axis='columns', inplace =True) # Renaming column A with 'new_a' and B with 'new_b'
Output:
new_a new_b
0 1 4
1 2 5
2 3 6
Renaming index/Row_Name using mapping:
df.rename({0: "x", 1: "y", 2: "z"}, axis='index', inplace =True) # Row name are getting replaced by 'x', 'y', and 'z'.
Output:
new_a new_b
x 1 4
y 2 5
z 3 6

Suppose your dataset name is df, and df has.
df = ['$a', '$b', '$c', '$d', '$e']`
So, to rename these, we would simply do.
df.columns = ['a','b','c','d','e']

Let's say this is your dataframe.
You can rename the columns using two methods.
Using dataframe.columns=[#list]
df.columns=['a','b','c','d','e']
The limitation of this method is that if one column has to be changed, full column list has to be passed. Also, this method is not applicable on index labels.
For example, if you passed this:
df.columns = ['a','b','c','d']
This will throw an error. Length mismatch: Expected axis has 5 elements, new values have 4 elements.
Another method is the Pandas rename() method which is used to rename any index, column or row
df = df.rename(columns={'$a':'a'})
Similarly, you can change any rows or columns.

If you've got the dataframe, df.columns dumps everything into a list you can manipulate and then reassign into your dataframe as the names of columns...
columns = df.columns
columns = [row.replace("$", "") for row in columns]
df.rename(columns=dict(zip(columns, things)), inplace=True)
df.head() # To validate the output
Best way? I don't know. A way - yes.
A better way of evaluating all the main techniques put forward in the answers to the question is below using cProfile to gage memory and execution time. #kadee, #kaitlyn, and #eumiro had the functions with the fastest execution times - though these functions are so fast we're comparing the rounding of 0.000 and 0.001 seconds for all the answers. Moral: my answer above likely isn't the 'best' way.
import pandas as pd
import cProfile, pstats, re
old_names = ['$a', '$b', '$c', '$d', '$e']
new_names = ['a', 'b', 'c', 'd', 'e']
col_dict = {'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}
df = pd.DataFrame({'$a':[1, 2], '$b': [10, 20], '$c': ['bleep', 'blorp'], '$d': [1, 2], '$e': ['texa$', '']})
df.head()
def eumiro(df, nn):
df.columns = nn
# This direct renaming approach is duplicated in methodology in several other answers:
return df
def lexual1(df):
return df.rename(columns=col_dict)
def lexual2(df, col_dict):
return df.rename(columns=col_dict, inplace=True)
def Panda_Master_Hayden(df):
return df.rename(columns=lambda x: x[1:], inplace=True)
def paulo1(df):
return df.rename(columns=lambda x: x.replace('$', ''))
def paulo2(df):
return df.rename(columns=lambda x: x.replace('$', ''), inplace=True)
def migloo(df, on, nn):
return df.rename(columns=dict(zip(on, nn)), inplace=True)
def kadee(df):
return df.columns.str.replace('$', '')
def awo(df):
columns = df.columns
columns = [row.replace("$", "") for row in columns]
return df.rename(columns=dict(zip(columns, '')), inplace=True)
def kaitlyn(df):
df.columns = [col.strip('$') for col in df.columns]
return df
print 'eumiro'
cProfile.run('eumiro(df, new_names)')
print 'lexual1'
cProfile.run('lexual1(df)')
print 'lexual2'
cProfile.run('lexual2(df, col_dict)')
print 'andy hayden'
cProfile.run('Panda_Master_Hayden(df)')
print 'paulo1'
cProfile.run('paulo1(df)')
print 'paulo2'
cProfile.run('paulo2(df)')
print 'migloo'
cProfile.run('migloo(df, old_names, new_names)')
print 'kadee'
cProfile.run('kadee(df)')
print 'awo'
cProfile.run('awo(df)')
print 'kaitlyn'
cProfile.run('kaitlyn(df)')

df = pd.DataFrame({'$a': [1], '$b': [1], '$c': [1], '$d': [1], '$e': [1]})
If your new list of columns is in the same order as the existing columns, the assignment is simple:
new_cols = ['a', 'b', 'c', 'd', 'e']
df.columns = new_cols
>>> df
a b c d e
0 1 1 1 1 1
If you had a dictionary keyed on old column names to new column names, you could do the following:
d = {'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}
df.columns = df.columns.map(lambda col: d[col]) # Or `.map(d.get)` as pointed out by #PiRSquared.
>>> df
a b c d e
0 1 1 1 1 1
If you don't have a list or dictionary mapping, you could strip the leading $ symbol via a list comprehension:
df.columns = [col[1:] if col[0] == '$' else col for col in df]

df.rename(index=str, columns={'A':'a', 'B':'b'})
pandas.DataFrame.rename

If you already have a list for the new column names, you can try this:
new_cols = ['a', 'b', 'c', 'd', 'e']
new_names_map = {df.columns[i]:new_cols[i] for i in range(len(new_cols))}
df.rename(new_names_map, axis=1, inplace=True)

Another way we could replace the original column labels is by stripping the unwanted characters (here '$') from the original column labels.
This could have been done by running a for loop over df.columns and appending the stripped columns to df.columns.
Instead, we can do this neatly in a single statement by using list comprehension like below:
df.columns = [col.strip('$') for col in df.columns]
(strip method in Python strips the given character from beginning and end of the string.)

It is real simple. Just use:
df.columns = ['Name1', 'Name2', 'Name3'...]
And it will assign the column names by the order you put them in.

# This way it will work
import pandas as pd
# Define a dictionary
rankings = {'test': ['a'],
'odi': ['E'],
't20': ['P']}
# Convert the dictionary into DataFrame
rankings_pd = pd.DataFrame(rankings)
# Before renaming the columns
print(rankings_pd)
rankings_pd.rename(columns = {'test':'TEST'}, inplace = True)

You could use str.slice for that:
df.columns = df.columns.str.slice(1)

Another option is to rename using a regular expression:
import pandas as pd
import re
df = pd.DataFrame({'$a':[1,2], '$b':[3,4], '$c':[5,6]})
df = df.rename(columns=lambda x: re.sub('\$','',x))
>>> df
a b c
0 1 3 5
1 2 4 6

My method is generic wherein you can add additional delimiters by comma separating delimiters= variable and future-proof it.
Working Code:
import pandas as pd
import re
df = pd.DataFrame({'$a':[1,2], '$b': [3,4],'$c':[5,6], '$d': [7,8], '$e': [9,10]})
delimiters = '$'
matchPattern = '|'.join(map(re.escape, delimiters))
df.columns = [re.split(matchPattern, i)[1] for i in df.columns ]
Output:
>>> df
$a $b $c $d $e
0 1 3 5 7 9
1 2 4 6 8 10
>>> df
a b c d e
0 1 3 5 7 9
1 2 4 6 8 10

Note that the approaches in previous answers do not work for a MultiIndex. For a MultiIndex, you need to do something like the following:
>>> df = pd.DataFrame({('$a','$x'):[1,2], ('$b','$y'): [3,4], ('e','f'):[5,6]})
>>> df
$a $b e
$x $y f
0 1 3 5
1 2 4 6
>>> rename = {('$a','$x'):('a','x'), ('$b','$y'):('b','y')}
>>> df.columns = pandas.MultiIndex.from_tuples([
rename.get(item, item) for item in df.columns.tolist()])
>>> df
a b e
x y f
0 1 3 5
1 2 4 6

If you have to deal with loads of columns named by the providing system out of your control, I came up with the following approach that is a combination of a general approach and specific replacements in one go.
First create a dictionary from the dataframe column names using regular expressions in order to throw away certain appendixes of column names and then add specific replacements to the dictionary to name core columns as expected later in the receiving database.
This is then applied to the dataframe in one go.
dict = dict(zip(df.columns, df.columns.str.replace('(:S$|:C1$|:L$|:D$|\.Serial:L$)', '')))
dict['brand_timeseries:C1'] = 'BTS'
dict['respid:L'] = 'RespID'
dict['country:C1'] = 'CountryID'
dict['pim1:D'] = 'pim_actual'
df.rename(columns=dict, inplace=True)

If you just want to remove the '$' sign then use the below code
df.columns = pd.Series(df.columns.str.replace("$", ""))

In addition to the solution already provided, you can replace all the columns while you are reading the file. We can use names and header=0 to do that.
First, we create a list of the names that we like to use as our column names:
import pandas as pd
ufo_cols = ['city', 'color reported', 'shape reported', 'state', 'time']
ufo.columns = ufo_cols
ufo = pd.read_csv('link to the file you are using', names = ufo_cols, header = 0)
In this case, all the column names will be replaced with the names you have in your list.

Here's a nifty little function I like to use to cut down on typing:
def rename(data, oldnames, newname):
if type(oldnames) == str: # Input can be a string or list of strings
oldnames = [oldnames] # When renaming multiple columns
newname = [newname] # Make sure you pass the corresponding list of new names
i = 0
for name in oldnames:
oldvar = [c for c in data.columns if name in c]
if len(oldvar) == 0:
raise ValueError("Sorry, couldn't find that column in the dataset")
if len(oldvar) > 1: # Doesn't have to be an exact match
print("Found multiple columns that matched " + str(name) + ": ")
for c in oldvar:
print(str(oldvar.index(c)) + ": " + str(c))
ind = input('Please enter the index of the column you would like to rename: ')
oldvar = oldvar[int(ind)]
if len(oldvar) == 1:
oldvar = oldvar[0]
data = data.rename(columns = {oldvar : newname[i]})
i += 1
return data
Here is an example of how it works:
In [2]: df = pd.DataFrame(np.random.randint(0, 10, size=(10, 4)), columns = ['col1', 'col2', 'omg', 'idk'])
# First list = existing variables
# Second list = new names for those variables
In [3]: df = rename(df, ['col', 'omg'],['first', 'ohmy'])
Found multiple columns that matched col:
0: col1
1: col2
Please enter the index of the column you would like to rename: 0
In [4]: df.columns
Out[5]: Index(['first', 'col2', 'ohmy', 'idk'], dtype='object')

move column in pandas dataframe

I have the following dataframe:
a b x y
0 1 2 3 -1
1 2 4 6 -2
2 3 6 9 -3
3 4 8 12 -4
How can I move columns b and x such that they are the last 2 columns in the dataframe? I would like to specify b and x by name, but not the other columns.

You can rearrange columns directly by specifying their order:
df = df[['a', 'y', 'b', 'x']]
In the case of larger dataframes where the column titles are dynamic, you can use a list comprehension to select every column not in your target set and then append the target set to the end.
>>> df[[c for c in df if c not in ['b', 'x']]
+ ['b', 'x']]
a y b x
0 1 -1 2 3
1 2 -2 4 6
2 3 -3 6 9
3 4 -4 8 12
To make it more bullet proof, you can ensure that your target columns are indeed in the dataframe:
cols_at_end = ['b', 'x']
df = df[[c for c in df if c not in cols_at_end]
+ [c for c in cols_at_end if c in df]]

cols = list(df.columns.values) #Make a list of all of the columns in the df
cols.pop(cols.index('b')) #Remove b from list
cols.pop(cols.index('x')) #Remove x from list
df = df[cols+['b','x']] #Create new dataframe with columns in the order you want

For example, to move column "name" to be the first column in df you can use insert:
column_to_move = df.pop("name")
# insert column with insert(location, column_name, column_value)
df.insert(0, "name", column_to_move)
similarly, if you want this column to be e.g. third column from the beginning:
df.insert(2, "name", column_to_move )

You can use to way below. It's very simple, but similar to the good answer given by Charlie Haley.
df1 = df.pop('b') # remove column b and store it in df1
df2 = df.pop('x') # remove column x and store it in df2
df['b']=df1 # add b series as a 'new' column.
df['x']=df2 # add b series as a 'new' column.
Now you have your dataframe with the columns 'b' and 'x' in the end. You can see this video from OSPY : https://youtu.be/RlbO27N3Xg4

similar to ROBBAT1's answer above, but hopefully a bit more robust:
df.insert(len(df.columns)-1, 'b', df.pop('b'))
df.insert(len(df.columns)-1, 'x', df.pop('x'))

This function will reorder your columns without losing data. Any omitted columns remain in the center of the data set:
def reorder_columns(columns, first_cols=[], last_cols=[], drop_cols=[]):
columns = list(set(columns) - set(first_cols))
columns = list(set(columns) - set(drop_cols))
columns = list(set(columns) - set(last_cols))
new_order = first_cols + columns + last_cols
return new_order
Example usage:
my_list = ['first', 'second', 'third', 'fourth', 'fifth', 'sixth']
reorder_columns(my_list, first_cols=['fourth', 'third'], last_cols=['second'], drop_cols=['fifth'])
# Output:
['fourth', 'third', 'first', 'sixth', 'second']
To assign to your dataframe, use:
my_list = df.columns.tolist()
reordered_cols = reorder_columns(my_list, first_cols=['fourth', 'third'], last_cols=['second'], drop_cols=['fifth'])
df = df[reordered_cols]

Simple solution:
old_cols = df.columns.values
new_cols= ['a', 'y', 'b', 'x']
df = df.reindex(columns=new_cols)

An alternative, more generic method;
from pandas import DataFrame
def move_columns(df: DataFrame, cols_to_move: list, new_index: int) -> DataFrame:
"""
This method re-arranges the columns in a dataframe to place the desired columns at the desired index.
ex Usage: df = move_columns(df, ['Rev'], 2)
:param df:
:param cols_to_move: The names of the columns to move. They must be a list
:param new_index: The 0-based location to place the columns.
:return: Return a dataframe with the columns re-arranged
"""
other = [c for c in df if c not in cols_to_move]
start = other[0:new_index]
end = other[new_index:]
return df[start + cols_to_move + end]

You can use pd.Index.difference with np.hstack, then reindex or use label-based indexing. In general, it's a good idea to avoid list comprehensions or other explicit loops with NumPy / Pandas objects.
cols_to_move = ['b', 'x']
new_cols = np.hstack((df.columns.difference(cols_to_move), cols_to_move))
# OPTION 1: reindex
df = df.reindex(columns=new_cols)
# OPTION 2: direct label-based indexing
df = df[new_cols]
# OPTION 3: loc label-based indexing
df = df.loc[:, new_cols]
print(df)
# a y b x
# 0 1 -1 2 3
# 1 2 -2 4 6
# 2 3 -3 6 9
# 3 4 -4 8 12

You can use movecolumn package in Python to move columns:
pip install movecolumn
Then you can write your code as:
import movecolumn as mc
mc.MoveToLast(df,'b')
mc.MoveToLast(df,'x')
Hope that helps.
P.S : The package can be found here. https://pypi.org/project/movecolumn/

You can also do this as a one-liner:
df.drop(columns=['b', 'x']).assign(b=df['b'], x=df['x'])

This will move any column to the last column :
Move any column to the last column of dataframe :
df= df[ [ col for col in df.columns if col != 'col_name_to_moved' ] + ['col_name_to_moved']]
Move any column to the first column of dataframe:
df= df[ ['col_name_to_moved'] + [ col for col in df.columns if col != 'col_name_to_moved' ]]
where col_name_to_moved is the column that you want to move.

I use Pokémon database as an example, the columns for my data base are
['Name', '#', 'Type 1', 'Type 2', 'Total', 'HP', 'Attack', 'Defense', 'Sp. Atk', 'Sp. Def', 'Speed', 'Generation', 'Legendary']
Here is the code:
import pandas as pd
df = pd.read_html('https://gist.github.com/armgilles/194bcff35001e7eb53a2a8b441e8b2c6')[0]
cols = df.columns.to_list()
cos_end= ["Name", "Total", "HP", "Defense"]
for i, j in enumerate(cos_end, start=(len(cols)-len(cos_end))):
cols.insert(i, cols.pop(cols.index(j)))
print(cols)
df = df.reindex(columns=cols)
print(df)

Renaming column names in Pandas

I want to change the column labels of a Pandas DataFrame from
['$a', '$b', '$c', '$d', '$e']
to
['a', 'b', 'c', 'd', 'e']

Rename Specific Columns
Use the df.rename() function and refer the columns to be renamed. Not all the columns have to be renamed:
df = df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'})
# Or rename the existing DataFrame (rather than creating a copy)
df.rename(columns={'oldName1': 'newName1', 'oldName2': 'newName2'}, inplace=True)
Minimal Code Example
df = pd.DataFrame('x', index=range(3), columns=list('abcde'))
df
a b c d e
0 x x x x x
1 x x x x x
2 x x x x x
The following methods all work and produce the same output:
df2 = df.rename({'a': 'X', 'b': 'Y'}, axis=1) # new method
df2 = df.rename({'a': 'X', 'b': 'Y'}, axis='columns')
df2 = df.rename(columns={'a': 'X', 'b': 'Y'}) # old method
df2
X Y c d e
0 x x x x x
1 x x x x x
2 x x x x x
Remember to assign the result back, as the modification is not-inplace. Alternatively, specify inplace=True:
df.rename({'a': 'X', 'b': 'Y'}, axis=1, inplace=True)
df
X Y c d e
0 x x x x x
1 x x x x x
2 x x x x x
From v0.25, you can also specify errors='raise' to raise errors if an invalid column-to-rename is specified. See v0.25 rename() docs.
Reassign Column Headers
Use df.set_axis() with axis=1 and inplace=False (to return a copy).
df2 = df.set_axis(['V', 'W', 'X', 'Y', 'Z'], axis=1, inplace=False)
df2
V W X Y Z
0 x x x x x
1 x x x x x
2 x x x x x
This returns a copy, but you can modify the DataFrame in-place by setting inplace=True (this is the default behaviour for versions <=0.24 but is likely to change in the future).
You can also assign headers directly:
df.columns = ['V', 'W', 'X', 'Y', 'Z']
df
V W X Y Z
0 x x x x x
1 x x x x x
2 x x x x x

Just assign it to the .columns attribute:
>>> df = pd.DataFrame({'$a':[1,2], '$b': [10,20]})
>>> df
$a $b
0 1 10
1 2 20
>>> df.columns = ['a', 'b']
>>> df
a b
0 1 10
1 2 20

The rename method can take a function, for example:
In [11]: df.columns
Out[11]: Index([u'$a', u'$b', u'$c', u'$d', u'$e'], dtype=object)
In [12]: df.rename(columns=lambda x: x[1:], inplace=True)
In [13]: df.columns
Out[13]: Index([u'a', u'b', u'c', u'd', u'e'], dtype=object)

As documented in Working with text data:
df.columns = df.columns.str.replace('$', '')

Pandas 0.21+ Answer
There have been some significant updates to column renaming in version 0.21.
The rename method has added the axis parameter which may be set to columns or 1. This update makes this method match the rest of the pandas API. It still has the index and columns parameters but you are no longer forced to use them.
The set_axis method with the inplace set to False enables you to rename all the index or column labels with a list.
Examples for Pandas 0.21+
Construct sample DataFrame:
df = pd.DataFrame({'$a':[1,2], '$b': [3,4],
'$c':[5,6], '$d':[7,8],
'$e':[9,10]})
$a $b $c $d $e
0 1 3 5 7 9
1 2 4 6 8 10
Using rename with axis='columns' or axis=1
df.rename({'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'}, axis='columns')
or
df.rename({'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'}, axis=1)
Both result in the following:
a b c d e
0 1 3 5 7 9
1 2 4 6 8 10
It is still possible to use the old method signature:
df.rename(columns={'$a':'a', '$b':'b', '$c':'c', '$d':'d', '$e':'e'})
The rename function also accepts functions that will be applied to each column name.
df.rename(lambda x: x[1:], axis='columns')
or
df.rename(lambda x: x[1:], axis=1)
Using set_axis with a list and inplace=False
You can supply a list to the set_axis method that is equal in length to the number of columns (or index). Currently, inplace defaults to True, but inplace will be defaulted to False in future releases.
df.set_axis(['a', 'b', 'c', 'd', 'e'], axis='columns', inplace=False)
or
df.set_axis(['a', 'b', 'c', 'd', 'e'], axis=1, inplace=False)
Why not use df.columns = ['a', 'b', 'c', 'd', 'e']?
There is nothing wrong with assigning columns directly like this. It is a perfectly good solution.
The advantage of using set_axis is that it can be used as part of a method chain and that it returns a new copy of the DataFrame. Without it, you would have to store your intermediate steps of the chain to another variable before reassigning the columns.
# new for pandas 0.21+
df.some_method1()
.some_method2()
.set_axis()
.some_method3()
# old way
df1 = df.some_method1()
.some_method2()
df1.columns = columns
df1.some_method3()

Since you only want to remove the $ sign in all column names, you could just do:
df = df.rename(columns=lambda x: x.replace('$', ''))
OR
df.rename(columns=lambda x: x.replace('$', ''), inplace=True)

Renaming columns in Pandas is an easy task.
df.rename(columns={'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}, inplace=True)

df.columns = ['a', 'b', 'c', 'd', 'e']
It will replace the existing names with the names you provide, in the order you provide.

Use:
old_names = ['$a', '$b', '$c', '$d', '$e']
new_names = ['a', 'b', 'c', 'd', 'e']
df.rename(columns=dict(zip(old_names, new_names)), inplace=True)
This way you can manually edit the new_names as you wish. It works great when you need to rename only a few columns to correct misspellings, accents, remove special characters, etc.

One line or Pipeline solutions
I'll focus on two things:
OP clearly states
I have the edited column names stored it in a list, but I don't know how to replace the column names.
I do not want to solve the problem of how to replace '$' or strip the first character off of each column header. OP has already done this step. Instead I want to focus on replacing the existing columns object with a new one given a list of replacement column names.
df.columns = new where new is the list of new columns names is as simple as it gets. The drawback of this approach is that it requires editing the existing dataframe's columns attribute and it isn't done inline. I'll show a few ways to perform this via pipelining without editing the existing dataframe.
Setup 1
To focus on the need to rename of replace column names with a pre-existing list, I'll create a new sample dataframe df with initial column names and unrelated new column names.
df = pd.DataFrame({'Jack': [1, 2], 'Mahesh': [3, 4], 'Xin': [5, 6]})
new = ['x098', 'y765', 'z432']
df
Jack Mahesh Xin
0 1 3 5
1 2 4 6
Solution 1
pd.DataFrame.rename
It has been said already that if you had a dictionary mapping the old column names to new column names, you could use pd.DataFrame.rename.
d = {'Jack': 'x098', 'Mahesh': 'y765', 'Xin': 'z432'}
df.rename(columns=d)
x098 y765 z432
0 1 3 5
1 2 4 6
However, you can easily create that dictionary and include it in the call to rename. The following takes advantage of the fact that when iterating over df, we iterate over each column name.
# Given just a list of new column names
df.rename(columns=dict(zip(df, new)))
x098 y765 z432
0 1 3 5
1 2 4 6
This works great if your original column names are unique. But if they are not, then this breaks down.
Setup 2
Non-unique columns
df = pd.DataFrame(
[[1, 3, 5], [2, 4, 6]],
columns=['Mahesh', 'Mahesh', 'Xin']
)
new = ['x098', 'y765', 'z432']
df
Mahesh Mahesh Xin
0 1 3 5
1 2 4 6
Solution 2
pd.concat using the keys argument
First, notice what happens when we attempt to use solution 1:
df.rename(columns=dict(zip(df, new)))
y765 y765 z432
0 1 3 5
1 2 4 6
We didn't map the new list as the column names. We ended up repeating y765. Instead, we can use the keys argument of the pd.concat function while iterating through the columns of df.
pd.concat([c for _, c in df.items()], axis=1, keys=new)
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 3
Reconstruct. This should only be used if you have a single dtype for all columns. Otherwise, you'll end up with dtype object for all columns and converting them back requires more dictionary work.
Single dtype
pd.DataFrame(df.values, df.index, new)
x098 y765 z432
0 1 3 5
1 2 4 6
Mixed dtype
pd.DataFrame(df.values, df.index, new).astype(dict(zip(new, df.dtypes)))
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 4
This is a gimmicky trick with transpose and set_index. pd.DataFrame.set_index allows us to set an index inline, but there is no corresponding set_columns. So we can transpose, then set_index, and transpose back. However, the same single dtype versus mixed dtype caveat from solution 3 applies here.
Single dtype
df.T.set_index(np.asarray(new)).T
x098 y765 z432
0 1 3 5
1 2 4 6
Mixed dtype
df.T.set_index(np.asarray(new)).T.astype(dict(zip(new, df.dtypes)))
x098 y765 z432
0 1 3 5
1 2 4 6
Solution 5
Use a lambda in pd.DataFrame.rename that cycles through each element of new.
In this solution, we pass a lambda that takes x but then ignores it. It also takes a y but doesn't expect it. Instead, an iterator is given as a default value and I can then use that to cycle through one at a time without regard to what the value of x is.
df.rename(columns=lambda x, y=iter(new): next(y))
x098 y765 z432
0 1 3 5
1 2 4 6
And as pointed out to me by the folks in sopython chat, if I add a * in between x and y, I can protect my y variable. Though, in this context I don't believe it needs protecting. It is still worth mentioning.
df.rename(columns=lambda x, *, y=iter(new): next(y))
x098 y765 z432
0 1 3 5
1 2 4 6

Column names vs Names of Series
I would like to explain a bit what happens behind the scenes.
Dataframes are a set of Series.
Series in turn are an extension of a numpy.array.
numpy.arrays have a property .name.
This is the name of the series. It is seldom that Pandas respects this attribute, but it lingers in places and can be used to hack some Pandas behaviors.
Naming the list of columns
A lot of answers here talks about the df.columns attribute being a list when in fact it is a Series. This means it has a .name attribute.
This is what happens if you decide to fill in the name of the columns Series:
df.columns = ['column_one', 'column_two']
df.columns.names = ['name of the list of columns']
df.index.names = ['name of the index']
name of the list of columns column_one column_two
name of the index
0 4 1
1 5 2
2 6 3
Note that the name of the index always comes one column lower.
Artefacts that linger
The .name attribute lingers on sometimes. If you set df.columns = ['one', 'two'] then the df.one.name will be 'one'.
If you set df.one.name = 'three' then df.columns will still give you ['one', 'two'], and df.one.name will give you 'three'.
BUT
pd.DataFrame(df.one) will return
three
0 1
1 2
2 3
Because Pandas reuses the .name of the already defined Series.
Multi-level column names
Pandas has ways of doing multi-layered column names. There is not so much magic involved, but I wanted to cover this in my answer too since I don't see anyone picking up on this here.
|one |
|one |two |
0 | 4 | 1 |
1 | 5 | 2 |
2 | 6 | 3 |
This is easily achievable by setting columns to lists, like this:
df.columns = [['one', 'one'], ['one', 'two']]

Many of pandas functions have an inplace parameter. When setting it True, the transformation applies directly to the dataframe that you are calling it on. For example:
df = pd.DataFrame({'$a':[1,2], '$b': [3,4]})
df.rename(columns={'$a': 'a'}, inplace=True)
df.columns
>>> Index(['a', '$b'], dtype='object')
Alternatively, there are cases where you want to preserve the original dataframe. I have often seen people fall into this case if creating the dataframe is an expensive task. For example, if creating the dataframe required querying a snowflake database. In this case, just make sure the the inplace parameter is set to False.
df = pd.DataFrame({'$a':[1,2], '$b': [3,4]})
df2 = df.rename(columns={'$a': 'a'}, inplace=False)
df.columns
>>> Index(['$a', '$b'], dtype='object')
df2.columns
>>> Index(['a', '$b'], dtype='object')
If these types of transformations are something that you do often, you could also look into a number of different pandas GUI tools. I'm the creator of one called Mito. It’s a spreadsheet that automatically converts your edits to python code.

Let's understand renaming by a small example...
Renaming columns using mapping:
df = pd.DataFrame({"A": [1, 2, 3], "B": [4, 5, 6]}) # Creating a df with column name A and B
df.rename({"A": "new_a", "B": "new_b"}, axis='columns', inplace =True) # Renaming column A with 'new_a' and B with 'new_b'
Output:
new_a new_b
0 1 4
1 2 5
2 3 6
Renaming index/Row_Name using mapping:
df.rename({0: "x", 1: "y", 2: "z"}, axis='index', inplace =True) # Row name are getting replaced by 'x', 'y', and 'z'.
Output:
new_a new_b
x 1 4
y 2 5
z 3 6

Suppose your dataset name is df, and df has.
df = ['$a', '$b', '$c', '$d', '$e']`
So, to rename these, we would simply do.
df.columns = ['a','b','c','d','e']

Let's say this is your dataframe.
You can rename the columns using two methods.
Using dataframe.columns=[#list]
df.columns=['a','b','c','d','e']
The limitation of this method is that if one column has to be changed, full column list has to be passed. Also, this method is not applicable on index labels.
For example, if you passed this:
df.columns = ['a','b','c','d']
This will throw an error. Length mismatch: Expected axis has 5 elements, new values have 4 elements.
Another method is the Pandas rename() method which is used to rename any index, column or row
df = df.rename(columns={'$a':'a'})
Similarly, you can change any rows or columns.

If you've got the dataframe, df.columns dumps everything into a list you can manipulate and then reassign into your dataframe as the names of columns...
columns = df.columns
columns = [row.replace("$", "") for row in columns]
df.rename(columns=dict(zip(columns, things)), inplace=True)
df.head() # To validate the output
Best way? I don't know. A way - yes.
A better way of evaluating all the main techniques put forward in the answers to the question is below using cProfile to gage memory and execution time. #kadee, #kaitlyn, and #eumiro had the functions with the fastest execution times - though these functions are so fast we're comparing the rounding of 0.000 and 0.001 seconds for all the answers. Moral: my answer above likely isn't the 'best' way.
import pandas as pd
import cProfile, pstats, re
old_names = ['$a', '$b', '$c', '$d', '$e']
new_names = ['a', 'b', 'c', 'd', 'e']
col_dict = {'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}
df = pd.DataFrame({'$a':[1, 2], '$b': [10, 20], '$c': ['bleep', 'blorp'], '$d': [1, 2], '$e': ['texa$', '']})
df.head()
def eumiro(df, nn):
df.columns = nn
# This direct renaming approach is duplicated in methodology in several other answers:
return df
def lexual1(df):
return df.rename(columns=col_dict)
def lexual2(df, col_dict):
return df.rename(columns=col_dict, inplace=True)
def Panda_Master_Hayden(df):
return df.rename(columns=lambda x: x[1:], inplace=True)
def paulo1(df):
return df.rename(columns=lambda x: x.replace('$', ''))
def paulo2(df):
return df.rename(columns=lambda x: x.replace('$', ''), inplace=True)
def migloo(df, on, nn):
return df.rename(columns=dict(zip(on, nn)), inplace=True)
def kadee(df):
return df.columns.str.replace('$', '')
def awo(df):
columns = df.columns
columns = [row.replace("$", "") for row in columns]
return df.rename(columns=dict(zip(columns, '')), inplace=True)
def kaitlyn(df):
df.columns = [col.strip('$') for col in df.columns]
return df
print 'eumiro'
cProfile.run('eumiro(df, new_names)')
print 'lexual1'
cProfile.run('lexual1(df)')
print 'lexual2'
cProfile.run('lexual2(df, col_dict)')
print 'andy hayden'
cProfile.run('Panda_Master_Hayden(df)')
print 'paulo1'
cProfile.run('paulo1(df)')
print 'paulo2'
cProfile.run('paulo2(df)')
print 'migloo'
cProfile.run('migloo(df, old_names, new_names)')
print 'kadee'
cProfile.run('kadee(df)')
print 'awo'
cProfile.run('awo(df)')
print 'kaitlyn'
cProfile.run('kaitlyn(df)')

df = pd.DataFrame({'$a': [1], '$b': [1], '$c': [1], '$d': [1], '$e': [1]})
If your new list of columns is in the same order as the existing columns, the assignment is simple:
new_cols = ['a', 'b', 'c', 'd', 'e']
df.columns = new_cols
>>> df
a b c d e
0 1 1 1 1 1
If you had a dictionary keyed on old column names to new column names, you could do the following:
d = {'$a': 'a', '$b': 'b', '$c': 'c', '$d': 'd', '$e': 'e'}
df.columns = df.columns.map(lambda col: d[col]) # Or `.map(d.get)` as pointed out by #PiRSquared.
>>> df
a b c d e
0 1 1 1 1 1
If you don't have a list or dictionary mapping, you could strip the leading $ symbol via a list comprehension:
df.columns = [col[1:] if col[0] == '$' else col for col in df]

df.rename(index=str, columns={'A':'a', 'B':'b'})
pandas.DataFrame.rename

If you already have a list for the new column names, you can try this:
new_cols = ['a', 'b', 'c', 'd', 'e']
new_names_map = {df.columns[i]:new_cols[i] for i in range(len(new_cols))}
df.rename(new_names_map, axis=1, inplace=True)

Another way we could replace the original column labels is by stripping the unwanted characters (here '$') from the original column labels.
This could have been done by running a for loop over df.columns and appending the stripped columns to df.columns.
Instead, we can do this neatly in a single statement by using list comprehension like below:
df.columns = [col.strip('$') for col in df.columns]
(strip method in Python strips the given character from beginning and end of the string.)

It is real simple. Just use:
df.columns = ['Name1', 'Name2', 'Name3'...]
And it will assign the column names by the order you put them in.

# This way it will work
import pandas as pd
# Define a dictionary
rankings = {'test': ['a'],
'odi': ['E'],
't20': ['P']}
# Convert the dictionary into DataFrame
rankings_pd = pd.DataFrame(rankings)
# Before renaming the columns
print(rankings_pd)
rankings_pd.rename(columns = {'test':'TEST'}, inplace = True)

You could use str.slice for that:
df.columns = df.columns.str.slice(1)

Another option is to rename using a regular expression:
import pandas as pd
import re
df = pd.DataFrame({'$a':[1,2], '$b':[3,4], '$c':[5,6]})
df = df.rename(columns=lambda x: re.sub('\$','',x))
>>> df
a b c
0 1 3 5
1 2 4 6

My method is generic wherein you can add additional delimiters by comma separating delimiters= variable and future-proof it.
Working Code:
import pandas as pd
import re
df = pd.DataFrame({'$a':[1,2], '$b': [3,4],'$c':[5,6], '$d': [7,8], '$e': [9,10]})
delimiters = '$'
matchPattern = '|'.join(map(re.escape, delimiters))
df.columns = [re.split(matchPattern, i)[1] for i in df.columns ]
Output:
>>> df
$a $b $c $d $e
0 1 3 5 7 9
1 2 4 6 8 10
>>> df
a b c d e
0 1 3 5 7 9
1 2 4 6 8 10

Note that the approaches in previous answers do not work for a MultiIndex. For a MultiIndex, you need to do something like the following:
>>> df = pd.DataFrame({('$a','$x'):[1,2], ('$b','$y'): [3,4], ('e','f'):[5,6]})
>>> df
$a $b e
$x $y f
0 1 3 5
1 2 4 6
>>> rename = {('$a','$x'):('a','x'), ('$b','$y'):('b','y')}
>>> df.columns = pandas.MultiIndex.from_tuples([
rename.get(item, item) for item in df.columns.tolist()])
>>> df
a b e
x y f
0 1 3 5
1 2 4 6

If you have to deal with loads of columns named by the providing system out of your control, I came up with the following approach that is a combination of a general approach and specific replacements in one go.
First create a dictionary from the dataframe column names using regular expressions in order to throw away certain appendixes of column names and then add specific replacements to the dictionary to name core columns as expected later in the receiving database.
This is then applied to the dataframe in one go.
dict = dict(zip(df.columns, df.columns.str.replace('(:S$|:C1$|:L$|:D$|\.Serial:L$)', '')))
dict['brand_timeseries:C1'] = 'BTS'
dict['respid:L'] = 'RespID'
dict['country:C1'] = 'CountryID'
dict['pim1:D'] = 'pim_actual'
df.rename(columns=dict, inplace=True)

If you just want to remove the '$' sign then use the below code
df.columns = pd.Series(df.columns.str.replace("$", ""))

In addition to the solution already provided, you can replace all the columns while you are reading the file. We can use names and header=0 to do that.
First, we create a list of the names that we like to use as our column names:
import pandas as pd
ufo_cols = ['city', 'color reported', 'shape reported', 'state', 'time']
ufo.columns = ufo_cols
ufo = pd.read_csv('link to the file you are using', names = ufo_cols, header = 0)
In this case, all the column names will be replaced with the names you have in your list.

Here's a nifty little function I like to use to cut down on typing:
def rename(data, oldnames, newname):
if type(oldnames) == str: # Input can be a string or list of strings
oldnames = [oldnames] # When renaming multiple columns
newname = [newname] # Make sure you pass the corresponding list of new names
i = 0
for name in oldnames:
oldvar = [c for c in data.columns if name in c]
if len(oldvar) == 0:
raise ValueError("Sorry, couldn't find that column in the dataset")
if len(oldvar) > 1: # Doesn't have to be an exact match
print("Found multiple columns that matched " + str(name) + ": ")
for c in oldvar:
print(str(oldvar.index(c)) + ": " + str(c))
ind = input('Please enter the index of the column you would like to rename: ')
oldvar = oldvar[int(ind)]
if len(oldvar) == 1:
oldvar = oldvar[0]
data = data.rename(columns = {oldvar : newname[i]})
i += 1
return data
Here is an example of how it works:
In [2]: df = pd.DataFrame(np.random.randint(0, 10, size=(10, 4)), columns = ['col1', 'col2', 'omg', 'idk'])
# First list = existing variables
# Second list = new names for those variables
In [3]: df = rename(df, ['col', 'omg'],['first', 'ohmy'])
Found multiple columns that matched col:
0: col1
1: col2
Please enter the index of the column you would like to rename: 0
In [4]: df.columns
Out[5]: Index(['first', 'col2', 'ohmy', 'idk'], dtype='object')