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How to replace data in one pandas df by the data of another one?

Want to replace some rows of some columns in a bigger pandas df by data in a smaller pandas df. The column names are same in both.
Tried using combine_first but it only updates the null values.
For example lets say df1.shape is 100, 25 and df2.shape is 10,5
df1
A B C D E F G ...Z Y Z
1 abc 10.20 0 pd.NaT
df2
A B C D E
1 abc 15.20 1 10
Now after replacing df1 should look like:
A B C D E F G ...Z Y Z
1 abc 15.20 1 10 ...
To replace values in df1 the condition is where df1.A = df2.A and df1.B = df2.B
How can it be achieved in the most pythonic way? Any help will be appreciated.

Don't know I really understood your question does this solves your problem ?
df1 = pd.DataFrame(data={'A':[1],'B':[2],'C':[3],'D':[4]})
df2 = pd.DataFrame(data={'A':[1],'B':[2],'C':[5],'D':[6]})
new_df=pd.concat([df1,df2]).drop_duplicates(['A','B'],keep='last')
print(new_df)
output:
A B C D
0 1 2 5 6

You could play with Multiindex.
First let us create those dataframe that you are working with:
cols = pd.Index(list(ascii_uppercase))
vals = np.arange(100*len(cols)).reshape(100, len(cols))
df = pd.DataFrame(vals, columns=cols)
df1 = pd.DataFrame(vals[:10,:5], columns=cols[:5])
Then transform A and B in indices:
df = df.set_index(["A","B"])
df1 = df1.set_index(["A","B"])*1.5 # multiply just to make the other values different
df.loc[df1.index, df1.columns] = df1
df = df.reset_index()

How to remove double quotes while assigning columns to dataframe

I have below list
ColumnName = 'Emp_id','Emp_Name','EmpAGe'
While i am trying to read above columns and assign inside dataframe i am getting extra double quotes
df = pd.dataframe(data,columns=[ColumnName])
columns=[ColumnName]
i am getting columns = ["'Emp_id','Emp_Name','EmpAGe'"]
how can i handle these extra double quotes and remove them while assigning header to data

This code
ColumnName = 'Emp_id','Emp_Name','EmpAGe'
Is a tuple and not a list.
In case you want three columns, each with values on the tuple above you gonna need
df = pd.dataframe(data,columns=list(ColumnName))

The problem is how you define the columns for pandas DataFrame.
The example below will build a correct data frame :
import pandas as pd
ColumnName1 = 'Emp_id','Emp_Name','EmpAGe'
df1 = [['A1','A1','A2'],['1','2','1'],['a0','a1','a3']]
df = pd.DataFrame(data=df1,columns=ColumnName1 )
df
Result :
Emp_id Emp_Name EmpAGe
0 A1 A1 A2
1 1 2 1
2 a0 a1 a3
A print screen of the code I wrote with the result, with no double quotations

Just for the shake of the understanding, where you can use col.replace to get the desired ..
Let take an example..
>>> df
col1" col2"
0 1 1
1 2 2
Result:
>>> df.columns = [col.replace('"', '') for col in df.columns]
# df.columns = df.columns.str.replace('"', '') <-- can use this as well
>>> df
col1 col2
0 1 1
1 2 2
OR
>>> df = pd.DataFrame({ '"col1"':[1, 2], '"col2"':[1,2]})
>>> df
"col1" "col2"
0 1 1
1 2 2
>>> df.columns = [col.replace('"', '') for col in df.columns]
>>> df
col1 col2
0 1 1
1 2 2

Your input is not quite right. ColumnName is already list-like and it should be passed on directly rather than wrapped in another list. In the latter case it would be interpreted as one single column.
df = pd.DataFrame(data, columns=ColumnName)

List of list of tuples to pandas dataframe

I have this array (it's a result from similarity calcul) it's a list of tuples like this:
example = [[(a,b), (c,d)], [(a1,b1), (c1,d2)] …]
In example there is 121044 list of 30 tuples each.
I want to have a pandas Dataframe like of just the second value of the tuples (i.e : b, d, b1, d2) without spending to much time compute it
Do you have any ideas ?

Use nested list comprehension:
df = pd.DataFrame([[y[1] for y in x] for x in example])
print (df)
0 1
0 b d
1 b1 d2
df = pd.DataFrame([[y[1] for y in x] for x in example], columns=['col1','col2'])
print (df)
col1 col2
0 b d
1 b1 d2

For numeric data, you can use numpy indexing directly. This should be more efficient than a list comprehension, as pandas uses numpy internally to store data in contiguous memory blocks.
import pandas as pd, numpy as np
example = [[(1,2), (3,4)], [(5,6), (7,8)]]
df = pd.DataFrame(np.array(example)[..., 1],
columns=['col1', 'col2'])
print(df)
col1 col2
0 2 4
1 6 8

Pandas - Interleave / Zip two DataFrames by row

Suppose I have two dataframes:
>> df1
0 1 2
0 a b c
1 d e f
>> df2
0 1 2
0 A B C
1 D E F
How can I interleave the rows? i.e. get this:
>> interleaved_df
0 1 2
0 a b c
1 A B C
2 d e f
3 D E F
(Note my real DFs have identical columns, but not the same number of rows).
What I've tried
inspired by this question (very similar, but asks on columns):
import pandas as pd
from itertools import chain, zip_longest
df1 = pd.DataFrame([['a','b','c'], ['d','e','f']])
df2 = pd.DataFrame([['A','B','C'], ['D','E','F']])
concat_df = pd.concat([df1,df2])
new_index = chain.from_iterable(zip_longest(df1.index, df2.index))
# new_index now holds the interleaved row indices
interleaved_df = concat_df.reindex(new_index)
ValueError: cannot reindex from a duplicate axis
The last call fails because df1 and df2 have some identical index values (which is also the case with my real DFs).
Any ideas?

You can sort the index after concatenating and then reset the index i.e
import pandas as pd
df1 = pd.DataFrame([['a','b','c'], ['d','e','f']])
df2 = pd.DataFrame([['A','B','C'], ['D','E','F']])
concat_df = pd.concat([df1,df2]).sort_index().reset_index(drop=True)
Output :
0 1 2
0 a b c
1 A B C
2 d e f
3 D E F
EDIT (OmerB) : Incase of keeping the order regardless of the index value then.
import pandas as pd
df1 = pd.DataFrame([['a','b','c'], ['d','e','f']]).reset_index()
df2 = pd.DataFrame([['A','B','C'], ['D','E','F']]).reset_index()
concat_df = pd.concat([df1,df2]).sort_index().set_index('index')

Use toolz.interleave
In [1024]: from toolz import interleave
In [1025]: pd.DataFrame(interleave([df1.values, df2.values]))
Out[1025]:
0 1 2
0 a b c
1 A B C
2 d e f
3 D E F

Here's an extension of #Bharath's answer that can be applied to DataFrames with user-defined indexes without losing them, using pd.MultiIndex.
Define Dataframes with the full set of column/ index labels and names:
df1 = pd.DataFrame([['a','b','c'], ['d','e','f']], index=['one', 'two'], columns=['col_a', 'col_b','col_c'])
df1.columns.name = 'cols'
df1.index.name = 'rows'
df2 = pd.DataFrame([['A','B','C'], ['D','E','F']], index=['one', 'two'], columns=['col_a', 'col_b','col_c'])
df2.columns.name = 'cols'
df2.index.name = 'rows'
Add DataFrame ID to MultiIndex:
df1.index = pd.MultiIndex.from_product([[1], df1.index], names=["df_id", df1.index.name])
df2.index = pd.MultiIndex.from_product([[2], df2.index], names=["df_id", df2.index.name])
Then use #Bharath's concat() and sort_index():
data = pd.concat([df1, df2], axis=0, sort=True)
data.sort_index(axis=0, level=data.index.names[::-1], inplace=True)
Output:
cols col_a col_b col_c
df_id rows
1 one a b c
2 one A B C
1 two d e f
2 two D E F

You could also preallocate a new DataFrame, and then fill it using a slice.
def interleave(dfs):
data = np.transpose(np.array([np.empty(dfs[0].shape[0]*len(dfs), dtype=dt) for dt in dfs[0].dtypes]))
out = pd.DataFrame(data, columns=dfs[0].columns)
for ix, df in enumerate(dfs):
out.iloc[ix::len(dfs),:] = df.values
return out
The preallocation code is taken from this question.
While there's a chance it could outperform the index method for certain data types / sizes, it won't behave gracefully if the DataFrames have different sizes.
Note - for ~200000 rows with 20 columns of mixed string, integer and floating types, the index method is around 5x faster.

You can try this way :
In [31]: import pandas as pd
...: from itertools import chain, zip_longest
...:
...: df1 = pd.DataFrame([['a','b','c'], ['d','e','f']])
...: df2 = pd.DataFrame([['A','B','C'], ['D','E','F']])
In [32]: concat_df = pd.concat([df1,df2]).sort_index()
...:
In [33]: interleaved_df = concat_df.reset_index(drop=1)
In [34]: interleaved_df
Out[34]:
0 1 2
0 a b c
1 A B C
2 d e f
3 D E F

Appending a list or series to a pandas DataFrame as a row?

So I have initialized an empty pandas DataFrame and I would like to iteratively append lists (or Series) as rows in this DataFrame. What is the best way of doing this?

df = pd.DataFrame(columns=list("ABC"))
df.loc[len(df)] = [1,2,3]

Sometimes it's easier to do all the appending outside of pandas, then, just create the DataFrame in one shot.
>>> import pandas as pd
>>> simple_list=[['a','b']]
>>> simple_list.append(['e','f'])
>>> df=pd.DataFrame(simple_list,columns=['col1','col2'])
col1 col2
0 a b
1 e f

Here's a simple and dumb solution:
>>> import pandas as pd
>>> df = pd.DataFrame()
>>> df = df.append({'foo':1, 'bar':2}, ignore_index=True)

Could you do something like this?
>>> import pandas as pd
>>> df = pd.DataFrame(columns=['col1', 'col2'])
>>> df = df.append(pd.Series(['a', 'b'], index=['col1','col2']), ignore_index=True)
>>> df = df.append(pd.Series(['d', 'e'], index=['col1','col2']), ignore_index=True)
>>> df
col1 col2
0 a b
1 d e
Does anyone have a more elegant solution?

Following onto Mike Chirico's answer... if you want to append a list after the dataframe is already populated...
>>> list = [['f','g']]
>>> df = df.append(pd.DataFrame(list, columns=['col1','col2']),ignore_index=True)
>>> df
col1 col2
0 a b
1 d e
2 f g

There are several ways to append a list to a Pandas Dataframe in Python. Let's consider the following dataframe and list:
import pandas as pd
# Dataframe
df = pd.DataFrame([[1, 2], [3, 4]], columns = ["col1", "col2"])
# List to append
list = [5, 6]
Option 1: append the list at the end of the dataframe with pandas.DataFrame.loc.
df.loc[len(df)] = list
Option 2: convert the list to dataframe and append with pandas.DataFrame.append().
df = df.append(pd.DataFrame([list], columns=df.columns), ignore_index=True)
Option 3: convert the list to series and append with pandas.DataFrame.append().
df = df.append(pd.Series(list, index = df.columns), ignore_index=True)
Each of the above options should output something like:
>>> print (df)
col1 col2
0 1 2
1 3 4
2 5 6
Reference : How to append a list as a row to a Pandas DataFrame in Python?

Converting the list to a data frame within the append function works, also when applied in a loop
import pandas as pd
mylist = [1,2,3]
df = pd.DataFrame()
df = df.append(pd.DataFrame(data[mylist]))

Here's a function that, given an already created dataframe, will append a list as a new row. This should probably have error catchers thrown in, but if you know exactly what you're adding then it shouldn't be an issue.
import pandas as pd
import numpy as np
def addRow(df,ls):
"""
Given a dataframe and a list, append the list as a new row to the dataframe.
:param df: <DataFrame> The original dataframe
:param ls: <list> The new row to be added
:return: <DataFrame> The dataframe with the newly appended row
"""
numEl = len(ls)
newRow = pd.DataFrame(np.array(ls).reshape(1,numEl), columns = list(df.columns))
df = df.append(newRow, ignore_index=True)
return df

If you want to add a Series and use the Series' index as columns of the DataFrame, you only need to append the Series between brackets:
In [1]: import pandas as pd
In [2]: df = pd.DataFrame()
In [3]: row=pd.Series([1,2,3],["A","B","C"])
In [4]: row
Out[4]:
A 1
B 2
C 3
dtype: int64
In [5]: df.append([row],ignore_index=True)
Out[5]:
A B C
0 1 2 3
[1 rows x 3 columns]
Whitout the ignore_index=True you don't get proper index.

simply use loc:
>>> df
A B C
one 1 2 3
>>> df.loc["two"] = [4,5,6]
>>> df
A B C
one 1 2 3
two 4 5 6

As mentioned here - https://kite.com/python/answers/how-to-append-a-list-as-a-row-to-a-pandas-dataframe-in-python, you'll need to first convert the list to a series then append the series to dataframe.
df = pd.DataFrame([[1, 2], [3, 4]], columns = ["a", "b"])
to_append = [5, 6]
a_series = pd.Series(to_append, index = df.columns)
df = df.append(a_series, ignore_index=True)

Consider an array A of N x 2 dimensions. To add one more row, use the following.
A.loc[A.shape[0]] = [3,4]

The simplest way:
my_list = [1,2,3,4,5]
df['new_column'] = pd.Series(my_list).values
Edit:
Don't forget that the length of the new list should be the same of the corresponding Dataframe.