I import a CSV as a DataFrame using:
import numpy as np
import pandas as pd
df = pd.read_csv("test.csv")
Then I'm trying to do a simple replace based on IDs:
df.loc[df.ID == 103, ['fname', 'lname']] = 'Michael', 'Johnson'
I get the following error:
AttributeError: 'list' object has no attribute 'loc'
Note, when I do print pd.version() I get 0.12.0, so it's not a problem (at least as far as I understand) with having pre-11 version. Any ideas?
To pickup from the comment: "I was doing this:"
df = [df.hc== 2]
What you create there is a "mask": an array with booleans that says which part of the index fulfilled your condition.
To filter your dataframe on your condition you want to do this:
df = df[df.hc == 2]
A bit more explicit is this:
mask = df.hc == 2
df = df[mask]
If you want to keep the entire dataframe and only want to replace specific values, there are methods such replace: Python pandas equivalent for replace. Also another (performance wise great) method would be creating a separate DataFrame with the from/to values as column and using pd.merge to combine it into the existing DataFrame. And using your index to set values is also possible:
df[mask]['fname'] = 'Johnson'
But for a larger set of replaces you would want to use one of the two other methods or use "apply" with a lambda function (for value transformations). Last but not least: you can use .fillna('bla') to rapidly fill up NA values.
The traceback indicates to you that df is a list and not a DataFrame as expected in your line of code.
It means that between df = pd.read_csv("test.csv") and df.loc[df.ID == 103, ['fname', 'lname']] = 'Michael', 'Johnson' you have other lines of codes that assigns a list object to df. Review that piece of code to find your bug
#Boud answer is correct. Loc assignment works fine if the right-hand-side list matches the number of replacing elements
In [56]: df = DataFrame(dict(A =[1,2,3], B = [4,5,6], C = [7,8,9]))
In [57]: df
Out[57]:
A B C
0 1 4 7
1 2 5 8
2 3 6 9
In [58]: df.loc[1,['A','B']] = -1,-2
In [59]: df
Out[59]:
A B C
0 1 4 7
1 -1 -2 8
2 3 6 9
Related
I'm trying to using pandas to append a blank row based on the values in the first column. When the first six characters in the first column don't match, I want an empty row between them (effectively creating groups). Here is an example of what the output could look like:
002446
002447-01
002447-02
002448
This is what I was able to put together thus far.
readie=pd.read_csv('title.csv')
i=0
for row in readie:
readie.append(row)
i+=1
if readie['column title'][i][0:5]!=readie['column title'][i+1][0:5]:
readie.append([])
When running this code, I get the following error message:
TypeError: cannot concatenate object of type '<class 'str'>'; only Series and DataFrame objs are valid
I believe there are other ways to do this, but I would like to use pandas if at all possible.
I'm using the approach from this answer.
Assuming that strings like '123456' and '123' are considered as not matching:
df_orig = pd.DataFrame(
{'col':['002446','002447-01','002447-02','002448','00244','002448']}
)
df = df_orig.reset_index(drop=True) # reset your index
first_6 = df['col'].str.slice(stop=6)
mask = first_6 != first_6.shift(fill_value=first_6[0])
df.index = df.index + mask.cumsum()
df = df.reindex(range(df.index[-1] + 1))
print(df)
col
0 002446
1 NaN
2 002447-01
3 002447-02
4 NaN
5 002448
6 NaN
7 00244
8 NaN
9 002448
I have a DataFrame object df. And I would like to modify job column so that all retired people are 1 and rest 0 (like shown here):
df['job'] = df['job'].apply(lambda x: 1 if x == "retired" else 0)
But I get a warning:
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
Why did I get it here though? From what I read it applies to situations where I take a slice of rows and then a column, but here I am just modyfing elements in a row. Is there a better way to do that?
Use:
df['job']=df['job'].eq('retired').astype(int)
or
df['job']=np.where(df['job'].eq('retired'),1,0)
So here's an example dataframe:
import pandas as pd
import numpy as np
data = {'job':['retired', 'a', 'b', 'retired']}
df = pd.DataFrame(data)
print(df)
job
0 retired
1 a
2 b
3 retired
Now, you can make use of numpy's where function:
df['job'] = np.where(df['job']=='retired', 1, 0)
print(df)
job
0 1
1 0
2 0
3 1
I would not suggest using apply here, as in the case of large data frame it could lower your performance.
I would prefer using numpy.select or numpy.where.
See This And This
I'm not used to pandas at all, thus the several question on my problem.
I have a function computing computing a list called solutions. This list can either be made of tuples of 3 values (a, b, c) or empty.
solutions = [(a,b,c), (d,e,f), (g,h,i)]
To save it, I first turn it into a numpy array, and then I save it with pandas after naming the columns.
solutions = np.asarray(solutions)
df = pd.DataFrame(solutions)
df.columns = ["Name1", "Name2", "Name3"]
df.to_pickle(path)
My issue is that I sometimes have a empty solutions list: solutions = []. Thus, the line df.columns raises an error. To bypass it, I currently check the size of solutions, and if it is empty, I do:
pickle.dump([], path, "wb")
I would like to be a more consistent between my data type, and to save the SAME format between both scenario.
=> If the list is empty, I would like to save the 3 columns name with an empty data frame. Ultimate goal, is to reopen the file with pd.read_pickle() and to access easily the data in it.
Second issue, I would like to reopen the files pickled, and to add a column. Could you show me the right way to do so?
And third question, how can I select a part of the dataframe. For instance, I want all lines in which the column Name1 value % 0.25 == 0.
Thanks
Create your dataframe using:
df = pandas.DataFrame(data=solutions, columns=['name1', 'name2', 'name3'])
If solutions is empty, it will nevertheless create a dataframe with 3 columns and 0 row.
In [2]: pd.DataFrame(data=[(1,2,3), (4,5,6)], columns=['a','b','c'])
Out[2]:
a b c
0 1 2 3
1 4 5 6
In [3]: pd.DataFrame(data=[], columns=['a','b','c'])
Out[3]:
Empty DataFrame
Columns: [a, b, c]
Index: []
For your third question:
df["Name1"] % 0.25 == 0
computes a series of booleans which are true where the value in the first column can be divided by 0.25. You can use it to select the rows of your dataframe:
df[ df["Name1"] % 0.25 == 0 ]
How can I find the row for which the value of a specific column is maximal?
df.max() will give me the maximal value for each column, I don't know how to get the corresponding row.
Use the pandas idxmax function. It's straightforward:
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].idxmax()
3
>>> df['B'].idxmax()
4
>>> df['C'].idxmax()
1
Alternatively you could also use numpy.argmax, such as numpy.argmax(df['A']) -- it provides the same thing, and appears at least as fast as idxmax in cursory observations.
idxmax() returns indices labels, not integers.
Example': if you have string values as your index labels, like rows 'a' through 'e', you might want to know that the max occurs in row 4 (not row 'd').
if you want the integer position of that label within the Index you have to get it manually (which can be tricky now that duplicate row labels are allowed).
HISTORICAL NOTES:
idxmax() used to be called argmax() prior to 0.11
argmax was deprecated prior to 1.0.0 and removed entirely in 1.0.0
back as of Pandas 0.16, argmax used to exist and perform the same function (though appeared to run more slowly than idxmax).
argmax function returned the integer position within the index of the row location of the maximum element.
pandas moved to using row labels instead of integer indices. Positional integer indices used to be very common, more common than labels, especially in applications where duplicate row labels are common.
For example, consider this toy DataFrame with a duplicate row label:
In [19]: dfrm
Out[19]:
A B C
a 0.143693 0.653810 0.586007
b 0.623582 0.312903 0.919076
c 0.165438 0.889809 0.000967
d 0.308245 0.787776 0.571195
e 0.870068 0.935626 0.606911
f 0.037602 0.855193 0.728495
g 0.605366 0.338105 0.696460
h 0.000000 0.090814 0.963927
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
In [20]: dfrm['A'].idxmax()
Out[20]: 'i'
In [21]: dfrm.iloc[dfrm['A'].idxmax()] # .ix instead of .iloc in older versions of pandas
Out[21]:
A B C
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
So here a naive use of idxmax is not sufficient, whereas the old form of argmax would correctly provide the positional location of the max row (in this case, position 9).
This is exactly one of those nasty kinds of bug-prone behaviors in dynamically typed languages that makes this sort of thing so unfortunate, and worth beating a dead horse over. If you are writing systems code and your system suddenly gets used on some data sets that are not cleaned properly before being joined, it's very easy to end up with duplicate row labels, especially string labels like a CUSIP or SEDOL identifier for financial assets. You can't easily use the type system to help you out, and you may not be able to enforce uniqueness on the index without running into unexpectedly missing data.
So you're left with hoping that your unit tests covered everything (they didn't, or more likely no one wrote any tests) -- otherwise (most likely) you're just left waiting to see if you happen to smack into this error at runtime, in which case you probably have to go drop many hours worth of work from the database you were outputting results to, bang your head against the wall in IPython trying to manually reproduce the problem, finally figuring out that it's because idxmax can only report the label of the max row, and then being disappointed that no standard function automatically gets the positions of the max row for you, writing a buggy implementation yourself, editing the code, and praying you don't run into the problem again.
You might also try idxmax:
In [5]: df = pandas.DataFrame(np.random.randn(10,3),columns=['A','B','C'])
In [6]: df
Out[6]:
A B C
0 2.001289 0.482561 1.579985
1 -0.991646 -0.387835 1.320236
2 0.143826 -1.096889 1.486508
3 -0.193056 -0.499020 1.536540
4 -2.083647 -3.074591 0.175772
5 -0.186138 -1.949731 0.287432
6 -0.480790 -1.771560 -0.930234
7 0.227383 -0.278253 2.102004
8 -0.002592 1.434192 -1.624915
9 0.404911 -2.167599 -0.452900
In [7]: df.idxmax()
Out[7]:
A 0
B 8
C 7
e.g.
In [8]: df.loc[df['A'].idxmax()]
Out[8]:
A 2.001289
B 0.482561
C 1.579985
Both above answers would only return one index if there are multiple rows that take the maximum value. If you want all the rows, there does not seem to have a function.
But it is not hard to do. Below is an example for Series; the same can be done for DataFrame:
In [1]: from pandas import Series, DataFrame
In [2]: s=Series([2,4,4,3],index=['a','b','c','d'])
In [3]: s.idxmax()
Out[3]: 'b'
In [4]: s[s==s.max()]
Out[4]:
b 4
c 4
dtype: int64
df.iloc[df['columnX'].argmax()]
argmax() would provide the index corresponding to the max value for the columnX. iloc can be used to get the row of the DataFrame df for this index.
A more compact and readable solution using query() is like this:
import pandas as pd
df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
print(df)
# find row with maximum A
df.query('A == A.max()')
It also returns a DataFrame instead of Series, which would be handy for some use cases.
Very simple: we have df as below and we want to print a row with max value in C:
A B C
x 1 4
y 2 10
z 5 9
In:
df.loc[df['C'] == df['C'].max()] # condition check
Out:
A B C
y 2 10
If you want the entire row instead of just the id, you can use df.nlargest and pass in how many 'top' rows you want and you can also pass in for which column/columns you want it for.
df.nlargest(2,['A'])
will give you the rows corresponding to the top 2 values of A.
use df.nsmallest for min values.
The direct ".argmax()" solution does not work for me.
The previous example provided by #ely
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].argmax()
3
>>> df['B'].argmax()
4
>>> df['C'].argmax()
1
returns the following message :
FutureWarning: 'argmax' is deprecated, use 'idxmax' instead. The behavior of 'argmax'
will be corrected to return the positional maximum in the future.
Use 'series.values.argmax' to get the position of the maximum now.
So that my solution is :
df['A'].values.argmax()
mx.iloc[0].idxmax()
This one line of code will give you how to find the maximum value from a row in dataframe, here mx is the dataframe and iloc[0] indicates the 0th index.
Considering this dataframe
[In]: df = pd.DataFrame(np.random.randn(4,3),columns=['A','B','C'])
[Out]:
A B C
0 -0.253233 0.226313 1.223688
1 0.472606 1.017674 1.520032
2 1.454875 1.066637 0.381890
3 -0.054181 0.234305 -0.557915
Assuming one want to know the rows where column "C" is max, the following will do the work
[In]: df[df['C']==df['C'].max()])
[Out]:
A B C
1 0.472606 1.017674 1.520032
The idmax of the DataFrame returns the label index of the row with the maximum value and the behavior of argmax depends on version of pandas (right now it returns a warning). If you want to use the positional index, you can do the following:
max_row = df['A'].values.argmax()
or
import numpy as np
max_row = np.argmax(df['A'].values)
Note that if you use np.argmax(df['A']) behaves the same as df['A'].argmax().
Use:
data.iloc[data['A'].idxmax()]
data['A'].idxmax() -finds max value location in terms of row
data.iloc() - returns the row
If there are ties in the maximum values, then idxmax returns the index of only the first max value. For example, in the following DataFrame:
A B C
0 1 0 1
1 0 0 1
2 0 0 0
3 0 1 1
4 1 0 0
idxmax returns
A 0
B 3
C 0
dtype: int64
Now, if we want all indices corresponding to max values, then we could use max + eq to create a boolean DataFrame, then use it on df.index to filter out indexes:
out = df.eq(df.max()).apply(lambda x: df.index[x].tolist())
Output:
A [0, 4]
B [3]
C [0, 1, 3]
dtype: object
what worked for me is:
df[df['colX'] == df['colX'].max()
You then get the row in your df with the maximum value of colX.
Then if you just want the index you can add .index at the end of the query.
How can I find the row for which the value of a specific column is maximal?
df.max() will give me the maximal value for each column, I don't know how to get the corresponding row.
Use the pandas idxmax function. It's straightforward:
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].idxmax()
3
>>> df['B'].idxmax()
4
>>> df['C'].idxmax()
1
Alternatively you could also use numpy.argmax, such as numpy.argmax(df['A']) -- it provides the same thing, and appears at least as fast as idxmax in cursory observations.
idxmax() returns indices labels, not integers.
Example': if you have string values as your index labels, like rows 'a' through 'e', you might want to know that the max occurs in row 4 (not row 'd').
if you want the integer position of that label within the Index you have to get it manually (which can be tricky now that duplicate row labels are allowed).
HISTORICAL NOTES:
idxmax() used to be called argmax() prior to 0.11
argmax was deprecated prior to 1.0.0 and removed entirely in 1.0.0
back as of Pandas 0.16, argmax used to exist and perform the same function (though appeared to run more slowly than idxmax).
argmax function returned the integer position within the index of the row location of the maximum element.
pandas moved to using row labels instead of integer indices. Positional integer indices used to be very common, more common than labels, especially in applications where duplicate row labels are common.
For example, consider this toy DataFrame with a duplicate row label:
In [19]: dfrm
Out[19]:
A B C
a 0.143693 0.653810 0.586007
b 0.623582 0.312903 0.919076
c 0.165438 0.889809 0.000967
d 0.308245 0.787776 0.571195
e 0.870068 0.935626 0.606911
f 0.037602 0.855193 0.728495
g 0.605366 0.338105 0.696460
h 0.000000 0.090814 0.963927
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
In [20]: dfrm['A'].idxmax()
Out[20]: 'i'
In [21]: dfrm.iloc[dfrm['A'].idxmax()] # .ix instead of .iloc in older versions of pandas
Out[21]:
A B C
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
So here a naive use of idxmax is not sufficient, whereas the old form of argmax would correctly provide the positional location of the max row (in this case, position 9).
This is exactly one of those nasty kinds of bug-prone behaviors in dynamically typed languages that makes this sort of thing so unfortunate, and worth beating a dead horse over. If you are writing systems code and your system suddenly gets used on some data sets that are not cleaned properly before being joined, it's very easy to end up with duplicate row labels, especially string labels like a CUSIP or SEDOL identifier for financial assets. You can't easily use the type system to help you out, and you may not be able to enforce uniqueness on the index without running into unexpectedly missing data.
So you're left with hoping that your unit tests covered everything (they didn't, or more likely no one wrote any tests) -- otherwise (most likely) you're just left waiting to see if you happen to smack into this error at runtime, in which case you probably have to go drop many hours worth of work from the database you were outputting results to, bang your head against the wall in IPython trying to manually reproduce the problem, finally figuring out that it's because idxmax can only report the label of the max row, and then being disappointed that no standard function automatically gets the positions of the max row for you, writing a buggy implementation yourself, editing the code, and praying you don't run into the problem again.
You might also try idxmax:
In [5]: df = pandas.DataFrame(np.random.randn(10,3),columns=['A','B','C'])
In [6]: df
Out[6]:
A B C
0 2.001289 0.482561 1.579985
1 -0.991646 -0.387835 1.320236
2 0.143826 -1.096889 1.486508
3 -0.193056 -0.499020 1.536540
4 -2.083647 -3.074591 0.175772
5 -0.186138 -1.949731 0.287432
6 -0.480790 -1.771560 -0.930234
7 0.227383 -0.278253 2.102004
8 -0.002592 1.434192 -1.624915
9 0.404911 -2.167599 -0.452900
In [7]: df.idxmax()
Out[7]:
A 0
B 8
C 7
e.g.
In [8]: df.loc[df['A'].idxmax()]
Out[8]:
A 2.001289
B 0.482561
C 1.579985
Both above answers would only return one index if there are multiple rows that take the maximum value. If you want all the rows, there does not seem to have a function.
But it is not hard to do. Below is an example for Series; the same can be done for DataFrame:
In [1]: from pandas import Series, DataFrame
In [2]: s=Series([2,4,4,3],index=['a','b','c','d'])
In [3]: s.idxmax()
Out[3]: 'b'
In [4]: s[s==s.max()]
Out[4]:
b 4
c 4
dtype: int64
df.iloc[df['columnX'].argmax()]
argmax() would provide the index corresponding to the max value for the columnX. iloc can be used to get the row of the DataFrame df for this index.
A more compact and readable solution using query() is like this:
import pandas as pd
df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
print(df)
# find row with maximum A
df.query('A == A.max()')
It also returns a DataFrame instead of Series, which would be handy for some use cases.
Very simple: we have df as below and we want to print a row with max value in C:
A B C
x 1 4
y 2 10
z 5 9
In:
df.loc[df['C'] == df['C'].max()] # condition check
Out:
A B C
y 2 10
If you want the entire row instead of just the id, you can use df.nlargest and pass in how many 'top' rows you want and you can also pass in for which column/columns you want it for.
df.nlargest(2,['A'])
will give you the rows corresponding to the top 2 values of A.
use df.nsmallest for min values.
The direct ".argmax()" solution does not work for me.
The previous example provided by #ely
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].argmax()
3
>>> df['B'].argmax()
4
>>> df['C'].argmax()
1
returns the following message :
FutureWarning: 'argmax' is deprecated, use 'idxmax' instead. The behavior of 'argmax'
will be corrected to return the positional maximum in the future.
Use 'series.values.argmax' to get the position of the maximum now.
So that my solution is :
df['A'].values.argmax()
mx.iloc[0].idxmax()
This one line of code will give you how to find the maximum value from a row in dataframe, here mx is the dataframe and iloc[0] indicates the 0th index.
Considering this dataframe
[In]: df = pd.DataFrame(np.random.randn(4,3),columns=['A','B','C'])
[Out]:
A B C
0 -0.253233 0.226313 1.223688
1 0.472606 1.017674 1.520032
2 1.454875 1.066637 0.381890
3 -0.054181 0.234305 -0.557915
Assuming one want to know the rows where column "C" is max, the following will do the work
[In]: df[df['C']==df['C'].max()])
[Out]:
A B C
1 0.472606 1.017674 1.520032
The idmax of the DataFrame returns the label index of the row with the maximum value and the behavior of argmax depends on version of pandas (right now it returns a warning). If you want to use the positional index, you can do the following:
max_row = df['A'].values.argmax()
or
import numpy as np
max_row = np.argmax(df['A'].values)
Note that if you use np.argmax(df['A']) behaves the same as df['A'].argmax().
Use:
data.iloc[data['A'].idxmax()]
data['A'].idxmax() -finds max value location in terms of row
data.iloc() - returns the row
If there are ties in the maximum values, then idxmax returns the index of only the first max value. For example, in the following DataFrame:
A B C
0 1 0 1
1 0 0 1
2 0 0 0
3 0 1 1
4 1 0 0
idxmax returns
A 0
B 3
C 0
dtype: int64
Now, if we want all indices corresponding to max values, then we could use max + eq to create a boolean DataFrame, then use it on df.index to filter out indexes:
out = df.eq(df.max()).apply(lambda x: df.index[x].tolist())
Output:
A [0, 4]
B [3]
C [0, 1, 3]
dtype: object
what worked for me is:
df[df['colX'] == df['colX'].max()
You then get the row in your df with the maximum value of colX.
Then if you just want the index you can add .index at the end of the query.