I have a pandas
data frame (v 0.20.3):
df = pd.DataFrame({'coname1': ['Apple','Yahoo'], 'coname2':['Apple', 'Google']})
df['eq'] = df.apply(lambda row: row['coname1'] == row['coname2'], axis=1).astype(bool)
coname1 coname2 eq
0 Apple Apple True
1 Yahoo Google False
If I would like to replace True/False
to 'Yes'/'No'
, I could run this:
df.replace({
True: 'Yes',
False: 'No'
})
coname1 coname2 eq
0 Apple Apple Yes
1 Yahoo Google No
Which seems to get the job done. However, if a data frame is just one row with a value of 0/1
in a column, it will be also replaced as it's being treated as Boolean.
df1 = pd.DataFrame({'coname1': [1], 'coname2':['Google'], 'coname3':[777]})
df1['eq'] = True
coname1 coname2 coname3 eq
0 1 Google 777 True
df1.replace({
True: 'Yes',
False: 'No'
})
coname1 coname2 coname3 eq
0 Yes Google 777 Yes
I would like to map True/False
to Yes/No
for all columns in the data frame that are of dtype
bool
.
How do I tell pandas
to run map True/False to arbitrary strings only for the columns that are of dtype
bool
without explicitly specifying the names of columns as I may not know them in advance?
Use the dtypes attribute to check if the column is boolean and filter based on that:
df = pd.DataFrame({'A': [0, 1], 'B': ['x', 'y'],
'C': [True, False], 'D': [False, True]})
df
Out:
A B C D
0 0 x True False
1 1 y False True
bool_cols = df.columns[df.dtypes == 'bool']
df[bool_cols] = df[bool_cols].replace({True: 'Yes', False: 'No'})
df
Out:
A B C D
0 0 x Yes No
1 1 y No Yes
I think the fastest way would be to use map in a loop though:
for col in df.columns[df.dtypes == 'bool']:
df[col] = df[col].map({True: 'Yes', False: 'No'})
A nice workaround is to create a function that first checks if the element is of type bool or not, and then use applymap
:
import pandas as pd
df1 = pd.DataFrame({'coname1': [1], 'coname2':['Google'], 'coname3':[777]})
df1['eq'] = True
def bool2yes(boolean):
if isinstance(boolean, bool):
if boolean == True:
return "Yes"
else:
return "No"
else:
return boolean
>>> df1.applymap(bool2yes)
coname1 coname2 coname3 eq
0 1 Google 777 Yes
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