Pandas filter columns of a DataFrame with bool

Question

For a DataFrame (df) with multiple columns and rows

     A   B  C  D
0    1   4  2  6
1    2   5  7  4
2    3   6  5  6

and another DataFrame (dfBool) containing dtype: bool

0  True
1  False
2  False
3  True

What is the easiest way to split this DataFrame by columns into two different DataFrames by transposing dfbool so you get the desired output

     A   D
0    1   6
1    2   4
2    3   6 

     B  C 
0    4  2  
1    5  7  
2    6  5  

I cannot understand, in my limited experience why dfTrue = df[dfBool.transpose() == True] does not work

Answer 1

I would like to modify EdChum's comment, because if dfBool is DataFrame, you have to first select column:

import pandas as pd

df = pd.DataFrame({'D': {0: 6, 1: 4, 2: 6},
                    'A': {0: 1, 1: 2, 2: 3},
                    'C': {0: 2, 1: 7, 2: 5},
                    'B': {0: 4, 1: 5, 2: 6}})
print (df)
   A  B  C  D
0  1  4  2  6
1  2  5  7  4
2  3  6  5  6

dfBool = pd.DataFrame({'a':[True, False, False, True]})
print (dfBool)
       a
0   True
1  False
2  False
3   True

#select first column in dfBool
df2 = (dfBool.iloc[:,0])
#or select column a in dfBool
#df2 = (dfBool.a)
print (df2)
0     True
1    False
2    False
3     True
Name: a, dtype: bool

print (df[df.columns[df2]])
   A  D
0  1  6
1  2  4
2  3  6

print (df[df.columns[~df2]])
   B  C
0  4  2
1  5  7
2  6  5

Another very nice solution from ayhan, thank you:

print (df.loc[:, dfBool.a.values])
   A  D
0  1  6
1  2  4
2  3  6

print (df.loc[:, ~dfBool.a.values])
   B  C
0  4  2
1  5  7
2  6  5

But if dfBool is Series, solution works very well:

dfBool = pd.Series([True, False, False, True])
print (dfBool)

0     True
1    False
2    False
3     True
dtype: bool

print (df[df.columns[dfBool]])
   A  D
0  1  6
1  2  4
2  3  6

print (df[df.columns[~dfBool]])
   B  C
0  4  2
1  5  7
2  6  5

And for Series:

print (df.loc[:, dfBool.values])
   A  D
0  1  6
1  2  4
2  3  6

print (df.loc[:, ~dfBool.values])
   B  C
0  4  2
1  5  7
2  6  5

Timings:

In [277]: %timeit (df[df.columns[dfBool.a]])
1000 loops, best of 3: 769 µs per loop

In [278]: %timeit (df.loc[:, dfBool1.a.values])
The slowest run took 9.08 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 380 µs per loop

In [279]: %timeit (df.transpose()[dfBool1.a.values].transpose())
The slowest run took 5.04 times longer than the fastest. This could mean that an intermediate result is being cached.
1000 loops, best of 3: 550 µs per loop

Code for timings:

import pandas as pd

df = pd.DataFrame({'D': {0: 6, 1: 4, 2: 6},
                    'A': {0: 1, 1: 2, 2: 3},
                    'C': {0: 2, 1: 7, 2: 5},
                    'B': {0: 4, 1: 5, 2: 6}})
print (df)
df = pd.concat([df]*1000, axis=1).reset_index(drop=True)

dfBool = pd.DataFrame({'a': [True, False, False, True]})
dfBool1 = pd.concat([dfBool]*1000).reset_index(drop=True)

Output is little different:

print (df[df.columns[dfBool.a]])
   A  A  A  A  A  A  A  A  A  A ...  D  D  D  D  D  D  D  D  D  D
0  1  1  1  1  1  1  1  1  1  1 ...  6  6  6  6  6  6  6  6  6  6
1  2  2  2  2  2  2  2  2  2  2 ...  4  4  4  4  4  4  4  4  4  4
2  3  3  3  3  3  3  3  3  3  3 ...  6  6  6  6  6  6  6  6  6  6

[3 rows x 2000 columns]

print (df.loc[:, dfBool1.a.values])
   A  D  A  D  A  D  A  D  A  D ...  A  D  A  D  A  D  A  D  A  D
0  1  6  1  6  1  6  1  6  1  6 ...  1  6  1  6  1  6  1  6  1  6
1  2  4  2  4  2  4  2  4  2  4 ...  2  4  2  4  2  4  2  4  2  4
2  3  6  3  6  3  6  3  6  3  6 ...  3  6  3  6  3  6  3  6  3  6

[3 rows x 2000 columns]

print (df.transpose()[dfBool1.a.values].transpose())
   A  D  A  D  A  D  A  D  A  D ...  A  D  A  D  A  D  A  D  A  D
0  1  6  1  6  1  6  1  6  1  6 ...  1  6  1  6  1  6  1  6  1  6
1  2  4  2  4  2  4  2  4  2  4 ...  2  4  2  4  2  4  2  4  2  4
2  3  6  3  6  3  6  3  6  3  6 ...  3  6  3  6  3  6  3  6  3  6

[3 rows x 2000 columns]