我想修改EdChum的评论,因为如果dfBool
是一个DataFrame
,你必须先选择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
df2 = (dfBool.iloc[:,0])
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
另一个非常好的解决方案来自ayhan,感谢您:
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
但是如果 dfBool
是一个 Series
,这个解决方案也能够很好地工作:
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
而对于 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
时间:
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
计时代码:
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)
输出略有不同:
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]
df[df.columns[dfBool]]
和df[df.columns[~dfBool]]
吗?请确认。 - EdChum