我有一个这样的数据框架:
import pandas as pd
import numpy as np
df = pd.DataFrame(np.random.rand(4,4), columns=list('abcd'))
df
a b c d
0 0.418762 0.042369 0.869203 0.972314
1 0.991058 0.510228 0.594784 0.534366
2 0.407472 0.259811 0.396664 0.894202
3 0.726168 0.139531 0.324932 0.906575
我如何得到除b之外的所有列?
当前回答
df[df.columns.difference(['b'])]
Out:
a c d
0 0.427809 0.459807 0.333869
1 0.678031 0.668346 0.645951
2 0.996573 0.673730 0.314911
3 0.786942 0.719665 0.330833
其他回答
类似于@Toms的答案，也可以选择除“b”以外的所有列，而不使用.loc，如下所示:
df[df.columns[~df.columns.isin(['b'])]]
你可以在索引中删除列:
df[df.columns.drop('b')]
or
df.loc[:, df.columns.drop('b')]
我认为一个很好的解决方案是使用pandas和regex的函数过滤器(匹配除“b”之外的所有内容):
df.filter(regex="^(?!b$)")
df[df.columns.difference(['b'])]
Out:
a c d
0 0.427809 0.459807 0.333869
1 0.678031 0.668346 0.645951
2 0.996573 0.673730 0.314911
3 0.786942 0.719665 0.330833
我测试了速度，发现对我来说。loc解决方案是最快的
df_working_1.loc[:, df_working_1.columns != "market_id"]
# 7.19 ms ± 201 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
df_working_1.drop("market_id", axis=1)
# 7.65 ms ± 136 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
df_working_1[df_working_1.columns.difference(['market_id'])]
# 7.58 ms ± 116 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
df_working_1[[i for i in list(df_working_1.columns) if i != 'market_id']]
# 7.57 ms ± 144 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
