我们说我有一个多索引的pandas数据框,看起来像是documentation。
import numpy as np
import pandas as pd
arrays = [np.array(['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux']),
np.array(['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two'])]
df = pd.DataFrame(np.random.randn(8, 4), index=arrays)
看起来像这样:
0 1 2 3
bar one -0.096648 -0.080298 0.859359 -0.030288
two 0.043107 -0.431791 1.923893 -1.544845
baz one 0.639951 -0.008833 -0.227000 0.042315
two 0.705281 0.446257 -1.108522 0.471676
foo one -0.579483 -2.261138 -0.826789 1.543524
two -0.358526 1.416211 1.589617 0.284130
qux one 0.498149 -0.296404 0.127512 -0.224526
two -0.286687 -0.040473 1.443701 1.025008
现在我只想要" ne"包含在MultiIndex的第二级。
有没有办法为包含(部分)的字符串切片MultiIndex?
答案 0 :(得分:11)
您可以应用以下掩码:
df = df.iloc[df.index.get_level_values(1).str.contains('ne')]
返回:
bar one -0.143200 0.523617 0.376458 -2.091154
baz one -0.198220 1.234587 -0.232862 -0.510039
foo one -0.426127 0.594426 0.457331 -0.459682
qux one -0.875160 -0.157073 -0.540459 -1.792235
编辑: 也可以在多个级别上应用逻辑掩码,例如:
df = df.iloc[(df.index.get_level_values(0).str.contains('ba')) | (df.index.get_level_values(1).str.contains('ne'))]
返回:
bar one 0.620279 1.525277 0.379649 -0.032608
two 0.465240 -0.190038 0.795730 1.720368
baz one 0.986828 -0.080394 -0.303319 0.747483
two 0.487534 1.597006 0.114551 0.299502
foo one -0.085700 0.112433 0.704043 0.264280
qux one -0.291758 -1.071669 0.794354 -1.805530