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问题:
This is an example DataFrame with multi index rows.
row_idx_arr = list(zip(['r0', 'r0', 'r0', 'r1', 'r1', 'r1', 'r2', 'r2', 'r2', 'r3', 'r3', 'r3'], ['r-00', 'r-01', 'r-02', 'r-00', 'r-01', 'r-02', 'r-00', 'r-01', 'r-02', 'r-00', 'r-01', 'r-02', ])) row_idx = pd.MultiIndex.from_tuples(row_idx_arr) d = pd.DataFrame((np.random.randn(36)*10).reshape(12,3), index=row_idx, columns=['c0', 'c1', 'returns']) c0 c1 returns r0 r-00 3.553446 5.434018 5.141394 r-01 10.045250 18.453873 13.170396 r-02 -7.231743 -11.695715 5.303477 r1 r-00 -1.302917 6.461693 15.016544 r-01 13.348552 -9.133629 -2.464875 r-02 11.157144 16.833344 -8.745151 r2 r-00 -10.937900 -14.829996 -8.457521 r-01 -7.495922 9.269724 -5.001560 r-02 -8.966551 11.063291 -2.420552 r3 r-00 -21.434668 -0.730560 5.550830 r-01 16.590447 -0.432384 -0.396881 r-02 -0.636957 -2.765959 2.591906
I'd like to create a new DataFrame where, for each row (r0, r1, r2, r3), I have the 2 entries (level 2 rows: r-00, r-01, r-02) with highest 'returns'.
Please note that this is an example, in my program I have thousands of rows.
回答1:
I think you can use nlargest with groupby:
import pandas as pd import numpy as np row_idx_arr = list(zip(['r0', 'r0', 'r0', 'r1', 'r1', 'r1', 'r2', 'r2', 'r2', 'r3', 'r3', 'r3'], ['r-00', 'r-01', 'r-02', 'r-00', 'r-01', 'r-02', 'r-00', 'r-01', 'r-02', 'r-00', 'r-01', 'r-02', ])) row_idx = pd.MultiIndex.from_tuples(row_idx_arr) d = pd.DataFrame((np.random.randn(36)*10).reshape(12,3), index=row_idx, columns=['c0', 'c1', 'returns']) print d c0 c1 returns r0 r-00 -13.417493 -14.758075 -3.650524 r-01 1.092054 -1.224499 -8.968738 r-02 4.793562 -9.958708 -16.554163 r1 r-00 -0.308835 -4.584725 -4.070714 r-01 -23.764872 0.240768 -24.110720 r-02 -4.054037 7.744689 12.762280 r2 r-00 9.160783 -16.041333 10.865837 r-01 -10.472071 -1.625311 17.091514 r-02 -13.009323 1.114351 -3.494279 r3 r-00 7.537877 -17.307256 -2.739447 r-01 -1.107766 1.458901 -19.214064 r-02 8.473581 -7.456646 1.427752 df = d.groupby(level=0, group_keys=False).apply(lambda x: x.nlargest(2, ['returns'])) print df c0 c1 returns r0 r-00 -13.417493 -14.758075 -3.650524 r-01 1.092054 -1.224499 -8.968738 r1 r-02 -4.054037 7.744689 12.762280 r-00 -0.308835 -4.584725 -4.070714 r2 r-01 -10.472071 -1.625311 17.091514 r-00 9.160783 -16.041333 10.865837 r3 r-02 8.473581 -7.456646 1.427752 r-00 7.537877 -17.307256 -2.739447
回答2:
The most elegant way would be the following:
d.groupby(axis=0, level=0, group_keys=False).nlargest(2, 'returns')
Unfortunately that doesn't work because DataFrameGroupBy (object returned by groupby) hasn't had nlargest method implemented yet in Pandas API.
But here is a workaround:
larg = d['returns'].groupby(level=0, group_keys=False).nlargest(2) d.ix[larg.index]
That works because groupby applied to a Series gives back a SeriesGroupBy object that has nlargest method implemented.
