python - 从数据框中删除反向重复项

Question

谁能提出一个好的解决方案来从数据框中删除反向重复项？

我的数据看起来像这样，其中第一列和第二列是颠倒的重复项。

TRINITY_DN16813_c0_g1_i3    TRINITY_DN16813_c0_g1_i4    96.491  228 8   0   202 429 417 190 3.049999999999999e-104  377
TRINITY_DN16813_c0_g1_i4    TRINITY_DN16813_c0_g1_i3    96.104  231 9   0   190 420 429 199 2.979999999999999e-104  377

我只需要保留一行，其中第三列具有较高的值

TRINITY_DN16813_c0_g1_i3    TRINITY_DN16813_c0_g1_i4    96.491  228 8   0   202 429 417 190 3.049999999999999e-104  377

这是我使用 series.isin() 时的结果。

TRINITY_DN28139_c0_g1_i2    TRINITY_DN28139_c0_g1_i5    99.971  3465    1   0   1   3465    1   3465    0.0 6394
TRINITY_DN28139_c0_g1_i5    TRINITY_DN28139_c0_g1_i2    99.971  3465    1   0   1   3465    1   3465    0.0 6394
TRINITY_DN25313_c0_g1_i6    TRINITY_DN25313_c0_g1_i5    99.97   3315    1   0   1   3315    1   3315    0.0 6117
TRINITY_DN25313_c0_g1_i5    TRINITY_DN25313_c0_g1_i6    99.97   3315    1   0   1   3315    1   3315    0.0 6117
TRINITY_DN25502_c0_g1_i3    TRINITY_DN25502_c0_g1_i4    99.96799999999999   3078    1   0   1   3078    1   3078    0.0 5679
TRINITY_DN25502_c0_g1_i4    TRINITY_DN25502_c0_g1_i3    99.96799999999999   3078    1   0   1   3078    1   3078    0.0 5679
TRINITY_DN28726_c0_g1_i2    TRINITY_DN28726_c0_g1_i1    99.96600000000001   5805    2   0   1   5805    1   5805    0.0 10709
TRINITY_DN28726_c0_g1_i1    TRINITY_DN28726_c0_g1_i2    99.96600000000001   5805    2   0   1   5805    1   5805    0.0 10709
TRINITY_DN27942_c0_g1_i7    TRINITY_DN27942_c0_g1_i6    99.964  2760    1   0   1   2760    1   2760    0.0 5092
TRINITY_DN25118_c0_g1_i1    TRINITY_DN25118_c0_g1_i2    99.964  2770    1   0   81  2850    204 2973    0.0 5110
TRINITY_DN27942_c0_g1_i6    TRINITY_DN27942_c0_g1_i7    99.964  2760    1   0   1   2760    1   2760    0.0 5092
TRINITY_DN25118_c0_g1_i2    TRINITY_DN25118_c0_g1_i1    99.964  2770    1   0   204 2973    81  2850    0.0 5110
TRINITY_DN28502_c1_g1_i9    TRINITY_DN28502_c1_g1_i7    99.963  2678    1   0   1928    4605    2021    4698    0.0 4940
TRINITY_DN28502_c1_g1_i7    TRINITY_DN28502_c1_g1_i9    99.963  2678    1   0   2021    4698    1928    4605    0.0 4940
TRINITY_DN25619_c0_g1_i1    TRINITY_DN25619_c0_g1_i8    99.963  2715    1   0   1   2715    1   2715    0.0 5009
TRINITY_DN25619_c0_g1_i8    TRINITY_DN25619_c0_g1_i1    99.963  2715    1   0   1   2715    1   2715    0.0 5009
TRINITY_DN23022_c0_g1_i5    TRINITY_DN23022_c0_g1_i1    99.962  2622    1   0   1   2622    1   2622    0.0 4837

Answer 1

试试这个。它完全在 Pandas 中(应该更快)这也纠正了我之前回答中的错误，但将标签作为一对的概念保持不变。

In [384]: df['pair'] = df[[0, 1]].apply(lambda x: '{}-{}'.format(*sorted((x[0], x[1]))), axis=1)

只获取每个重复结果的最大值:

In [385]: dfd = df.loc[df.groupby('pair')[2].idxmax()]

如果您需要将名称放在单独的列中:

In [398]: dfd[0] = dfd['pair'].transform(lambda x: x.split('-')[0])
In [399]: dfd[1] = dfd['pair'].transform(lambda x: x.split('-')[1])