基于列标签在pandas中重塑数据框。

Question

基于列标签在pandas中重塑数据框。

12

什么是在pandas中重新塑造以下数据框的最佳方法？该数据框df对于每个样本（在这种情况下为s1和s2）都具有x，y值，看起来像这样：

In [23]: df = pandas.DataFrame({"s1_x": scipy.randn(10), "s1_y": scipy.randn(10), "s2_x": scipy.randn(10), "s2_y": scipy.randn(10)})
In [24]: df
Out[24]: 
       s1_x      s1_y      s2_x      s2_y
0  0.913462  0.525590 -0.377640  0.700720
1  0.723288 -0.691715  0.127153  0.180836
2  0.181631 -1.090529 -1.392552  1.530669
3  0.997414 -1.486094  1.207012  0.376120
4 -0.319841  0.195289 -1.034683  0.286073
5  1.085154 -0.619635  0.396867  0.623482
6  1.867816 -0.928101 -0.491929 -0.955295
7  0.920658 -1.132057  1.701582 -0.110299
8 -0.241853 -0.129702 -0.809852  0.014802
9 -0.019523 -0.578930  0.803688 -0.881875

s1_x和s1_y是样本1的x/y值，s2_x，s2_y是样本2的样本值，等等。如何将其重新整形为一个仅包含x、y列但包含附加列sample的DataFrame，该列表示DataFrame中的每一行是来自s1还是s2？例如：

          x         y      sample
0  0.913462  0.525590          s1
1  0.723288 -0.691715          s1
2  0.181631 -1.090529          s1
3  0.997414 -1.486094          s1
...
5  0.396867  0.623482          s2
...

这对于后续使用Rpy2绘制图形非常有用，因为许多R绘图功能可以利用此分组变量，所以这就是我重新整理数据框的动机。

我认为Chang She给出的答案不适用于具有唯一索引（如此示例）的数据框：

In [636]: df = pandas.DataFrame({"s1_x": scipy.randn(10), "s1_y": scipy.randn(10), "s2_x": scipy.randn(10), "s2_y": scipy.randn(10), "names": range(10)})
In [637]: df
Out[637]: 
   names      s1_x      s1_y      s2_x      s2_y
0      0  0.672298  0.415366  1.034770  0.556209
1      1  0.067087 -0.851028  0.053608 -0.276461
2      2 -0.674174 -0.099015  0.864148 -0.067240
3      3  0.542996 -0.813018  2.283530  2.793727
4      4  0.216633 -0.091870 -0.746411 -0.421852
5      5  0.141301 -1.537721 -0.371601 -1.594634
6      6  1.267148 -0.833120  0.369516 -0.671627
7      7 -0.231163 -0.557398  1.123155  0.865140
8      8  1.790570 -0.428563  0.668987  0.632409
9      9 -0.820315 -0.894855  0.673247 -1.195831
In [638]: df.columns = pandas.MultiIndex.from_tuples([tuple(c.split('_')) for c in df.columns])

In [639]: df.stack(0).reset_index(1)
Out[639]: 
  level_1         x         y
0      s1  0.672298  0.415366
0      s2  1.034770  0.556209
1      s1  0.067087 -0.851028
1      s2  0.053608 -0.276461
2      s1 -0.674174 -0.099015
2      s2  0.864148 -0.067240
3      s1  0.542996 -0.813018
3      s2  2.283530  2.793727
4      s1  0.216633 -0.091870
4      s2 -0.746411 -0.421852
5      s1  0.141301 -1.537721
5      s2 -0.371601 -1.594634
6      s1  1.267148 -0.833120
6      s2  0.369516 -0.671627
7      s1 -0.231163 -0.557398
7      s2  1.123155  0.865140
8      s1  1.790570 -0.428563
8      s2  0.668987  0.632409
9      s1 -0.820315 -0.894855
9      s2  0.673247 -1.195831

转换成功了，但在此过程中列"names"丢失了。我该如何在对具有名称中包含_的列进行融合转换的同时保留df中的"names"列？"names"列仅为数据帧中的每行分配一个唯一名称。本例中是数字类型，但在我的数据中它们是字符串标识符。谢谢。

- user248237

1个回答

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可以查看英文原文，
原文链接

- Chang She · Accepted Answer

我假设您已经有了数据框。在这种情况下，您可以将列转换为MultiIndex，然后使用stack和reset_index。请注意，您还需要重命名和重新排序列，并按样本排序，才能获得您在问题中发布的确切内容：

In [4]: df = pandas.DataFrame({"s1_x": scipy.randn(10), "s1_y": scipy.randn(10), "s2_x": scipy.randn(10), "s2_y": scipy.randn(10)})

In [5]: df.columns = pandas.MultiIndex.from_tuples([tuple(c.split('_')) for c in df.columns])

In [6]: df.stack(0).reset_index(1)
Out[6]: 
  level_1         x         y
0      s1  0.897994 -0.278357
0      s2 -0.008126 -1.701865
1      s1 -1.354633 -0.890960
1      s2 -0.773428  0.003501
2      s1 -1.499422 -1.518993
2      s2  0.240226  1.773427
3      s1 -1.090921  0.847064
3      s2 -1.061303  1.557871
4      s1 -1.697340 -0.160952
4      s2 -0.930642  0.182060
5      s1 -0.356076 -0.661811
5      s2  0.539875 -1.033523
6      s1 -0.687861 -1.450762
6      s2  0.700193  0.658959
7      s1 -0.130422 -0.826465
7      s2 -0.423473 -1.281856
8      s1  0.306983  0.433856
8      s2  0.097279 -0.256159
9      s1  0.498057  0.147243
9      s2  1.312578  0.111837

你可以通过直接创建具有MultiIndex的DataFrame来避免MultiIndex转换的开销。

编辑：使用merge将原始id加入

In [59]: df
Out[59]: 
   names      s1_x      s1_y      s2_x      s2_y
0      0  0.732099  0.018387  0.299856  0.737142
1      1  0.914755 -0.798159 -0.732868 -1.279311
2      2 -1.063558  0.161779 -0.115751 -0.251157
3      3 -1.185501  0.095147 -1.343139 -0.003084
4      4  0.622400 -0.299726  0.198710 -0.383060
5      5  0.179318  0.066029 -0.635507  1.366786
6      6 -0.820099  0.066067  1.113402  0.002872
7      7  0.711627 -0.182925  1.391194 -2.788434
8      8 -1.124092  1.303375  0.202691 -0.225993
9      9 -0.179026  0.847466 -1.480708 -0.497067

In [60]: id = df.ix[:, ['names']]

In [61]: df.columns = pandas.MultiIndex.from_tuples([tuple(c.split('_')) for c in df.columns])

In [62]: pandas.merge(df.stack(0).reset_index(1), id, left_index=True, right_index=True)
Out[62]: 
  level_1         x         y  names
0      s1  0.732099  0.018387      0
0      s2  0.299856  0.737142      0
1      s1  0.914755 -0.798159      1
1      s2 -0.732868 -1.279311      1
2      s1 -1.063558  0.161779      2
2      s2 -0.115751 -0.251157      2
3      s1 -1.185501  0.095147      3
3      s2 -1.343139 -0.003084      3
4      s1  0.622400 -0.299726      4
4      s2  0.198710 -0.383060      4
5      s1  0.179318  0.066029      5
5      s2 -0.635507  1.366786      5
6      s1 -0.820099  0.066067      6
6      s2  1.113402  0.002872      6
7      s1  0.711627 -0.182925      7
7      s2  1.391194 -2.788434      7
8      s1 -1.124092  1.303375      8
8      s2  0.202691 -0.225993      8
9      s1 -0.179026  0.847466      9
9      s2 -1.480708 -0.497067      9

或者：

    In [64]: df
Out[64]: 
   names      s1_x      s1_y      s2_x      s2_y
0      0  0.744742 -1.123403  0.212736  0.005440
1      1  0.465075 -0.673491  1.467156 -0.176298
2      2 -1.111566  0.168043 -0.102142 -1.072461
3      3  1.226537 -1.147357 -1.583762 -1.236582
4      4  1.137675  0.224422  0.738988  1.528416
5      5 -0.237014 -1.110303 -0.770221  1.389714
6      6 -0.659213  2.305374 -0.326253  1.416778
7      7  1.524214 -0.395451 -1.884197  0.524606
8      8  0.375112 -0.622555  0.295336  0.927208
9      9  1.168386 -0.291899 -1.462098  0.250889

In [65]: df = df.set_index('names')

In [66]: df.columns = pandas.MultiIndex.from_tuples([tuple(c.split('_')) for c in df.columns])

In [67]: df.stack(0).reset_index(1)
Out[67]: 
      level_1         x         y
names                            
0          s1  0.744742 -1.123403
0          s2  0.212736  0.005440
1          s1  0.465075 -0.673491
1          s2  1.467156 -0.176298
2          s1 -1.111566  0.168043
2          s2 -0.102142 -1.072461
3          s1  1.226537 -1.147357
3          s2 -1.583762 -1.236582
4          s1  1.137675  0.224422
4          s2  0.738988  1.528416
5          s1 -0.237014 -1.110303
5          s2 -0.770221  1.389714
6          s1 -0.659213  2.305374
6          s2 -0.326253  1.416778
7          s1  1.524214 -0.395451
7          s2 -1.884197  0.524606
8          s1  0.375112 -0.622555
8          s2  0.295336  0.927208
9          s1  1.168386 -0.291899
9          s2 -1.462098  0.250889