5

我有一个在行上DataFrame使用 a的熊猫：MultiIndex

index = pandas.MultiIndex.from_tuples(list(itertools.product(range(3), range(3))))
df = pandas.DataFrame(numpy.random.randn(9,3), index=index, columns=['A', 'B', 'C'])

            A         B         C
0 0  2.400417  0.698638  1.231540
  1 -0.023154 -2.110450  0.774964
  2 -1.282392 -0.062794  1.471655
1 0 -1.081853  0.261876 -1.771075
  1 -2.013747 -0.377957 -0.393802
  2  1.711172 -0.552468  1.018727
2 0  0.155821 -0.222691  0.496586
  1  0.563638 -0.756709  1.050212
  2 -1.446159 -0.891549  0.256695

我想在索引的第一级对这个 DataFrame 进行洗牌，所以可能的结果是：

            A         B         C
1 0 -1.081853  0.261876 -1.771075
  1 -2.013747 -0.377957 -0.393802
  2  1.711172 -0.552468  1.018727
0 0  2.400417  0.698638  1.231540
  1 -0.023154 -2.110450  0.774964
  2 -1.282392 -0.062794  1.471655
2 0  0.155821 -0.222691  0.496586
  1  0.563638 -0.756709  1.050212
  2 -1.446159 -0.891549  0.256695
4

3 回答 3

6

reindex当传递与所需顺序匹配的重新排序的元组数组时，该方法可以完成此操作。此时，可以根据您的问题进行重新排序。例如：

In [38]: df
Out[38]: 
            A         B         C
0 0 -1.725337  0.111493  0.178294
  1 -1.809003 -0.614219 -0.931909
  2  0.621427 -0.186233  0.254727
1 0 -1.322863  1.242415  1.375579
  1  0.249738 -1.280204  0.356491
  2 -0.743671  0.325841 -0.167772
2 0 -0.070937  0.401172 -1.790801
  1  1.433794  2.257198  1.848435
  2 -1.021557 -1.054363 -1.485536

In [39]: neworder = [1, 0, 2]

In [41]: newindex = sorted(df.index, key=lambda x: neworder.index(x[0]))

In [42]: newindex
Out[42]: 
[(1L, 0L),
 (1L, 1L),
 (1L, 2L),
 (0L, 0L),
 (0L, 1L),
 (0L, 2L),
 (2L, 0L),
 (2L, 1L),
 (2L, 2L)]

In [43]: df.reindex(newindex)
Out[43]: 
            A         B         C
1 0 -1.322863  1.242415  1.375579
  1  0.249738 -1.280204  0.356491
  2 -0.743671  0.325841 -0.167772
0 0 -1.725337  0.111493  0.178294
  1 -1.809003 -0.614219 -0.931909
  2  0.621427 -0.186233  0.254727
2 0 -0.070937  0.401172 -1.790801
  1  1.433794  2.257198  1.848435
  2 -1.021557 -1.054363 -1.485536
于 2012-05-05T13:49:37.423 回答
0

如果以下工作会容易得多，但没有：

df.ix[[1, 0, 2]]

以下是更多的解决方法。也许有更好的方法，但我想不通。这只DataFrame是以正确的顺序创建一个“切片”列表并将它们与pandas.concat.

In : df
Out:
            A         B         C
0 0  1.202098 -0.031121  1.417629
  1 -0.895862  0.697531 -0.572411
  2  1.179101 -0.008602  1.583385
1 0  1.969477 -0.968004 -0.567695
  1 -1.504443 -0.002264 -0.413091
  2 -1.412457  0.310518  0.267475
2 0 -0.385933 -0.471800 -0.598141
  1 -0.105032  0.443437 -0.615566
  2 -1.035326 -0.282289 -0.042762

In : shuffled = [2,0,1]

In : df2 = pandas.concat([df.ix[i:i] for i in shuffled])

In : df2
Out:
            A         B         C
2 0 -0.385933 -0.471800 -0.598141
  1 -0.105032  0.443437 -0.615566
  2 -1.035326 -0.282289 -0.042762
0 0  1.202098 -0.031121  1.417629
  1 -0.895862  0.697531 -0.572411
  2  1.179101 -0.008602  1.583385
1 0  1.969477 -0.968004 -0.567695
  1 -1.504443 -0.002264 -0.413091
  2 -1.412457  0.310518  0.267475
于 2012-05-05T03:16:40.570 回答
0

一个简单，简洁的解决方案是

icol = df.index.levels[k]
df.loc[icol[permutation]]

这适用于最外层。要获得一般级别，请使用swaplevel或IndexSlice

完整的 MWE：

import pandas
from io import StringIO

table = """   
i j  A         B         C
0 0  2.400417  0.698638  1.231540
  1 -0.023154 -2.110450  0.774964
  2 -1.282392 -0.062794  1.471655
1 0 -1.081853  0.261876 -1.771075
  1 -2.013747 -0.377957 -0.393802
  2  1.711172 -0.552468  1.018727
2 0  0.155821 -0.222691  0.496586
  1  0.563638 -0.756709  1.050212
  2 -1.446159 -0.891549  0.256695
"""

df = pandas.read_fwf(StringIO(table), dtype={"i":"UInt8", "j":"UInt8"})
df = df.ffill().set_index(["i", "j"])

i, j = df.index.levels
idx = pandas.IndexSlice
perm = [2,0,1]
df.loc[i[perm]]

退货

            A         B         C
i j                              
2 0  0.155821 -0.222691  0.496586
  1  0.563638 -0.756709  1.050212
  2 -1.446159 -0.891549  0.256695
0 0  2.400417  0.698638  1.231540
  1 -0.023154 -2.110450  0.774964
  2 -1.282392 -0.062794  1.471655
1 0 -1.081853  0.261876 -1.771075
  1 -2.013747 -0.377957 -0.393802
  2  1.711172 -0.552468  1.018727

并df.loc[idx[i, j[perm]], :]给出

            A         B         C
i j                              
0 2 -1.282392 -0.062794  1.471655
  0  2.400417  0.698638  1.231540
  1 -0.023154 -2.110450  0.774964
1 2  1.711172 -0.552468  1.018727
  0 -1.081853  0.261876 -1.771075
  1 -2.013747 -0.377957 -0.393802
2 2 -1.446159 -0.891549  0.256695
  0  0.155821 -0.222691  0.496586
  1  0.563638 -0.756709  1.050212

不过要小心！尽管df.loc[idx[i[perm], :], :]并df.loc[idx[i[[perm]], j], :]给出相同的结果，但上面df.loc[idx[i, j[perm]], :]返回的结果与（可能）意外结果不同df.loc[idx[:, j[perm]], :]：

            A         B         C
i j                              
0 2 -1.282392 -0.062794  1.471655
1 2  1.711172 -0.552468  1.018727
2 2 -1.446159 -0.891549  0.256695
0 0  2.400417  0.698638  1.231540
1 0 -1.081853  0.261876 -1.771075
2 0  0.155821 -0.222691  0.496586
0 1 -0.023154 -2.110450  0.774964
1 1 -2.013747 -0.377957 -0.393802
2 1  0.563638 -0.756709  1.050212
于 2021-10-05T15:33:28.603 回答