在多索引数据框中重命名索引值

时间:2013-12-11 20:49:33

标签: python-2.7 pandas

创建我的数据框:

from pandas import *
arrays = [['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
          ['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']]

tuples = zip(*arrays)

index = MultiIndex.from_tuples(tuples, names=['first','second'])
data = DataFrame(randn(8,2),index=index,columns=['c1','c2'])

data
Out[68]: 
                    c1        c2
first second                    
bar   one     0.833816 -1.529639
      two     0.340150 -1.818052
baz   one    -1.605051 -0.917619
      two    -0.021386 -0.222951
foo   one     0.143949 -0.406376
      two     1.208358 -2.469746
qux   one    -0.345265 -0.505282
      two     0.158928  1.088826

我想重命名“第一个”索引值,例如“bar” - >“cat”,“baz” - >“dog”等。但是,我读过的每个例子都在单级索引和/或循环遍历整个索引,从头开始有效地重新创建它。我想的是:

data = data.reindex(index={'bar':'cat','baz':'dog'})

但是这不起作用,我也不希望它在多个索引上工作。我可以在不循环整个数据框索引的情况下进行这样的替换吗?

开始修改

我对在发布之前更新到0.13犹豫不决,所以我使用了以下解决方法:

index = data.index.tolist()
for r in xrange( len(index) ):
    index[r] = (codes[index[r][0]],index[r][1])

index = pd.MultiIndex.from_tuples(index,names=data.index.names)
data.index = index

以前定义的代码字典在哪里:字符串对。这实际上并没有像我期望的那样大的性能(需要几秒钟来操作~110万行)。它不像单线,但确实有效。

结束修改

2 个答案:

答案 0 :(得分:16)

使用set_levels方法(new in version 0.13.0):

data.index.set_levels([[u'cat', u'dog', u'foo', u'qux'], 
                       [u'one', u'two']], inplace=True)

产量

                    c1        c2
first second                    
cat   one    -0.289649 -0.870716
      two    -0.062014 -0.410274
dog   one     0.030171 -1.091150
      two     0.505408  1.531108
foo   one     1.375653 -1.377876
      two    -1.478615  1.351428
qux   one     1.075802  0.532416
      two     0.865931 -0.765292

要根据dict重新映射级别,可以使用如下函数:

def map_level(df, dct, level=0):
    index = df.index
    index.set_levels([[dct.get(item, item) for item in names] if i==level else names
                      for i, names in enumerate(index.levels)], inplace=True)

dct = {'bar':'cat', 'baz':'dog'}
map_level(data, dct, level=0)

这是一个可运行的例子:

import numpy as np
import pandas as pd

arrays = [['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
          ['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']]
tuples = zip(*arrays)
index = pd.MultiIndex.from_tuples(tuples, names=['first','second'])
data = pd.DataFrame(np.random.randn(8,2),index=index,columns=['c1','c2'])
data2 = data.copy()

data.index.set_levels([[u'cat', u'dog', u'foo', u'qux'], 
                       [u'one', u'two']], inplace=True)
print(data)
#                     c1        c2
# first second                    
# cat   one     0.939040 -0.748100
#       two    -0.497006 -1.185966
# dog   one    -0.368161  0.050339
#       two    -2.356879 -0.291206
# foo   one    -0.556261  0.474297
#       two     0.647973  0.755983
# qux   one    -0.017722  1.364244
#       two     1.007303  0.004337

def map_level(df, dct, level=0):
    index = df.index
    index.set_levels([[dct.get(item, item) for item in names] if i==level else names
                      for i, names in enumerate(index.levels)], inplace=True)
dct = {'bar':'wolf', 'baz':'rabbit'}
map_level(data2, dct, level=0)
print(data2)
#                      c1        c2
# first  second                    
# wolf   one     0.939040 -0.748100
#        two    -0.497006 -1.185966
# rabbit one    -0.368161  0.050339
#        two    -2.356879 -0.291206
# foo    one    -0.556261  0.474297
#        two     0.647973  0.755983
# qux    one    -0.017722  1.364244
#        two     1.007303  0.004337

答案 1 :(得分:1)

set_levels方法导致我的新列名无序。所以我找到了一个不太干净的不同解决方案,但效果很好。方法是print df.index(或等效df.columns),然后复制并粘贴输出,并更改所需的值。例如:

print data.index
  

MultiIndex(等级= [[' bar',' baz',' foo',' qux'],[&#39 ;一个','两个']],标签= [[0,0,1,1,2,2,3,3],[0,1,0,1,0,1] ,0,1]],              names = [' first',' second'])

data.index = MultiIndex(levels=[['new_bar', 'new_baz', 'new_foo', 'new_qux'],
                                ['new_one', 'new_two']],
                        labels=[[0, 0, 1, 1, 2, 2, 3, 3], [0, 1, 0, 1, 0, 1, 0, 1]],
                        names=['first', 'second'])

我们也可以通过编辑标签来完全控制名称。例如:

data.index = MultiIndex(levels=[['bar', 'baz', 'foo', 'qux'],
                                ['one', 'twooo', 'three', 'four',
                                 'five', 'siz', 'seven', 'eit']],
                        labels=[[0, 0, 1, 1, 2, 2, 3, 3], [0, 1, 2, 3, 4, 5, 6, 7]],
                        names=['first', 'second'])

请注意,在此示例中,我们已经完成了from pandas import MultiIndexfrom pandas import *之类的操作。