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Python 数据帧多索引合并_Python_Pandas_Merge_Multi Index - Fatal编程技术网

Python 数据帧多索引合并

Python 数据帧多索引合并,python,pandas,merge,multi-index,Python,Pandas,Merge,Multi Index,我想问一个关于在pandas中合并多索引数据帧的问题,下面是一个假设场景: arrays = [['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'], ['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']] tuples = list(zip(*arrays)) index1 = pd.MultiIndex.from_tuples(tuples, nam

我想问一个关于在pandas中合并多索引数据帧的问题,下面是一个假设场景:

arrays = [['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
            ['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']]
tuples = list(zip(*arrays))
index1 = pd.MultiIndex.from_tuples(tuples, names=['first', 'second'])
index2 = pd.MultiIndex.from_tuples(tuples, names=['third', 'fourth'])

s1 = pd.DataFrame(np.random.randn(8), index=index1, columns=['s1'])
s2 = pd.DataFrame(np.random.randn(8), index=index2, columns=['s2'])
那么

s1.merge(s2, how='left', left_index=True, right_index=True)

将导致错误

我是否必须在s1/s2中的任何一个上重置_index(),才能使其正常工作


谢谢

看起来您需要将它们结合使用

s1.merge(s2, left_index=True, right_on=['third', 'fourth'])
#s1.merge(s2, right_index=True, left_on=['first', 'second'])
输出:
除了使用@ALollz所指出的索引名称外,您只需使用
loc
,它将自动匹配索引

s1.loc[:, 's2'] = s2   # Or explicitly, s2['s2']

                s1           s2
first   second      
bar     one     -0.111384   -2.341803
        two     -1.226569    1.308240
baz     one      1.880835    0.697946
        two     -0.008979   -0.247896
foo     one      0.103864   -1.039990
        two      0.836931    0.000811
qux     one     -0.859005   -1.199615
        two     -0.321341   -1.098691
一般的公式是

s1.loc[:, s2.columns] = s2
rename_轴
您可以重命名一的索引级别,并让
join
执行它的操作

s1.join(s2.rename_axis(s1.index.names))

                    s1        s2
first second                    
bar   one    -0.696420 -1.040463
      two     0.640891  1.483262
baz   one     1.598837  0.097424
      two     0.003994 -0.948419
foo   one    -0.717401  1.190019
      two    -1.201237 -0.000738
qux   one     0.559684 -0.505640
      two     1.979700  0.186013

concat

首先通过
combine\u分配它

s1.combine_first(s2)
Out[19]: 
                    s1        s2
first second                    
bar   one     0.039203  0.795963
      two     0.454782 -0.222806
baz   one     3.101120 -0.645474
      two    -1.174929 -0.875561
foo   one    -0.887226  1.078218
      two     1.507546 -1.078564
qux   one     0.028048  0.042462
      two     0.826544 -0.375351

# s2.combine_first(s1)

这是许多新用户/编码人员感到沮丧的事情之一,有太多不同的方法来做同样的事情。我喜欢这一点,因为根据数据集或您为什么要首先这样做,您可以选择易于编码和理解的路线,也可以优化以获得更快的运行时间路线。
s1.join(s2.rename_axis(s1.index.names))

                    s1        s2
first second                    
bar   one    -0.696420 -1.040463
      two     0.640891  1.483262
baz   one     1.598837  0.097424
      two     0.003994 -0.948419
foo   one    -0.717401  1.190019
      two    -1.201237 -0.000738
qux   one     0.559684 -0.505640
      two     1.979700  0.186013
pd.concat([s1, s2], axis=1)

                    s1        s2
first second                    
bar   one    -0.696420 -1.040463
      two     0.640891  1.483262
baz   one     1.598837  0.097424
      two     0.003994 -0.948419
foo   one    -0.717401  1.190019
      two    -1.201237 -0.000738
qux   one     0.559684 -0.505640
      two     1.979700  0.186013
s1.combine_first(s2)
Out[19]: 
                    s1        s2
first second                    
bar   one     0.039203  0.795963
      two     0.454782 -0.222806
baz   one     3.101120 -0.645474
      two    -1.174929 -0.875561
foo   one    -0.887226  1.078218
      two     1.507546 -1.078564
qux   one     0.028048  0.042462
      two     0.826544 -0.375351

# s2.combine_first(s1)