熊猫数据多指标合并

Question

我想问一个关于在熊猫中合并多索引数据的问题，下面是一个假设的场景：

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'])

那你也可以

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

或

s1.merge(s2, how='left', left_on=['first', 'second'], right_on=['third', 'fourth'])

会导致错误。

我是否必须在reset_index() / s1/s2上完成这项工作？

Answer 1

看来你需要使用它们的组合。

s1.merge(s2, left_index=True, right_on=['third', 'fourth'])
#s1.merge(s2, right_index=True, left_on=['first', 'second'])

输出：

               s1        s2
bar one  0.765385 -0.365508
    two  1.462860  0.751862
baz one  0.304163  0.761663
    two -0.816658 -1.810634
foo one  1.891434  1.450081
    two  0.571294  1.116862
qux one  1.056516 -0.052927
    two -0.574916 -1.197596

Answer 2

除了使用@ALollz指出的索引名称之外，您可以简单地使用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

Answer 3

按combine_first分配

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)