Pandas Dataframe Multiindex Merge

Question

I wanted to ask a questions regarding merging multiindex dataframe in pandas, here is a hypothetical scenario:

arrays = [[\'bar\', \'bar\', \'baz\', \'baz\'         


        

Answer 1

rename_axis

You can rename the index levels of one and let join do its thing

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

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

Answer 2

Seems like you need to use a combination of them.

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

Output:

               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 3

Assign it by combine_first

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)

Answer 4

Other than using the indexes names as pointed by @ALollz, you can simply use loc, which will match indexes automatically

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

A general formula would be

s1.loc[:, s2.columns] = s2