Pandas 数据帧多索引合并
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Pandas Dataframe Multiindex Merge
提问by learningToCode
I wanted to ask a questions regarding merging multiindex dataframe in pandas, here is a hypothetical scenario:
我想问一个关于在Pandas中合并多索引数据框的问题，这是一个假设的场景：
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'])
Then either
那么要么
s1.merge(s2, how='left', left_index=True, right_index=True)
or
或者
s1.merge(s2, how='left', left_on=['first', 'second'], right_on=['third', 'fourth'])
will result in error.
会导致错误。
Do I have to do reset_index() on either s1/s2 to make this work?
我是否必须在 s1/s2 上执行 reset_index() 才能使其工作？
Thanks
谢谢
回答by ALollz
Seems like you need to use a combination of them.
似乎您需要结合使用它们。
s1.merge(s2, left_index=True, right_on=['third', 'fourth'])
#s1.merge(s2, right_index=True, left_on=['first', 'second'])
Output:
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
回答by YOBEN_S
Assign it by combine_first
分配给 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)
回答by rafaelc
Other than using the indexes names as pointed by @ALollz, you can simply use loc, which will match indexes automatically
除了使用@ALollz 指向的索引名称之外，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
A general formula would be
一个通用的公式是
s1.loc[:, s2.columns] = s2
回答by piRSquared
rename_axis
rename_axis
You can rename the index levels of one and let joindo its thing
您可以重命名一个的索引级别并让它join做它的事情
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
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
