问题
Say I have a dictionary that looks like this:
dictionary = {\'A\' : {\'a\': [1,2,3,4,5],
\'b\': [6,7,8,9,1]},
\'B\' : {\'a\': [2,3,4,5,6],
\'b\': [7,8,9,1,2]}}
and I want a dataframe that looks something like this:
A B
a b a b
0 1 6 2 7
1 2 7 3 8
2 3 8 4 9
3 4 9 5 1
4 5 1 6 2
Is there a convenient way to do this? If I try:
In [99]:
DataFrame(dictionary)
Out[99]:
A B
a [1, 2, 3, 4, 5] [2, 3, 4, 5, 6]
b [6, 7, 8, 9, 1] [7, 8, 9, 1, 2]
I get a dataframe where each element is a list. What I need is a multiindex where each level corresponds to the keys in the nested dict and the rows corresponding to each element in the list as shown above. I think I can work a very crude solution but I\'m hoping there might be something a bit simpler.
回答1:
Pandas wants the MultiIndex values as tuples, not nested dicts. The simplest thing is to convert your dictionary to the right format before trying to pass it to DataFrame:
>>> reform = {(outerKey, innerKey): values for outerKey, innerDict in dictionary.iteritems() for innerKey, values in innerDict.iteritems()}
>>> reform
{('A', 'a'): [1, 2, 3, 4, 5],
('A', 'b'): [6, 7, 8, 9, 1],
('B', 'a'): [2, 3, 4, 5, 6],
('B', 'b'): [7, 8, 9, 1, 2]}
>>> pandas.DataFrame(reform)
A B
a b a b
0 1 6 2 7
1 2 7 3 8
2 3 8 4 9
3 4 9 5 1
4 5 1 6 2
[5 rows x 4 columns]
回答2:
dict_of_df = {k: pd.DataFrame(v) for k,v in dictionary.items()}
df = pd.concat(dict_of_df, axis=1)
Note that the order of columns is lost for python < 3.6
来源:https://stackoverflow.com/questions/24988131/nested-dictionary-to-multiindex-dataframe-where-dictionary-keys-are-column-label