entityMap|blocks|key|9io08|text|Pandas希望将MultiIndex值作为元组，而不是嵌套的字典。最简单的做法是在尝试将字典传递给DataFrame之前将其转换为正确的格式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|8rq0i|>>>+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]|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

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:

<pre><code>&gt;&gt;&gt; reform = {(outerKey, innerKey): values for outerKey, innerDict in dictionary.iteritems() for innerKey, values in innerDict.iteritems()}
&gt;&gt;&gt; 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]}
&gt;&gt;&gt; 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]
</code></pre>

entityMap|blocks|key|7hl7o|text|dict_of_df+=+{k:+pd.DataFrame(v)+for+k,v+in+dictionary.items()}
df+=+pd.concat(dict_of_df,+axis=1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|73igt|请注意，对于低于3.6的python，列的顺序会丢失|unstyled^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|I|9|@]|A|@]|B|$]]]]

<pre><code>dict_of_df = {k: pd.DataFrame(v) for k,v in dictionary.items()}
df = pd.concat(dict_of_df, axis=1)
</code></pre>

Note that the order of columns is lost for python &lt; 3.6

entityMap|blocks|key|f6ebb|text|这个答案有点晚了，但是...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|c73pu|您正在寻找的功能是.stack|offset|length|style|CODE|b5o3q|df+=+pandas.DataFrame.from_dict(dictionary,+orient="index").stack().to_frame()
#+to+break+out+the+lists+into+columns
df+=+pd.DataFrame(df[0].values.tolist(),+index=df.index)|code-block|syntax|javascript^0|0|9|6|0^^$0|$]|1|@$2|3|4|5|6|7|8|N|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|O|9|@$E|P|F|Q|G|H]]|A|@]|B|$]]|$2|I|4|J|6|K|8|R|9|@]|A|@]|B|$L|M]]]]

This answer is a little late to the game, but...

You're looking for the functionality in <code>.stack</code>:

<pre><code>df = pandas.DataFrame.from_dict(dictionary, orient="index").stack().to_frame()
# to break out the lists into columns
df = pd.DataFrame(df[0].values.tolist(), index=df.index)
</code></pre>

entityMap|blocks|key|f0b3p|text|这个递归函数应该可以工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2ugpi|def+reform_dict(dictionary,+t=tuple(),+reform={}):
++++for+key,+val+in+dictionary.items():
++++++++t+=+t+%2B+(key,)
++++++++if+isinstance(val,+dict):
++++++++++++reform_dict(val,+t,+reform)
++++++++else:
++++++++++++reform.update({t:+val})
++++++++t+=+t[:-1]
++++return+reform|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

This recursive function should work:
<pre><code>def reform_dict(dictionary, t=tuple(), reform={}):
 for key, val in dictionary.items():
 t = t + (key,)
 if isinstance(val, dict):
 reform_dict(val, t, reform)
 else:
 reform.update({t: val})
 t = t[:-1]
 return reform
</code></pre>

entityMap|blocks|key|3l0mj|text|如果字典中的列表长度不同，可以采用BrenBarn的方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7k26r|>>>+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]}}

>>>+reform+=+{(outerKey,+innerKey):+values+for+outerKey,+innerDict+in+dictionary.items()+for+innerKey,+values+in+innerDict.items()}
>>>+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]}

>>>+pandas.DataFrame.from_dict(reform,+orient='index').transpose()
>>>+df.columns+=+pd.MultiIndex.from_tuples(df.columns)
+++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++NaN
[5+rows+x+4+columns]|code-block|syntax|javascript^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|I|9|@]|A|@]|B|$F|G]]]]

If lists in the dictionary are not of the same lenght, you can adapte the method of BrenBarn.
<pre><code>&gt;&gt;&gt; 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]}}

&gt;&gt;&gt; reform = {(outerKey, innerKey): values for outerKey, innerDict in dictionary.items() for innerKey, values in innerDict.items()}
&gt;&gt;&gt; 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]}

&gt;&gt;&gt; pandas.DataFrame.from_dict(reform, orient='index').transpose()
&gt;&gt;&gt; df.columns = pd.MultiIndex.from_tuples(df.columns)
 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 NaN
[5 rows x 4 columns]
</code></pre>

Say I have a dictionary that looks like this:

<pre><code>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]}}
</code></pre>

and I want a dataframe that looks something like this:

<pre><code> 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
</code></pre>

Is there a convenient way to do this? If I try:

<pre><code>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]
</code></pre>

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.

Nested dictionary to multiindex dataframe where dictionary keys are column labels

 假设我有一个字典，如下所示： dictionary = {'A' : {'a': [1,2,3,4,5],                     'b': [6,7,8,9,1]},              'B' : {'a': [2,3,4,5,6],                     'b': [7,...

问从字典到多索引数据帧的嵌套字典，其中字典键是列标签
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