我有一个DataFrames字典，其中键被认为是有意义的：
In [32]: x = pd.DataFrame(dict(foo=[1,2,3], bar=[4,5,6])).set_index('foo') In [33]: y = pd.DataFrame(dict(foo=[7,8,9], bar=[10,11,12])).set_index('foo') In [34]: z = dict(x=x, y=y)看起来像：
In [43]: x Out[43]: bar foo 1 4 2 5 3 6 In [44]: y Out[44]: bar foo 7 10 8 11 9 12有一个很好的方法来获取以下DataFrame：
foo bar x 1 4 2 5 3 6 y 7 10 8 11 9 12I have a dictionary of DataFrames, where the keys are assumed to be meaningful:
In [32]: x = pd.DataFrame(dict(foo=[1,2,3], bar=[4,5,6])).set_index('foo') In [33]: y = pd.DataFrame(dict(foo=[7,8,9], bar=[10,11,12])).set_index('foo') In [34]: z = dict(x=x, y=y)Which looks like:
In [43]: x Out[43]: bar foo 1 4 2 5 3 6 In [44]: y Out[44]: bar foo 7 10 8 11 9 12Is there a nice way to get the following DataFrame:
foo bar x 1 4 2 5 3 6 y 7 10 8 11 9 12最满意答案
您可以使用concat ，字典的键将自动用于新的索引级别：
In [6]: z = dict(x=x, y=y) In [7]: pd.concat(z) Out[7]: bar foo x 1 4 2 5 3 6 y 7 10 8 11 9 12您还可以使用concat的names参数为此新索引级别指定名称（例如names=['key'] ）。
You can use concat for this, and the keys of the dictionary will automatically be used for a new index level:
In [6]: z = dict(x=x, y=y) In [7]: pd.concat(z) Out[7]: bar foo x 1 4 2 5 3 6 y 7 10 8 11 9 12You can also give this new index level a name using the names argument of concat (eg names=['key']).
从DataFrames字典创建DataFrame(Create a DataFrame from a dictionary of DataFrames)我有一个DataFrames字典，其中键被认为是有意义的：
In [32]: x = pd.DataFrame(dict(foo=[1,2,3], bar=[4,5,6])).set_index('foo') In [33]: y = pd.DataFrame(dict(foo=[7,8,9], bar=[10,11,12])).set_index('foo') In [34]: z = dict(x=x, y=y)看起来像：
In [43]: x Out[43]: bar foo 1 4 2 5 3 6 In [44]: y Out[44]: bar foo 7 10 8 11 9 12有一个很好的方法来获取以下DataFrame：
foo bar x 1 4 2 5 3 6 y 7 10 8 11 9 12I have a dictionary of DataFrames, where the keys are assumed to be meaningful:
In [32]: x = pd.DataFrame(dict(foo=[1,2,3], bar=[4,5,6])).set_index('foo') In [33]: y = pd.DataFrame(dict(foo=[7,8,9], bar=[10,11,12])).set_index('foo') In [34]: z = dict(x=x, y=y)Which looks like:
In [43]: x Out[43]: bar foo 1 4 2 5 3 6 In [44]: y Out[44]: bar foo 7 10 8 11 9 12Is there a nice way to get the following DataFrame:
foo bar x 1 4 2 5 3 6 y 7 10 8 11 9 12最满意答案
您可以使用concat ，字典的键将自动用于新的索引级别：
In [6]: z = dict(x=x, y=y) In [7]: pd.concat(z) Out[7]: bar foo x 1 4 2 5 3 6 y 7 10 8 11 9 12您还可以使用concat的names参数为此新索引级别指定名称（例如names=['key'] ）。
You can use concat for this, and the keys of the dictionary will automatically be used for a new index level:
In [6]: z = dict(x=x, y=y) In [7]: pd.concat(z) Out[7]: bar foo x 1 4 2 5 3 6 y 7 10 8 11 9 12You can also give this new index level a name using the names argument of concat (eg names=['key']).
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