创建一个空的MultiIndex [英] Creating an empty MultiIndex

问题描述

我想在为其分配行之前使用 MultiIndex 创建一个空 DataFrame .我已经发现空的DataFrame不喜欢动态分配MultiIndex，因此我在创建过程中设置了MultiIndex 名称.但是，我不想分配级别，因为这将在以后完成.这是到目前为止我得到的最好的代码:

I would like to create an empty DataFrame with a MultiIndex before assigning rows to it. I already found that empty DataFrames don't like to be assigned MultiIndexes on the fly, so I'm setting the MultiIndex names during creation. However, I don't want to assign levels, as this will be done later. This is the best code I got to so far:

def empty_multiindex(names):
    """
    Creates empty MultiIndex from a list of level names.
    """
    return MultiIndex.from_tuples(tuples=[(None,) * len(names)], names=names)

哪个给我

In [2]:

empty_multiindex(['one','two', 'three'])

Out[2]:

MultiIndex(levels=[[], [], []],
           labels=[[-1, -1, -1], [-1, -1, -1], [-1, -1, -1]],
           names=[u'one', u'two', u'three'])

和

In [3]:
DataFrame(index=empty_multiindex(['one','two', 'three']))

Out[3]:
one two three
NaN NaN NaN

好吧，我对这些NaN毫无用处.我稍后可以轻松删除它们，但这显然是一个棘手的解决方案.任何人都有更好的吗?

Well, I have no use for these NaNs. I can easily drop them later, but this is obviously a hackish solution. Anyone has a better one?

推荐答案

解决方案是省略标签.这对我来说很好:

The solution is to leave out the labels. This works fine for me:

>>> my_index = pd.MultiIndex(levels=[[],[],[]],
                             labels=[[],[],[]],
                             names=[u'one', u'two', u'three'])
>>> my_index
MultiIndex(levels=[[], [], []],
           labels=[[], [], []],
           names=[u'one', u'two', u'three'])
>>> my_columns = [u'alpha', u'beta']
>>> df = pd.DataFrame(index=my_index, columns=my_columns)
>>> df
Empty DataFrame
Columns: [alpha, beta]
Index: []
>>> df.loc[('apple','banana','cherry'),:] = [0.1, 0.2]
>>> df
                    alpha beta
one   two    three            
apple banana cherry   0.1  0.2

希望有帮助！

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