pandas 创建一个空的 MultiIndex
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Creating an empty MultiIndex
提问by dmvianna
I would like to create an emptyDataFramewith a MultiIndexbefore 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 namesduring 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:
我想在为其分配行之前创建一个带有MultiIndex的空DataFrame。我已经发现空的 DataFrame 不喜欢动态分配 MultiIndex，所以我在创建过程中设置了 MultiIndex名称。但是，我不想分配levels，因为这将在以后完成。这是迄今为止我得到的最好的代码：
def empty_multiindex(names):
"""
Creates empty MultiIndex from a list of level names.
"""
return MultiIndex.from_tuples(tuples=[(None,) * len(names)], names=names)
Which gives me
这给了我
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'])
and
和
In [3]:
DataFrame(index=empty_multiindex(['one','two', 'three']))
Out[3]:
one two three
NaN NaN NaN
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?
好吧，我对这些 NaN 没有用。稍后我可以轻松地删除它们，但这显然是一种骇人听闻的解决方案。有人有更好的吗？
回答by RoG
The solution is to leave out the labels. This works fine for me:
解决办法是去掉标签。这对我来说很好用：
>>> 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
Hope that helps!
希望有帮助！
回答by Jean Paul
Another solution which is maybe a little simpler is to use the function set_index:
另一个可能更简单的解决方案是使用该函数set_index：
>>> import pandas as pd
>>> df = pd.DataFrame(columns=['one', 'two', 'three', 'alpha', 'beta'])
>>> df = df.set_index(['one', 'two', 'three'])
>>> 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
回答by mcsoini
Using pd.MultiIndex.from_arraysallows for a slightly more concise solution when defining the index explicitly:
pd.MultiIndex.from_arrays在显式定义索引时，使用允许稍微更简洁的解决方案：
import pandas as pd
ind = pd.MultiIndex.from_arrays([[]] * 3, names=(u'one', u'two', u'three'))
df = pd.DataFrame(columns=['alpha', 'beta'], index=ind)
df.loc[('apple','banana','cherry'), :] = [4, 3]
alpha beta
one two three
apple banana cherry 4 3
