按MultiIndex级别或子级别对pandas DataFrame进行切片 [英] Slice pandas DataFrame by MultiIndex level or sublevel

问题描述

受此答案的启发，并且缺少对

Inspired by this answer and the lack of an easy answer to this question I found myself writing a little syntactic sugar to make life easier to filter by MultiIndex level.

def _filter_series(x, level_name, filter_by):
    """
    Filter a pd.Series or pd.DataFrame x by `filter_by` on the MultiIndex level
    `level_name`

    Uses `pd.Index.get_level_values()` in the background. `filter_by` is either
    a string or an iterable.
    """
    if isinstance(x, pd.Series) or isinstance(x, pd.DataFrame):
        if type(filter_by) is str:
            filter_by = [filter_by]

        index = x.index.get_level_values(level_name).isin(filter_by)
        return x[index]
    else:
        print "Not a pandas object"

但是，如果我认识熊猫开发团队(而且我正在慢慢地开始！)，已经有一种不错的方法可以做到这一点，而我只是不知道它是什么！

But if I know the pandas development team (and I'm starting to, slowly!) there's already a nice way to do this, and I just don't know what it is yet!

我说得对吗?

推荐答案

使用master/0.14中的新多索引切片器非常容易(即将发布)，请参见

This is very easy using the new multi-index slicers in master/0.14 (releasing soon), see here

存在一个开放的问题，使其在语法上更容易(不难做到)，请参见此处 例如这样的内容:df.loc[{ 'third' : ['C1','C3'] }]我认为是合理的

There is an open issue to make this syntatically easier (its not hard to do), see here e.g something like this: df.loc[{ 'third' : ['C1','C3'] }] I think is reasonable

这是您的操作方法(需要master/0.14):

Here's how you can do it (requires master/0.14):

In [2]: def mklbl(prefix,n):
   ...:     return ["%s%s" % (prefix,i)  for i in range(n)]
   ...: 


In [11]: index = MultiIndex.from_product([mklbl('A',4),
mklbl('B',2),
mklbl('C',4),
mklbl('D',2)],names=['first','second','third','fourth'])

In [12]: columns = ['value']

In [13]: df = DataFrame(np.arange(len(index)*len(columns)).reshape((len(index),len(columns))),index=index,columns=columns).sortlevel()

In [14]: df
Out[14]: 
                           value
first second third fourth       
A0    B0     C0    D0          0
                   D1          1
             C1    D0          2
                   D1          3
             C2    D0          4
                   D1          5
             C3    D0          6
                   D1          7
      B1     C0    D0          8
                   D1          9
             C1    D0         10
                   D1         11
             C2    D0         12
                   D1         13
             C3    D0         14
                   D1         15
A1    B0     C0    D0         16
                   D1         17
             C1    D0         18
                   D1         19
             C2    D0         20
                   D1         21
             C3    D0         22
                   D1         23
      B1     C0    D0         24
                   D1         25
             C1    D0         26
                   D1         27
             C2    D0         28
                   D1         29
             C3    D0         30
                   D1         31
A2    B0     C0    D0         32
                   D1         33
             C1    D0         34
                   D1         35
             C2    D0         36
                   D1         37
             C3    D0         38
                   D1         39
      B1     C0    D0         40
                   D1         41
             C1    D0         42
                   D1         43
             C2    D0         44
                   D1         45
             C3    D0         46
                   D1         47
A3    B0     C0    D0         48
                   D1         49
             C1    D0         50
                   D1         51
             C2    D0         52
                   D1         53
             C3    D0         54
                   D1         55
      B1     C0    D0         56
                   D1         57
             C1    D0         58
                   D1         59
                             ...

[64 rows x 1 columns]

在所有级别上创建索引器，选择所有条目

Create an indexer across all of the levels, selecting all entries

In [15]: indexer = [slice(None)]*len(df.index.names)

使我们关注的级别只有我们关注的条目

Make the level we care about only have the entries we care about

In [16]: indexer[df.index.names.index('third')] = ['C1','C3']

选择它(重要的是这是一个元组！)

Select it (its important that this is a tuple!)

In [18]: df.loc[tuple(indexer),:]
Out[18]: 
                           value
first second third fourth       
A0    B0     C1    D0          2
                   D1          3
             C3    D0          6
                   D1          7
      B1     C1    D0         10
                   D1         11
             C3    D0         14
                   D1         15
A1    B0     C1    D0         18
                   D1         19
             C3    D0         22
                   D1         23
      B1     C1    D0         26
                   D1         27
             C3    D0         30
                   D1         31
A2    B0     C1    D0         34
                   D1         35
             C3    D0         38
                   D1         39
      B1     C1    D0         42
                   D1         43
             C3    D0         46
                   D1         47
A3    B0     C1    D0         50
                   D1         51
             C3    D0         54
                   D1         55
      B1     C1    D0         58
                   D1         59
             C3    D0         62
                   D1         63

[32 rows x 1 columns]

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