问题描述
受此答案的启发,并且缺少对我发现自己写了一些语法糖,使生活更容易按MultiIndex级别进行过滤. /p>
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),请参见此处
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 heree.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]
这篇关于按MultiIndex级别或子级别对pandas DataFrame进行切片的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!