给出两个pandas系列对象A和Matches。匹配包含A的索引的子集并具有布尔条目。如何做一个逻辑索引的等效?
如果匹配的长度与A相同,则可以使用:
A[Matches] = 5.*Matches
匹配时间短于A:
error: Unalignable boolean Series key provided
In [15]: A = pd.Series(range(10))
In [16]: A
Out[16]: 0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
dtype: int64
In [17]: Matches = (A<3)[:5]
In [18]: Matches
Out[18]: 0 True
1 True
2 True
3 False
4 False
dtype: bool
In [19]: A[Matches] = None
---------------------------------------------------------------------------
IndexingError Traceback (most recent call last)
<ipython-input-19-7a04f32ce860> in <module>()
----> 1 A[Matches] = None
C:\Anaconda\lib\site-packages\pandas\core\series.py in __setitem__(self, key, value)
631
632 if _is_bool_indexer(key):
--> 633 key = _check_bool_indexer(self.index, key)
634 try:
635 self.where(~key, value, inplace=True)
C:\Anaconda\lib\site-packages\pandas\core\indexing.py in _check_bool_indexer(ax, key)
1379 mask = com.isnull(result.values)
1380 if mask.any():
-> 1381 raise IndexingError('Unalignable boolean Series key provided')
1382
1383 result = result.astype(bool).values
IndexingError: Unalignable boolean Series key provided
In [20]:
我要找的结果是:
In [16]: A
Out[16]: 0 None
1 None
2 None
3 3
4 4
5 5
6 6
7 7
8 8
9 9
dtype: int64
匹配系列的构造是人为的,仅供参考。另外,在我的例子中,行索引显然是非数字的,不等于元素值......
答案 0 :(得分:4)
嗯,你不能拥有你想要的东西,因为int64
不是包含None的系列的可能dtype。 None不是整数。但你可以近距离接触:
>>> A = pd.Series(range(10))
>>> Matches = (A<3)[:5]
>>> A[Matches[Matches].index] = None
>>> A
0 None
1 None
2 None
3 3
4 4
5 5
6 6
7 7
8 8
9 9
dtype: object
哪个有效,因为Matches[Matches]
选择Matches
的元素为真。