9

我无法对熊猫系列对象进行分组。DataFrames 很好，但我似乎无法对 Series 进行 groupby。有没有人能让这个工作？

>>> import pandas as pd
>>> a = pd.Series([1,2,3,4], index=[4,3,2,1])
>>> a
4    1
3    2
2    3
1    4
dtype: int64
>>> a.groupby()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/share/apps/install/anaconda/lib/python2.7/site-packages/pandas/core/generic.py", line 153, in groupby
    sort=sort, group_keys=group_keys)
  File "/share/apps/install/anaconda/lib/python2.7/site-packages/pandas/core/groupby.py", line 537, in groupby
    return klass(obj, by, **kwds)
  File "/share/apps/install/anaconda/lib/python2.7/site-packages/pandas/core/groupby.py", line 195, in __init__
    level=level, sort=sort)
  File "/share/apps/install/anaconda/lib/python2.7/site-packages/pandas/core/groupby.py", line 1326, in _get_grouper
    ping = Grouping(group_axis, gpr, name=name, level=level, sort=sort)
  File "/share/apps/install/anaconda/lib/python2.7/site-packages/pandas/core/groupby.py", line 1203, in __init__
    self.grouper = self.index.map(self.grouper)
  File "/share/apps/install/anaconda/lib/python2.7/site-packages/pandas/core/index.py", line 878, in map
    return self._arrmap(self.values, mapper)
  File "generated.pyx", line 2200, in pandas.algos.arrmap_int64 (pandas/algos.c:61221)
TypeError: 'NoneType' object is not callable
4

3 回答 3

13

您需要传递某种映射（可能是字典/函数/索引）

In [6]: a
Out[6]: 
4    1
3    2
2    3
1    4
dtype: int64

In [7]: a.groupby(a.index).sum()
Out[7]: 
1    4
2    3
3    2
4    1
dtype: int64

In [3]: a.groupby(lambda x: x % 2 == 0).sum()
Out[3]: 
False    6
True     4
dtype: int64
于 2013-07-29T16:45:20.060 回答
6

如果您需要按系列的值分组：

grouped = a.groupby(a)

或者

grouped = a.groupby(lambda x: a[x])
于 2016-08-31T11:08:17.703 回答
0

不要太认真地回答这个问题；）我并不是说这是一个好主意。

如果你真的想内联，或者以“流利”的方式，你可以做这样的事情。

def smart_groupby(self, by=None, *args, **kwargs):
    if by is None:
        return self.groupby(self, *args, **kwargs)
    return self.groupby(by, *args, **kwargs)

import pandas as pd
ps.Series.groupby = smart_groupby

pd.Series(['a', 'a', 'a', 'b', 'b']).groupby().count()

结果是

a    3
b    2
dtype: int64

它应该像往常一样运行，但如果您省略by基于自身的 it 组，它还有一个额外的好处。

于 2019-06-12T14:51:27.050 回答