我正在尝试在熊猫系列上进行Group_by，然后对其进行排名 . 奇怪的是，这在以前版本的熊猫中有效，但自从我们将熊猫版本升级到0.14.0后，现在已停止工作了
这是一个例子
该系列
i1 = pd.MultiIndex(levels=[[0, 1, 2, 3], [u'A', u'B'], [u'Spar', u'PnP', 'Checkers', 'Woolworths']],
labels=[[0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3],
[0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1],
[0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3]],
names=[u'respondent', u'survey', u'brand'])
s1 = pd.Series.from_array([1, 2, 3, 4, 2, 3, 4, 1, 3, 4, 2, 1, 4, 1, 2, 3, 1, 2, 3, 4, 3, 2, 1, 4, 2, 3, 4, 1, 1, 4, 3, 2], index = i1, name='usage')
s1
respondent survey brand
0 A Spar 1
PnP 2
Checkers 3
Woolworths 4
B Spar 2
PnP 3
Checkers 4
Woolworths 1
1 A Spar 3
PnP 4
Checkers 2
Woolworths 1
B Spar 4
PnP 1
Checkers 2
Woolworths 3
2 A Spar 1
PnP 2
Checkers 3
Woolworths 4
B Spar 3
PnP 2
Checkers 1
Woolworths 4
3 A Spar 2
PnP 3
Checkers 4
Woolworths 1
B Spar 1
PnP 4
Checkers 3
Woolworths 2
Name: usage, dtype: int64
当我尝试按照以下方式进行分组时
s1.groupby(['respondent']).rank()
我收到以下错误
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
<ipython-input-41-14bf5be195e8> in <module>()
----> 1 s1.groupby(['respondent']).mean()
/Users/donovanthomson/anaconda/lib/python2.7/site-packages/pandas/core/generic.pyc in groupby(self, by, axis, level, as_index, sort, group_keys, squeeze)
2727 axis = self._get_axis_number(axis)
2728 return groupby(self, by, axis=axis, level=level, as_index=as_index,
-> 2729 sort=sort, group_keys=group_keys, squeeze=squeeze)
2730
2731 def asfreq(self, freq, method=None, how=None, normalize=False):
/Users/donovanthomson/anaconda/lib/python2.7/site-packages/pandas/core/groupby.pyc in groupby(obj, by, **kwds)
1098 raise TypeError('invalid type: %s' % type(obj))
1099
-> 1100 return klass(obj, by, **kwds)
1101
1102
/Users/donovanthomson/anaconda/lib/python2.7/site-packages/pandas/core/groupby.pyc in __init__(self, obj, keys, axis, level, grouper, exclusions, selection, as_index, sort, group_keys, squeeze)
384 if grouper is None:
385 grouper, exclusions, obj = _get_grouper(obj, keys, axis=axis,
--> 386 level=level, sort=sort)
387
388 self.obj = obj
/Users/donovanthomson/anaconda/lib/python2.7/site-packages/pandas/core/groupby.pyc in _get_grouper(obj, key, axis, level, sort)
1978 exclusions.append(gpr)
1979 name = gpr
-> 1980 gpr = obj[gpr]
1981
1982 if isinstance(gpr, Categorical) and len(gpr) != len(obj):
/Users/donovanthomson/anaconda/lib/python2.7/site-packages/pandas/core/series.pyc in __getitem__(self, key)
477 def __getitem__(self, key):
478 try:
--> 479 result = self.index.get_value(self, key)
480
481 if not np.isscalar(result):
/Users/donovanthomson/anaconda/lib/python2.7/site-packages/pandas/core/index.pyc in get_value(self, series, key)
2554 raise InvalidIndexError(key)
2555 else:
-> 2556 raise e1
2557 except Exception: # pragma: no cover
2558 raise e1
KeyError: 'respondent'
1 回答
您需要将索引级别分组，而不是不存在的列：
如果您愿意，也可以使用
s1.groupby(level='respondent').rank()从技术上讲，我认为它不应该在以前的版本中工作，因为它在索引级别与列上的语义不同分组