Pandas - Groupby多列

时间:2015-11-13 22:33:29

标签: python pandas

我正在尝试按多列进行分组,并将它们聚合在一起,以便在分组后成为列表。

目前,DataFrame看起来像这样:

enter image description here

我试过用这个:

grouped = DataFrame.groupby(['jobname', 'block'], axis=0)
DataFrame= grouped.aggregate(lambda x: list(x))

但是,当我在IPython中应用它时,它给了我这个错误:

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-221-97113b757fa1> in <module>()
----> 1 cassandraFrame_2 = grouped.aggregate(lambda x: list(x))
      2 cassandraFrame_2

/usr/local/lib/python2.7/dist-packages/pandas/core/groupby.pyc in aggregate(self, arg, *args, **kwargs)
   2867 
   2868             if self.grouper.nkeys > 1:
-> 2869                 return self._python_agg_general(arg, *args, **kwargs)
   2870             else:
   2871 

/usr/local/lib/python2.7/dist-packages/pandas/core/groupby.pyc in _python_agg_general(self, func, *args, **kwargs)
   1166         for name, obj in self._iterate_slices():
   1167             try:
-> 1168                 result, counts = self.grouper.agg_series(obj, f)
   1169                 output[name] = self._try_cast(result, obj)
   1170             except TypeError:

/usr/local/lib/python2.7/dist-packages/pandas/core/groupby.pyc in agg_series(self, obj, func)
   1633             return self._aggregate_series_fast(obj, func)
   1634         except Exception:
-> 1635             return self._aggregate_series_pure_python(obj, func)
   1636 
   1637     def _aggregate_series_fast(self, obj, func):

/usr/local/lib/python2.7/dist-packages/pandas/core/groupby.pyc in _aggregate_series_pure_python(self, obj, func)
   1667                 if (isinstance(res, (Series, Index, np.ndarray)) or
   1668                         isinstance(res, list)):
-> 1669                     raise ValueError('Function does not reduce')
   1670                 result = np.empty(ngroups, dtype='O')
   1671 

ValueError: Function does not reduce

最终,我想将相同的作业名分组,并阻塞在一起,但数据是一个元组列表,现在它是一个3项元组。

例如:

jobname       block         data
Complete-Test Simple_buff   (tuple_1)
Complete-Test Simple_buff   (tuple_2)

骨料:

jobname       block         data
Complete-Test Simple_buff   [(tuple_1),(tuple_2)]

我可以按jobname分组,但是,这会将block聚合在一起,但我想保持blocks分开。

有人能指出我正确的方向吗?

由于

1 个答案:

答案 0 :(得分:6)

看起来显式检查聚合函数返回的值不是SeriesIndexnp.ndarraylist

因此,以下内容应该有效:

grouped = df.groupby(['jobname', 'block'])
aggregated = grouped.aggregate(lambda x: tuple(x))