为什么在聚合不存在的列时,pandas会为列值提供NaN?

时间:2014-05-27 17:16:24

标签: python pandas

我想在下面的DataFrame中按字母数字求和:

In [10]: df
Out[10]:
  letter  number
0      A       1
1      A       2
2      B       3
3      B       4
4      C       5
5      C       6

[6 rows x 2 columns]

这很容易实现:

In [11]: df.groupby('letter')[['number']].sum()
Out[11]:
        number
letter
A            3
B            7
C           11

[3 rows x 1 columns]

但如果我拼错了我的专栏,我会得到NaN个值:

In [12]: df.groupby('letter')[['numberrrrr']].sum()
Out[12]:
        numberrrrr
letter
A              NaN
B              NaN
C              NaN

[3 rows x 1 columns]

这导致我们的团队非常追逐以确定错误的位置。相反,我们想要一个错误陈述,如:

In [13]: df.groupby('letter')['numberrrrr'].sum()
---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)
<ipython-input-13-8ebcdeee8710> in <module>()
----> 1 df.groupby('letter')['numberrrrr'].sum()

/usr/local/Anaconda/lib/python2.7/site-packages/pandas/core/groupby.pyc in __getitem__(self, key)
   2475         else:
   2476             if key not in self.obj:  # pragma: no cover
-> 2477                 raise KeyError(str(key))
   2478             # kind of a kludge
   2479             return SeriesGroupBy(self.obj[key], selection=key,

KeyError: 'numberrrrr'

当请求的列丢失时,是否有任何特殊原因从聚合返回DataFrame不会导致错误?

这是关于pandas 0.13.1。

1 个答案:

答案 0 :(得分:3)

这在master / 0.14.0(本周结束)中修复;如果您想尝试

,则rc1为here
In [7]: df.groupby('letter')[['number']].sum()
Out[7]: 
        number
letter        
A            3
B            7
C           11

In [8]: df.groupby('letter')[['numberrrr']].sum()
KeyError: "Columns not found: 'numberrrr'"

In [9]: pd.__version__
Out[9]: '0.14.0rc1-43-g0dec048'