实现`df [m] = df [x] + df [y] + df [z]`的更好方法

QM.*_*.py 2 python nan apply pandas

我想得到三列的总和,我采取的方法如下:

In [14]:

a_pd = pd.DataFrame({'a': np.arange(3),
                     'b': [5, 7, np.NAN],
                     'c': [2, 9, 0]})
a_pd
Out[14]:
a   b   c
0   0   5.0 2
1   1   7.0 9
2   2   NaN 0
In [18]:

b_pd = a_pd['a'] + a_pd['b'] + a_pd['c']
b_pd
Out[18]:
0     7.0
1    17.0
2     NaN
dtype: float64
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但正如您所看到的,NaN不能被排除在外.所以我试过np.add(),但有些不对劲:

In [19]:

b_pd = a_pd[['a', 'b', 'c']].apply(np.add, axis=1)
b_pd
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-19-f52f400573b4> in <module>()
----> 1 b_pd = a_pd[['a', 'b', 'c']].apply(np.add, axis=1)
      2 b_pd

F:\anaconda\lib\site-packages\pandas\core\frame.pyc in apply(self, func, axis, broadcast, raw, reduce, args, **kwds)
   4045 
   4046         if isinstance(f, np.ufunc):
-> 4047             results = f(self.values)
   4048             return self._constructor(data=results, index=self.index,
   4049                                      columns=self.columns, copy=False)

ValueError: invalid number of arguments
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所以,我想知道你是如何修复这个bug的.

ayh*_*han 5

您可以使用DataFrame的sum方法:

a_pd.sum(axis=1)
Out: 
0     7.0
1    17.0
2     2.0
dtype: float64
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如果要指定列:

a_pd[['a', 'b', 'c']].sum(axis=1)
Out: 
0     7.0
1    17.0
2     2.0
dtype: float64
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