如何解决与 Series.fillna() 相关的 Pandas 问题？
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How to solve the Pandas issue related to Series.fillna()?
提问by bigbug
I just upgrade from Pandas 0.11 to 0.13.0rc1. The upgration caused one error related to Series.fillna().
我只是从 Pandas 0.11 升级到 0.13.0rc1。升级导致了一个与 Series.fillna() 相关的错误。
>>> df
sales net_pft
STK_ID RPT_Date
600809 20060331 5.8951 1.1241
20060630 8.3031 1.5464
20060930 11.9084 2.2990
20061231 NaN 2.6060
20070331 5.9129 1.3334
[5 rows x 2 columns]
>>> type(df['sales'])
<class 'pandas.core.series.Series'>
>>> df['sales'] = df['sales'].fillna(df['net_pft'])
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "D:\Python27\lib\site-packages\pandas\core\generic.py", line 1912, in fillna
obj.fillna(v, inplace=True)
AttributeError: 'numpy.float64' object has no attribute 'fillna'
>>>
Why df['sales']become 'numpy.float64'object when it is used in fillna()? How to correctly do "fill the NaN of one column with the other column's value" ?
为什么在使用时df['sales']成为'numpy.float64'对象fillna()？如何正确地“用另一列的值填充一列的 NaN”？
采纳答案by joris
There was a recent discussion on this, and it is fixed in pandas master: https://github.com/pydata/pandas/issues/5703(after the release of 0.13rc1, so it will be fixed in final 0.13).
最近有一个讨论，已经在pandas master中修复了：https: //github.com/pydata/pandas/issues/5703（0.13rc1发布后，所以会在final 0.13中修复）。
Note: the behaviour changed! This was not supported behaviour in pandas <= 0.12, as @behzad.nouri points out (using a Series as input to fillna). However it did work but was apparantly based on the location, which was wrong. But as long as both serieses (df['sales']and df['net_pft']in you case) have the same index, this will not matter.
In pandas 0.13, it will be supported but based on the index of the Series. See comment here: https://github.com/pydata/pandas/issues/5703#issuecomment-30663525
注意：行为改变了！正如@behzad.nouri 指出的那样（使用 Series 作为 的输入fillna），这在 pandas <= 0.12 中不受支持。然而，它确实有效，但显然是基于位置，这是错误的。但只要双方个系列（df['sales']和df['net_pft']你的情况下）具有相同的索引，因此关系不大。
在 pandas 0.13 中，它将被支持，但基于系列的索引。请参阅此处的评论：https: //github.com/pydata/pandas/issues/5703#issuecomment-30663525
回答by behzad.nouri
it seems more like what you are trying to do is:
看起来更像是你想要做的是：
idx = df['sales'].isnull( )
df['sales'][ idx ] = df['net_pft'][ idx ]
because what you are providing as valueargument to fillnais a series, the code goes into the bellow branch which calls fillnafor every index item of the provided series. If selfwas a DataFrame this would have worked correctly, that is it would fillnaeach column using the provided series, but since selfhere is a Series it will break.
因为您提供的value参数fillna是一个系列，所以代码进入波纹管分支，该分支调用fillna所提供系列的每个索引项。如果self是一个 DataFrame 这将正常工作，也就是说它会fillna使用提供的系列的每一列，但由于self这里是一个系列，它会中断。
As in the documentationto fillnaa DataFrame the parameter value can be
在DataFrame的文档中，fillna参数值可以是
alternately a dict of values specifying which value to use for each column (columns not in the dict will not be filled).
或者是一个值字典，指定用于每列的值（不在字典中的列将不会被填充）。
from the source code below, if valueis a Series it will work the same way as a dict using the Series' index as keys to fillnacorresponding columns.
从下面的源代码中，如果value是一个系列，它将以与使用系列索引作为fillna对应列的键的字典相同的方式工作。
else: # value is not None
if method is not None:
raise ValueError('cannot specify both a fill method and value')
if len(self._get_axis(axis)) == 0:
return self
if isinstance(value, (dict, com.ABCSeries)):
if axis == 1:
raise NotImplementedError('Currently only can fill '
'with dict/Series column '
'by column')
result = self if inplace else self.copy()
for k, v in compat.iteritems(value):
if k not in result:
continue
obj = result[k]
obj.fillna(v, inplace=True)
return result
else:
new_data = self._data.fillna(value, inplace=inplace,
downcast=downcast)
