如果另一列是NaN，如何替换列中的值? [英] How to replace values in a column if another column is a NaN?

问题描述

因此，这应该是地球上最简单的事情.伪代码:

So this should be the easiest thing on earth. Pseudocode:

Replace column C with NaN if column E is NaN

我知道我可以通过以下方式来做到这一点:拉出E列为NaN的所有数据帧行，替换所有C列，然后将其合并到原始数据集中，但这对于一个简单的操作来说似乎需要做很多工作.为什么这样行不通:

I know I can do this by pulling out all dataframe rows where column E is NaN, replacing all of Column C, and then merging that on the original dataset, but that seems like a lot of work for a simple operation. Why doesn't this work:

样本数据:

dfz = pd.DataFrame({'A' : [1,0,0,1,0,0],
               'B' : [1,0,0,1,0,1],
               'C' : [1,0,0,1,3,1],
               'D' : [1,0,0,1,0,0],
               'E' : [22.0,15.0,None,10.,None,557.0]})

替换功能:

def NaNfunc(dfz):
  if dfz['E'] == None:
    return None
  else:
    return dfz['C']

dfz['C'] = dfz.apply(NaNfunc, axis=1)

以及如何在一行中做到这一点?

And how to do this in one line?

推荐答案

使用np.where:

In [34]:
dfz['C'] = np.where(dfz['E'].isnull(), dfz['E'], dfz['C'])
dfz

Out[34]:
   A  B   C  D    E
0  1  1   1  1   22
1  0  0   0  0   15
2  0  0 NaN  0  NaN
3  1  1   1  1   10
4  0  0 NaN  0  NaN
5  0  1   1  0  557

或者只是掩盖df:

In [38]:
dfz.loc[dfz['E'].isnull(), 'C'] = dfz['E']
dfz

Out[38]:
   A  B   C  D    E
0  1  1   1  1   22
1  0  0   0  0   15
2  0  0 NaN  0  NaN
3  1  1   1  1   10
4  0  0 NaN  0  NaN
5  0  1   1  0  557

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