python - 基于忽略缺失值的多个条件的新列

Question

我有以下带有一些缺失值的数据框:

       A     B
0  63.0   9.0
1   NaN  35.0
2  51.0  95.0
3  25.0  11.0
4  91.0   NaN
5   2.0  47.0
6  37.0  10.0
7   NaN  88.0
8  75.0  87.0
9  92.0  21.0

我想根据以上两列的条件创建一个新列:

df['C'] = numpy.where((df['A']>55) | (df['B']>55), "Yes", "No")

这有效但没有考虑缺失值:

      A     B    C
0  63.0   9.0  Yes
1   NaN  35.0   No
2  51.0  95.0  Yes
3  25.0  11.0   No
4  91.0   NaN  Yes
5   2.0  47.0   No
6  37.0  10.0   No
7   NaN  88.0  Yes
8  75.0  87.0  Yes
9  92.0  21.0  Yes

为了纠正缺失值，我必须运行以下代码:

df['C'] = numpy.where((df['A'].isnull()) | (df['B'].isnull()), numpy.nan, df['C'])

然后我得到适当的新专栏:

      A     B    C
0  63.0   9.0  Yes
1   NaN  35.0  NaN
2  51.0  95.0  Yes
3  25.0  11.0   No
4  91.0   NaN  NaN
5   2.0  47.0   No
6  37.0  10.0   No
7   NaN  88.0  NaN
8  75.0  87.0  Yes
9  92.0  21.0  Yes

有没有更好的方法可以让我仅在单行代码中更正缺失值？

Answer 1

使用 np.select。条件的顺序在这里很重要，因为 np.select 选择第一个有效选项，所以您必须先进行 null 检查。

c1 = df.isnull().any(1)
c2 = df.gt(55).any(1)

df['C'] = np.select([c1, c2], [np.nan, 'Yes'], 'No')

      A     B    C
0  63.0   9.0  Yes
1   NaN  35.0  nan
2  51.0  95.0  Yes
3  25.0  11.0   No
4  91.0   NaN  nan
5   2.0  47.0   No
6  37.0  10.0   No
7   NaN  88.0  nan
8  75.0  87.0  Yes
9  92.0  21.0  Yes