我有类似如下的数据:
A B C
0 M M M
1 Y M M
2 Y NaN NaN
3 Y Y etc
我需要的是:
A B C F
0 M M M 3
1 Y M M 4
2 Y NaN NaN 0
3 Y Y etc 5
我不知道如何处理 rows[2, 3],这里我列出我用过但不起作用的代码如下:
df.loc[df['A'] == 'M', 'F'] = '3'
df.loc[((df.A != 'M') & (df.B == 'M')), 'F'] = '4'
df.loc[(df.A != 'M') & (df.B != 'M') & (df.C != ''), 'F'] = '5'
df.loc[(df.A != 'M') & (df.B != 'M') & (df.C == ''), 'F'] = '0'
使用numpy.select
与 Series.notna
:
m1 = df['A'] == 'M'
m2 = df['B'] == 'M'
m3 = df['C'].notna()
df['F'] = np.select([m1, m2, m3], ['3','4','5'], default='0')
print (df)
A B C F
0 M M M 3
1 Y M M 4
2 Y NaN NaN 0
3 Y Y etc 5
如有必要,添加更多条件使用~
通过按位与
- &
反转掩码和链:
m1 = df['A'] == 'M'
m2 = df['B'] == 'M'
m3 = df['C'].notna()
m11 = ~m1
m22 = ~m2
m33 = ~m3
df['F'] = np.select([m1, m2 & m11, m3 & m11 & m22], ['3','4','5'], default='0')
编辑:
您的解决方案可能会被 Series.isna
更改和 Series.notna
:
df.loc[df['A'] == 'M', 'F'] = '3'
df.loc[((df.A != 'M') & (df.B == 'M')), 'F'] = '4'
df.loc[(df.A != 'M') & (df.B != 'M') & (df.C.notna()), 'F'] = '5'
df.loc[(df.A != 'M') & (df.B != 'M') & (df.C.isna()), 'F'] = '0'
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