如何 select NaN 行并将 value_A 分配给新列？代码 df.loc['condition_1' & 'condition_2' & 'df.column_a == ''] = 'value_a'

Question

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

我不知道如何处理行[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'

Answer 1

使用numpy.select with 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 ~ 反转掩码并按 bitwise AND - &:

链接

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因 Series.isna 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'

如何 select NaN 行并将 value_A 分配给新列？代码 df.loc['condition_1' & 'condition_2' & 'df.column_a == ''] = 'value_a'

how to select NaN rows and assign value_A to a new column? code df.loc['condition_1' & 'condition_2' & 'df.column_a == ''] = 'value_a'

python

nan

pandas

pandas-loc