Question

            A         B       C    D
0  2002-01-13  15:00:00  Joseph  3.9
1  2002-01-13  15:00:00    Emma  1.9
2  2002-01-13  16:00:00  Joseph  8.0
3  2002-01-13  16:00:00    Emma  9.0
4  2002-01-14  15:00:00  Joseph  0.2
5  2002-01-14  15:00:00    Emma  7.0
6  2002-01-14  16:00:00  Joseph  1.6
7  2002-01-14  16:00:00    Emma  3.4

我希望获得一个新的df["E"]列，用于修复＆＃34; Joseph＆＃34;和＃34;艾玛＆＃34; D值在15:00:00，每天剩下的时间。

输出应为：

            A         B       C    D     E
0  2002-01-13  15:00:00  Joseph  3.9   3.9
1  2002-01-13  15:00:00    Emma  1.9   1.9
2  2002-01-13  16:00:00  Joseph  8.0   3.9
3  2002-01-13  16:00:00    Emma  9.0   3.9
4  2002-01-14  15:00:00  Joseph  0.2   0.2
5  2002-01-14  15:00:00    Emma  7.0   7.0
6  2002-01-14  16:00:00  Joseph  1.6   0.2
7  2002-01-14  16:00:00    Emma  3.4   7.0

Answer 1

据推测，您希望在groupby和A上C，然后在transform上使用first + D。

df['E'] = df.groupby(['A', 'C']).D.transform('first')
df

            A         B       C    D    E
0  2002-01-13  15:00:00  Joseph  3.9  3.9
1  2002-01-13  15:00:00    Emma  1.9  1.9
2  2002-01-13  16:00:00  Joseph  8.0  3.9
3  2002-01-13  16:00:00    Emma  9.0  1.9
4  2002-01-14  15:00:00  Joseph  0.2  0.2
5  2002-01-14  15:00:00    Emma  7.0  7.0
6  2002-01-14  16:00:00  Joseph  1.6  0.2
7  2002-01-14  16:00:00    Emma  3.4  7.0

如果条目的开始时间早于15:00:00，请先屏蔽D然后transform：

df['E'] = df.assign(
        D=df.D.mask(df.B.ne('15:00:00'))
 ).groupby(['A', 'C']).D.transform('first')

df
            A         B       C    D    E
0  2002-01-13  15:00:00  Joseph  3.9  3.9
1  2002-01-13  15:00:00    Emma  1.9  1.9
2  2002-01-13  16:00:00  Joseph  8.0  3.9
3  2002-01-13  16:00:00    Emma  9.0  1.9
4  2002-01-14  15:00:00  Joseph  0.2  0.2
5  2002-01-14  15:00:00    Emma  7.0  7.0
6  2002-01-14  16:00:00  Joseph  1.6  0.2
7  2002-01-14  16:00:00    Emma  3.4  7.0

修复多列条件的值

1 个答案: