Python在满足条件之前从行复制值

Question

我确实有以下数据：

             A    B
Date
2019-11-04   0    0
2019-11-05   0    0
2019-11-06   1   -2.197387
2019-11-07   1   -2.203836
2019-11-08   1   -2.208839
2019-11-09  -1   -2.207335
2022-11-10  -1   -0.968055
2022-11-11  -1   -0.971627
2022-11-12   0    0
2022-11-13   1   -0.974535
2022-11-14   1   -0.972994

我期待的是以下几点：

             A    B
Date
2019-11-04   0    0
2019-11-05   0    0
2019-11-06   1   -2.197387
2019-11-07   1   -2.197387
2019-11-08   1   -2.197387
2019-11-09  -1   -2.207335
2022-11-10  -1   -2.207335
2022-11-11  -1   -2.207335
2022-11-12   0    0
2022-11-13   1   -0.974535
2022-11-14   1   -0.974535

我尝试了..loc 1和..loc 1，条件是.iloc或.iloc必须是!=0，但在这两种情况下，0只是复制到下一行。

Answer 1

使用GroupBy.pipe和GroupBy.first，将A的连续值与Series.cumsum的移位值进行比较

df['B'] = df.groupby(df.A.ne(df.A.shift()).cumsum())['B'].transform('first')
print (df)
            A         B
Date                   
2019-11-04  0  0.000000
2019-11-05  0  0.000000
2019-11-06  1 -2.197387
2019-11-07  1 -2.197387
2019-11-08  1 -2.197387
2019-11-09 -1 -2.207335
2022-11-10 -1 -2.207335
2022-11-11 -1 -2.207335
2022-11-12  0  0.000000
2022-11-13  1 -0.974535
2022-11-14  1 -0.974535

另一种方法是使用Series.where替换每个连续值的非第一个值，并向前填充缺失的值(如果列中没有缺失值，则使用此方法)：

df['B'] = df['B'].where(df.A.ne(df.A.shift())).ffill()
print (df)
            A         B
Date                   
2019-11-04  0  0.000000
2019-11-05  0  0.000000
2019-11-06  1 -2.197387
2019-11-07  1 -2.197387
2019-11-08  1 -2.197387
2019-11-09 -1 -2.207335
2022-11-10 -1 -2.207335
2022-11-11 -1 -2.207335
2022-11-12  0  0.000000
2022-11-13  1 -0.974535
2022-11-14  1 -0.974535