基于现有列的熊猫数据栏创建

Question

我需要actual_new栏从熊猫的实际数据。

start time      end time        actual      actual_new
4/1/2022 20:00  4/1/2022 21:00  0.749123    0.749123
4/1/2022 21:00  4/1/2022 22:00  0.749123    0.770175
4/1/2022 22:00  4/1/2022 23:00  0.749123    0.725439
4/1/2022 23:00  4/2/2022 0:00   0.749123    0.659649
4/2/2022 0:00   4/2/2022 1:00   0.749123    0.245614
4/2/2022 1:00   4/2/2022 2:00   0.749123    0.078947
4/1/2022 21:00  4/1/2022 22:00  0.770175    0.749123
4/1/2022 22:00  4/1/2022 23:00  0.770175    0.770175
4/1/2022 23:00  4/2/2022 0:00   0.770175    0.725439
4/2/2022 0:00   4/2/2022 1:00   0.770175    0.659649
4/2/2022 1:00   4/2/2022 2:00   0.770175    0.245614
4/2/2022 2:00   4/2/2022 3:00   0.770175    0.078947
4/1/2022 22:00  4/1/2022 23:00  0.725439    0.749123
4/1/2022 23:00  4/2/2022 0:00   0.725439    0.770175
4/2/2022 0:00   4/2/2022 1:00   0.725439    0.725439
4/2/2022 1:00   4/2/2022 2:00   0.725439    0.659649
4/2/2022 2:00   4/2/2022 3:00   0.725439    0.245614
4/2/2022 3:00   4/2/2022 4:00   0.725439    0.078947
4/1/2022 23:00  4/2/2022 0:00   0.659649    0.749123
4/2/2022 0:00   4/2/2022 1:00   0.659649    0.770175
4/2/2022 1:00   4/2/2022 2:00   0.659649    0.725439
4/2/2022 2:00   4/2/2022 3:00   0.659649    0.659649
4/2/2022 3:00   4/2/2022 4:00   0.659649    0.245614
4/2/2022 4:00   4/2/2022 5:00   0.659649    0.078947
4/2/2022 0:00   4/2/2022 1:00   0.245614    0.749123
4/2/2022 1:00   4/2/2022 2:00   0.245614    0.770175
4/2/2022 2:00   4/2/2022 3:00   0.245614    0.725439
4/2/2022 3:00   4/2/2022 4:00   0.245614    0.659649
4/2/2022 4:00   4/2/2022 5:00   0.245614    0.245614
4/2/2022 5:00   4/2/2022 6:00   0.245614    0.078947
4/2/2022 1:00   4/2/2022 2:00   0.078947    0.749123
4/2/2022 2:00   4/2/2022 3:00   0.078947    0.770175
4/2/2022 3:00   4/2/2022 4:00   0.078947    0.725439
4/2/2022 4:00   4/2/2022 5:00   0.078947    0.659649
4/2/2022 5:00   4/2/2022 6:00   0.078947    0.245614
4/2/2022 6:00   4/2/2022 7:00   0.078947    0.078947

Answer 1

df['actual_new'] = list(df['actual'].unique())*int(df.shape[0]/len(uniques))

Answer 2

试着这样做：

df['actual_new'] = df['actual']

或

df['actual_new'] = df.apply(lambda x: x['actual'], axis = 1)