将DateTimeIndex的time元素分配给新列

时间:2016-06-01 15:23:06

标签: python pandas dataframe datetimeindex

我正在使用以下DataFrame,并且只想访问DateTimeIndex的时间(不是日期):

                    idle wheel  Induce wheel axial  radial
tiempo          
5/30/2016 19:37:46  -1,099.12   -1,048.78   -477.13
5/30/2016 19:37:47  -1,099.12   -1,048.78   -476.98
5/30/2016 19:37:48  -1,099.12   -1,048.78   -477.21
5/30/2016 19:37:49  -1,099.12   -1,048.78   -477.13
5/30/2016 19:37:50  -1,099.12   -1,048.78   -477.21
5/30/2016 19:37:51  -1,099.12   -1,048.78   -477.35
5/30/2016 19:37:52  -1,099.12   -1,048.78   -477.13
5/30/2016 19:37:53  -1,099.12   -1,048.78   -476.98
5/30/2016 19:37:54  -1,099.12   -1,048.78   -476.98
5/30/2016 19:37:55  -1,099.12   -1,048.78   -476.98
5/30/2016 19:37:56  -1,099.12   -1,048.78   -476.98
5/30/2016 19:37:57  -1,099.12   -1,048.78   -476.98

我想只保留19:..:..,而不是日期部分。 我一直在寻找,但无法找到任何解决方案。

1 个答案:

答案 0 :(得分:2)

.time index上使用'tiempo'

df['time'] = df.index.time

或,如果'tiempo'column使用

df['time'] = df.tiempo.dt.time

得到:

                tiempo  idle wheel  Induce wheel axial    radial      time
0  2016-05-30 19:37:46    -1099.12    -1048.78           -477.13  19:37:46
1  2016-05-30 19:37:47    -1099.12    -1048.78           -476.98  19:37:47
2  2016-05-30 19:37:48    -1099.12    -1048.78           -477.21  19:37:48
3  2016-05-30 19:37:49    -1099.12    -1048.78           -477.13  19:37:49
4  2016-05-30 19:37:50    -1099.12    -1048.78           -477.21  19:37:50
5  2016-05-30 19:37:51    -1099.12    -1048.78           -477.35  19:37:51
6  2016-05-30 19:37:52    -1099.12    -1048.78           -477.13  19:37:52
7  2016-05-30 19:37:53    -1099.12    -1048.78           -476.98  19:37:53
8  2016-05-30 19:37:54    -1099.12    -1048.78           -476.98  19:37:54
9  2016-05-30 19:37:55    -1099.12    -1048.78           -476.98  19:37:55
10 2016-05-30 19:37:56    -1099.12    -1048.78           -476.98  19:37:56
11 2016-05-30 19:37:57    -1099.12    -1048.78           -476.98  19:37:57