我正在使用下面的DataFrame并且只想访问我的DateTimeIndex的时间(而不是日期):
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
我只想保留 19:..:..，而不是日期部分。
我一直在寻找但找不到任何解决方案。
最佳答案
在您的索引上使用.time 'tiempo':
df['time'] = df.index.time
或者，如果'tiempo'是列使用
df['time'] = df.tiempo.dt.time
获取:
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
关于python - 将 DateTimeIndex 的时间元素分配给新列，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/37572827/