我想计算特定时间范围内两列之间的时差。
我尝试df.between_time
,但它仅适用于索引。
例如时间范围:18.00-8.00
数据:
start stop
0 2018-07-16 16:00:00 2018-07-16 20:00:00
1 2018-07-11 08:03:00 2018-07-11 12:03:00
2 2018-07-13 17:54:00 2018-07-13 21:54:00
3 2018-07-14 13:09:00 2018-07-14 17:09:00
4 2018-07-20 00:21:00 2018-07-20 04:21:00
5 2018-07-20 17:00:00 2018-07-21 09:00:00
预期结果:
start stop time_diff
0 2018-07-16 16:00:00 2018-07-16 20:00:00 02:00:00
1 2018-07-11 08:03:00 2018-07-11 12:03:00 0
2 2018-07-13 17:54:00 2018-07-13 21:54:00 03:54:00
3 2018-07-14 13:09:00 2018-07-14 17:09:00 0
4 2018-07-20 00:21:00 2018-07-20 04:21:00 04:00:00
5 2018-07-20 17:00:00 2018-07-21 09:00:00 14:00:00
注意:如果time_diff> 1天,我已经处理了该案。
问题:我应该构建一个函数来执行此操作,还是应该有pandas内置函数来执行此操作?任何帮助或指导将不胜感激。
答案 0 :(得分:1)
我认为这可以解决
tmp = pd.DataFrame({'time1': pd.to_datetime(['2018-07-16 16:00:00', '2018-07-11 08:03:00',
'2018-07-13 17:54:00', '2018-07-14 13:09:00',
'2018-07-20 00:21:00', '2018-07-20 17:00:00']),
'time2': pd.to_datetime(['2018-07-16 20:00:00', '2018-07-11 12:03:00',
'2018-07-13 21:54:00', '2018-07-14 17:09:00',
'2018-07-20 04:21:00', '2018-07-21 09:00:00'])})
time1_date = tmp.time1.dt.date.astype(str)
tmp['rule18'], tmp['rule08'] = pd.to_datetime(time1_date + ' 18:00:00'), pd.to_datetime(time1_date + ' 08:00:00')
# if stop exceeds 18:00:00, compute time difference from this hour
tmp['time_diff_rule1'] = np.where(tmp.time2 > tmp.rule18, (tmp.time2 - tmp.rule18), (tmp.time2 - tmp.time1))
# rearrange the dataframe with your second rule
tmp['time_diff_rule2'] = np.where((tmp.time2 < tmp.rule18) & (tmp.time1 > tmp.rule08), 0, tmp['time_diff_rule1'])
time_diff_rule1 time_diff_rule2
0 02:00:00 02:00:00
1 04:00:00 00:00:00
2 03:54:00 03:54:00
3 04:00:00 00:00:00
4 04:00:00 04:00:00
5 15:00:00 15:00:00