请参阅此Manipulating pandas columns帖子
我分享了此数据框：
+----------+------------+-------+-----+------+
| Location | Date | Event | Key | Time |
+----------+------------+-------+-----+------+
| i2 | 2019-03-02 | 1 | a | |
| i2 | 2019-03-02 | 1 | a | |
| i2 | 2019-03-02 | 1 | a | |
| i2 | 2019-03-04 | 1 | a | 2 |
| i2 | 2019-03-15 | 2 | b | 0 |
| i9 | 2019-02-22 | 2 | c | 0 |
| i9 | 2019-03-10 | 3 | d | |
| i9 | 2019-03-10 | 3 | d | 0 |
| s8 | 2019-04-22 | 1 | e | |
| s8 | 2019-04-25 | 1 | e | |
| s8 | 2019-04-28 | 1 | e | 6 |
| t14 | 2019-05-13 | 3 | f | |
+----------+------------+-------+-----+------+
这是一个后续问题。如下所示，在“日期”之后再考虑两列。
+-----------------------+----------------------+
| Start Time (hh:mm:ss) | Stop Time (hh:mm:ss) |
+-----------------------+----------------------+
| 13:24:38 | 14:17:39 |
| 03:48:36 | 04:17:20 |
| 04:55:05 | 05:23:48 |
| 08:44:34 | 09:13:15 |
| 19:21:05 | 20:18:57 |
| 21:05:06 | 22:01:50 |
| 14:24:43 | 14:59:37 |
| 07:57:32 | 09:46:21
| 19:21:05 | 20:18:57 |
| 21:05:06 | 22:01:50 |
| 14:24:43 | 14:59:37 |
| 07:57:32 | 09:46:21 |
+-----------------------+----------------------+
任务保持不变-获得时间差，但以小时为单位，对应于第一行的停止时间和最后一行的开始时间 每个键。
基于答案，我正在尝试类似的事情：
df['Time']=df.groupby(['Location','Event']).Date.\
transform(lambda x : (x.iloc[-1]-x.iloc[0]))[~df.duplicated(['Location','Event'],keep='last')]
df['Time_h']=df.groupby(['Location','Event'])['Start Time (hh:mm:ss)','Stop Time (hh:mm:ss)'].\
transform(lambda x,y : (x.iloc[-1]-y.iloc[0]))[~df.duplicated(['Location','Event'],keep='last')] # This gives an error on transform
分别计算天数和小时数之间的差异，然后合并。有没有更好的办法？