我有一个像这样的数据集
user-id date-time msg
1 2016-12-09 10:25:00 1
2 2016-12-09 10:26:00 0
3 2016-12-09 10:26:00 1
2 2016-12-09 10:27:00 1
1 2016-12-09 10:28:00 2
2 2016-12-09 10:28:00 1
3 2016-12-09 10:29:00 2
2 2016-12-09 10:29:00 1
1 2016-12-09 10:30:00 3
我想拥有一个新的列来计算每个记录与消息与该记录相似的第一次时差。这样的东西：
user-id date-time msg time-diffrence
1 2016-12-09 10:25:00 1 00:00
2 2016-12-09 10:26:00 0 00:00
3 2016-12-09 10:26:00 1 01:00
2 2016-12-09 10:27:00 1 02:00
1 2016-12-09 10:28:00 2 00:00
2 2016-12-09 10:28:00 1 03:00
3 2016-12-09 10:29:00 2 01:00
2 2016-12-09 10:29:00 1 04:00
1 2016-12-09 10:30:00 3 00:00
我找到了仅考虑日期时间或使用LOC或ILOC的解决方案，但它们不起作用。
选项＃1
使用groupby和iloc：
df['time-difference'] = df.groupby('msg')['date-time'].apply(lambda x: x - x.iloc[0])
输出：
user-id date-time msg time-difference
0 1 2016-12-09 10:25:00 1 00:00:00
1 2 2016-12-09 10:26:00 0 00:00:00
2 3 2016-12-09 10:26:00 1 00:01:00
3 2 2016-12-09 10:27:00 1 00:02:00
4 1 2016-12-09 10:28:00 2 00:00:00
5 2 2016-12-09 10:28:00 1 00:03:00
6 3 2016-12-09 10:29:00 2 00:01:00
7 2 2016-12-09 10:29:00 1 00:04:00
8 1 2016-12-09 10:30:00 3 00:00:00
选项＃2
与transform和first或min使用groupby：
df['time-difference'] = df['date-time'] - df.groupby('msg')['date-time'].transform('first')
输出：
user-id date-time msg time-difference
0 1 2016-12-09 10:25:00 1 00:00:00
1 2 2016-12-09 10:26:00 0 00:00:00
2 3 2016-12-09 10:26:00 1 00:01:00
3 2 2016-12-09 10:27:00 1 00:02:00
4 1 2016-12-09 10:28:00 2 00:00:00
5 2 2016-12-09 10:28:00 1 00:03:00
6 3 2016-12-09 10:29:00 2 00:01:00
7 2 2016-12-09 10:29:00 1 00:04:00
8 1 2016-12-09 10:30:00 3 00:00:00