如何根据与前一行的差异对行进行分组？

Question

我有以下数据帧：

    | start_time          | end_time            | id  |
    |---------------------|---------------------|-----|
    | 2017-03-30 01:00:00 | 2017-03-30 01:15:30 |1    |
    | 2017-03-30 02:02:00 | 2017-03-30 03:30:00 |4    |
    | 2017-03-30 03:37:00 | 2017-03-30 03:39:00 |7    |
    | 2017-03-30 03:41:30 | 2017-03-30 04:50:00 |8    |
    | 2017-03-30 07:10:00 | 2017-03-30 07:10:30 |10   |
    | 2017-03-30 07:11:00 | 2017-03-30 07:20:00 |13   |
    | 2017-03-30 07:22:00 | 2017-03-30 08:00:00 |15   |
    | 2017-03-30 10:00:00 | 2017-03-30 10:03:00 |20   |

当行“I-1”的完成时间最多比行“I”的开始时间早900秒时，我想将同一id下的行分组。
基本上，上述示例的输出是： 结果是：

    | start_time          | end_time            | id  |
    |---------------------|---------------------|-----|
    | 2017-03-30 01:00:00 | 2017-03-30 01:15:30 |1    |
    | 2017-03-30 02:02:00 | 2017-03-30 03:30:00 |4    |
    | 2017-03-30 03:37:00 | 2017-03-30 03:39:00 |4    |
    | 2017-03-30 03:41:30 | 2017-03-30 04:50:00 |4    |
    | 2017-03-30 07:10:00 | 2017-03-30 07:10:30 |10   |
    | 2017-03-30 07:11:00 | 2017-03-30 07:20:00 |10   |
    | 2017-03-30 07:22:00 | 2017-03-30 08:00:00 |10   |
    | 2017-03-30 10:00:00 | 2017-03-30 10:03:00 |20   |

我通过以下代码实现了这一点，但我确信有一种更优雅（更高效）的方法：

df['endTime_delayed'] = df.end_time.shift(1)
df['id_delayed'] = df['id'].shift(1)
for (i,row) in df.iterrows():
    if (row.start_time-row.endTime_delayed).seconds <= 900 :
        df.id.iloc[i] = df.id_delayed.iloc[i]
        try :
            df.id_delayed.iloc[i+1] = df.id.iloc[i]
        except : 
            break

Answer 1

mask和ffill

diff = df.start_time.sub(df.end_time.shift())
mask = diff < pd.Timedelta(900, unit='s')
df.id.mask(mask).ffill().astype(df.id.dtype)

0     1
1     4
2     4
3     4
4    10
5    10
6    10
7    20
Name: id, dtype: int64