在一天中的每个小时拆分时间范围答案

【问题标题】：Split time ranges on each hour of the day在一天中的每个小时拆分时间范围
【发布时间】：2021-08-17 03:01:33
【问题描述】：

我试图将一天中多个小时重叠的时间范围分成两行，每行代表每小时的时间（并留下开始和结束时间都在一天中的同一小时内的时间范围)。

示例数据框：

df1 = pd.DataFrame({'id':['A','B','C'],
                   'start_time':['2021-03-01 11:52:11+00:00','2021-03-01 11:53:10+00:00','2021-03-01 12:01:54+00:00'],
                   'end_time':['2021-03-01 12:14:45+00:00','2021-03-01 12:09:40+00:00','2021-03-01 12:57:03+00:00']})
df1['start_time'] = pd.to_datetime(df1['start_time']) 
df1['end_time'] = pd.to_datetime(df1['end_time'])

看起来像：

--------------------------------------------------------------------------
index    id   start_time                     end_time
0        A    2021-03-01 11:52:11+00:00      2021-03-01 12:14:45+00:00
1        B    2021-03-01 11:53:10+00:00      2021-03-01 12:09:40+00:00
2        C    2021-03-01 12:01:54+00:00      2021-03-01 12:57:03+00:00
--------------------------------------------------------------------------

我希望得到如下结果：

--------------------------------------------------------------------------
index    id   start_time                     end_time
0        A    2021-03-01 11:52:11+00:00      2021-03-01 12:00:00+00:00
1        A    2021-03-01 12:00:00+00:00      2021-03-01 12:14:45+00:00
2        B    2021-03-01 11:53:10+00:00      2021-03-01 12:00:00+00:00
3        B    2021-03-01 12:00:00+00:00      2021-03-01 12:09:40+00:00
4        C    2021-03-01 12:01:54+00:00      2021-03-01 12:57:03+00:00
--------------------------------------------------------------------------

【问题讨论】：

标签： python pandas datetime time

【解决方案1】：

def myfunc(sr):
    """Return all pairwise hours between start_time and end_time."""
    dtl = pd.date_range(sr["start_time"].floor("H"),
                        sr["end_time"].ceil("H"),
                        freq="H").tolist()
    dtl[0] = sr["start_time"]
    dtl[-1] = sr["end_time"]
    return list(zip(dtl[0:-1], dtl[1:]))

# One row -> (start, next_hour) or (curr_hour, next_hour) or (curr_hour, end)
pairs = df1.apply(myfunc, axis="columns").explode()

# Build output dataframe from list 'pairs'
out = pd.DataFrame(pairs.tolist(), 
                   columns=["start_time", "end_time"],
                   index=pairs.index)

# Merge with 'df1'
out = df1[["id"]].join(out).reset_index(drop=True)

>>> out
  id                start_time                  end_time
0  A 2021-03-01 11:52:11+00:00 2021-03-01 12:00:00+00:00
1  A 2021-03-01 12:00:00+00:00 2021-03-01 12:14:45+00:00
2  B 2021-03-01 11:53:10+00:00 2021-03-01 12:00:00+00:00
3  B 2021-03-01 12:00:00+00:00 2021-03-01 12:09:40+00:00
4  C 2021-03-01 12:01:54+00:00 2021-03-01 12:57:03+00:00

【讨论】：