Python 熊猫：如果满足条件，则在数据帧中包含新的时间戳行

我有一个如下所示的数据帧：

    value                 timestamp
18.832939   2019-03-04 12:37:26 UTC
18.832939   2019-03-04 12:38:26 UTC
18.832939   2019-03-04 12:39:27 UTC
18.955200   2019-03-04 12:40:28 UTC
18.784912   2019-03-04 12:44:32 UTC
18.784912   2019-03-04 12:45:33 UTC
20.713936   2019-03-04 17:59:36 UTC
20.871742   2019-03-04 18:08:31 UTC
20.871742   2019-03-04 18:09:32 UTC
20.873871   2019-03-04 18:10:32 UTC

我希望得到以下结果，其中我确定了大于2分钟但小于15分钟2

    value                 timestamp
18.832939   2019-03-04 12:37:26 UTC
18.832939   2019-03-04 12:38:26 UTC
18.832939   2019-03-04 12:39:27 UTC
18.955200   2019-03-04 12:40:28 UTC
      NaN   2019-03-04 12:41:28 UTC
      NaN   2019-03-04 12:42:28 UTC
      NaN   2019-03-04 12:43:28 UTC
18.784912   2019-03-04 12:44:32 UTC
18.784912   2019-03-04 12:45:33 UTC
20.713936   2019-03-04 17:59:36 UTC
      NaN   2019-03-04 18:00:36 UTC
      NaN   2019-03-04 18:01:36 UTC
      NaN   2019-03-04 18:02:36 UTC
      NaN   2019-03-04 18:03:36 UTC
      NaN   2019-03-04 18:04:36 UTC
      NaN   2019-03-04 18:05:36 UTC
      NaN   2019-03-04 18:06:36 UTC
      NaN   2019-03-04 18:07:36 UTC
20.871742   2019-03-04 18:08:31 UTC
20.871742   2019-03-04 18:09:32 UTC
20.873871   2019-03-04 18:10:32 UTC

这意味着，为了实现这一目标，我必须做两件事：

确定间隙在哪里符合我想要的条件。因为我们可能会有超过15分钟的差距，我对此不感兴趣。 一旦确定，用1分钟的增量或带有时间戳的均匀间隔值创建新行。 我可以用这个做第一个：

df['aux_1'] = ((df['timestamp'].diff() > '0 days 00:02:00') & (df['timestamp'].diff() < '0 days 00:15:00')).astype(int) #get ending of the gap.
df['aux_2'] = df['aux_1'].shift(-1) #beginning of the gap.
df['intervals'] = df['aux_1'] + df['aux_2'] #both beginning and end with numeric consecutive flags contained in a single column.

但是，我不知道如何做第二部分，至少熊猫不喜欢。以某种方式确定我想要填充的时间戳间隔的开始和结束，然后应用asfreq'1m'，并使用该向量来填充我想要的间隔，这将是理想的。只是我不知道该怎么做

---

有人能帮我吗？提前谢谢

熊猫不太喜欢，但我会做以下事情

new_timestamp = []
for i, row in df.iterrows():
    if row['aux_2']==0:
        new_timestamp.append(row['timestamp'])
    elif row['aux_2']==1:
        new_timestamp += pd.date_range(row['timestamp'], df.iloc[i+1]['timestamp'], freq='min').to_list()

new_df = df.set_index('timestamp')
new_df = new_df.loc[new_timestamp]

这导致

print(new_df)
timestamp                   value       aux_1   aux_2   intervals
2019-03-04 12:37:26+00:00   18.832939   0.0     0.0     0.0
2019-03-04 12:38:26+00:00   18.832939   0.0     0.0     0.0
2019-03-04 12:39:27+00:00   18.832939   0.0     0.0     0.0
2019-03-04 12:40:28+00:00   18.955200   0.0     1.0     1.0
2019-03-04 12:41:28+00:00   NaN     NaN     NaN     NaN
2019-03-04 12:42:28+00:00   NaN     NaN     NaN     NaN
2019-03-04 12:43:28+00:00   NaN     NaN     NaN     NaN
2019-03-04 12:44:28+00:00   NaN     NaN     NaN     NaN
2019-03-04 12:44:32+00:00   18.784912   1.0     0.0     1.0
2019-03-04 12:45:33+00:00   18.784912   0.0     0.0     0.0
2019-03-04 17:59:36+00:00   20.713936   0.0     1.0     1.0
2019-03-04 18:00:36+00:00   NaN     NaN     NaN     NaN
2019-03-04 18:01:36+00:00   NaN     NaN     NaN     NaN
2019-03-04 18:02:36+00:00   NaN     NaN     NaN     NaN
2019-03-04 18:03:36+00:00   NaN     NaN     NaN     NaN
2019-03-04 18:04:36+00:00   NaN     NaN     NaN     NaN
2019-03-04 18:05:36+00:00   NaN     NaN     NaN     NaN
2019-03-04 18:06:36+00:00   NaN     NaN     NaN     NaN
2019-03-04 18:07:36+00:00   NaN     NaN     NaN     NaN
2019-03-04 18:08:31+00:00   20.871742   1.0     0.0     1.0
2019-03-04 18:09:32+00:00   20.871742   0.0     0.0     0.0 

---

我的建议是：1生成一个带有时间戳列的数据帧，间隔为1分钟。2将数据连接回新创建的数据帧中，使用key:timestamp，精确到分钟级别。这很好，我的意思是，我希望有一种方法可以做到这一点，而不必将时间戳设置为索引。但是如果我想的话，我可以改变列的位置，并设置一个新的数字升序索引。谢谢你的支持！