熊猫：如果满足条件，则在数据框中添加新的时间戳记行

Question

我有一个看起来像这样的数据框：

    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

我想要以下结果，在这里我确定所有大于2分钟但小于15分钟（2 <t <15）的时间戳间隔：

    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，我必须做两件事才能实现这个目标：

1. 确定差距在哪里满足我想要的条件。 因为我们之间的差距可能甚至超过15分钟，而我对此并不感兴趣。
2. 确定后，使用“时间戳记”增加1分钟或以均匀间隔的值创建新行。

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，我不确定如何做第二部分，至少不是“类似熊猫”。 最好以某种方式标识我打算填充的时间戳间隔的开始-结束，然后应用asfreq（'1m'），然后使用该向量填充我想要的间隔。 只是不确定如何正确地执行操作。

有人能帮我吗？ 提前致谢。

Answer 1

熊猫不是很喜欢，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