计算 pandas 事件之间的时差 [英] Calculating time difference between events in pandas

问题描述

+--------------------------------------------------------------+
| 2014-08-12T10:30:14.6938893+10:00     Reading received START |
| 2014-08-12T10:30:14.6938893+10:00       Reading received ADD |
| 2014-08-12T10:30:14.7094893+10:00    Reading received UPDATE |
| 2014-08-12T10:30:14.7094893+10:00    Reading received COMMIT |
| 2014-08-12T10:30:14.7094893+10:00               Commit start |
| 2014-08-12T10:30:14.7406893+10:00                 Commit end |
| 2014-08-12T10:30:14.7406893+10:00    Reading received FINISH |
| 2014-08-12T10:30:23.3206893+10:00     Reading received START |
| 2014-08-12T10:30:23.3206893+10:00       Reading received ADD |
| 2014-08-12T10:30:23.3362893+10:00    Reading received UPDATE |
| 2014-08-12T10:30:23.3362893+10:00    Reading received COMMIT |
| 2014-08-12T10:30:23.3362893+10:00               Commit start |
| 2014-08-12T10:30:23.3674893+10:00                 Commit end |
| 2014-08-12T10:30:23.3674893+10:00    Reading received FINISH |
+--------------------------------------------------------------+

鉴于该值描述一个事件的时间序列，如何计算重复发生的事件之间的时间差，例如阅读已收到START 和随后的阅读已完成之间的平均差异?

Given a time series where the value describes an event, how can I calculate delta times between recurring events, e.g. the average difference between Reading received START and the subsequent Reading received FINISH?

有没有比这更好的方法了?

Is there a better way than then e.g.

left = df[df.Event == 'Reading received START']
right = df[df.Event == 'Reading received FINISH']
left.index = range(len(left))
right.index = range(len(right))
delta = (right.Time - left.Time)

推荐答案

为明确起见，我假设您正在从较大的数据框中显示索引和一列(称为事件").那是对的吗? 怎么样:

To be explicit, I'm assuming that you are showing the index and one column (called 'Event') from a larger dataframe. Is that correct? How about the following:

relevant_df = df[df.Event.isin(['Reading received START','Reading received START'])
relevant_ts_as_series = pd.Series(relevant_df.index)
diff = relevant_ts_as_series - relevant_ts_as_series.shift()

然后，您可以根据需要选择diff.mean().

Then you can take diff.mean() if you like.

我敢打赌，除了将索引转换为系列以外，还有一种更优雅的方法，但这应该对您有用.

I bet there's a more elegant way than turning the index into a Series, but this should work for you.

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