计算pandas中事件之间的时差

Question

+--------------------------------------------------------------+
| 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 |
+--------------------------------------------------------------+

给定值描述事件的时间序列，如何计算重复事件之间的增量时间，例如： 阅读收到START 和随后阅读收到FINISH 之间的平均差异？

有没有比这更好的方法，例如。

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)

Answer 1

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“＃39;事件＆＃39;”）。那是对的吗？ 如下：

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()

如果您愿意，可以diff.mean()。

我敢打赌，除了将索引转换为系列之外，还有一种更优雅的方式，但这应该适合你。