python - 搜索缺少的时间戳并在 python 中显示？

Question

这是我的一些数据集，其中包含时间、温度 1、温度 2

Timestamp.             Temperature1.        Temperature2
09/01/2016 00:00:08          53.4.                       45.5
09/01/2016 00:00:38.         53.5.                       45.2
09/01/2016 00:01:08.         54.6.                        43.2
09/01/2016 00:01:38.         55.2.                        46.3
09/01/2016 00:02:08.         54.5.                        45.5
09/01/2016 00:04:08.         54.2.                       35.5
09/01/2016 00:05:08.         52.4.                        45.7
09/01/2016 00:05:38.         53.4.                         45.2

我的数据每 30 秒接收一次..

这是我的数据集..缺少一些时间戳..bcoz。每隔 30 秒我的数据就来一次……所以有些数据点丢失了……如何找到该数据点……并将数据作为 NAN 插入到那里……请帮帮我..

Answer 1

你可以使用resample('30S', base=8)方法:

In [20]: x.resample('30S', base=8).mean()
Out[20]:
                                 Temperature1    Temperature2
Timestamp
2016-09-01 00:00:08                      53.4            45.5
2016-09-01 00:00:38                      53.5            45.2
2016-09-01 00:01:08                      54.6            43.2
2016-09-01 00:01:38                      55.2            46.3
2016-09-01 00:02:08                      54.5            45.5
2016-09-01 00:02:38                       NaN             NaN
2016-09-01 00:03:08                       NaN             NaN
2016-09-01 00:03:38                       NaN             NaN
2016-09-01 00:04:08                      54.2            35.5
2016-09-01 00:04:38                       NaN             NaN
2016-09-01 00:05:08                      52.4            45.7
2016-09-01 00:05:38                      53.4            45.2

上面的解决方案假设 Timestamp 是 datetime 数据类型并且它已被设置为索引。如果 Timestamp 是常规列(不是索引)，那么从 Pandas 0.19.0 开始我们可以在常规列(它必须是 datetime dtype)上重新采样，使用 on='column_name' 参数:

In [26]: x.resample('30S', on='Timestamp', base=8).mean()
Out[26]:
                                 Temperature1    Temperature2
Timestamp
2016-09-01 00:00:08                      53.4            45.5
2016-09-01 00:00:38                      53.5            45.2
2016-09-01 00:01:08                      54.6            43.2
2016-09-01 00:01:38                      55.2            46.3
2016-09-01 00:02:08                      54.5            45.5
2016-09-01 00:02:38                       NaN             NaN
2016-09-01 00:03:08                       NaN             NaN
2016-09-01 00:03:38                       NaN             NaN
2016-09-01 00:04:08                      54.2            35.5
2016-09-01 00:04:38                       NaN             NaN
2016-09-01 00:05:08                      52.4            45.7
2016-09-01 00:05:38                      53.4            45.2

如果您需要动态找到您的base值，您可以这样做:

In [21]: x.index[0].second
Out[21]: 8

来自 docs :

base : int, default 0

For frequencies that evenly subdivide 1 day, the “origin” of the aggregated intervals. For example, for 5min frequency, base could range from 0 through 4.

Defaults to 0