如何更改 R 中的时间序列(XTS 或 ZOO)? [英] How can I alter a time series (XTS or ZOO) in R?

问题描述

我是 stackoverflow 的新手，也是 R 的新手，但我已经搜索了很长时间，但找不到以下问题的答案.

I am new to stackoverflow and fairly new to R but have searched long and hard and cannot find an answer to the following question.

我有许多数据文件是温度与时间序列的关系.我将 CSV 作为 ZOO 对象导入，然后转换为 XTS.一个正确的文件看起来像这样，一个小时和一个半小时的读数:

I have a number of data files that are temperature against a time series. I am importing the CSV as a ZOO object then converting to XTS. A correct file looks like this, with readings on the hour and the half hour:

>head(master1)
                       S_1
2010-03-03 00:00:00 2.8520
2010-03-03 00:30:00 2.6945
2010-03-03 01:00:00 2.5685
2010-03-03 01:30:00 2.3800
2010-03-03 02:00:00 2.2225
2010-03-03 02:30:00 2.0650

但有些时间值略有出入 - 即 23:59:00 不是 00:00:00，或 00:29:00 而不是 00:30:00.

But the time value on some are slightly out - i.e. 23:59:00 not 00:00:00, or 00:29:00 instead of 00:30:00.

>head(master21)
                       S_21
2010-03-04 23:59:00  -0.593
2010-03-05 00:29:00  -0.908
2010-03-05 00:59:00  -1.034
2010-03-05 01:29:00  -1.223
2010-03-05 01:59:00  -1.349
2010-03-05 02:29:00  -1.538

我想更正这些时间序列，因为微小的差异对我的分析并不重要，我最终想合并文件，所以每个时间序列都需要有相同的时间.

I want to correct these time series, as the minute difference is not important for my analysis and I ultimately want to merge the files, so each timeseries needs to have the same timing.

我想要一个可以说将时间序列向前移动 1 分钟，但不要更改数据列(例如 S_21)"的命令.我对 gsub() 进行了一些更简单的更改，并且在将数据转换为 ZOO 或 XTS 之前考虑了一个复杂的正则表达式来更改数据.我已经阅读了 lag() 和 diff() 但它们似乎移动了相对于时间序列的数据值；如果我错了，请纠正我.

I want a command that can just say "shift the time series forward by 1 minute, but don't alter the data column (e.g. S_21). I have had some luck with gsub() on easier changes, and contemplated a complex regex to change the data before it is converted to ZOO or XTS. I have read about lag() and diff() but they seem to move the data values relative to the time series; please correct me if I am wrong.

如果能帮助解决这个问题，我们将不胜感激.

Any help solving this issue would be much appreciated.

推荐答案

尝试

index(master21) <- index(master21) + 60    # adds a minute

这将在时间索引上增加一分钟.然后您可以使用 merge() 作为时间戳对齐.

which will add a minute to the time index. You can then use merge() as the timestamps align.

更一般地说，zoo 包的小插图对您也很有用.

More generally, the vignettes of the zoo package will be useful for you too.

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