如何更改 R 中的时间序列（XTS 或 ZOO）？
我是 stackoverflow 的新手，对 R 也相当陌生，但经过长时间的艰苦搜索，找不到以下问题的答案。
我有许多数据文件，它们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。正确的文件如下所示，其中包含整点和半点的读数：
>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
但有些文件的时间值稍有偏差 - 即 23:59:00 而不是 00:00:00，或者 00:29:00 而不是 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
我想纠正这些时间序列，因为微小的差异对我的分析并不重要，而且我最终想要合并文件，因此每个时间序列需要具有相同的时间。
我想要一个可以只说“将时间序列向前移动 1 分钟，但不更改数据列（例如 S_21）”的命令。 我在 gsub() 上获得了一些运气，可以更轻松地进行更改，并考虑使用复杂的正则表达式来更改数据，然后再将其转换为 ZOO 或 XTS。我已经阅读过有关 lag() 和 diff() 的内容，但它们似乎相对于时间序列移动了数据值；如果我错了，请纠正我。
任何解决此问题的帮助将不胜感激。
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.
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
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
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.
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.
Any help solving this issue would be much appreciated.
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尝试
将时间索引增加一分钟。然后，您可以使用
merge()作为时间戳对齐。更一般地说，
zoo包的小插图对您也很有用。Try
which will add a minute to the time index. You can then use
merge()as the timestamps align.More generally, the vignettes of the
zoopackage will be useful for you too.