剥离日期并保持时间

问题描述

很多人问如何剥夺时间和保持日期，但是呢呢呢？给定：

  myDateTime<  - 11/02/2014 14:22:45
  

我想看看：

  myTime 
 [1]14:22:45
  

时区不是必需的。

我已经尝试了（从其他答案）

  as .POSIXct（substr（myDateTime，12,19），format =％H：％M：％S）
  

[1]2013-04-13 14:22:45 NZST

目的是分析几天内按时间记录的事件

感谢

编辑

原来没有纯粹的时间对象，所以每次也必须有一个日期。

最后我使用了

  as.POSIXct（as.numeric（as.POSIXct（myDateTime））%% 86400，origin =2000-01-01）
  

而不是字符解决方案，因为我需要对结果进行算术。这个解决方案类似于我的原来的一个，除了日期可以一致地被控制 - 2000-01-01在这种情况下，而我的尝试只是使用当前日期在运行时。

解决方案

如果GMT日期内的时间对您的问题有用，可以使用 %% 余数运算符，取余数mod $ 86400 （一天中的秒数）。

 code>邮票<  -  c（2013-04-12 19:00:00，2010-04-01 19:00:01，2018-06-18 19:00:02） 
 as.numeric（as.POSIXct（stamps））%% 86400 
 ## [1] 0 1 2 
  

Lots of people ask how to strip the time and keep the date, but what about the other way around? Given:

myDateTime <- "11/02/2014 14:22:45"

I would like to see:

myTime
[1] "14:22:45"

Time zone not necessary.

I've already tried (from other answers)

as.POSIXct(substr(myDateTime, 12,19),format="%H:%M:%S")

[1] "2013-04-13 14:22:45 NZST"

The purpose is to analyse events recorded over several days by time of day only.

Thanks

Edit:

It turns out there's no pure "time" object, so every time must also have a date.

In the end I used

as.POSIXct(as.numeric(as.POSIXct(myDateTime)) %% 86400, origin = "2000-01-01")

rather than the character solution, because I need to do arithmetic on the results. This solution is similar to my original one, except that the date can be controlled consistently - "2000-01-01" in this case, whereas my attempt just used the current date at runtime.

解决方案

If the time within a GMT day is useful for your problem, you can get this with %%, the remainder operator, taking the remainder modulo 86400 (the number of seconds in a day).

stamps <- c("2013-04-12 19:00:00", "2010-04-01 19:00:01", "2018-06-18 19:00:02")
as.numeric(as.POSIXct(stamps)) %% 86400
## [1] 0 1 2

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