将"hr"转换为"00:00:00"；到日期时间"00:00:00" [英] Converting chr "00:00:00" to date-time "00:00:00"

问题描述

我的问题来自

My question comes from this question. The question had the following character string.

x <- "2007-02-01 00:00:00"
y <- "02/01/2007 00:06:10"

如果您尝试将此字符串转换为日期类对象，则会发生一些有趣的事情.

If you try to convert this string to date-class object, something funny happens.

这是@nrusell的答案中的一个示例.

This is a sample from @nrusell's answer.

as.POSIXct(x,tz=Sys.timezone())
[1] "2007-02-01 EST"

as.POSIXct(y,format="%m/%d/%Y %H:%M:%S",tz=Sys.timezone())
[1] "2007-02-01 00:06:10 EST" 

如您所见，00:00:00在第一个示例中消失了. @Richard Scriven在我们的讨论中使用lubridate留下了以下示例.

As you see, 00:00:00 disappears from the first example. @Richard Scriven left the following example in our discussion using lubridate.

dt <- as.POSIXct("2007-02-01 00:00:00")
hour(dt) <- hour(dt)+1
dt
[1] "2007-02-01 01:00:00 EST"
hour(dt) <- hour(dt)-1
dt
[1] "2007-02-01 EST"

再次，00:00:00消失.为什么R在转换后避免在日期类对象中保留00:00:00?我们如何保留00:00:00?

Once again, 00:00:00 disappears. Why does R avoid keeping 00:00:00 in date-class object after conversion? How can we keep 00:00:00?

推荐答案

如果日期的时间部分是午夜，则只是print会删除精度.这是??strftime帮助中的文字，特别是format参数:

It is just the print that remove the precision if the time part of a date is a midnight. This is literlay explained in ??strftime help, specially the format parameter:

一个字符串.默认值为％Y-％m-％d％H:％M:％S"(如果有) 组件的时间组件不是午夜，而％Y-％m-％d" 否则

A character string. The default is "%Y-%m-%d %H:%M:%S" if any component has a time component which is not midnight, and "%Y-%m-%d" otherwise

一个想法是重新定义POSIXct对象的S3方法print:

One idea is to redefine the S3 method print for POSIXct object:

print.POSIXct <- function(x,...)print(format(x,"%Y-%m-%d %H:%M:%S"))

现在以您的示例为例，如果您打印x日期(带有午夜部分)，您将得到:

Now for your example if your print your x date(with midnight part) you get:

x <- "2007-02-01 00:00:00"
x <- as.POSIXct(x,tz=Sys.timezone())
x
[1] "2007-02-01 00:00:00"

这篇关于将"hr"转换为"00:00:00"；到日期时间"00:00:00"的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！