在R中如何将格式化的本地时间转换为历元?

I have a local time like "2013-08-27 10:01:22", how do I convert it to epoch time?

我有一个本地时间，比如“2013-08-27 10:01:22”，我该如何将它转换为*时间呢?

basically I need the opposite of as.POSIXct, I searched on google and surprisingly didn't find any.

基本上我需要a的反义词。POSIXct，我在谷歌上搜索，却没有找到。

2 个解决方案

#1

You can use as.integer to get seconds since epoch began...

你可以使用as.integer来获取秒，从纪元开始……

x <- as.POSIXct( Sys.time() )
#[1] "2013-08-27 12:37:17 BST"

class(x)
#[1] "POSIXct" "POSIXt" 

as.integer( x )
#[1] 1377603437

Using a vector of strings called times:

用一个叫做时间的弦向量:

times
#[1] "2013-08-27 12:39:32" "2013-08-27 12:39:33" "2013-08-27 12:39:34"
#[4] "2013-08-27 12:39:35" "2013-08-27 12:39:36" "2013-08-27 12:39:37"
#[7] "2013-08-27 12:39:38" "2013-08-27 12:39:39" "2013-08-27 12:39:40"
#[10] "2013-08-27 12:39:41"

as.integer( as.POSIXct( times ) )
#[1] 1377603609 1377603610 1377603611 1377603612 1377603613 1377603614
#[7] 1377603615 1377603616 1377603617 1377603618

Without the timezone in the strings you will probably have to specify the tz argument to as.POSIXct, e.g. as.integer( as.POSIXct( times ) , tz = "BST" ) for British Summertime.

如果没有字符串中的时区，您可能需要将tz参数指定为。POSIXct,例如as.integer(如。英国夏令时的POSIXct(times)， tz =“BST”。

#2

The accepted answer truncates time to the whole second. POSIXct actually provides sub-second resolution, though. As mentioned in the comments by “statquant”, you can use as.numeric to obtain the exact epoch:

被接受的答案将时间缩短到整个秒。然而，POSIXct实际上提供了亚秒分辨率。正如“statquant”评论中提到的，您可以使用As。数值以获得确切的历元:

result = as.numeric(as.POSIXct(Sys.time()))

Beware that with the default options for digit display in R this will look like it has no digits behind the decimal point:

请注意，在R中显示数字的默认选项将会看起来在小数点后没有数字:

> result
[1] 1480599768

However, these are simply truncated in the display. To make them visible, use:

然而，这些只是在显示中被截断。要使它们可见，请使用:

> dput(result)
1480599767.58447

… or set options('digits') to a higher value.

或将选项(“位数”)设置为更高的值。

#1