将POSIXct对象传递给函数返回数字向量

I'm trying to do some manipulation on the POSIXct vector, but when I pass it to a function, the vector changes into a numeric vector, instead of retaining POSIXct class, even when the function itself only returns the object:

我想对POSIXct向量做一些操作，但是当我将它传递给一个函数时，这个向量会变成一个数字向量，而不是保持POSIXct类，即使函数本身只返回对象:

# Sample dates from vector and it's class.
> dates <- as.POSIXct(c("2012-02-01 12:32:00", "2012-10-24 17:25:56", "2008-09-26 17:13:31", "2011-08-23 11:11:17,", "2015-09-19 22:28:33"), tz = "America/Los_Angeles")
> dates
[1] "2012-02-01 12:32:00 PST" "2012-10-24 17:25:56 PDT" "2008-09-26 17:13:31 PDT" "2011-08-23 11:11:17 PDT" "2015-09-19 22:28:33 PDT"
> class(dates)
[1] "POSIXct" "POSIXt" 
# Simple subset is retaining original class.
> qq <- dates[1:5]
> qq
[1] "2012-02-01 12:32:00 PST" "2012-10-24 17:25:56 PDT" "2008-09-26 17:13:31 PDT" "2011-08-23 11:11:17 PDT" "2015-09-19 22:28:33 PDT"
> class(qq)
[1] "POSIXct" "POSIXt" 
# sapply on the same subset using simple "return" function changes class to "numeric" - why? How to retain "POSIXct"?
> qq2 <- sapply(dates[1:5], function(x) x)
> qq2
[1] 1328128320 1351124756 1222474411 1314123077 1442726913
> class(qq2)
[1] "numeric"

Why it happens? How can I retain the POSIXct class of the original vector? I know that POSIXct is numeric under the hood, but I want to retain the original class for readability.

为什么会发生?如何保留原向量的POSIXct类?我知道POSIXct在外壳下是数字的，但是我想保留原来的类以使其具有可读性。

1 个解决方案

#1

We can use lapply instead of sapply as sapply by default has the option simplify = TRUE. So, if the list elements are of the same length, it will simplify it to vector or matrix depending on the length of the list elements and POSIXct is stored as numeric.

我们可以使用lapply代替sapply，因为默认情况下，sapply的选项是simplify = TRUE。因此，如果列表元素的长度相同，它将根据列表元素的长度将其简化为向量或矩阵，POSIXct将存储为数字。

lst <- lapply(dates, function(x) x)

If we need to use sapply, then an option would simplify = FALSE

如果我们需要使用sapply，那么一个选项将简单化= FALSE

lst <- sapply(dates, function(x) x, simplify=FALSE)

After applying the function, if we need as a vector output,

在应用这个函数之后，如果我们需要一个矢量输出，

do.call("c", lst)

Regarding the change of timezone, it is documented in the ?DateTimeClasses

关于时区的变更，在?DateTimeClasses中有记载

Using c on "POSIXlt" objects converts them to the current time zone, and on "POSIXct" objects drops any "tzone" attributes (even if they are all marked with the same time zone).

在“POSIXlt”对象上使用c将它们转换为当前时区，在“POSIXct”对象上删除任何“tzone”属性(即使它们都标记为相同的时区)。

So, the possible option would be (as mentioned in the comments by @kmo)

因此，可能的选项是(如@kmo在评论中提到的)

.POSIXct(lst, tz = "America/Los_Angeles")
#[1] "2012-02-01 12:32:00 PST" "2012-10-24 17:25:56 PDT" "2008-09-26 17:13:31 PDT" "2011-08-23 11:11:17 PDT" "2015-09-19 22:28:33 PDT"

Or as @thelatemail mentioned in the comments

或者如评论中提到的@thelatemail

.POSIXct(sapply(dates,I), attr(dates,"tzone") )

#1

lst <- lapply(dates, function(x) x)