Question

我有一个日期的字符向量，我想将其转换为不同的时区。

 pubdate
[1] "Fri, 10 Jul 2015 03:21:23 +0000" "Fri, 10 Jul 2015 03:04:55 +0000"
[3] "Thu, 09 Jul 2015 23:49:01 +0000" "Thu, 09 Jul 2015 23:30:37 +0000"
 [5] "Thu, 09 Jul 2015 23:27:44 +0000" "Thu, 09 Jul 2015 23:16:46 +0000"
 [7] "Thu, 09 Jul 2015 23:14:06 +0000" "Thu, 09 Jul 2015 23:10:20 +0000"
 [9] "Thu, 09 Jul 2015 23:07:52 +0000" "Thu, 09 Jul 2015 22:37:41 +0000"
[11] "Thu, 09 Jul 2015 22:35:06 +0000"

我为此创建了一个函数。

temp <- as.matrix(0)

    for (i in 1:length(pubdate)){
            tmp_dta <- strptime(pubdate[[i]],format="%a, %d %b %Y %H:%M:%S", tz="GMT")
            tmp_dta$hour <- tmp_dta$hour - 1
            tmp_dta <- as.POSIXct(tmp_dta)
            attributes(tmp_dta)$tzone <- "Asia/Manila" 
            temp[i] <- tmp_dta

    }

然而，当我试图打印临时数据时，它似乎返回了秒数。这里

> temp
 [1] 1436494883 1436493895 1436482141 1436481037 1436480864 1436480206 1436480046 1436479820
 [9] 1436479672 1436477861 1436477706

我可以知道如何更改它作为日期返回吗？例如：＆＃34; 2015-07-10 10:21:23 PHT＆＃34;

谢谢！

更新：正如Nicola在下面所建议的那样，我删除了循环并添加了他建议的代码。下面的代码有效：

tmp_dta <- strptime(pubdate,format="%a, %d %b %Y %H:%M:%S", tz="GMT")

    x <- as.POSIXct(tmp_dta)

    attributes(x)$tzone <- "Asia/Manila" 

    newpubdate <- x - 3600

Answer 1

您可以在一行中完成整个操作：

pubdate <- c( "Fri, 10 Jul 2015 03:21:23 +0000" ,"Fri, 10 Jul 2015 03:04:55 +0000","Thu, 09 Jul 2015 23:49:01 +0000", "Thu, 09 Jul 2015 23:30:37 +0000","Thu, 09 Jul 2015 23:27:44 +0000", "Thu, 09 Jul 2015 23:16:46 +0000","Thu, 09 Jul 2015 23:14:06 +0000","Thu, 09 Jul 2015 23:10:20 +0000", "Thu, 09 Jul 2015 23:07:52 +0000", "Thu, 09 Jul 2015 22:37:41 +0000","Thu, 09 Jul 2015 22:35:06 +0000")
strptime(pubdate,"%a, %d %b %Y %H:%M:%S %z",tz="Asia/Manila")
# [1] "2015-07-10 11:21:23 PHT" "2015-07-10 11:04:55 PHT" "2015-07-10 07:49:01 PHT" "2015-07-10 07:30:37 PHT" "2015-07-10 07:27:44 PHT" "2015-07-10 07:16:46 PHT" "2015-07-10 07:14:06 PHT" "2015-07-10 07:10:20 PHT"
# [9] "2015-07-10 07:07:52 PHT" "2015-07-10 06:37:41 PHT" "2015-07-10 06:35:06 PHT"

如果您认为存在一小时的差异（我不认为有：您是否考虑过夏令时？），那么，正如@nicola建议的那样：

strptime(pubdate,"%a, %d %b %Y %H:%M:%S %z",tz="Asia/Manila") - as.difftime(1,unit="hours")
# [1] "2015-07-10 10:21:23 PHT" "2015-07-10 10:04:55 PHT" "2015-07-10 06:49:01 PHT" "2015-07-10 06:30:37 PHT" "2015-07-10 06:27:44 PHT" "2015-07-10 06:16:46 PHT" "2015-07-10 06:14:06 PHT" "2015-07-10 06:10:20 PHT"
# [9] "2015-07-10 06:07:52 PHT" "2015-07-10 05:37:41 PHT" "2015-07-10 05:35:06 PHT"

日期以秒为单位返回

1 个答案: