将dd/mm/yy和dd/mm/yyyy转换为日期
我有一个字符向量,带有像这样的各种格式的日期
I have some a character vector with dates in various formats like this
dates <- c("23/11/12", "20/10/2012", "22/10/2012" ,"23/11/12")
我想将它们转换为日期.我已经从lubridate包中尝试了很好的dmy,但这是行不通的:
I want to convert these to Dates. I have tried the very good dmy from the lubridate package, but this does not work:
dmy(dates)
[1] "0012-11-23 UTC" "2012-10-20 UTC" "2012-10-22 UTC" "0012-11-23 UTC"
它将/12年视为0012.
It is treating the /12 year as if it is 0012.
因此,我现在尝试使用正则表达式选择每种类型,并使用as.Date()分别转换为日期.但是,我尝试选择dd/mm/yy的正则表达式仅不起作用.
So I now am trying regular expression to select each type and individually convert to dates using as.Date(). However the regular expression I have tried to select the dd/mm/yy only does not work.
dates[grep('[0-9]{2}/[0-9]{2}/[0-9]{2,2}', dates)]
返回
[1] "23/11/12" "20/10/2012" "22/10/2012" "23/11/12"
我认为{2,2}应该得到正好2个数字,而不是全部.我不太擅长使用正则表达式,因此我们将不胜感激.
I thought that the {2,2} should get a exactly 2 numbers and not all of them. I'm not very good at regular expression so any help will be appreciated.
谢谢
编辑
我实际上拥有的是以下三种不同类型的日期
What I actually have are three different types of date as below
dates <- c("23-Jul-2013", "23/11/12", "20/10/2012", "22/10/2012" ,"23/11/12")
我想将它们转换为日期
parse_date_time(dates,c('dmy'))
给我
[1] "2013-07-23" "0012-11-23" "2012-10-20" "2012-10-22" "0012-11-23"
但是,这是错误的,0012应该是2012.我想(一个非常简单的)解决方案.
However, this is wrong and 0012 should be 2012. I would like (a fairly simple) solution to this.
我现在有一个解决方案(感谢@plannapus)是使用正则表达式 我实际上最终创建了此函数,因为我仍然遇到一些情况,即润滑方法将12变成0012
One solution I now have (thanks to @plannapus)is to use regular expressions I actually ended up creating this function as I was still getting some cases where the lubridate approach was turning 12 into 0012
asDateRegex <- function(dates,
#selects strings from the vector dates using regexes and converts these to Dates
regexes = c('[0-9]{2}/[0-9]{2}/[0-9]{4}', #dd/mm/yyyy
'[0-9]{2}/[0-9]{2}/[0-9]{2}$', #dd/mm/yy
'[0-9]{2}-[[:alpha:]]{3}-[0-9]{4}'), #dd-mon-yyyy
orders = 'dmy',
...){
require(lubridate)
new_dates <- as.Date(rep(NA, length(dates)))
for(reg in regexes){
new_dates[grep(reg, dates)] <- as.Date(parse_date_time(dates[grep(reg, dates)], order = orders))
}
new_dates
}
asDateRegex (dates)
[1] "2012-10-20" "2013-07-23" "2012-11-23" "2012-10-22" "2012-11-23"
但这不是很优雅.有更好的解决方案吗?
But this is not very elegant. Any better solutions?
您可以在lubridate
中使用parse_date_time
:
some.dates <- c("23/11/12", "20/10/2012", "22/10/2012" ,"23/11/12")
parse_date_time(some.dates,c('dmy'))
[1] "2012-11-23 UTC" "2012-10-20 UTC" "2012-10-22 UTC" "2012-11-23 UTC"
但是,请注意格式的顺序很重要:
But , Note that the order of format is important :
some.dates <- c("20/10/2012","23/11/12", "22/10/2012" ,"23/11/12")
parse_date_time(some.dates,c('dmY','dmy'))
[1] "2012-10-20 UTC" "2012-11-23 UTC" "2012-10-22 UTC" "2012-11-23 UTC"
编辑
内部parse_date_time
使用的是guess_formats
(我猜它使用了一些正则表达式):
Internally parse_date_time
is using guess_formats
(which I guess uses some regular expressions):
guess_formats(some.dates,c('dmy'))
dmy dmy dmy dmy
"%d/%m/%Y" "%d/%m/%y" "%d/%m/%Y" "%d/%m/%y"
如评论中所述,您可以像这样使用parse_date_time
:
As mentioned in the comment you can use parse_date_time
like this:
as.Date(dates, format = guess_formats(dates,c('dmy')))