hi: i tried regular expressions and , as usual, failed. but below does
do the job uglily using strsplit. i'd still be curious and appreciate if
someone could do the regular expression method. thanks.
dts <- c("1990m12", "1992m8") #March 1990 and Aug 1992
#SPLIT IT USING m
temp <- strsplit(dts,"m")
# PASTE THE #'S BUT CHECK
# FOR GREATER THAN 9 UGLIFIES IT
moddates <- lapply(temp,function(.str) {
if ( as.numeric(.str[2]) > 9 ) {
paste(.str[1],as.numeric(.str[2]),"01",sep="-")
} else {
paste(.str[1],"-0",as.numeric(.str[2]),"-01",sep="")
}
})
# MAKE THE DATE
datelist<-lapply(moddates,as.Date ,format="%Y-%m-%d")
On Thu, Dec 18, 2008 at 1:14 AM, Shruthi Jayaram wrote:
Hello,
If I have a character string like
d <- c("1990m3", "1992m8") #March 1990 and Aug 1992
what is the easiest way to convert it into any standard date form; for
example,
d <- c("01/03/1990", "01/08/1992")
I looked at as.Date but it doesn't seem to address my problem as I
have an
"m" stuck in the middle of my character string which R does not
recognise.
Would be very grateful for any help on this.
Shruthi
--
View this message in context:
http://www.nabble.com/Parsing-unusual-date-format-tp21067562p21067562.html
Sent from the R help mailing list archive at Nabble.com.
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
