You can use regular expressions:
as.Date(sub('(\\d+)m(\\d+)','\\1-\\2-01',dts,perl=TRUE))
but as.Date isn't as inflexible as you think:
as.Date(paste(dts,'m01',sep=''),'%Ym%mm%d')
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Thu, 18 Dec 2008, markle...@verizon.net wrote:
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
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