Don't (can't) use paste0 with sep='' -- that's redundant, and I'm surprised not
an error.
Michael
On Jun 29, 2012, at 7:18 PM, arun <smartpink...@yahoo.com> wrote:
> Hi,
>
> I knew that you already got many replies.
>
> Here is my contribution.
> dat1<-paste0(expand.grid(d,e)$Var1,expand.grid(d,e)$Var2,sep="")
>
> or
> dat2<-as.vector(rbind(paste0(d,e[1],sep=""),paste0(d,e[2],sep=""),paste0(d,e[3],sep="")))
>
>
> #Still Bert's one line is the shortest
>
>
> A.K.
>
>
>
> ----- Original Message -----
> From: arun.gurubaramurugeshan <arun.gurubaramuruges...@autozone.com>
> To: r-help@r-project.org
> Cc:
> Sent: Thursday, June 28, 2012 5:03 PM
> Subject: [R] Nested For Loop
>
> I am creating a nested for loop and following are the codes I'm using, but I
> am not acheiving what I want.
>
> I have a vector
>
> d<-151:159
>
> I have another vector
>
> e<-e<-c("apple", "orange", "banana")
>
> I need to create f as
> 151apple
> 151orange
> 151banana
> .
> .
> 159apple
> 159orange
> 159banana
>
> Here is how I wrote nested for loop...
>
> for (i in 1:length(d))
> { for (j in 1:length(e))
> {
> x[j]<-paste(d[i],e[j],sep="")
> print(x[j])
> }
> }
>
> The result of the above codes is....
>
>> for (i in 1:length(d))
> + { for (j in 1:length(e))
> + {
> + x[j]<-paste(d[i],e[j],sep="")
> + print(x[j])
> + }
> + }
> [1] "151apple"
> [1] "151orange"
> [1] "151banana"
>>
>
>
> What do I need to do this looping produce the desired result.
>
> Thanks
> Arun
> ----
>
>
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Nested-For-Loop-tp4634804.html
> Sent from the R help mailing list archive at Nabble.com.
>
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>
> ______________________________________________
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
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