Re: [R] sample

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From: Sundar Dorai-Raj (sundar.dorai-raj@pdf.com)
Date: Thu 06 May 2004 - 04:58:29 EST


Message-id: <40993955.9000907@pdf.com>


Rogério Rosa da Silva wrote:
> Dear List:
>
> I have the following simple program:
>
> x<- sample(site)
> VarGuilda1<- var(tapply(x,site,func1))
> VarGuilda2<- var(tapply(x,site,func2))
> VarGuilda3<- var(tapply(x,site,func3))
> VarGuilda4<- var(tapply(x,site,func4))
> VarGuilda5<- var(tapply(x,site,func5))
> VarGuilda6<- var(tapply(x,site,func6))
> VarGuilda7<- var(tapply(x,site,func7))
> VarGuilda8<- var(tapply(x,site,func8))
> VarGuilda9<- var(tapply(x,site,func9))
> Var<-cbind(VarGuilda1,VarGuilda2,VarGuilda3,VarGuilda4,VarGuilda5,VarGuilda6,VarGuilda7,VarGuilda8,VarGuilda9)
> write(Var,file="LAU_Var_01.txt", ncol=9)
>
> Every time I want to repeat this I have to change the name of *.txt
> file manually. How can I automate this, so it could be done for all the
> *.txt files (1000) I have to generate.
>
>
> Thanks in advance, Rogério
>

Use ?paste.

for(i in 1:1000) {
   ...
   file <- paste("LAU_Var_", i, ".txt", sep = "")
   write(Var, file = file, ncol = 9)
}

or if you want something like "LAU_VAR_xxxx.txt", you can use ?sprintf:

file <- "LAU_Var_%04d.txt"
for(i in 1:1000) {
   ...
   write(Var, file = sprintf(file, i), ncol = 9)
}

--sundar

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