Re: [R] Assignments inside lapply

From: Kenn Konstabel <lebatsnok_at_gmail.com>
Date: Wed, 27 Apr 2011 13:12:39 +0300

On Wed, Apr 27, 2011 at 12:58 PM, Nick Sabbe <nick.sabbe_at_ugent.be> wrote:
> No, that does not work.
> You cannot do assignment within (l)apply.
> Nor in any other function for that matter.

Yes that may work if you want to.
You can do non-local assignment within lapply using <<- (and, for that matter, within any other function) but there is no one-word answer to the question whether this is a good idea.

a <- list()
lapply(1:5, function(x) a[[x]] <<- x)
print(a)

KK

>
>
> Nick Sabbe
> --
> ping: nick.sabbe@ugent.be
> link: http://biomath.ugent.be
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>
> -- Do Not Disapprove
>
>
>
> -----Original Message-----
> From: r-help-bounces_at_r-project.org [mailto:r-help-bounces_at_r-project.org] On
> Behalf Of Alaios
> Sent: woensdag 27 april 2011 11:37
> To: R-help_at_r-project.org
> Subject: [R] Assignments inside lapply
>
> Dear all I would like to ask you if an assignment can be done inside a
> lapply statement.
>
> For example
>
> I would like to covert a double nested for loop
>
> for (i in c(1:dimx)){
>  for (j in c(1:dimy)){
>      Powermap[i,j] <- Pr(c(i,j),c(PRX,PRY),f)
>   }
> }
>
> to something like that:
>
>
> ij<-expand.grid(i=seq(1:dimx),j=(1:dimy))
>
> unlist(lapply(1:nrow(ij),function(rowId) { return
> (Powermap[i,j]<-Pr(c(ij$i[rowId],ij$j[rowId]),c(PRX,PRY),f))   }))
>
>
> as you can see lapply does not return nothing as the assignment is done
> inside the function. Would that work correctly? What are the cases such a
> statement will misfunction?
>
> I would like to thank you in advace for your help.
>
> Best Regards
> Alex
>
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>



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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. Received on Wed 27 Apr 2011 - 10:16:33 GMT

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