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Message-ID: <CAAxdm-7Pm7R1LDTrkLVzx-iEhAtzvSZNr0P14ti=-o9ma4zNtA@mail.gmail.com>
Date: 2011-11-03T17:44:17Z
From: jim holtman
Subject: optimising a loop
In-Reply-To: <161DC602615F6943A19BAEA3DBF2E4B90177CC8DD55A@HARFANG.intranet.mrn.gouv>

try this:

> nom.plac2 <- c("102", "103", "301", "303","304", "403")
> poids2 <- matrix(NA, 6,6, dimnames=list(nom.plac2,nom.plac2))
> poids2
    102 103 301 303 304 403
102  NA  NA  NA  NA  NA  NA
103  NA  NA  NA  NA  NA  NA
301  NA  NA  NA  NA  NA  NA
303  NA  NA  NA  NA  NA  NA
304  NA  NA  NA  NA  NA  NA
403  NA  NA  NA  NA  NA  NA
> wei2 <- matrix(c(.6,.4,.5,.5,.9,.1,.8,.2,.7,.3,.6,.4),6,2,dimnames=list(nom.plac2, c("p1","p2")),byrow=T)
> voisin <- matrix(c("103","304", "303", "102", "103" ,"303","403","304","303","102","103" ,"303"),
+          6,2,dimnames=list(nom.plac2, c("v1","v2")),byrow=T)
>
> # do matrix addressing by converting names to numbers
> indx <- rbind(
+     cbind(match(rownames(voisin), rownames(poids2))
+            , match(voisin[, 1], colnames(poids2))
+            )
+   , cbind(match(rownames(voisin), rownames(poids2))
+            , match(voisin[, 2], colnames(poids2))
+            )
+   )
> indx
      [,1] [,2]
 [1,]    1    2
 [2,]    2    4
 [3,]    3    2
 [4,]    4    6
 [5,]    5    4
 [6,]    6    2
 [7,]    1    5
 [8,]    2    1
 [9,]    3    4
[10,]    4    5
[11,]    5    1
[12,]    6    4
>
> # change the data
> poids2[indx] <- c(wei2[,1], wei2[,2])
> poids2
    102 103 301 303 304 403
102  NA 0.6  NA  NA 0.4  NA
103 0.5  NA  NA 0.5  NA  NA
301  NA 0.9  NA 0.1  NA  NA
303  NA  NA  NA  NA 0.2 0.8
304 0.3  NA  NA 0.7  NA  NA
403  NA 0.6  NA 0.4  NA  NA
>
>
>

On Thu, Nov 3, 2011 at 1:25 PM,
<Bastien.Ferland-Raymond at mrnf.gouv.qc.ca> wrote:
> Dear R community,
>
> I'm trying to remove a loop from my code but I'm stock and I can't find a good way to do it. ?Hopefully one of you will have something clever to propose.
>
> Here is a simplified example:
>
> I have a squared matrix:
>
>> nom.plac2 <- c("102", "103", "301", "303","304", "403")
>> poids2 <- matrix(NA, 6,6, dimnames=list(nom.plac2,nom.plac2))
>> poids2
> ? ?102 103 301 303 304 403
> 102 ?NA ?NA ?NA ?NA ?NA ?NA
> 103 ?NA ?NA ?NA ?NA ?NA ?NA
> 301 ?NA ?NA ?NA ?NA ?NA ?NA
> 303 ?NA ?NA ?NA ?NA ?NA ?NA
> 304 ?NA ?NA ?NA ?NA ?NA ?NA
> 403 ?NA ?NA ?NA ?NA ?NA ?NA
>
> I want to replace some of the NAs following specific criterion included in 2 others matrix:
>
>> wei2 <- matrix(c(.6,.4,.5,.5,.9,.1,.8,.2,.7,.3,.6,.4),6,2,dimnames=list(nom.plac2, c("p1","p2")),byrow=T)
>> wei2
> ? ? p1 ?p2
> 102 0.6 0.4
> 103 0.5 0.5
> 301 0.9 0.1
> 303 0.8 0.2
> 304 0.7 0.3
> 403 0.6 0.4
>> voisin <- matrix(c("103","304", "303", "102", "103" ,"303","403","304","303","102","103" ,"303"),
> ? ? ? ? ?6,2,dimnames=list(nom.plac2, c("v1","v2")),byrow=T)
>> voisin
> ? ?v1 ? ?v2
> 102 "103" "304"
> 103 "303" "102"
> 301 "103" "303"
> 303 "403" "304"
> 304 "303" "102"
> 403 "103" "303"
>
> So my final result is:
>
> ? ?102 103 301 303 304 403
> 102 ?NA 0.6 ?NA ?NA 0.4 ?NA
> 103 0.5 ?NA ?NA 0.5 ?NA ?NA
> 301 ?NA 0.9 ?NA 0.1 ?NA ?NA
> 303 ?NA ?NA ?NA ?NA 0.2 0.8
> 304 0.3 ?NA ?NA 0.7 ?NA ?NA
> 403 ?NA 0.6 ?NA 0.4 ?NA ?NA
>
>
> So, globally I want to fill for each line of "poids2" data from "wei2" associated with the good the good identifier found in "voisin".
>
> This can easily be done by a loop:
>
>> loop <- poids2
>> for(i in 1:6){
> + loop[i,voisin[i,]] <- wei2[i,]
> + }
>
> But I expect it to be quite slow with my larger dataset.
>
> Does any of you has an idea how I could remove the loop and speed up the operation?
>
> Best regards,
>
>
> Bastien Ferland-Raymond, M.Sc. Stat., M.Sc. Biol.
> Division des orientations et projets sp?ciaux
> Direction des inventaires forestiers
> Minist?re des Ressources naturelles et de la Faune du Qu?bec
>
> ______________________________________________
> R-help at 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.
>



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.