Here is a faster way of doing the replacement: (provide reproducible
data next time)
> x <- matrix(sample(6:9, 64, TRUE), 8)
> x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 8 7 7 6 7 8 7 9
[2,] 7 7 8 6 7 6 7 7
[3,] 7 7 7 6 9 6 6 7
[4,] 9 9 7 6 8 7 6 6
[5,] 6 9 9 8 8 9 8 9
[6,] 9 7 6 9 7 8 6 7
[7,] 7 9 8 9 7 9 7 8
[8,] 9 9 6 9 9 8 8 6> x.f <- 1:8 # replacement values based on column
> x.ind <- which(x == 9, arr.ind=TRUE)
> x.ind
row col
[1,] 4 1
[2,] 6 1
[3,] 8 1
[4,] 4 2
[5,] 5 2
[6,] 7 2
[7,] 8 2
[8,] 5 3
[9,] 6 4
[10,] 7 4
[11,] 8 4
[12,] 3 5
[13,] 8 5
[14,] 5 6
[15,] 7 6
[16,] 1 8
[17,] 5 8> x[x.ind] <- x.f[x.ind[,'col']]
> x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 8 7 7 6 7 8 7 8
[2,] 7 7 8 6 7 6 7 7
[3,] 7 7 7 6 5 6 6 7
[4,] 1 2 7 6 8 7 6 6
[5,] 6 2 3 8 8 6 8 8
[6,] 1 7 6 4 7 8 6 7
[7,] 7 2 8 4 7 6 7 8
[8,] 1 2 6 4 5 8 8 6
On Wed, Oct 28, 2009 at 12:55 PM, Marcio Resende
<mresendeufv at yahoo.com.br> wrote:>
> #Mdarts is a matrix 2343x788
> #frequencia is a vector 2343x1
> # 9 in Mdarts[fri,frj] stands for my missing values which i want to replace
> by the value in the vector frequencia
>
>
>
Mdarts<-t(matrix(scan("C:/GWS/CNB/dartg.txt"),ncol=nindT,nrow=nm,
byrow=T))
> frequencia <- matrix(scan("C:/GWS/CNB/freq.txt"),ncol=1)
> for (fri in 1:nindT){
> for (frj in 1:nm){
> Mdarts[fri,frj] <- if (Mdarts[fri,frj] == 9) frequencia[frj] else
> Mdarts[fri,frj]
> Mdarts[fri,frj] <- Mdarts[fri,frj]/1-(frequencia[frj]^2)
> }
> }
>
> Is there a faster way to it?
> Maybe using any apply function?
> Thanks in advance
> --
> View this message in context:
http://www.nabble.com/Is-there-a-faster-way-to-do-it--tp26098223p26098223.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
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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.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?