Hi,
I have two matricies a and x:
a<-matrix(c(3,4,5,2,3,4,1,1,2), nrow=3, ncol=3)
[,1] [,2] [,3]
[1,] 3 2 1
[2,] 4 3 1
[3,] 5 4 2
x<-matrix(c(3, NA, NA, NA, 2, 5, NA, 2, 2), nrow=3, ncol=3)
[,1] [,2] [,3]
[1,] 3 NA NA
[2,] NA 2 2
[3,] NA 5 2
I wish to combine these two into one matrix using the values from x where x has
values, and values from a where x has NA's, giving a new matrix which would
look like this:
ax<-matrix(c(3,4,5,2,2,5,1,2,2), nrow=3, ncol=3)
[,1] [,2] [,3]
[1,] 3 2 1
[2,] 4 2 2
[3,] 5 5 2
I want an automatic way of doing this as my actual application is a much larger
matrix.
Thanks in advance
Tom
_________________________________________________________________
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try this: a <- matrix(c(3,4,5,2,3,4,1,1,2), nrow=3, ncol=3) x <- matrix(c(3, NA, NA, NA, 2, 5, NA, 2, 2), nrow=3, ncol=3) ind <- is.na(x) x[ind] <- a[ind] x I hope it helps. Best, Dimitris Tom Liptrot wrote:> Hi, > > I have two matricies a and x: > > a<-matrix(c(3,4,5,2,3,4,1,1,2), nrow=3, ncol=3) > > [,1] [,2] [,3] > [1,] 3 2 1 > [2,] 4 3 1 > [3,] 5 4 2 > > x<-matrix(c(3, NA, NA, NA, 2, 5, NA, 2, 2), nrow=3, ncol=3) > > [,1] [,2] [,3] > [1,] 3 NA NA > [2,] NA 2 2 > [3,] NA 5 2 > > I wish to combine these two into one matrix using the values from x where x has values, and values from a where x has NA's, giving a new matrix which would look like this: > > ax<-matrix(c(3,4,5,2,2,5,1,2,2), nrow=3, ncol=3) > > [,1] [,2] [,3] > [1,] 3 2 1 > [2,] 4 2 2 > [3,] 5 5 2 > > I want an automatic way of doing this as my actual application is a much larger matrix. > > Thanks in advance > > Tom > > > _________________________________________________________________ > > [[elided Hotmail spam]] > > > [[alternative HTML version deleted]] > > ______________________________________________ > 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. >-- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014
On Mon, Jul 13, 2009 at 11:31 AM, Tom Liptrot <tomliptrot at hotmail.com> wrote:> I wish to combine these two into one matrix using the values from x where x has values, and values from a where x has NA's, giving a new matrix which would look like this:This should do the trick: x[which(is.na(x))]=a[which(is.na(x))] -- Michael Knudsen micknudsen at gmail.com http://www.google.com/profiles/micknudsen
Hi
matrix is virtually a vector so you can find index of values of x and add
those values to propper places in ax
ax <- a
idx <- which(!is.na(x))
ax[idx] <- x[idx]
ax
[,1] [,2] [,3]
[1,] 3 2 1
[2,] 4 2 2
[3,] 5 5 2
Regards
Petr
r-help-bounces at r-project.org napsal dne 13.07.2009 11:31:12:
>
> Hi,
>
> I have two matricies a and x:
>
> a<-matrix(c(3,4,5,2,3,4,1,1,2), nrow=3, ncol=3)
>
> [,1] [,2] [,3]
> [1,] 3 2 1
> [2,] 4 3 1
> [3,] 5 4 2
>
> x<-matrix(c(3, NA, NA, NA, 2, 5, NA, 2, 2), nrow=3, ncol=3)
>
> [,1] [,2] [,3]
> [1,] 3 NA NA
> [2,] NA 2 2
> [3,] NA 5 2
>
> I wish to combine these two into one matrix using the values from x
where x > has values, and values from a where x has NA's, giving a new matrix
which > would look like this:
>
> ax<-matrix(c(3,4,5,2,2,5,1,2,2), nrow=3, ncol=3)
>
> [,1] [,2] [,3]
> [1,] 3 2 1
> [2,] 4 2 2
> [3,] 5 5 2
>
> I want an automatic way of doing this as my actual application is a much
larger matrix.>
> Thanks in advance
>
> Tom
>
>
> _________________________________________________________________
>
> [[elided Hotmail spam]]
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> 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.
The following code should do it. This assumes matrices a and x are of the same dimension which is why you can index a as below x[is.na(x)==TRUE]<-a[is.na(x)==TRUE] -- View this message in context: http://www.nabble.com/Combine-two-matricies-tp24458609p24458797.html Sent from the R help mailing list archive at Nabble.com.