On Thu, Jul 15, 2010 at 4:21 PM, Dennis Murphy wrote:
> Hi:
>
> Tal was on the right track, but the function in package zoo that applies
> here is rollapply():
>
> library(zoo)
> m<-matrix(seq(1,80),ncol=20, nrow=4)
> t(apply(m, 1, function(x) rollapply(zoo(x), width = 5, FUN = mean, by = 5)))
>
Hi:
Tal was on the right track, but the function in package zoo that applies
here is rollapply():
library(zoo)
m<-matrix(seq(1,80),ncol=20, nrow=4)
t(apply(m, 1, function(x) rollapply(zoo(x), width = 5, FUN = mean, by = 5)))
[,1] [,2] [,3] [,4]
[1,]9 29 49 69
[2,] 10 30 50
Eik and Patrick,
Thanks I will give those a try.
On Thu, Jul 15, 2010 at 8:15 AM, Patrick J Rogers wrote:
> Hi Steven,
>
> You can just cut the matrix up into a 5 column matrix and use apply as
> normal
>
> m2<-matrix(as.vector(t(m)), ncol=5, byrow=TRUE)
> result<-matrix(apply(m2, 1, mean), nc
Hi Steven,
You can just cut the matrix up into a 5 column matrix and use apply as normal
m2<-matrix(as.vector(t(m)), ncol=5, byrow=TRUE)
result<-matrix(apply(m2, 1, mean), ncol=ncol(m)/ncol(m2), byrow=TRUE)
result
--
Patrick Rogers
Dept. of Political Science
University of California, San Diego
Hi Steven,
how about this (using simple matrix algebra)
wind<-5 #size of window to be aggregated
if (ncol(m)%%wind!=0) stop('dimension of m does not fit') #crude check
tmat<-apply(diag(ncol(m)/wind),1,function(x)rep(x,each=wind))/wind
m%*%tmat
hth.
Am 15.07.2010 11:39, schrieb steven mosher:
>
You might find it useful to look at
rollmean
from the
zoo
package.
Tal
Contact
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www.r-stati
Given a matrix of MxN
want to take the means of rows in the following fashion
m<-matrix(seq(1,80),ncol=20, nrow=4)
result<-matrix(NA,nrow=4,ncol=20/5)
result[,1]<-apply(m[,1:5],1,mean)
result[,2]<-apply(m[,6:10],1,mean)
result[,3]<-apply(m[,11:15],1,mean)
result[,4]<-apply(m[,16:20],1,mean)
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