Dear John,
Perhaps is possible to do the job using apply or any of its friends. For
now, here is a solution using for() :
# Data sets
set.seed(123)
sample1<-matrix(data=c(1:20),byrow=TRUE,ncol=5,nrow=4)
sample2<-sample1+rnorm(20)
# t-test p-values
k=nrow(sample1)
res=NULL
for(i in 1:k) res[i]=t.t
How about the following?
sapply(1:nrow(sample1), function(x) t.test(sample1[x,], sample2[x,])$p.value)
Gabor
On Thu, Oct 16, 2008 at 6:07 PM, John Sorkin
<[EMAIL PROTECTED]> wrote:
> Although I am grateful to Jorge and David for their suggestions, their
> solution will not solve my problem as s
Although I am grateful to Jorge and David for their suggestions, their solution
will not solve my problem as sample2 is not fixed. Let me demonstrate what I
want to do:
# Define two matricies
sample1<-matrix(data=c(1:20),byrow=TRUE,ncol=5,nrow=4)
sample2<-sample1+rnorm(20)
These are the comput
3 matches
Mail list logo
