Thanks, Simon. This is very useful.
Brian
On Mon, Mar 28, 2011 at 11:39 PM, Simon Blomberg wrote:
> ?cophenetic
>
> Cheers,
>
> Simon.
>
> On 29/03/11 10:27, Ben Bolker wrote:
>
>> Brian Pellerin gmail.com> writes:
>>
>>
>>
>>
>>
Hello R users,
If I generate a random tree with n=10 tips as rtree(n=10) say, is there a
way to have the distances between all tips put into a n by n matrix?
Sincerely,
Brian
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R-help@r-project.org mailing
Thanks, David. This works splendidly. Thanks for your help.
Sincerely,
Brian
>>> David Winsemius 03/26/11 3:06 PM >>>
On Mar 26, 2011, at 9:44 AM, Brian Pellerin wrote:
> Hello,
>
> I would like to take advantage of the upper.tri() function here but
> I don&
Hello,
I would like to take advantage of the upper.tri() function here but I don't
know exactly. Here is some working code...
i<-5
fi<-matrix(0,nrow=i,ncol=i)
for(r in 1:i){
for(c in 1:i){
if(r==c){
fi[r,c]<-1
}else if(rhttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Thanks, we're almost there. The 3rd statement needs to
satisfy fi_2[r,c]<-fi_2[c,r] where rwrote:
> Try this:
>
> fi_2 <- diag(1, i)
> fi_2[lower.tri(fi_2)] <- 1 - runif(sum(lower.tri(fi_2))) ^ .5
> fi_2[upper.tri(fi_2)] <- fi_2[lower.tri(fi_2)]
>
> On Tue, M
Hello R users,
I would like to reduce the number of for loops in my code. I build these
matrices (5 times). The main diagonal are 1s and the two sides along the
main diagonal mirror each other as follows:
i<-5
fi<-matrix(0,nrow=i,ncol=i)#floral inhibition matrix for(r in 1:i){ for(c in
1:i){
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