On 2011-04-26 05:26, Dimitris Rizopoulos wrote:
one way is the following:
a<- matrix(rnorm(9), 3 ,3)
aa<- a[order(row(a), -a)]
matrix(aa, nrow(a), byrow = TRUE)[, 2]
That's clever, Dmitris. And very fast!
Lars: my apologies; I didn't read your request
carefully. You asked for more _efficient_ and
I just thought 'shorter code'. I should know
that whenver I think 'apply', I should think
'matrix'.
Peter Ehlers
I hope it helps.
Best,
Dimitris
On 4/26/2011 2:01 PM, Lars Bishop wrote:
Hi,
I need to extract the second largest element from each row of a
matrix. Below is my solution, but I think there should be a more efficient
way to accomplish the same, or not?
set.seed(1)
a<- matrix(rnorm(9), 3 ,3)
sec.large<- as.vector(apply(a, 1, order, decreasing=T)[2,])
ans<- sapply(1:length(sec.large), function(i) a[i, sec.large[i]])
ans
Thanks in advance for your help,
Lars.
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