Hello,
Better yet is function aggregate.
# Create some data matrix
Y <- matrix(c(sample(80, 507, TRUE), sample(-1:1, 4*507, TRUE)), ncol=5)
X <- aggregate(Y[, 5]~Y[, 1], Y, sum) # In your case Y[, 4] not Y[, 1]
Hope this helps,
Rui Barradas
Em 05-06-2012 09:53, ?zg?r Asar escreveu:> Hi,
>
> x<-matrix(0,80,ncol=1)
>
> will create x matrix with all elements 0 (to be filled by the sums that you
> need)
>
> sum(y[y[,4]==1,5])
>
> will calculate the sum of 5th column of y with 4th column=1
>
> Similarly,
>
> sum(y[y[,4]==80,5])
>
> will calculate the sum of 5th column of y with 4th column=80.
>
> You can adapt this to your case, with simple loops, etc.
>
> Hope this helps.
> Ozgur
>
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