On 25-01-2013, at 10:25, marcoguerzoni <marco.guerz...@unito.it> wrote:
> dear all,
>
> thank you for reading.
>
> I have a dataset of artists and where and when they had an exhibition.
> I'd like to create an affiliation network in the form of matrix, telling me
> which aritist have been in the same at the same time.
> I manage to do it, but given that I have 96000 observation the program takes
> 30 months to complete.
> her what i have done.
> the data look like this
>
> Artist <-c(1,2,3,2,4,4,5)
> Begin <- as.Date(c('2006-08-23', '2006-03-21', '2006-03-06', '2006-01-13',
> '2006-05-20', '2006-07-13', '2006-07-20'))
> End <- as.Date(c('2006-10-23', '2006-11-30', '2006-05-06', '2006-12-13',
> '2006-09-20', '2006-08-13', '2006-09-20'))
> Istitution <- c(1, 2, 2, 1, 1, 2, 1)
>
> artist is the name of the artist, Begin and End is the when and Istitutionis
> the where.
>
> my IF is working,
>
>
> #number of unique artist
> c <- unique(Artist)
> d <- length(c)
> a <-length(Artist)
>
> B <- mat.or.vec(d,d)
>
> for(i in 1:d) {
> for(j in 1:d) {
> if (Istitution[i] == Istitution[j]) {
> if (Begin[i] <= End[j])
> {
> if (End[i]-Begin[j] >= 0) {
> B[i,j] <- B[i,j]+1
> B[i,i] <- 0
> }
> }
> else{
> if (End[j]-Begin[i] >= 0) {
> B[i,j] <- B[i,j]+1
> B[i,i] <- 0
> }
> }
> }
> }
> print(i)
> }
> do you have a way to make the programm simpler and faster?
It is not clear why you are only using the unique artists.
You shouldn't be using "c" as variable name. It is a builtin function.
Since the result is symmetric you can change the j-loop to for(j in (i+1):d).
After the loop you can do
B[lower.tri(B)] <- t(B)[lower.tri(B)]
to fill the remainder of the matrix B. This would certainly be more efficient.
But I don't quite understand what you are trying to do.
With you example you could compute the result you desire.
Gerrit's answer is concise.
Berend
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