If cycle takes to much time...

On 25-01-2013, at 10:25, marcoguerzoni <marco.guerzoni at 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