Yes, this is sort of what I was looking for, there is one problem though. f(i,j) will be called K times but there is only need for one call. Maybe this is a little better (no redundant calls to f()), but there still is a lot of unnecessary copying, maybe it will be optimized away?
tmp = [f(i, j) for i in 1:n, j in i:m] [tmp[i,j][k] for k in 1:K, i in 1:n, j in 1:m] Are there better ways? Den måndagen den 24:e mars 2014 kl. 23:12:03 UTC+1 skrev Gunnar Farnebäck: > > [f(i,j)[k] for k in 1:K, i in 1:n, j in i:m] > > Den måndagen den 24:e mars 2014 kl. 15:07:49 UTC+1 skrev Linus Mellberg: >> >> Hi! >> >> I'm trying to construct a 3 dimensional array from a number of 1 >> dimensional arrays. Essentially what i would like to do is >> >> a = [f(i, j) for i in 1:n, j in 1:m] >> >> where f(i, j) is a function that returns an array (note, f has to >> construct the entire array at the same time). The code above creates a >> 2-dimensional array of arrays, but I would like to get a 3-dimensional >> array with the arrays returned by f in the first dimension with i and j in >> the second and third dimension, hope you understand >> >> a[:,:,1] = [f(1,1) f(2,1) ... f(n,1)] >> a[:,:,2] = [f(1,2) f(2,2) ... f(n,2)] >> . >> . >> . >> a[:,:,m] = [f(1,m) f(2,m) ... f(n,m)] >> >> f(i,j) are column arrays above. >> >> It can be achieved by first creating the large matrix and then filling it >> >> a = zeros(Int64, k, n, m) >> for i in 1:n, j in 1:m >> a[:,i,j] = f(i,j) >> end >> >> Is this the only way? I find it sort of ugly when its usually possible to >> do nice construction using comprehensions in other cases. >> >>
