In order to be a little more specific I wanted to add, that it seems weird that I can use the variables for the if clause, but not for creating the other ranges, it's just that I don't know how to express myself correctly, I hope you can understand me.
El miércoles, 10 de agosto de 2016, 11:56:00 (UTC-5), Ismael Venegas Castelló escribió: > > Is there a way to make reference of the internal variables of an array > comprehension? I'm trying to improve this Rosetta Code task: > > > - https://rosettacode.org/wiki/List_comprehensions#Julia > > > const n = 20 > sort(filter(x -> x[1] < x[2] && x[1]^2 + x[2]^2 == x[3]^2, [(a, b, c) for > a=1:n, b=1:n, c=1:n])) > > > In Python it's: > > In [2]: n = 20 > > In [3]: [(x,y,z) for x in xrange(1,n+1) for y in xrange(x,n+1) for z in > xrange(y,n+1) if x**2 + y**2 == z**2] > Out[3]: [(3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15), (12 > , 16, 20)] > > > I'll update the task with: > > julia> [(x, y, z) for x = 1:n, y = 1:n, z = 1:n if x < y && x^2 + y^2 == z > ^2] |> sort > 6-element Array{Tuple{Int64,Int64,Int64},1}: > (3,4,5) > (5,12,13) > (6,8,10) > (8,15,17) > (9,12,15) > (12,16,20) > > But I tried this and it doesn't work, I wonder why? > > julia> [(x, y, z) for x = 1:n, y = x:n, z = y:n if x^2 + y^2 == z^2] > ERROR: UndefVarError: x not defined > > Is there a way to do this? Thanks in advance! >
