Júlio Hoffimann (Jul 22 2023 at 21:14): Júlio Hoffimann (Jul 22 2023 at 21:14):Given the following generator:
xs = range(0, stop=1, length=10)
ys = range(0, stop=1, length=10)
((x, y) for (x, y) in Iterators.product(xs, ys))
How to "vec" it so that the result is 1-dimensional? I am currently relying on IterTools.ivec, but would like to drop this dependency.
Is there an easy workaround that doesn't rely on defining a whole new IVec struct?
Mason Protter (Jul 22 2023 at 21:21):I’m pretty sure you can just put an if true at the end of the generator to turn it into a filtered generator and then it will lose all shape information
Jakob Nybo Nissen (Jul 22 2023 at 21:22):Yes but it also loses its length information (making the collect significantly slower). You can call collect, then call vec on it.
Júlio Hoffimann (Jul 22 2023 at 21:35):In my use case I actually managed to do a simpler trick:
https://github.com/JuliaGeometry/Meshes.jl/commit/16b340e5123dccbac98c742f142a987eb6b93cfd#
Júlio Hoffimann (Jul 22 2023 at 21:36):Because I have the number of iterators a priori, I can manually add two "for" keywords to flatten the vector. This is different than a single "for" with commas.
Notification Bot (Jul 22 2023 at 21:36):Júlio Hoffimann has marked this topic as resolved.
Júlio Hoffimann (Jul 22 2023 at 21:37):julia> ((x, y) for (x, y) in Iterators.product(xs, ys)) |> collect
10×10 Matrix{Tuple{Float64, Float64}}:
(0.0, 0.0) (0.0, 0.111111) (0.0, 0.222222) (0.0, 0.333333) … (0.0, 0.666667) (0.0, 0.777778) (0.0, 0.888889) (0.0, 1.0)
(0.111111, 0.0) (0.111111, 0.111111) (0.111111, 0.222222) (0.111111, 0.333333) (0.111111, 0.666667) (0.111111, 0.777778) (0.111111, 0.888889) (0.111111, 1.0)
(0.222222, 0.0) (0.222222, 0.111111) (0.222222, 0.222222) (0.222222, 0.333333) (0.222222, 0.666667) (0.222222, 0.777778) (0.222222, 0.888889) (0.222222, 1.0)
(0.333333, 0.0) (0.333333, 0.111111) (0.333333, 0.222222) (0.333333, 0.333333) (0.333333, 0.666667) (0.333333, 0.777778) (0.333333, 0.888889) (0.333333, 1.0)
(0.444444, 0.0) (0.444444, 0.111111) (0.444444, 0.222222) (0.444444, 0.333333) (0.444444, 0.666667) (0.444444, 0.777778) (0.444444, 0.888889) (0.444444, 1.0)
(0.555556, 0.0) (0.555556, 0.111111) (0.555556, 0.222222) (0.555556, 0.333333) … (0.555556, 0.666667) (0.555556, 0.777778) (0.555556, 0.888889) (0.555556, 1.0)
(0.666667, 0.0) (0.666667, 0.111111) (0.666667, 0.222222) (0.666667, 0.333333) (0.666667, 0.666667) (0.666667, 0.777778) (0.666667, 0.888889) (0.666667, 1.0)
(0.777778, 0.0) (0.777778, 0.111111) (0.777778, 0.222222) (0.777778, 0.333333) (0.777778, 0.666667) (0.777778, 0.777778) (0.777778, 0.888889) (0.777778, 1.0)
(0.888889, 0.0) (0.888889, 0.111111) (0.888889, 0.222222) (0.888889, 0.333333) (0.888889, 0.666667) (0.888889, 0.777778) (0.888889, 0.888889) (0.888889, 1.0)
(1.0, 0.0) (1.0, 0.111111) (1.0, 0.222222) (1.0, 0.333333) (1.0, 0.666667) (1.0, 0.777778) (1.0, 0.888889) (1.0, 1.0)
julia> ((x, y) for y in ys for x in xs) |> collect
100-element Vector{Tuple{Float64, Float64}}:
(0.0, 0.0)
(0.1111111111111111, 0.0)
(0.2222222222222222, 0.0)
(0.3333333333333333, 0.0)
(0.4444444444444444, 0.0)
(0.5555555555555556, 0.0)
⋮
(0.4444444444444444, 1.0)
(0.5555555555555556, 1.0)
(0.6666666666666666, 1.0)
(0.7777777777777778, 1.0)
(0.8888888888888888, 1.0)
(1.0, 1.0)
Out of curiosity does calling Vector on it work? That actually sounds like the nicest way of solving it in general
Júlio Hoffimann (Jul 22 2023 at 23:26):You mean replacing collect with Vector @Jakob Nybo Nissen?
Júlio Hoffimann (Jul 22 2023 at 23:26):I think we need to collect the generator before feeding it to the Vector constructor.
Jakob Nybo Nissen (Jul 23 2023 at 07:22):Yeah... I think the Vector constructor ought to take arbitrary iterables
Last updated: Oct 02 2023 at 04:34 UTC