Mesh grids and position array tranform

New to Julia
1

I am working on the mesh grids, I want to ask whether there is quick transformation and inverse transformation between array_x and flat_x?

array_x = [ [1, 1, 1, 1],
            [2, 2, 2, 2],
            [3, 3, 3, 3]]

and
flat_x = [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3]

2

Going to flat_x you can use

julia> reduce(vcat, array_x)
12-element Vector{Int64}:
 1
 1
 1
 1
 2
 2
 2
 2
 3
 3
 3
 3

which will allocate a new (contiguous) array. There might be non-allocating approaches, but then you would be working with non-contiguous memory, which might also not be ideal.
For the converse direction you can use

julia> eachrow(reshape(flat_x, 4, 3)')
3-element RowSlices{LinearAlgebra.Adjoint{Int64, Matrix{Int64}}, Tuple{Base.OneTo{Int64}}, SubArray{Int64, 1, LinearAlgebra.Adjoint{Int64, Matrix{Int64}}, Tuple{Int64, Base.Slice{Base.OneTo{Int64}}}, false}}:
 [1, 1, 1, 1]
 [2, 2, 2, 2]
 [3, 3, 3, 3]

or if the number of columns is fixed, statically known, and relatively small

julia> using StaticArrays

julia> reinterpret(SVector{4, Int64}, flat_x)
3-element reinterpret(SVector{4, Int64}, ::Vector{Int64}):
 [1, 1, 1, 1]
 [2, 2, 2, 2]
 [3, 3, 3, 3]
3

In my experience, you almost never need to create mesh grids like that, with lots of redundant values. What do you need it for?

4

Note that Julia has 2d arrays — you don’t generally use arrays of arrays for this purpose like you would in Python syntax. (Also, you can often avoid “mesh-grid” arrays using broadcasting etc., as @DNF alluded to.)

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