I have: a = rand(Float32, h*w) b = reshape(a, h, w) c = similar(a) I would like to iterate over array b, calculate a result based on element value and its 2D coordinates and put it in c. The order of elements in c is irrelevant and it can be constructed in a different way than shown above. The options I see are: Use push(c, result) to initially empty array c, which is very inefficient, unless there is a way to reserve memory for empty array. Use two iterators of different shape in parallel, which I don’t know how to. Use 2D iterator over array b and calculate 1D index (or iterator) from it to put the result in array c What is a clear and concise way to do it? Performance is not critical here.

How about: Use option (3), but create a new 2D array b_new of the same shape as b, and then do c = vec(b_new). EDIT: vec does not allocate a new array; it just gives a view that looks like a vector.

I don’t think the performance of push! is so bad (I don’t believe it naively allocates on every push but does something smarter to reduce the churn), especially for something that is not performance critical. I think it often makes for simpler code.

Also, are you aware that you can index into an array using both 2D and 1D indices? For example julia> a = randn(3,5) 3×5 Array{Float64,2}: 0.180949 1.04227 0.59108 -1.29255 0.766903 -1.05146 0.878143 -0.201572 1.16771 -0.603789 -0.763806 -1.60095 -0.759488 1.00734 0.089463 julia> a[2,3] -0.20157212679471334 julia> a[8] -0.20157212679471334

sizehint!(v, 100) reserves memory. But indeed push! is (very) clever about memory use.

You could do this: a = rand(Float32, h*w) b = reshape(a, h, w) c = similar(a) foo(x) = 2 * x foo(x, i) = x + i[2] # this one doesn't care about the 2D index of b, but uses a linear index for both for i in eachindex(b, c) c[i] = foo(b[i]) end # if you need the 2D array indices ind = CartesianIndices(b) for i in eachindex(b, c) c[i] = foo(b[i], ind[i]) end

Thank you all for suggestions. Thinking my problem through, I noticed that array dimensionality in Julia is very fungible through reshape, which seems to be computationally cheap. I was stuck in a different language mindset where this was not a simple and cheap task. As a consequence, my initial requirement about iterating simultaneously over two arrays with different dimensions, but the same number of elements, is not necessary. I can restate it as: a = rand(Float32, h*w) b = reshape(a, h, w) c = similar(b) so I will be iterating over arrays b and c with the same dimensions, which makes it simpler. When different dimensions are required by a consuming function, I can just reshape resultant array c : reshape(c, h*w)

[image] pauljurczak: reshape(c, h*w) Or use vec(c), which is more readable and doesn’t require you to keep track of the sizes.

Thanks. vec(c) is much better than reshape(c, h*w).

Here is my first stab at this with f0 using a plain integer indexes and f1 using CartesianIndices depth2d = reshape(depth1d, width, height) depth4d = similar(depth2d, Point3f) function f0(depth2d, depth4d) @inbounds for i = 1:width, j = 1:height d = min(rangeinf, depth2d[i, j]) depth4d[i, j] = Point3f(d*(i-cx)/fl, d*(j-cy)/fl, d) end end function f1(depth2d, depth4d) for i in CartesianIndices(depth2d) d = min(rangeinf, depth2d[i]) depth4d[i] = Point3f(d*(i[1]-cx)/fl, d*(i[2]-cy)/fl, d) end end julia> judge(median(@benchmark f0(depth2d, depth4d)), median(@benchmark f1(depth2d, depth4d))) BenchmarkTools.TrialJudgement: time: -16.82% => improvement (5.00% tolerance) Employing CartesianIndices doesn’t make this simple, tiny function any clearer, and is less performant. Was this expected?

This is not a self-contained example, so I can’t run it for proper comparison. But, at least, you are using @inbounds in one function and not in the other. Also, remember to interpolate variables with $ when benchmarking with BenchmarkTools. (I think nearly a third of my posts on this forum contain the phrase “remember to interpolate”.)

Iterating over 2D array and placing results in 1D array with the same number of elements

New to Julia

Elrod July 3, 2020, 2:12pm 20

Sounds like this.

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Iterating over 2D array and placing results in 1D array with the same number of elements

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