How to "localize" slicing an array?

General Usage
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1

How can I “localize” slicing on dimension relative to “available” indices? In the example below, I would need L[:, j] to mean “get all elements along dimension 1 relative to elements present in skipzeros”. How can I do that?

L = rand(10, 10)
idx0 = rand(0:1, size(L))
L .= L .* idx0

skipzeros = filter(x -> x ∉ findall(L .== 0), CartesianIndices((1:10, 1:10))

for idx in skipzeros
    i, j = Tuple(idx)
    t1 = L[:, j] .+ 100
end
2

Is the following what you want? (taking the liberty to simplify the code a bit)

L = rand(10, 10) .* rand.(Bool)

skipzeros = findall(!=(0), L)

for idx in skipzeros
    same_2nd = getindex.(skipzeros, 2) .== idx[2]
    t1 = L[skipzeros[same_2nd]] .+ 100
end

This makes a vector skipzeros[same_2nd] which contains all the Cartesian indices that satisfy the condition. Then instead of slicing with [:, j] you can use this list of Cartesian indices to extract the corresponding values from L.

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3

This does it, thank you. Could there be a more streamlined way?

4

It depends what you really want to do. The given example finds the nonzero values in the same column for each nonzero value in L, and adds 100. So if column 3 has 5 nonzero values, it will calculate 5 times the same vector v .+ 100 (where v is the list of nonzero values in column 3). A simpler way to do this is

L = rand(10, 10) .* rand.(Bool)
cols_nozero = filter.(!iszero, eachcol(L))
[v .+ 100 for v in cols_nozero for _ in v]
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