给定两个数组

julia> a = [30, 14, 10, 21] 
julia> b = [2, 8, 24, 1]

我想获得排列索引 ( perm_ixs) b，使得 的元素大小排序b对应于 的元素大小排序a。

julia> perm_ixs =  [3, 1, 4, 2] 
julia> b[perm_ixs]
4-element Vector{Int64}:
 24
  2
  1
  8

从上面可以看出，中的perm_ixs元素是第三大元素，并且 中 的第三大元素位于索引 处。因此，在 permuted 中，第二个元素应该有一个 value 。我当前的方法如下所示，看起来相当幼稚。有没有更有效的方法来做到这一点？2``b``a``14``2``b``2

julia> function get_size_based_permutation(a, b)
    @assert length(a) == length(b)

    # Get vectors ranking the size of elements relative to others in the vector
    # where size 1 is the largest element and size n is the smallest 
    @show sizes_a = invperm(sortperm(a; rev=true))
    @show sizes_b = invperm(sortperm(b; rev=true))

    # Map the ix of the matching size element in one array to the 
    # linearly ordered ixs of the other array
    p = collect(1:length(sizes_a))
    for (ix, size_of_ele_in_b) in enumerate(sizes_b)
        matching_size_ix = findfirst(ele -> ele == size_of_ele_in_b, sizes_a)
        p[matching_size_ix] = ix 
    end

    return p
end
julia> a = [30, 14, 10, 21] 
julia> b = [2, 8, 24, 1]
julia> perm_ixs = get_size_based_permutation(a, b)
sizes_a = invperm(sortperm(a; rev = true)) = [1, 3, 4, 2]
sizes_b = invperm(sortperm(b; rev = true)) = [3, 2, 1, 4]
4-element Vector{Int64}:
 3
 1
 4
 2

julia> b[perm_ixs]
4-element Vector{Int64}:
 24
  2
  1
  8