I have a vector A. I need a vector RA which ranks each element of A. Is there a better way to do this than what I have below? A = [4;2;5;1;6;7] RA = sum((A .== sort(A)') .* cumsum(ones(length(A)))', dims=2) display([A RA])

I think they mean the inverse of sortperm, the solution would be invperm(sortperm(A)).

julia> sortperm(A) 6-element Vector{Int64}: 4 2 1 3 5 6 julia> A[sortperm(A)] 6-element Vector{Int64}: 1 2 4 5 6 7 do you want some kind of sortperm?

If I understand the problem correctly, the result should be vector with indices of the elements after sorting, for A it would be [ 3 2 4 1 5 6]. Command sortperm(A, rev = true) does not return it, invperm(sortperm(A)) does.

Thanks invperm(sortperm(A)) solves it

just as an alternative to the first proposal :grinning: [findfirst(sa->sa==a, sort(A)) for a in A]

This function built on a simple loop, could be an extra argument in favor of the thesis (which is not mine: I prefer the invperm (...)) solution) that sometimes a for loop may be preferable to the use of Base library functions function rankfy1(A) rank=Int[] for a in A r=1 for i in eachind…

If we are playing with the code this algorithm can be also expressed as [count(<=(a),A) for a in A]; if we do not want Base one can also write the same thing like this map(A) do a S = 0 for b in A b <= a && (S += 1) end S end All are O(n^2), so not very good.

OK. I had done some quick tests on the vector A proposed by the OP. On larger vectors invperm () proves to be much more efficient. This confirms me in the idea that, whenever possible, it is better to use functions written by “expert” developers rather than relying on your own loops. This, for e…

See Rankings and Rank Correlations · StatsBase.jl for general ranking: it lets you disambiguate ranks when ties are present.

It seems to do something like this, obviously in a much more general way. function rank1(A) sp=sortperm(A) invsp=Array{Int}(undef, length(A)) foreach(i->invsp[sp[i]]=i, eachindex(A)) return invsp end

Ranking of elements of a vector

General Usage

jling October 5, 2022, 3:05pm 4

ehh, sortperm(; rev = true)?

Topic		Replies	Views
Efficient way to apply monotonous rank transformation to an array Performance	4	558	April 16, 2020
Rank with ties New to Julia	5	429	July 21, 2022
Cant find order General Usage sortperm	2	497	December 26, 2021
Sort by index an array New to Julia sortperm	1	3670	December 12, 2018
Question about sort? General Usage	3	311	May 22, 2022

Ranking of elements of a vector

Related topics