Maximal value - product of two lists

General Usage
1

Concerning the following situation: max \forall i { h_i * X_{mij} }

Is there a function to obtain the maximal value between the product of two lists?

2

This?

julia> h = rand(10); x = rand(10,10,10);

julia> m = 2; j = 3;

julia> maximum(h .* x[m,:,j]) # note the dot in .* : element-wise multiplication
0.6705150233027053

julia> maximum(h[i] * x[m,i,j] for i in eachindex(h))
0.6705150233027053

The second won´t allocate intermediate arrays which are allocated in the first.

1 Like
3

I’d also like to point out the generic version of this, reduce, which I think is easier to remember than the difference between max and maximum. reduce(max, (h[i] * x[m, i, j] for i in eachindex(h))) does the same thing, but works for generic “reducing” functions (functions that can take a collection and summarize it into one number). For example, reduce(+, array) will take an array and return the sum.

4

TensorCast.jl is a friend:

using TensorCast
maximum(@cast _[i] := h[i] * x[$m,i,$j])
5

TensorCast is still a friend, with the @reduce macro, if you want to make the calculation for all m and j:

using TensorCast
@reduce S[m,j] := maximum(i) h[i] * x[m,i,j]

For example:

julia> h = rand(3)
3-element Vector{Float64}:
 0.8601614537887764
 0.9085586312619703
 0.5942727699394317

julia> x = rand(4,3,2)
4Ă—3Ă—2 Array{Float64, 3}:
[:, :, 1] =
 0.350826  0.739942  0.565042
 0.751966  0.430443  0.293777
 0.714827  0.974151  0.680083
 0.752213  0.289802  0.784696

[:, :, 2] =
 0.95478   0.0218679  0.652646 
 0.308032  0.499596   0.724832 
 0.426453  0.865667   0.790571 
 0.370235  0.494717   0.53362

julia> @reduce S[m,j] := maximum(i) h[i] * x[m,i,j]
4Ă—2 Matrix{Float64}:
 0.672281  0.821265
 0.646812  0.453912
 0.885073  0.786509
 0.647025  0.44948

julia> S[1,1]
0.672280575421645

julia> S[1,1] == h[2]*x[1,2,1]
true
3 Likes
6

The computation for single (m,j) could be also written as:

@reduce _ := maximum(i) h[i] * x[$m,i,$j]
7

Dear colleagues, many thanks to you for you support.