The consensus on this forum seems to be that Interpolations.jl is extremely fast, so I must not be using it in a smart way. Here’s an MWE of the Matlab code I’m trying to adapt to Julia. If some stuff looks weird (e.g. generating random numbers inside a loop), it’s because I’m using that stuff as a…

MATLAB: pts(:,2) = rand(Npt,1) * 6 - 3; Julia: pts[2,:] = rand(1,Npt)*6 .- 3 Given that pts is 2 x 10_000, assigning to 10_000 elements on each information is going to hurt performance (and give you the wrong answer).

Ugh duh, sorry. The vector vs. one-row matrix always confuses me when I’m going back and forth between languages. Fixed: pts[2,:] = rand(Npt)*6 .- 3 With no splatting and with @inbounds, it’s now 650 ms. If I further replace = with .=, I can shave off another 2000 allocations and 50 ms. But we’re …

Ok, try replacing the inner loop with result[:,j] .= fct.(eachrow(pts)...)

What are the dimensions of pts in Julia and MATLAB? If in MATLAB it is 10_000 x 2, and in Julia it is 2 x 10_000, I’d like to point out that: julia> x1 = rand(2,10_000); julia> x2 = rand(10_000,2); julia> @benchmark $x1[1,:] .= 0.5 BenchmarkTools.Trial: memory estimate: 0 bytes allocs estim…

Why would x1 and x2 have different performance? Seems like that shouldnt be a thing, and if it is a thing then it should be documented somewhere.

This is general (other than requiring at least 2 dimensions, which the MATLAB code also does), and fairly fast: using Interpolations, Random, BenchmarkTools f(a,b) = a^2 + b^2 const xs = range(0,5,length=100); const ys = range(-3,3,length=200); const vals = [f(x,y) for x in xs, y in ys]; const fct …

[image] chakravala: Why would x1 and x2 have different performance? Seems like that shouldnt be a thing, and if it is a thing then it should be documented somewhere. It’s a cpu-cache issue. Memory is transferred between the cpu and memory in chunks of 64 consecutive bytes or so (a cache line)…

It’s also a vectorization issue. If you have AVX2, you can do: @fastmath begin x[1,1] = 6x[1,1] - 3 x[2,1] = 6x[2,1] - 3 x[3,1] = 6x[3,1] - 3 x[4,1] = 6x[4,1] - 3 end with 1 load instruction, 1 fused multiply add, and 1 store (you’ll also have to load the constants, but that’s only necessary once…

Man, this is a powerful demonstration of how important it is to store the dimensions in a type rather than have it be determined at runtime. Since your code has a couple of different optimizations (including the mutating rand, which I hadn’t thought of), I decided to go from my code to yours step by…

Why are my linear interpolations 10x faster in MATLAB?

General Usage Performance

tbeason June 11, 2020, 11:43pm 9

Row-major vs column-major storage. It is faster to access items in the array in the sequence that they are stored.

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Why are my linear interpolations 10x faster in MATLAB?

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