Hi! I have some operation of summing two arrays (vd, bd) of different lengths ((vl + 1)^2 and ((bd + 1)^2). I can't do summing as d = vd + bd (error: different lengths). How can I do this operations faster? I compile MEX, but this did not give the required increase in speed. I need speed up adding because it using around 20 million
function [l, d] = add(vl, vd, bl, bd, lmax)
l = max([vl, bl, lmax]);
d = zeros((l + 1)^2, 1);
i = 1:(vl + 1)^2;
j = 1:(bl + 1)^2;
d(i) = d(i) + vd(i);
d(j) = d(j) + bd(j);
end

2 commentaires

Can you share the detail example? What would be the result in the following case?
vd=[1 2 3]
bd=[4 5 6 7]
Igor Arkhandeev
Igor Arkhandeev le 27 Mai 2021
Of course!
d = [5 7 9 7]; % 1 + 4, 2 + 5, 3 + 6, 7

Connectez-vous pour commenter.

 Réponse acceptée

Jan
Jan le 27 Mai 2021
Modifié(e) : Jan le 27 Mai 2021

0 votes

It would save time to work with the real sizes instead of providing vl, if you actually need (vl+1)^2. This would save 60 mllion squaring operations.
function [l, d] = add(vl, vd, bl, bd, lmax)
l = max(max(vl, bl), lmax);
d = zeros((l + 1)^2, 1);
iv = (vl + 1)^2;
id = (bl + 1)^2;
d(1:iv) = d(1:iv) + vd(1:iv); % Or: d(1:iv) + vd; ?
d(1:id) = d(1:id) + bd(1:id); % Or: d(1:id) + bd; ?
end
The JIT acceleration profits from using x(a:b) instead of creating the index vector v=a:b; x(v).
If the inputs are uniquely explained, it would be possible to modify the function without too much guessing. Maybe this would be better already, maybe not:
function [len, d] = add(a, b, len)
na = numel(a);
nb = numel(b);
len = max(max(na, nb), len);
d = zeros(len, 1);
if na == nb
d = a + b;
elseif na > nb
d(nb+1:na) = a(nb+1:na);
d(1:nb) = a(1:nb) + b;
else
d(na+1:nb) = b(na+1:nb);
d(1:na) = a + b(1:na);
end
end

Plus de réponses (1)

Matt J
Matt J le 27 Mai 2021
Modifié(e) : Matt J le 27 Mai 2021
vd=[1 2 3];
bd=[4 5 6 7];
N=max(numel(vd), numel(bd));
vd(end+1:N)=0;
bd(end+1:N)=0;
d=vd+bd
d = 1×4
5 7 9 7

Catégories

En savoir plus sur Write C Functions Callable from MATLAB (MEX Files) dans Centre d'aide et File Exchange

Produits

Version

R2018b

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by