Add operation is too slow

7 次查看（过去 30 天）
Igor Arkhandeev
Igor Arkhandeev 2021-5-27
编辑： Jan 2021-5-27
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 个评论
KALYAN ACHARJYA
KALYAN ACHARJYA 2021-5-27
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 2021-5-27
Of course!
d = [5 7 9 7]; % 1 + 4, 2 + 5, 3 + 6, 7

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Jan
Jan 2021-5-27
编辑：Jan 2021-5-27
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

更多回答（1 个）

Matt J
Matt J 2021-5-27
编辑：Matt J 2021-5-27
Ran in:
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

类别

在 Help Center 和 File Exchange 中查找有关 MATLAB Compiler 的更多信息

标签

产品


版本

R2018b

Community Treasure Hunt

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

Start Hunting!

Translated by