# Add operation is too slow

1 ビュー (過去 30 日間)
Igor Arkhandeev 2021 年 5 月 27 日

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 件のコメント表示非表示 1 件の古いコメント
Igor Arkhandeev 2021 年 5 月 27 日
Of course!
d = [5 7 9 7]; % 1 + 4, 2 + 5, 3 + 6, 7

サインインしてコメントする。

### 採用された回答

Jan 2021 年 5 月 27 日

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

サインインしてコメントする。

### その他の回答 (1 件)

Matt J 2021 年 5 月 27 日

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

サインインしてコメントする。

R2018b

### Community Treasure Hunt

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

Start Hunting!