How to accelerate the running speed of this code?
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function [ S ] = fun(X,A,B,C)
%% obtain S from X
%% the size of X is A*B*C
S = zeros(A,B,C);
for a = 1:A
for b = 1:B
Z = X - X(a,b,:);
Z(a,b,:) = X(a,b,:);
S(a,b,:) = prod(Z,[1,2]);
end
end
end
I have a three dimensional matrix X (with dimension A*B*C) and I want to obtain a matrix S of the same dimension. This code is really time-consuming, but i don't know how to accelerate it. Is there any way possible to handle this problem?
1 commentaire
Hancheng Zhu
le 9 Oct 2024
Réponses (1)
Here's an approach that may or may not be faster (depending on the size of your array X) and may or may not run (depending on the size of X and how much RAM your machine has) but gives the same result as your original code.
X = rand(50,100,10);
S_old = fun_old(X);
S_new = fun_new(X);
isequal(S_old,S_new)
timeit(@()fun_old(X))
timeit(@()fun_new(X))
function S = fun_new(X)
[A,B,C] = size(X);
Z = X-reshape(permute(X,[3 1 2]),[1,1,C,A*B]);
[ii,jj,kk] = ind2sub([A,B,C],1:A*B*C);
mm = repmat(1:A*B,1,C);
idx = sub2ind([A,B,C,A*B],ii,jj,kk,mm);
Z(idx) = X;
S = reshape(permute(prod(Z,[1 2]),[4 3 1 2]),[A,B,C]);
end
function S = fun_old(X)
[A,B,C] = size(X);
S = zeros(A,B,C);
for a = 1:A
for b = 1:B
Z = X - X(a,b,:);
Z(a,b,:) = X(a,b,:);
S(a,b,:) = prod(Z,[1,2]);
end
end
end
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