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Possible Duplicate:
How can I divide each row of a matrix by a fixed row?
我正在寻找一种优雅的方法来从矩阵的每一行中减去相同的向量。这是一种不优雅的方式。
a = [1 2 3];
b = rand(7,3);
c(:,1) = b(:,1) - a(1);
c(:,2) = b(:,2) - a(2);
c(:,3) = b(:,3) - a(3);
而且，优雅的方法不能比这个方法慢。
我试过了
c = b-repmat(a,size(b,1),1);
而且它看起来更慢。
编辑:获胜者是这种方法。
c(:,1) = b(:,1) - a(1);
c(:,2) = b(:,2) - a(2);
c(:,3) = b(:,3) - a(3);
编辑:更多方法和 tic toc 结果:
n = 1e6;
m = 3;
iter = 100;
a = rand(1,m);
b = rand(n,m);
tic
c = zeros(size(b));
for i = 1:iter
c(:,1) = b(:,1) - a(1);
c(:,2) = b(:,2) - a(2);
c(:,3) = b(:,3) - a(3);
end
toc
tic
c = zeros(size(b));
for i = 1:iter
c(:,1) = b(:,1) - a(1);
c(:,2) = b(:,2) - a(2);
c(:,3) = b(:,3) - a(3);
end
toc
tic
c = zeros(size(b));
for i = 1:iter
for j = 1:3
c(:,j) = b(:,j) - a(j);
end
end
toc
tic
for i = 1:iter
c = b-repmat(a,size(b,1),1);
end
toc
tic
for i = 1:iter
c = bsxfun(@minus,b,a);
end
toc
tic
c = zeros(size(b));
for i = 1:iter
for j = 1:size(b,1)
c(j,:) = b(j,:) - a;
end
end
toc
结果
Elapsed time is 0.622730 seconds.
Elapsed time is 0.627321 seconds.
Elapsed time is 0.713384 seconds.
Elapsed time is 2.621642 seconds.
Elapsed time is 1.323490 seconds.
Elapsed time is 17.269901 seconds.