如何从矩阵的每一行中减去一个向量? [英] How to subtract a vector from each row of a matrix?

问题描述

可能的重复:
如何划分每一行一个矩阵的固定行?

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法.

I'm looking for an elegant way to subtract the same vector from each row of a matrix. Here is a non elegant way of doing it.

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);

另外，优雅的方式不能比这个方法慢.

Also, the elegant way can't be slower than this method.

我试过了

c = b-repmat(a,size(b,1),1); 

而且看起来更慢.

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);

更多方法和 tic toc 结果:

More methods, and tic toc results:

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.

推荐答案

这是我的贡献:

c = b -ones(size(b))*diag(a)

现在对其进行速度测试:

Now speed testing it:

tic
for i = 1:10000
    c = zeros(size(b));
    b = rand(7,3);
    c = b - ones(size(b))*diag(a);
end
toc

结果:

经过的时间是 0.099979 秒.

没有那么快，但很干净.

Not quite as fast, but it is clean.

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