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

问题描述

可能重复:
如何划分每一行固定行的矩阵的大小?

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.这是一种不太优雅的方法.

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);

此外，优雅的方法也不能比这种方法慢.

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

我尝试过

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

而且似乎慢了.

赢家就是这种方法.

c(:,1) = b(:,1) - a(1);
c(:,2) = b(:,2) - a(2);
c(:,3) = b(:,3) - a(3);

更多方法和抽查结果:

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.

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:

Elapsed time is 0.099979 seconds.

不是那么快，但是很干净.

Not quite as fast, but it is clean.

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