如何从矩阵的每一行中减去一个向量？ [复制]

可能的重复：
如何将矩阵的每一行除以固定行？ https://stackoverflow.com/questions/4723824/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法。

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，优雅的方式不可能比这种方法慢。

我试过了

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

而且看起来更慢。

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

results

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.

这是我的贡献：

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

现在对其进行速度测试：

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.

速度不是那么快，但很干净。