将一个矩阵与另一个矩阵排序

更清楚的方法是使用循环

A = rand(3,4);
B = rand(3,4);
[sortedA,ind] = sort(A,2);

for r = 1:size(A,1)
   B(r,:) = B(r,ind(r,:));
end

下面是我很快编出来测试的代码:

siz = 10:100:1010;
tt = zeros(100,2,length(siz));

for s = siz
    for k = 1:100

        A = rand(s,1*s);
        B = rand(s,1*s);
        [sortedA,ind] = sort(A,2);

        tic;
        for r = 1:size(A,1)
            B(r,:) = B(r,ind(r,:));
        end,tt(k,1,s==siz) = toc;

        tic;
        m = size(A,1);
        B = B(bsxfun(@plus,(ind-1)*m,(1:m).'));
        tt(k,2,s==siz) = toc;

    end
end

m = squeeze(mean(tt,1));

m(1,:)./m(2,:)

ans =

    0.7149    2.1508    1.2203    1.4684    1.2339    1.1855    1.0212    1.0201    0.8770       0.8584    0.8405

ans =

    0.8431    1.2874    1.3550    1.1311    0.9979    0.9921    0.8263    0.7697    0.6856    0.7004    0.7314

Sort()返回按维度排序的索引。可以显式地为其他维度构造索引,使行保持稳定,然后使用线性索引重新排列整个数组。

A = rand(3,4);
B = A; %// Start with same values so we can programmatically check result

[A2 ix2] = sort(A,2);
%// ix2 is the index along dimension 2, and we want dimension 1 to remain unchanged
ix1 = repmat([1:size(A,1)]', [1 size(A,2)]); %//'
%// Convert to linear index equivalent of the reordering of the sort() call
ix = sub2ind(size(A), ix1, ix2) 
%// And apply it
B2 = B(ix)
ok = isequal(A2, B2) %// confirm reordering

你就不能这么做吗?

[val ind]=sort(A);
B=B(ind);

它对我有用,除非我对你的问题理解错误。