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我正在训练我自己的自组织地图来聚类颜色值。现在我想制作某种U 矩阵来显示节点与其直接邻居之间的欧几里德距离。我现在的问题是，我的算法效率很低！！肯定有一种方法可以更有效地计算它吗？

function displayUmatrix(dims,weights) %#dims is [30 30], size(weights) = [900 3], 
                                      %#consisting of values between 1 and 0

hold on; 
axis off;
A = zeros(dims(1), dims(2), 3);
B = reshape(weights',[dims(1) dims(2) size(weights,1)]);
if size(weights,1)==3
    for i=1:dims(1)
        for j=1:dims(2)
            if i~=1
                if j~=1
                    A(i,j,:)=A(i,j,:)+(B(i,j,:)-B(i-1,j-1,:)).^2;
                end
                A(i,j,:)=A(i,j,:)+(B(i,j,:)-B(i-1,j,:)).^2;
                if j~=dims(2)
                    A(i,j,:)=A(i,j,:)+(B(i,j,:)-B(i-1,j+1,:)).^2;
                end
            end
            if i~=dims(1)
                if j~=1
                    A(i,j,:)=A(i,j,:)+(B(i,j,:)-B(i+1,j-1,:)).^2;
                end
                A(i,j,:)=A(i,j,:)+(B(i,j,:)-B(i+1,j,:)).^2;
                if j~=dims(2)
                    A(i,j,:)=A(i,j,:)+(B(i,j,:)-B(i+1,j+1,:)).^2;
                end
            end 
            if j~=1
                A(i,j,:)=A(i,j,:)+(B(i,j,:)-B(i,j-1,:)).^2;
            end
            if j~=dims(2)
                A(i,j,:)=A(i,j,:)+(B(i,j,:)-B(i,j+1,:)).^2;
            end
            C(i,j)=sum(A(i,j,:));
        end
    end
    D = flipud(C);
    maximum = max(max(D));
    D = D./maximum;
    imagesc(D)
else
    error('display function does only work on 3D input');
end
hold off;
drawnow;

结尾

谢谢，马克斯

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1 回答 1

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您可以通过以下方式计算每个点到其右侧邻居的（平方）距离：

sum((B(:,1:end-1,:) - B(:,2:end,:)).^2, 3)

同样，您计算每个点到下面点的距离，以及两条对角线上的距离。对于边界上的点，您没有所有这些值，因此您用零填充它们。然后将距离相加并将它们除以一个点必须获得的与所有邻居的平均距离。

这是我的代码：

%calculate distances to neighbors
right = sum((B(:,1:end-1,:)- B(:,2:end,:)).^2, 3);
bottom = sum((B(1:end-1,:,:)- B(2:end,:,:)).^2, 3); zeros();
diag1 = sum((B(1:end-1,1:end-1,:)- B(2:end,2:end,:)).^2, 3);
diag2 = sum((B(2:end,2:end,:)- B(1:end-1,1:end-1,:)).^2, 3);

%pad them with zeros to the correct size
rightPadded = [right zeros(dim(1) , 1)];
leftPadded = [zeros(dim(1) , 1) right];

botomPadded = [bottom; zeros(1,dim(2))];
upPadded = [zeros(1,dim(2));bottom];

bottomRight = zeros(dim(1), dim(2));
bottomRight(1:end-1,1:end-1) = diag1;
upLeft = zeros(dim(1), dim(2));
upLeft(2:end,2:end) = diag1;

bottomLeft = zeros(dim(1), dim(2));
bottomLeft(1:end-1,2:end) = diag2;
upRight = zeros(dim(1), dim(2));
upRight(2:end,1:end-1) = diag2;

%add distances to all neighbors
sumDist = rightPadded + leftPadded + bottomRight + upLeft + bottomLeft + upRight;

%number of neighbors a point has
neighborNum = zeros(dim(1), dim(2)) + 8;
neighborNum([1 end],:) = 5;
neighborNum(:,[1 end]) = 5;
neighborNum([1 end],[1 end]) = 3;

%divide summed distance by number of neighbors
avgDist = sumDist./neighborNum;

It is all vectorized, so it should be faster than your version. If you want the exact U-matrix, you can interleave the average distances with the neighboring distances.

于 2013-03-22T18:47:37.633 回答