计算多个欧几里德有效的方法Matlab的的距离 [英] Efficient way to compute multiple euclidean distances Matlab

问题描述

我正在训练我自己的自组织映射集群colorvalues​​。现在，我想打一些一 U-矩阵显示节点及其直接邻国之间的欧氏距离。我现在的问题是，我的算法是非常效率不高！当然是有办法更有效地计算呢？

I am training my own self organizing map to cluster colorvalues. Now I want to make some sort of a U-matrix to show euclidean distances between the nodes and their direct neighbors. My problem now is, that my algorithm is quite unefficient!! There is certainly a way to compute this more efficiently?

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;

结束

谢谢，最大

推荐答案

您可以通过计算其neigbor各点的（平方）距离右：

You can calculate the (squared) distance of each point to its neigbor to the right by:

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

类似地，计算各点的距离的点的下方，并在两个对角线。你不必对，所以你垫他们用零边框点的所有这些值。然后添加的距离，由点必须得到所有邻居的平均距离的邻居数除以他们。

Similarly, you calculate the distance of each point to the point below, and on both diagonals. You don't have all those values for points on borders so you pad them with zeros. Then you add the distances and divide them by the number of neighbors a point has to get the average distance to all the neighbors.

下面是我的code：

%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;

这完全是矢量，所以它应该比你的版本速度更快。
如果你想确切的U-矩阵，可以与邻近的距离交错的平均距离。

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.

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