计算多个欧氏距离的有效方法Matlab

时间:2013-03-22 17:02:45

标签: arrays matlab euclidean-distance

我正在训练自己的自组织地图来聚类颜色值。现在我想制作某种U-matrix来显示节点与其直接邻居之间的欧氏距离。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;

谢谢,Max

1 个答案:

答案 0 :(得分:2)

您可以通过以下方式计算每个点到其右边的(平方)距离:

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

同样,您可以计算每个点到下方点以及两个对角线的距离。你没有为边界上的点设置所有这些值,所以你用零填充它们。然后你添加距离并将它们除以一个点的邻居数量,以获得到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;

它全部是矢量化的,所以它应该比你的版本更快。 如果你想要精确的U矩阵,你可以将平均距离与相邻距离交错。