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;
端
谢谢,Max
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。