Matlab：排序二维矩阵并保留三角形组中的节点

Question

我正在尝试优化一个 Matlab 脚本（下方），该脚本找到二维 space 中所有左下三角形函数值的边界框。代码遍历所有三角形，然后根据功能值按升序对节点进行排序。这似乎效率低下。

有没有什么方法可以在循环之前对函数值进行排序，但仍保留三角形组中的节点？或者其他一些加快速度的聪明方法？

clear;

x = (1:600)';
y = (1:500);
z = 2 * x.^2 + y;

zGrid = linspace(min(z, [], 'all') - 1, max(z, [], 'all') + 1, 200);

for iX = 1:length(x) - 1
    for iY = 1:length(y) - 1
        % Node indices
        xIndices = [iX, iX, iX + 1];
        yIndices = [iY, iY + 1, iY];

        % Node values
        xTmp = x(xIndices);
        yTmp = y(yIndices);
        zTmp = z(sub2ind(size(z), xIndices, yIndices));

        % Node sorted according to z
        [zSorted, indicesSorted] = sort(zTmp);
        xSorted = xTmp(indicesSorted);
        ySorted = yTmp(indicesSorted);

        % Get bounding box on zGrid
        iMin = find(zGrid <= zSorted(1), 1, 'last');
        iMax = find(zGrid(iMin:end) >= zSorted(end), 1, 'first') + (iMin - 1);
    end
end

Answer 1

您可以使用 meshgrid 生成所有索引，然后只需修改代码即可获得所需的输出：

x = (1:600).';
y = (1:500);
z = 2 * x.^2 + y;

zGrid = linspace(min(z(:)) - 1, max(z(:)) + 1, 200);
[X,Y] = meshgrid(x(1:end-1),y(1:end-1));
X = X(:);
Y = Y(:);

% Node indices
xIndices = [X, X, X + 1];
yIndices = [Y, Y + 1, Y];

% Node values
xTmp = x(xIndices);
yTmp = y(yIndices);
zTmp = z(sub2ind(size(z), xIndices, yIndices));

% Node sorted according to z
[zSorted, indicesSorted] = sort(zTmp,2);
xSorted = xTmp(indicesSorted+[0:3:(length(X)-1)*3].');
ySorted = yTmp(indicesSorted+[0:3:(length(Y)-1)*3].');

% Get bounding box on zGrid
% I use fliplr to get the last value and not the first one
[~,iMin] = max(fliplr(zGrid <= zSorted(:,1)),[], 2);
iMin = 200-iMin+1; %first to last value conversion
[~,iMax] = max(zGrid(iMin:end) >= zSorted(:,end),[],2);
iMax = iMax + iMin -1;

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