Matlab：更快地找到 ND 矩阵中每个元素的一维线性插值节点和权重

Question

在我现在正在处理的一个问题中，我计算矩阵 x 中的一些值，然后我需要为 x 中的每个元素找到下面最接近的元素的索引单调递增向量 X 以及 x 元素与其任一侧的第一个元素的相对接近度。（这本质上是线性插值，没有进行实际的插值。）我做了很多次，所以我对它尽可能快非常感兴趣。

我写了一个函数 locate，我可以用一些示例数据调用它：

X = linspace(5, 300, 40)';
x = randi(310, 5, 6, 7);

[ii, weights] = locate(x, X);

我写了两个版本的locate。第一个用于说明，第二个是我加速计算的最佳尝试。对于如何进一步提高性能，您有任何建议或替代方法吗？

1.博览会

function [ii, weights] = locate(x, X)
    % LOCATE Locate first node on grid below a given value.
    %
    %   [ii, weights] = locate(x, X) returns the first node in X that is below
    %   each element in x and the relative proximities to the two closest nodes.
    %
    %   X must be a monotonically increasing vector. x is a matrix (of any
    %   order).

    % Preallocate
    ii = ones(size(x));  % Indices of first node below (or 1 if no nodes below)
    weights = zeros([2, size(x)]);  % Relative proximity of the two closest nodes

    % Find indices and compute weights
    for ix = 1:numel(x)
        if x(ix) <= X(1)
            ii(ix) = 1;
            weights(:, ix) = [1; 0];
        elseif x(ix) >= X(end)
            ii(ix) = length(X) - 1;
            weights(:, ix) = [0; 1];
        else
            ii(ix) = find(X <= x(ix), 1, 'last');
            weights(:, ix) = ...
                [X(ii(ix) + 1) - x(ix); x(ix) - X(ii(ix))] / (X(ii(ix) + 1) - X(ii(ix)));
        end
    end
end

2。最佳尝试

function [ii, weights] = locate(x, X)
    % LOCATE Locate first node on grid below a given value.
    %
    %   [ii, weights] = locate(x, X) returns the first node in X that is below
    %   each element in x and the relative proximities to the two closest nodes.
    %
    %   X must be a monotonically increasing vector. x is a matrix (of any
    %   order).

    % Preallocate
    ii = ones(size(x));  % Indices of first node below (or 1 if no nodes below)
    weights = zeros([2, size(x)]);  % Relative proximity of the two closest nodes

    % Find indices
    for iX = 1:length(X) - 1
        ii(X(iX) <= x) = iX;
    end

    % Find weights
    below = x <= X(1);
    weights(1, below) = 1;  % All mass on the first node
    weights(2, below) = 0;

    above = x >= X(end);
    weights(1, above) = 0;
    weights(2, above) = 1;  % All mass on the last node

    interior = ~below & ~above;
    xInterior = x(interior)';
    iiInterior = ii(interior);
    XBelow = X(iiInterior)';
    XAbove = X(iiInterior + 1)';
    weights(:, interior) = ...
        [XAbove - xInterior; xInterior - XBelow] ./ (XAbove - XBelow);
end

Answer 1

在 Brain2Mesh toolbox.

中查看我的 polylineinterp 函数

https://github.com/fangq/brain2mesh/blob/master/polylineinterp.m

几乎完全一样，除了 polylen 输入就像你的 X 的差异。

一般来说，vectorizing这种操作就是用histc()，像这行

https://github.com/fangq/brain2mesh/blob/master/polylineinterp.m#L52

Matlab：更快地找到 ND 矩阵中每个元素的一维线性插值节点和权重

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