要完成此任务，除了计算插值，还要完成插值的整个过程.这是从 Octave c ++源代码翻译而来的解决方案.输入格式与 interpn 的第一签名相同函数，除了不需要v数组.同样，X应该是向量，并且不应该是ndgrid格式.输出W(权重)和I(位置)都具有(a ,b)的大小，其中a是网格上点的邻居数，而b是要内插的请求点数

To do the task the whole process of interpolation ,except computing the interpolated values, should be done. Here is a solution translated from the Octave c++ source. Format of the input is the same as the frst signature of the interpn function except that there is no need to the v array. Also Xs should be vectors and should not be of the ndgrid format. Both the outputs W (weights) and I (positions) have the size (a ,b) that a is the number of neighbors of a points on the grid and b is the number of requested points to be interpolated.

function [W , I] = lininterpnw(varargin)
% [W I] = lininterpnw(X1,X2,...,Xn,Xq1,Xq2,...,Xqn)
    n     = numel(varargin)/2;
    x     = varargin(1:n);
    y     = varargin(n+1:end);
    sz    = cellfun(@numel,x);
    scale = [1 cumprod(sz(1:end-1))];
    Ni    = numel(y{1});
    index = zeros(n,Ni);
    x_before = zeros(n,Ni);
    x_after = zeros(n,Ni);
    for ii = 1:n
        jj = interp1(x{ii},1:sz(ii),y{ii},'previous');
        index(ii,:) = jj-1;
        x_before(ii,:) = x{ii}(jj);
        x_after(ii,:) = x{ii}(jj+1);
    end
    coef(2:2:2*n,1:Ni) = (vertcat(y{:}) - x_before) ./ (x_after - x_before);
    coef(1:2:end,:)    = 1 - coef(2:2:2*n,:);
    bit = permute(dec2bin(0:2^n-1)=='1', [2,3,1]);
    %I = reshape(1+scale*bsxfun(@plus,index,bit), Ni, []).';  %Octave
    I = reshape(1+sum(bsxfun(@times,scale(:),bsxfun(@plus,index,bit))), Ni, []).';
    W = squeeze(prod(reshape(coef(bsxfun(@plus,(1:2:2*n).',bit),:).',Ni,n,[]),2)).';
end

测试:

x={[1 3 8 9],[2 12 13 17 25]};
v = rand(4,5);
y={[1.5 1.6 1.3 3.5,8.1,8.3],[8.4,13.5,14.4,23,23.9,24.2]};

[W I]=lininterpnw(x{:},y{:});

sum(W.*v(I))
interpn(x{:},v,y{:})

感谢@SardarUsama的测试和他的有用评论.

Thanks to @SardarUsama for testing and his useful comments.