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我有兴趣在一堆参数值的排列上计算一个函数。我想保持它对 N 维的通用性，但让我从 3 维开始写出来。使用 生成排列很容易meshgrid，但我不知道如何将生成的数组重新整形为多维？这是一个起点：

%These are the 3 variations of parameters, with some values.  
params1 = [100, 200, 300];%Picking these so it is easy to correlate to the function
params2 = [10, 20];
params3 = [1, 2];

%This generates parameter_values as the cartesian productpermutations.
[vec1, vec2, vec3] = meshgrid(params1, params2, params3);
parameter_values = [vec1(:) vec2(:) vec3(:)];

%Calculates functions on the set of parameters.
%Would have a fancier function, of course, this just makes it easy to see the results.
raw_vals = parameter_values(:,1) + parameter_values(:,2) + parameter_values(:,3); 

%Rearrange into a multiarray to access by parameter indices.
f_vals = reshape(raw_vals, [length(params1), length(params2), length(params3)]) %WRONG? 

%THE FOLLOWING FAIL BUT WOULD BE EXPECTED WITH THESE PARAMETERS AND THE FUNCTION.
assert(f_vals(2,1,1) == 211)
assert(f_vals(3,2,2) == 322)
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1 回答 1

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你想要ndgrid而不是meshgrid在这种情况下。

meshgrid的语法是[X,Y] = meshgrid(xgv,ygv)导致Y(:)变化最快的，而不是X(:). 有关详细信息，请参阅网格数据表示。换句话说，你得到

>> [vec1, vec2, vec3] = meshgrid(params1, params2, params3)
vec1(:,:,1) =
   100   200   300
   100   200   300
vec1(:,:,2) =
   100   200   300
   100   200   300
vec2(:,:,1) =
    10    10    10
    20    20    20
vec2(:,:,2) =
    10    10    10
    20    20    20
...

但你想得到：

>> [vec1, vec2, vec3] = ndgrid(params1, params2, params3)
vec1(:,:,1) =
   100   100
   200   200
   300   300
vec1(:,:,2) =
   100   100
   200   200
   300   300
vec2(:,:,1) =
    10    20
    10    20
    10    20
vec2(:,:,2) =
    10    20
    10    20
    10    20
...

如果你切换到ndgrid，那么你会得到f_vals(2,1,1) == 211预期的结果。

推广到 N 维可以这样完成：

    params = {[100, 200, 300],[10, 20],[1, 2]};
    vecs = cell(numel(params),1);
    [vecs{:}] = ndgrid(params{:});
    parameter_values = reshape(cat(numel(vecs)+1,vecs{:}),[],numel(vecs));
    raw_vals = sum(parameter_values,2);
    f_vals = reshape(raw_vals,cellfun(@numel,params))
于 2013-11-15T01:24:06.977 回答