MATLAB: Accessing an element of a multidimensional array with a list

匆匆过客 提交于 2019-11-28 00:11:59

One solution is to create a comma-separated list out of your vector of subscripted indices inds. You can do this by converting it to a cell array using NUM2CELL, then using the {:} syntax when indexing A:

inds = num2cell(inds);
value = A(inds{:});

I think this tool might help you:

if you have an ND matrix like R = rand(5,10,15,20), and you want to access elements in a particular pattern, you can use spindex to access the output in the same shape as the input access. So if you have size(i1) = [5,5,5], size(i2) = [5,5,5], etc. Then size(spindex(R,i1,i2,i3,i4)) also equals [5,5,5].

%#example:

z = reshape(1:(5^4),[5,5,5,5]);
zid1 = [1,1,5];
zid2 = [1,2,5];
zid3 = [1,3,5];
zid4 = [1,4,5];
zOut = spindex(z,zid1,zid2,zid3,zid4)
%#   should be like [1,431,625]
zid1 = [1,2;3,4];
zid2 = [1,1;1,1];
zid3 = [1,1;1,1];
zid4 = [1,1;1,1];
zOut = spindex(z,zid1,zid2,zid3,zid4)
%%#    should be like [[1,2];[3,4]]

you will need to add the code below as spindex.m to a location in your MATLAB path.

 function outM = spindex(inM,varargin)
 %function outM = spindex(inM,varargin)
 %
 %returns a matrix indexed from inM via index variables contained in varargin
 %useful for retreiving multiple values from a large multidimensional matrix
 %
 %
 %inM is an N-d matrix
 %the index variables stored in varargin must be as numerous as the number of dimensions in inM
 %each index variable must be identical in size
 %
 %example:
 %
 %z = reshape(1:(5^4),[5,5,5,5]);
 %zid1 = [1,1,5];
 %zid2 = [1,2,5];
 %zid3 = [1,3,5];
 %zid4 = [1,4,5];
 %zOut = spindex(z,zid1,zid2,zid3,zid4)
 %%   should be like [1,431,625]
 %zid1 = [1,2;3,4];
 %zid2 = [1,1;1,1];
 %zid3 = [1,1;1,1];
 %zid4 = [1,1;1,1];
 %zOut = spindex(z,zid1,zid2,zid3,zid4)
 %%    should be like [[1,2];[3,4]]
 sz = size(inM);
 ndim = length(sz);
 if((ndim == 2) & (sz(2) ==1)) % ndim always returns at least 2
   ndim =1;
 end
 if(nargin ~= (ndim +1))
    extraDims = setdiff(1:(nargin - 1),1:ndim);
    for iExtraDim = extraDims
       if(any(varargin{iExtraDim}~=1))
          error('must have as many indicies as dimensions\n');
       end
    end
 end
 szid = size(varargin{1});
 for i = 1:ndim
    szid2 = size(varargin{i});
    if(any(szid2 ~= szid))
       error('indicies must have identical shape');
    end
    ndIdxs(:,i) = varargin{i}(:);
 end
 if(ndim == 1)
    idxs = ndIdxs(:,1);
 else
    idxs = myNDsub2ind(size(inM),ndIdxs);
 end
 outM = nan(1,length(idxs));
 outM(find(not(isnan(idxs)))) = inM(idxs(find(not(isnan(idxs)))));
 outM = reshape(outM,size(varargin{1}));





 function ndx = myNDsub2ind(siz,subs)
 %function ndx = NDsub2ind(siz,subs)
 %-------------------------------
 %works more smoothly when the dimensionality of the mtrx is unknown
 %siz should be like [10 10 4 5] if subs is like
 % 9 8 3 5
 % 1 1 1 1
 % 10 10 4 5
 % 5 8 3 3
 %
 % siz will be rotated for you if submit a row vec instead a col vector
 % example: NDsub2ind([10 10 4 5],[[9,8,3,5];[1,1,1,1]])
 %----------------------------------------------
 if(size(siz,1) > 1) && (size(siz,2) > 1)
    error('the siz variable must be a vector');
 end

 if((size(subs,1) ~= 1) && (size(subs,2) == 1))
    subs = subs';
 end
 siz = siz(:)';
 if length(siz)<2
         error('MATLAB:sub2ind:InvalidSize',...
             'Size vector must have at least 2 elements.');
 end

 if ((length(siz) ~= size(subs,2)))
     error('NDsub2ind: length(siz) must = size(subs,2)');
 end

 nPoints = size(subs,1);


 %Compute linear indices
 k = [1 cumprod(siz(1:end-1))];
 ndx = ones(nPoints,1);
 s = size(subs); %For size comparison
 for i = 1:length(siz),
     v = subs;
     fNaN = find(   (v(:,i) < 1) | (v(:,i) > siz(i))   );
     %Verify subscripts are within range
     v(fNaN,i) = nan;
     ndx = ndx + (v(:,i)-1)*k(i);
 end
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!