Find the elements of a ND matrix given the index of the elements

[b,idx]=min(a,[],2) -- gives the value as well as the index of the min value of each row of matrix a.

Now the ques is - how to get b if i've 'a' & 'idx'.

For 2D,i can use a for loop to do it.

But how to do it for ND scenarios?

For example,

a=[0.8147    0.9134    0.2785
   0.9058    0.6324    0.5469
   0.1270    0.0975    0.9575]
b=[0.2785
   0.5469
   0.0975]
idx=[3
       3
       2]

for i=1:length(idx)
    b(i)=a(i,idx(i));
end

-- it'll do for 2D.

[darova]

Can you show you calculate idx for 3d dimension?

[Asif Newaz]

sorry.i dont understand ur question.

For a 3D matrix a,

[b,idx]=min(a,[],2) -will generate a 3D array of idx.

Answer 1

[Stephen23]

Surprisingly there is no trivial way to do this, but one general solution is to generate linear indices, e.g.:

% input data (any ND array and the corresponding output index from MIN or MAX):
a = [0.8147,0.9134,0.2785;0.9058,0.6324,0.5469;0.1270,0.0975,0.9575]; % ND array.
idx = [3;3;2] % indices, must be the same orientation as returned by MIN or MAX.
% sizes of input data:
idx = double(idx);
sza = size(a);
szn = size(idx);
szn(end+1:numel(sza)) = 1;
idd = szn~=sza;
% linear indices:
tmp = arrayfun(@(n)1:n,szn,'uni',0);
[tmp{:}] = ndgrid(tmp{:});
tmp{idd} = idx;
ndx = sub2ind(sza,tmp{:});

You can use the linear indices very simply:

>> a(ndx)
ans =
   0.2785
   0.5469
   0.0975