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

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Asif Newaz
Asif Newaz 2020 年 4 月 29 日
コメント済み: Asif Newaz 2020 年 4 月 29 日
[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.
  2 件のコメント
Asif Newaz
Asif Newaz 2020 年 4 月 29 日
sorry.i dont understand ur question.
For a 3D matrix a,
[b,idx]=min(a,[],2) -will generate a 3D array of idx.

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Stephen
Stephen 2020 年 4 月 29 日
編集済み: Stephen 2020 年 4 月 29 日
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

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