How to find indices of a rectangular region inside big matrix? | Efficiently

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How can indices of elements belong to a rectangular region can be found?
A = [4 4 4 4 4 4 4
4 1 1 1 1 3 0
4 1 3 3 1 3 0
4 1 3 3 1 3 0
4 1 1 1 1 3 0
4 4 4 4 4 4 4];
Input: Matrix's size[height, width] , [Row_start Row_end], [Col_start Col_end]
Output: [21 22 23 27 28 29 33 34 35]
Why efficiently : to do same for multiple combinations of rows & columns
Thank you
  4 个评论
Caglar
Caglar 2018-11-2
I am not sure If I understood what you are asking for. If you want to get a part of a matrix you can do it like A(3:5,4:6) .
JAI PRAKASH
JAI PRAKASH 2018-11-2
Sorry if I am not clear in my question.
I want linear indices of all the elements present in rectangular region.
e.g

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采纳的回答

Bruno Luong
Bruno Luong 2018-11-2
编辑:Bruno Luong 2018-11-2
recidx = (Row_start:Row_End)' + height*(Col_start-1:Col_end-1)
  6 个评论

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更多回答(2 个)

Geoffrey Schivre
Geoffrey Schivre 2018-11-2
Hello,
I'm not sure if it's efficient enough but try :
p = nchoosek([Row_start:Row_end,Col_start:Col_end],2);
idx = unique(sub2ind(size(A),p(:,1),p(:,2)));
Geoffrey
  5 个评论
Geoffrey Schivre
Geoffrey Schivre 2018-11-2
Sorry, I didn't refresh this page so I didn't see that your question was answered. I'm glad you find what you ask for !

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Caglar
Caglar 2018-11-2
编辑:Caglar 2018-11-2
function result = stack (A,row_start,row_end,col_start,col_end)
% A = [4 4 4 4 4 4 4
% 4 1 1 1 1 3 0
% 4 1 3 3 1 3 0
% 4 1 3 3 1 3 0
% 4 1 1 1 1 3 0
% 4 4 4 4 4 4 4];
% row_start=3; col_start=4;
% row_end=5; col_end=6;
height=(size(A,1));
result=(row_start:row_end)+(height)*((col_start:col_end)'-1);
result=transpose(result); result=result(:);
end

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