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

Question

JAI PRAKASH 2018-11-2

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https://ww2.mathworks.cn/matlabcentral/answers/427676-how-to-find-indices-of-a-rectangular-region-inside-big-matrix-efficiently

编辑： JAI PRAKASH 2018-11-24

How can indices of elements belong to a rectangular region can be found?

A = [4     4     4     4     4     4     4
   1     1     1     1     3     0
   1     3     3     1     3     0
   1     3     3     1     3     0
   1     1     1     1     3     0
   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

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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 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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Answer 1

Bruno Luong 2018-11-2

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https://ww2.mathworks.cn/matlabcentral/answers/427676-how-to-find-indices-of-a-rectangular-region-inside-big-matrix-efficiently#answer_344888

编辑：Bruno Luong 2018-11-2

recidx = (Row_start:Row_End)' + height*(Col_start-1:Col_end-1)

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JAI PRAKASH 2018-11-3

Hi @Bruno

Do you have some idea for below question also.

How can already used elements be eliminated one by one in 2D matrix while moving upwards? | Efficiently

Which is nothing but continuation of the problem, that you solved just now.

Bruno Luong 2018-11-3

James's solution is very well. I doubt I can do better.

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Answer 2

Geoffrey Schivre 2018-11-2

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https://ww2.mathworks.cn/matlabcentral/answers/427676-how-to-find-indices-of-a-rectangular-region-inside-big-matrix-efficiently#answer_344889

在 MATLAB Online 中打开

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

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Geoffrey Schivre 2018-11-2

在 MATLAB Online 中打开

When you are looking for big rectangle use :

p1 = nchoosek([Row_start:Col_end],2);
p2 = nchoosek([Col_start:Row_end],2);
p3 = intersect([Row_start:Col_end],[Col_start:Row_end]);
p = [p1;p2;[p3',p3']];
idx = sub2ind(size(A),p(:,1),p(:,2));

This methode is less efficient on small rectangle but more efficient on large one due to the use of nchoosek on smaller vectors :

small rectangle (100 by 100) :

tic
N = 10000;
n1 = 10;
n2 = 10;
s1 = 99;
s2 = 99;
p = nchoosek([n1:(n1+s1),n2:(n2+s2)],2);
idx = unique(sub2ind([N,N],p(:,1),p(:,2)));
toc
Elapsed time is 0.016082 seconds.
tic
N = 10000;
n1 = 10;
n2 = 10;
s1 = 99;
s2 = 99;
p1 = nchoosek([n1:(n2+s2)],2);
p2 = nchoosek([n2:(n1+s1)],2);
p3 = intersect([n1:(n2+s2)],[n2:(n1+s1)]);
p = [p1;p2;[p3',p3']];
idx = sub2ind([N,N],p(:,1),p(:,2));
toc
Elapsed time is 0.020043 seconds.

large rectangle (1000 by 1000) :

tic
N = 10000;
n1 = 10;
n2 = 10;
s1 = 999;
s2 = 999;
p = nchoosek([n1:(n1+s1),n2:(n2+s2)],2);
idx = unique(sub2ind([N,N],p(:,1),p(:,2)));
toc
Elapsed time is 21.936548 seconds.
tic
N = 10000;
n1 = 10;
n2 = 10;
s1 = 999;
s2 = 999;
p1 = nchoosek([n1:(n2+s2)],2);
p2 = nchoosek([n2:(n1+s1)],2);
p3 = intersect([n1:(n2+s2)],[n2:(n1+s1)]);
p = [p1;p2;[p3',p3']];
idx = sub2ind([N,N],p(:,1),p(:,2));
toc
Elapsed time is 5.138268 seconds.

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 !

JAI PRAKASH 2018-11-2

Ya, But I learn interesting things from your comments.

'nchoosek' is new to me

Thanks

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Answer 3

Caglar 2018-11-2

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https://ww2.mathworks.cn/matlabcentral/answers/427676-how-to-find-indices-of-a-rectangular-region-inside-big-matrix-efficiently#answer_344890

编辑：Caglar 2018-11-2

在 MATLAB Online 中打开

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