hi every body !! my question is how to get the position of the similar elements in same matrix??
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Rutika Titre
el 27 de Oct. de 2015
Comentada: Image Analyst
el 29 de Oct. de 2015
suppose if i have matrix
M=[0 0 0 0 0;0 0 -1 0 0; 0 0 -1 0 -1;0 0 -1 0 -1; -1 -1 0 -1 -1];
i have to compare each row as in..1st row to 2nd row then 1st row to third row, 1st row to 4th one and 1st to 5th one and then again 2nd row to 3rd ans so on....I m trying to compare first and 2nd row row where
ft=[0 0 0 0 0];
st=[0 0 -1 0 0];
now i want that output should be pos=1,2,4,5. Where 1,2,4,,5 are the position of the similar elemnts in ft and st. I have tried many ways but -1 is the barrier here.
M stuck here plz help me out of this...any kind of help will b appreciated. my email id:- titre.rutika25@gmail.com
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Eng. Fredius Magige
el 27 de Oct. de 2015
Editada: Eng. Fredius Magige
el 27 de Oct. de 2015
Hi write down your code, which more preferable than abc....
Respuesta aceptada
dpb
el 27 de Oct. de 2015
Editada: dpb
el 27 de Oct. de 2015
Basic idea is
>> find(M(1,:)==M(2,:))
ans =
1 2 4 5
>>
Apply above for each row pairwise; note you'll have to save results in a cell array as there won't necessarily be the same length output vector for each row (and could, theoretically, have an empty row as well).
Implementation is a nested loop;
R=size(M,1); % # rows
k=0; % output counter
for i=1:R
for j=i+1:R
k=k+1;
out{k}=find(M(j,:)==M(i,:));
end
end
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Image Analyst
el 27 de Oct. de 2015
What does similar mean to you? Is 3.0001 similar to 3.0004? Or do you mean equal rather than similar? And, if you mean equal, are you using integers, or floating point numbers with fractional parts? Why? See the FAQ: http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F
And why not just use a for loop:
[rows, columns] = size(M)
for row = 1 : rows
thisRow = M(row, :);
newM = repmat(thisRow, [rows, 1]);
differences = M - newM;
% etc.
end
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Rutika Titre
el 29 de Oct. de 2015
1 comentario
Image Analyst
el 29 de Oct. de 2015
VIrtually all the Image Processing articles are logged here: http://www.visionbib.com/bibliography/contents.html There is a search capability at the bottom of the page. Good luck.
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