Efficiently find indices in matrix

Question

Ryan Magee 2013-8-29

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https://ww2.mathworks.cn/matlabcentral/answers/85960-efficiently-find-indices-in-matrix

评论： Ryan Magee 2013-9-26

Massive edit/complete rewording by example.

if true
  A = [1:10];
  for i = 1:10,
      for j = 1:10,
          c(i,j) = a(i)-a(j);
      end
  end
  map = c == %%%;
  init = a;
  initval(~map) = nan;
  initval(isnan(initval)) = 0;
  uval = unique(initval);
end

Where %%% is a placeholder for "if that element in c is divisible by 5", a condition I'm not sure how to put in efficiently yet. Oversimplified, but gets the idea across. I have a large vector (not all consecutive integers, a bunch of random numbers) that I'm essentially subtracting from itself element by element and am looking for every value in that resulting matrix that is a multiple of 5. Then, I want it to tell me the values from A that "generated it", ie 1 6 2 7 3 8 4 9 etc.

Any suggestions?

Thanks for looking.

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Azzi Abdelmalek 2013-8-29

Post a short example with expected result

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Answer 1

Image Analyst 2013-9-26

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Regarding your new question

map = mod(c,5) == 0
valuesThatGeneratedIt = c(map)

But make sure they're all integers or you may not get 0 exactly. See the FAQ.

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Ryan Magee 2013-9-26

Awesome, thank you so much. Sorry it took so long to get back to you...I was away from the internet for quite some time.

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Answer 2

Image Analyst 2013-8-29

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Why don't you just make a logical (binary) "map" of where C meets your criteria? For a simple example, let's find out where C is -10:

binaryMap = C == -10; % map of where C = -10.

(If you want to take a histogram and find negative values of C that occur at least twice, then you can do that with a different criteria, a little more complicated.) Now you had to create 2D matrices to add A and B (using repmat or whatever) to get C. So you can just use the same indexes to get the A Values or B values. for example mask A like this:

AmatchingConditions = A; % Initialize.
AmatchingConditions(~binaryMap) = nan; % Set non-matching locations to nan

Now when you look at A, only the elements in A that match your criteria will appear and those that don't contribute to matching your criteria will appear as nan.