MATLAB Answers

Return row and col of ismembertol comparisons?

2 views (last 30 days)
Andrew Poissant
Andrew Poissant on 4 Sep 2017
Edited: Andrew Poissant on 4 Sep 2017
I have a layered matrix, N, that is 834x834x3 and I want to scan that matrix using ismembertol to see which ordered pairs from F are in N and then return the row and col for each match. Matrix F is 93x2. I have my attempt at it below but it does not work properly and I do not think find is the function I need. I did not define the matrices N and F because they are defined using previous code. I know I need to use something like ismembertol(...) & ismembertol(...) to scan matrix N for common values between it and F but I am not sure how to return the row and col. Any suggestions?
[r, c] = size(F);
tol = 1;
for i = 1:r
[row(i), col(i)] = find(ismembertol(F(i,1), N(:,:,2), tol) & ismembertol(F(i,2), N(:,:,3), tol))
end

  0 Comments

Sign in to comment.

Answers (1)

KSSV
KSSV on 4 Sep 2017
A = rand(3,3,3) ;
B = rand(2,2,3) ;
%%find values equal to 0.5
[LIA,LocAllB] = ismembertol(B, A, 0.1)
LIA returns an array containing logical 1 (true) where the elements of B are within tolerance 0.1 of the elements in A.
LocB contains the index location in A for each element in B that is a member of A.

  3 Comments

Andrew Poissant
Andrew Poissant on 4 Sep 2017
Yes but I need to find the common row and col such that the values in both columns of F are members of N. How can I do that?
KSSV
KSSV on 4 Sep 2017
Yes....that's what LIA and LOCB gives you...
Andrew Poissant
Andrew Poissant on 4 Sep 2017
I am confused. I have the matrix F that is being used for ismembertol versus two layers of matrix N. I do not see where the matrix F (not layered) is set up and compared to to the two different layers of N. I see that you set up two 3 layered matrices and compared the two but I am comparing one column of matrix F to second layer of N and the second column of F to the third layer of N. Am I missing something in your solution?

Sign in to comment.

Sign in to answer this question.

Tags