How to find array elements that meet a condition defined by an index vector?

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Let A be an n by n matrix of 0's. Let L be an n by 1 matrix of integers, which will be used as an index vector. I want to modify the entries of A based on the following rules:
If L(i)=5 and L(j)=5, then A(i,j)=1;
If L(i)=5 xor L(j)=5 (i.e., exatly one of L(i) and L(j) is 5), then A(i,j)=2;
What is the fastest/simplest code to achieve this? I can use a nested for loop for i and j and modify A(i,j) entry by entry. Is there any built in function that I can use?

Accepted Answer

Guillaume
Guillaume on 21 Jul 2019
It's not clear to me why L is a matrix of integers when the only value you care about is 5. Shouldn't it be a binary matrix?
We're going to transform it in a binary matrix anyway:
%demo data
n = 20;
A = zeros(n);
L = randi([3 6], n, 1);
is5 = L == 5; %binary matrix indicated where 5 are in L
A(is5 & is5') = 1; %assign 1 when L(i) and L(j) are 5
A(xor(is5, is5')) = 2; %assign 2 when L(i) xor L(j) is 5
and looking at the result made me realise that another way to obtain the same is:
A(is5, :) = 2;
A(:, is5) = 2;
A(is5, is5) = 1;
  3 Comments
Guillaume
Guillaume on 21 Jul 2019
bsxfun would be more efficient than repmat, but it's been 3 years now that it's not been needed. The bsxfun version:
A(bsxfun(@and, is5, is5')) = 1;
A(bsxfun(@xor, is5, is5')) = 2;
The 2nd version does an or but then overwrite the elements when both are equal, effectively turning it in an xor. I actually prefer the 2nd version, the outcome is more obvious.
Rik
Rik on 21 Jul 2019
You're right, I overlooked the implicit xor. And about bsxfun: a fair number of people still use older releases than 3 years back, and since we weren't provided with explicit information, I thought it would be better to include the remark.

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