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

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?

[KALYAN ACHARJYA]

Can you explain the difference between L(i) & L(j)?

[Dong Dong]

They impose conditions on the first and second index of the matrix.

Answer 1

[Guillaume]

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;

[Rik]

Note that on old releases you will need repmat (or bsxfun?) to expand the vector to an array. The last three lines will work on any release, although those don't take the xor into account.

[Guillaume]

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]

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.