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

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Dong Dong 2019-7-21

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评论： Dong Dong 2019-7-21

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?

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KALYAN ACHARJYA 2019-7-21

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

Dong Dong 2019-7-21

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

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Answer 1

Guillaume 2019-7-21

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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;

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Guillaume 2019-7-21

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 2019-7-21

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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