# Why do logical indices implicitly reshape? Is there a workaround?

조회 수: 2(최근 30일)
Joel Lynch 2021년 8월 3일
편집: Joel Lynch 2021년 8월 3일
I've encountered this several times, and it always bothers me. Say for example I want to sum each row of matrix A, for values less than 0.5:
A = rand(3);
B = zeros( size(A) );
B(A<0.5) = A(A<0.5);
row_sum = sum( B, 2 );
But it feels like I should be able to simplify this, with something like:
row_sum = sum( A(A<0.5), 2 );
but this obviously fails because while A<0.5 preserves shape, A(A<0.5) returns a vector where the A<0.5 matrix is implicitly linearized.
I get why this happens (A>0.5 elements would be undefined in a matrix), but it seems incongruous with how logical indices are presented to the user as shape-preserving operations. After all, A<0.5 returns a logical matrix and you would need to call find(A<0.5) to acess the linear indices used to generate A(A<0.5).
In my humble opinion, it seems like this is exactly what NaN should exist for. If the default behavior of A(A<0.5) was to return NaN for A(A>0.5), then having functions like sum with 'omitnan' enabled by default would produce more intuitive results. Am I missing something?
Getting back to earth, are there any practical ways to simplify the above process?
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Joel Lynch 2021년 8월 3일
Can you please give a reference to the MATLAB documentation where...
I certainly didn't mean to imply there was an error or inconsistency in the MATLAB documentation; I'm aware this is the normal documented MATLAB behavior. I should have said "logical operations are returned to the user as shape preserving boolean matrices"
My point being that this rule seems counter-intuitive when logical indexing is the primary use of logical operations (or maybe they aren't?). Why not return the linear indices (e.g. find(A<0.5)) that are actually used in logical indexing, and use an inverse function to find(), when the matrix is needed?

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### 채택된 답변

Jakeb Chouinard 2021년 8월 3일
편집: Jakeb Chouinard 2021년 8월 3일
Assuming that modifying A is not allowed, I think you're close to the most simple way to do what you need.
You could also do something like:
A = rand(3);
B = A;
B(B<0.5) = 0; %or NaN, if that's preferred
sum(B,2)
Or:
A = rand(3);
cmpMatrix = A>=0.5; %CoMParison Matrix
sum(A.*cmpMatrix,2)
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Jakeb Chouinard 2021년 8월 3일
That's the spirit! At least then you're using up less space, though if you'll need to call the comparison matrix again, you may be better off keeping it stored. And just since I know the mods prefer to have questions be closed once answered, could you select my response as an accepted answer?
Cheers!

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R2021a

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