Compare two vectors for similarity

2012 12 9
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更新时间：2022 2 2
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How to compare two vectors quickly. Right now I print out each in a loop and examine them by eye, is there a way i can find if two are almost similar.

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I am comparing A to B and then A to C, I need a single number that will allow me to quickly judge A resembles B or C.
maxanto
maxanto 2022-2-2
isequal(a, b)
Returns true if each element of vector a is equal to each element of vector b. If some element of a are different from b returns false.

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 采纳的回答

Matt Fig
Matt Fig 2012-12-9

0 个投票

What is the criteria for 'almost similar' in your application? 90% same exact values? 90% of the values in one vector within 95% of some other value in the other vector? Do the values have to be in the same positions? Do the vectors have to be the same length? Perhaps a few short examples would help...

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Souparno Bandyopadhyay
编辑：Walter Roberson 2012-12-9
A, B, C are vectors of length 4. They are coefficients of a linear prediction filter.
say A = [1 2 3 4]
and B = [1.1 2.2 3.3 4.4]
and C = [1 2 3 4.5]
so if
i = 1:4
delta = (A[i] - B[i]);
sum = sum + delta;
end
% i want the minimum value of sum for 512 iterations. but I want to know if there exists a function that would allow me to replace the for loop.
Matt Fig
Matt Fig 2012-12-9
编辑：Matt Fig 2012-12-9
Souparno, that is not valid MATLAB code. But if it were transformed to valid code such that you are after variable S. (You should not name a variable sum because this will mask the MATLAB function SUM. Same goes for i.)
S = 0;
for ii = 1:4
delta = (A(ii) - B(ii));
S = S + delta;
end
This can be replaced by:
S = sum(A-B)
I have no idea what you mean by 512 iterations. You mean the real vectors are 512 elements instead of 4?
I am using lpc(x,4), x is a set of discrete data to get 4 coeff vector of a linear prediction filter. Then I am comparing these coeff to another set that i get from executing lpc over another part of the same x(n) to find which two parts resemble the closest.
Jan
Jan 2012-12-9
@Souparno: Accepting an answer means that the problem is solved. Is this true here?
yes, S = sum(A-B), is what I was looking for.

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Greg Heath
Greg Heath 2012-12-10

14 个投票

S = sum(A-B) is NOT a useful function for quantifying similarity because positive and negative terms will cancel.
The most common are
mae(A-B) % mean(abs(A-B))
sae(A-B) % sum(abs(A-B))
norm(A-B,1) % sum(abs(A-B))
norm(A-B,inf) % max(abs(A-B))
mse(A-B) % mean((A-B).^2)
sse(A-B) % sum((A-B).^2)
norm(A-B) % sqrt(sse(A-B))
Hope this helps.
Thank you for formally accepting my answer
Greg

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