Compare two vectors for similarity

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.

[Souparno Bandyopadhyay]

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]

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.

Answer 1

[Matt Fig]

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...

[Souparno Bandyopadhyay]

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]

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?

[Souparno Bandyopadhyay]

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]

@Souparno: Accepting an answer means that the problem is solved. Is this true here?

[Souparno Bandyopadhyay]

yes, S = sum(A-B), is what I was looking for.

Answer 2

[Greg Heath]

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