Find closest matrix from list

I've got a matrix A(3x4) and a list of similar matrix M(i).x, where i=1:100. I need to find the matrix from list M which will be closest to my matrix A. How can I do that?

[Azzi Abdelmalek]

What is your criterion ? What should we compare to tell that a is closer then b to c?

[Iain]

closest in what sense?

if:

A = [0 1];

which would be closer

 M(1).x = [0 1000];
 M(2).x = [500 501];
 M(3).x = [200 -200];

[Jan]

@lain: On first sight M(1).x is closest, because it is found in the topmost line. But if A is defined after M, M(3).x is closest. ;-)

Answer 1

[Jan]

bestValue = -Inf;
bestIndex = 0;
for k = 1:numel(M)
  Value = getDistance(A, M(k).x);
  if Value > bestValue
    bestValue = Value;
    bestIndex = k;
  end
end

Now insert your criterion to determine the distance as you like. Perhaps you are looking for "Value < bestValue" and want to start with +Inf.

[Andrew]

Sorry. but I can't find function getDistance in a Matlab Help

Answer 2

[Azzi Abdelmalek]

I propose this criterion sum(abs(a-b)) to be the smallest

for k=1:100
    s(k)=sum(sum(abs(A-M(k).x)))
end
[~,idx]=min(s);
Res=M(idx).m

Answer 3

[Andrew]

Sorry for incomplete question.

For example I've got matrix A[ 1 2 2; 1 2 3; 1 2 4] and in a list is present matrix M(3).x=[ 1 2 3; 1 2 3; 1 2 4] and M(4).x=[ 1 2 4; 1 2 3; 1 2 4]. Than matrix M(3).x will be closest. A can't use mean or sum of values in matrix to compare.

[Azzi Abdelmalek]

Why M(3) is the closest? Are you comparing just by eyes?

[Andrew]

Because the difference is only 1 with M(3) and 2 with M(4).

A[ 1 2 2; 1 2 3; 1 2 4]

M(3).x=[ 1 2 3; 1 2 3; 1 2 4]

M(4).x=[ 1 2 4; 1 2 3; 1 2 4]

[Azzi Abdelmalek]

What about this case

M(3).x=[ 5 22 3; 10 21 3; 1 52 4]
M(4).x=[ 1 2 44; 11 2 13; 15 2 4]

[Andrew]

yes, here is a problem. I have to find the closest value, but there should be some level of difference. For example, if the mean value of the first 3 numbers in M(3) differ from the first three values in A for 10, second three values for 5 and the third one for 5, and in M(4) differs for 7,7,7 respectively, it better to choose M(4) because the mean difference of each value is less. Somehow in that way. Sorry for English, I hope you understand what I mean.

[Iain]

The question is what definition are you using for "distance" all these are valid options...

 distance = max(abs(A(:)-M(i).x(:)));
 distance = sum(abs(A(:)-M(i).x(:)));
 distance = sum((A(:)-M(i).x(:)).^2);
 distance = max(abs(A(:)-M(i).x(:)+mean(M(i).x(:))-mean(A(:)) ));

[Andrew]

Oh, I've just realized that I can use

 distance = sum(abs(A(:)-M(i).x(:)));

to find the closest and use

distance = max(abs(A(:)-M(i).x(:)));

to cut off that values which are too "far". Okay, thanks a lot for your help!! I will try to do that.