Find closest matrix from list

2013 7 30
3 个回答
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0 个投票

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?

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Azzi Abdelmalek
Azzi Abdelmalek 2013-7-30
What is your criterion ? What should we compare to tell that a is closer then b to c?
Iain
Iain 2013-7-30
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
Jan 2013-7-30
编辑：Jan 2013-7-30
@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. ;-)

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回答（3 个）

Jan
Jan 2013-7-30

0 个投票

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.

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Andrew
Andrew 2013-7-30
Sorry. but I can't find function getDistance in a Matlab Help

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Azzi Abdelmalek
Azzi Abdelmalek 2013-7-30
编辑：Azzi Abdelmalek 2013-7-30

0 个投票

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
Andrew
Andrew 2013-7-30
编辑：Andrew 2013-7-30

0 个投票

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.

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Azzi Abdelmalek
Azzi Abdelmalek 2013-7-30
编辑：Azzi Abdelmalek 2013-7-30
Why M(3) is the closest? Are you comparing just by eyes?
Andrew
Andrew 2013-7-30
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
Azzi Abdelmalek 2013-7-30
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
Andrew 2013-7-30
编辑：Andrew 2013-7-30
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
Iain 2013-7-30
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
Andrew 2013-7-30
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.

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Andrew
2013-7-30

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