How to find a unique indicator of a specific matrix?

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M 2023-2-8

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https://ww2.mathworks.cn/matlabcentral/answers/1909110-how-to-find-a-unique-indicator-of-a-specific-matrix

评论： Walter Roberson 2023-2-8

Can we find a unique indicator of a specific matrix?

for example, if there is the following matrix (generated by a specific case) , how can I find one indicator of it?

I tried to take the sumation of the absolute values of the Matrix, but it didnt work because it was similer or too close for other cases

M1 and M2 is the range that can the matrix M varies between them, they are generated from a one case but with a little variation in some parameters (I want to find indicator for each other and compare it with other cases)

in this case the indicator should be too close for these two matrices (they are generated from the same case)

M1 = [-0.321306015516281	-0.274347013364667	-0.244382392248534	-0.288238228952784;
0.0558620817276004	0.372558615192237	0.572777978587165	0.503058540034827;
0.879663610530521	-0.136530080441011	-0.146669343761813	-0.155051653631985;
-0.176098379345313	0.737213426291489	-0.317456953062300	-0.314726433029229;
-0.211259054671114	-0.342511568755319	0.574666860881602	-0.415568275258076;
-0.210220781888802	-0.326336514970413	-0.399587060803839	0.606679564783506]
M2 = [-0.329410894346382	-0.282164376523392	-0.250593904107229	-0.300770395163912;
0.0391745398039566	0.367487954611743	0.578801050109399	0.490318907952166;
0.880000126456464	-0.133241897165618	-0.141354513303187	-0.151019721814831;
-0.173510462331790	0.738010210780251	-0.318373231341420	-0.312738411276027;
-0.207557875879909	-0.344447577135199	0.568108475231932	-0.416369292422166;
-0.205833532679448	-0.322922994018591	-0.397620657310764	0.612493983612021]

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M 2023-2-8

any indicator, like one number represents a matrix ...

M 2023-2-8

@Les Beckham comparing by indicator is easier in this problem, because for each case there is a range

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Answer 1

Walter Roberson 2023-2-8

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https://ww2.mathworks.cn/matlabcentral/answers/1909110-how-to-find-a-unique-indicator-of-a-specific-matrix#answer_1166920

As of R2022b you can use keyHash . Before that,

https://www.mathworks.com/matlabcentral/fileexchange/31272-datahash but see https://stackoverflow.com/questions/48667340/matlab-fast-hash-of-a-matrix for faster Java based code.

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Walter Roberson 2023-2-8

@M

No. All cases involving summarizing multiple values as a single value are mathematically classed as "hashing". Some hashing methods are better for particular purposes than other methods are, but all deterministic functions

are classified as hash functions.

M 2023-2-8

编辑：M 2023-2-8

@Walter Roberson ok, but this method didnt work with my problem for example there is only slight difference between the above matrices and they should indicat for a case, as range for exapmle.. and any number between this range should indicate for the same case...

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Answer 2

dpb 2023-2-8

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移动：dpb 2023-2-8

在 MATLAB Online 中打开

Along @Les Beckham's thought process, perhaps <ismembertol> might be of some use...let's see

M1 = [-0.321306015516281 -0.274347013364667 -0.244382392248534 -0.288238228952784;

0.0558620817276004 0.372558615192237 0.572777978587165 0.503058540034827;

0.879663610530521 -0.136530080441011 -0.146669343761813 -0.155051653631985;

-0.176098379345313 0.737213426291489 -0.317456953062300 -0.314726433029229;

-0.211259054671114 -0.342511568755319 0.574666860881602 -0.415568275258076;

-0.210220781888802 -0.326336514970413 -0.399587060803839 0.606679564783506];

M2 = [-0.329410894346382 -0.282164376523392 -0.250593904107229 -0.300770395163912;

0.0391745398039566 0.367487954611743 0.578801050109399 0.490318907952166;

0.880000126456464 -0.133241897165618 -0.141354513303187 -0.151019721814831;

-0.173510462331790 0.738010210780251 -0.318373231341420 -0.312738411276027;

-0.207557875879909 -0.344447577135199 0.568108475231932 -0.416369292422166;

-0.205833532679448 -0.322922994018591 -0.397620657310764 0.612493983612021];

100*(M2-M1)./M1 % percentage difference

ans = 6×4

2.5225 2.8494 2.5417 4.3479 -29.8728 -1.3610 1.0516 -2.5324 0.0383 -2.4084 -3.6237 -2.6004 -1.4696 0.1081 0.2886 -0.6317 -1.7520 0.5652 -1.1413 0.1928 -2.0870 -1.0460 -0.4921 0.9584

tol0=1E-4; % set initial tolerance tightly...

N=numel(M1); % check for when all are outside tol

i=1;

t{i,1}=tol0;

n{i,1}=sum(ismembertol(M1,M2,t{i}),'all');

while n{i}<N

i=i+1;

t{i}=2*t{i-1};

n{i}=sum(ismembertol(M1,M2,t{i}),'all');

if i>10, break, end

end

semilogx(cell2mat(t),cell2mat(n),'x-','linewidth',1.5)

grid

You can 'spearmint with alternatives on max points or all or whatever suits....

We have no basis for any idea of what is/is not important here from which to judge.

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dpb 2023-2-8

编辑：dpb 2023-2-8

Old version of MATLAB then; you'll have to use the colon expansion idiom or the nested-call to sum() method if can't upgrade to a release that supports the 'all' keyword -- and I'm not sure just which release introduced it, but it's been a while by now.

As for what you compare, that's your call depending upon what your requirements are and how you determine what is "close enough". The above just illustrates how many elements of your sample arrays fail the tolerance test depending upon the size of the tolerance you've chosen. if judging is based on how many elements are differerent, that's the count -- the Q? only you can answer is just what defines "too much". Is one element being different by a huge amount a problem if all the rest are identical? If that's a failure, then the above isn't going to be much help. If they all can be somewhat different as is the example that shouldn't be according to the post; then perphaps the level would be say, 5, elements faill the test. The big unknown is whether a similar result would hold at a given tolerance value or not between other datasets besides this one...we dunno; just a thought of one thing you could look at in trying to create a metric.

dpb 2023-2-8

Yeah, @Walter Roberson; I don't know why I tend to forget about nnz -- just 30-yr old habits die slowly, I guess, mostly. It's the better idiom; I should remember using it.

Walter Roberson 2023-2-8

在 MATLAB Online 中打开

A = [1 2 3]
A = 1×3
     1     2     3
B = [18 42 17 3 14 11 (1+eps)]
B = 1×7
   18.0000   42.0000   17.0000    3.0000   14.0000   11.0000    1.0000
[found, idx] = ismembertol(A,B)
found = 1×3 logical array
   1   0   1
idx = 1×3
     7     0     4

You can see from this that ismembertol() does not operate element-wise. Each member of the first matrix is compared to each element of the second matrix.

If you want to compare element-by-element then if all of the elements are in the same range, then

num_matches = nnz(abs(A-B) <= tolerance)

ismembertol() is for the case where each element of A is to be compared to all elements of B

When you do not pass a tolerance factor into ismembertol() then it uses 1e-12 (double) or 1e-6 (single) . If you use the datascale option you can specify a tolerance per column when ByRows option is used.

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