\-operator to perform least squares fit on multiple sample-sets

 Is there a matrix expansion of the \-operator that enables least squares fit of y = M\x where M is an m-by-n-by-p matrix and x is an m-length vector m=[3 .. 10]; n =3 for order 2?

[Andrew Newell]

What length does y have and how many sets of x do you have?

Answer 1

[bym]

The \ operator can do that already. See my answer here

http://www.mathworks.com/matlabcentral/answers/24086-ols-regression-for-multiplr-ys-xs

[Eugene]

Thanks. I'll look at the code.

[Eugene]

It appears that this code has 3 sets of x and 3 sets of y. My issues is many-sets of x, but only one set of y. If I understand the code correctly it requires that the y's not overlap each other in order to get the 3-distinct curves. If I try replicating the y-values then I only get 1 set of coeffs.

Somehow I need to get the operator to take the data n-values at a time without a for-loop. Any thoughts would be appreciated

Thanks

[Eugene]

Since I'm trying to fit multiple samples to multiple targets, the fact that y can be a matrix is very helpful but not everything I would like. If the operator could parse the data as a function of the order and the size of y that might work.

thanks

Answer 2

[Andrew Newell]

If I understand your question, you are trying to estimate y given multiple measurements of x. Note that if M were a matrix, the correct least squares estimate would be

yhat = (M'*M)\(M'*x);

(see also this answer).

You would be well advised to recast your problem. Instead of several vectors x, make one vector X containing all the data. Then reshape M so

y = M*X;

and then use the above expression.

[Eugene]

Thank you for the suggestion. Let me see if I can get there.

[Eugene]

Thanks. I tried recasting per your suggestion but was unsuccessful in attaining my desired result. I believe I have not

posed my question well so here is a numeric example.

y is the target to which all x's need to be adjusted. In my original post I discribed y as being an m x 3 matrix, but y as

a single vector would also be acceptable.

{

y = 2360; 6805; 11877; 15238;

}

Here are 4 examples of x-values; each row is an x.

{

X = [ ...

3313.272 7651.640 12685.088 16030.168;

3260.816 7591.840 12632.344 15971.112;

3225.880 7548.640 12567.560 15900.824;

3205.088 7562.496 12626.608 15983.656];

}

Putting these into Vondermonde (V) matricies (done in parallel of course):

{

V1 = [ ...

10977771.3459840 3313.27200000000 1;

58547594.6896000 7651.64000000000 1;

160911457.567744 12685.0880000000 1;

256966286.108224 16030.1680000000 1];

V2 = [ ...

10632920.9858560 3260.81600000000 1;

57636034.5856000 7591.84000000000 1;

159576114.934336 12632.3440000000 1;

255076418.516544 15971.1120000000 1];

V3 = [ ...

10406301.7744000 3225.88000000000 1;

56981965.8496000 7548.64000000000 1;

157943564.353600 12567.5600000000 1;

252836203.878976 15900.8240000000 1];

V4 = [ ...

10272589.0877440 3205.08800000000 1;

57191345.7500160 7562.49600000000 1;

159431229.585664 12626.6080000000 1;

255477259.126336 15983.6560000000 1];

}

Using p = V\y gives four sets of p-values (this is where I have used a for-loop):

p1 = [ ...

-1.19794168760304e-06;

1.03531698605336;

-1054.41785901502;

];

p2 = [ ...

-1.22990697430811e-06;

1.03605999963114;

-1001.23771919361;

];

p3 = [ ...

-1.21854840069794e-06;

1.03874536729461;

-975.178458457347;

];

p4 = [ ...

-1.16835324217934e-06

1.02950561368923

-924.033546502190

];

so the four results where Vn*pn:

yo =

2362.71819206764 6797.31839976274 11885.9066739332 15234.0567342364

2364.08580088584 6793.47706749778 11890.3648240961 15232.0723075203

2363.00886460663 6796.50108756282 11886.8543919961 15233.6356558344

2363.62052909954 6794.77884476849 11888.8582773600 15232.7423487719

compared to

y = 2360; 6805; 11877; 15238;

This 'fits' the x's to y.

I need to accomplish the same thing on thousands of x's.

Hopefully this explains my question better. Any further suggestions would be much appreciated.

[Andrew Newell]

So what you're really trying to do is to fit second order polynomials y=a*x^2+b*x+c to the data?

[Eugene]

I'm really trying to fit the data, using individual second order polynomials to the same (desired) target-curve. To me, this means there is a different a,b, & c for each data-set. If that is what you meant, I apologize for being so dense.

[Andrew Newell]

I'm not sure what you mean by "curve". To me, a set of (x,y) values represents points on a curve - in the example above, you would have four curves, one for each row of X.

[Eugene]

Yes an (x,y) set are points on a curve. In this case y is also a curve (a set of data with respect to (wrt), say z) and all four rows of X are also wrt z and we are trying to find the quadratic coefficients that when applied to each row in X, that will cause each row in X to closely match y.