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Re: fixed points piecewise-linear fitting


From: Sergei Steshenko
Subject: Re: fixed points piecewise-linear fitting
Date: Sat, 17 Mar 2012 09:32:08 -0700 (PDT)




----- Original Message -----
> From: Ben Abbott <address@hidden>
> To: Sergei Steshenko <address@hidden>
> Cc: help-octave Octave <address@hidden>
> Sent: Saturday, March 17, 2012 5:03 PM
> Subject: Re: fixed points piecewise-linear fitting
> 
> On Mar 17, 2012, at 7:43 AM, Sergei Steshenko wrote:
> 
>>  Hello,
>> 
>>  strictly speaking, it's not an Octave-specific question, but an 
> algorithmic one.
>> 
>>  Suppose there is a measured function Y(X). In Octave terms X is a vector 
> with N elements.
>> 
>>  Suppose there are fixed points Xf such that
>> 
>>  X(1) <= Xf(1)
>>  Xf(end) <= X(end).
>> 
>>  The Xf points are more sparse than X.
>> 
>> 
>>  The Xf points are fixed, i.e. one can't change them as he/she pleases.
>> 
>>  The goal is to find piecewise-linear function Yf(Xf) which best fits Y(X).
>> 
>>  I.e. for each two Xf(k), Xf(k+1) pair of points to find a piece of straight 
> line defined by Yf(k), Yf(k+1)pair of points such that the whole Yf fits Y 
> pretty well.
>> 
>>  Best fitting I'm interested in is according to minimum of sum(abs(Y - 
> Yf_interpolated)). The Yf_interpolated is linear interpolated Yf on X, so 
> dimensions of Y and Yf_interpolated match.
>> 
>> 
>>  I did some quick web searching and my impression is that there is no 
> universally adopted algorithm for this task, but there is a number solutions, 
> including some for R-language.
>> 
>>  I myself wrote a straightforward brute force implementation which works 
> pretty well and acceptably fast for me.
>> 
>>  Anyway, I'm writing this Email in the hope to be educated by the 
> community - maybe there are already more elegant wheels than the one I've 
> invented.
>> 
>>  Thanks,
>>    Sergei.
> 
> 
> I don't have an answer, but if I understand what you're looking for, 
> I'd be interested in a solution as well.
> 
> You desire solution for Yf that produces a least squares error between Y and 
> Yf_interpolated?
> 
> Where ...
> 
>     Yf_interpolated = interp1 (Xf, Yf, X, "linear");
> 
> ... and you'd like the solution that minimizes ...
> 
>     sum ((Yf_interpolated - Y).^2)
> 
> A linear solution would work. It would also be nice to extend the solution to 
> allow for a piece-wise continuous quadratic solution. It would also be 
> beneficial to allow the order of the polynomial pieces to be specified.
> 
> I think this would make a nice addition to piece-wise polynomial functions 
> already included in Octave.
> 
> If you like the idea, please enter this on the task list.
> 
>     https://savannah.gnu.org/task/?group=octave
> 
> ... and attached your current version so that those interested can suggest 
> changes.
> 
> Ben

Well, in my case it's just sum(abs(Yf_interpolated - Y)), not 
sum((Yf_interpolated - Y).^2).

I don't understand your "A linear solution would work".


My solution is poor man's brute force one.

I.e. I have a pretty good Yf initial approximation, and I have Y_step.

I have an outer loop which is iterations and an inner loop on 'k'

For each Yf(k) I try (Yf(k) + Y_step) and (Yf(k) - Y_step) and check whether my 
sum(abs(Yf_interpolated - Y)) becomes smaller or not.

If it is smaller, Yf(k) is replaced with Yf(k) +/- Y_step - depending on which 
of them gives smaller sum(abs(Yf_interpolated - Y)).

The outer loop is run until there is no improvement in sum(abs(Yf_interpolated 
- Y)) or until number of iterations is exhausted - in my case very good fitting 
is not critical. In my case I never reach iterations limit - I intentionally 
make it hight; for my data I think I never have more than 200 iterations.

In my case length(Yf) == length(Xf) == 64.

Of course, I have no mathematical proof that I am reaching the global minimum 
of sum(abs(Yf_interpolated - Y)). Typically sum(abs(Yf_interpolated - Y)) 
becomes about 2 times smaller than it was for initial Yf, i.e. in practice I 
see that my algorithm improves fitting.

My dirty little secret is that probably in my case initial Yf approximation 
would do :).

Regards,
  Sergei.




>


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