Is it possible to do an algebraic curve fit with just a single pass of the sample data?

Question

I would like to do an algebraic curve fit of 2D data points, but for various reasons - it isn't really possible to have much of the sample data in memory at once, and iterating through all of it is an expensive process.

(The reason for this is that actually I need to fit thousands of curves simultaneously based on gigabytes of data which I'm reading off disk, and which is therefore sloooooow).

Note that the number of polynomial coefficients will be limited (perhaps 5-10), so an exact fit will be extremely unlikely, but this is ok as I'm trying to find an underlying pattern in data with a lot of random noise. I understand how one can use a genetic algorithm to fit a curve to a dataset, but this requires many passes through the sample data, and thus isn't practical for my application.

Is there a way to fit a curve with a single pass of the data, where the state that must be maintained from sample to sample is minimal?

I should add that the nature of the data is that the points may lie anywhere on the X axis between 0.0 and 1.0, but the Y values will always be either 1.0 or 0.0.

So, in Java, I'm looking for a class with the following interface:

public interface CurveFit {
   public void addData(double x, double y);
   public List<Double> getBestFit(); // Returns the polynomial coefficients
}

The class that implements this must not need to keep much data in its instance fields, no more than a kilobyte even for millions of data points. This means that you can't just store the data as you get it to do multiple passes through it later.

edit: Some have suggested that finding an optimal curve in a single pass may be impossible, however an optimal fit is not required, just as close as we can get it in a single pass.

The bare bones of an approach might be if we have a way to start with a curve, and then a way to modify it to get it slightly closer to new data points as they come in - effectively a form of gradient descent. It is hoped that with sufficient data (and the data will be plentiful), we get a pretty good curve. Perhaps this inspires someone to a solution.

Answer 1

Yes, it is a projection. For

y = X beta + error

where lowercased terms are vectors, and X is a matrix, you have the solution vector

\hat{beta} = inverse(X'X) X' y

as per the OLS page. You almost never want to compute this directly but rather use LR, QR or SVD decompositions. References are plentiful in the statistics literature.

If your problem has only one parameter (and x is hence a vector as well) then this reduces to just summation of cross-products between y and x.

Answer 2

If you don't mind that you'll get a straight line "curve", then you only need six variables for any amount of data. Here's the source code that's going into my upcoming book; I'm sure that you can figure out how the DataPoint class works:

Interpolation.h:

#ifndef __INTERPOLATION_H
#define __INTERPOLATION_H

#include "DataPoint.h"

class Interpolation
{
private:
  int m_count;
  double m_sumX;
  double m_sumXX;  /* sum of X*X */
  double m_sumXY;  /* sum of X*Y */
  double m_sumY;
  double m_sumYY;  /* sum of Y*Y */

public:
  Interpolation();

  void addData(const DataPoint& dp);

  double slope() const;
  double intercept() const;

  double interpolate(double x) const;
  double correlate() const;
};

#endif // __INTERPOLATION_H

Interpolation.cpp:

#include <cmath>

#include "Interpolation.h"

Interpolation::Interpolation()
{
  m_count = 0;
  m_sumX = 0.0;
  m_sumXX = 0.0;
  m_sumXY = 0.0;
  m_sumY = 0.0;
  m_sumYY = 0.0;
}

void Interpolation::addData(const DataPoint& dp)
{
  m_count++;
  m_sumX += dp.getX();
  m_sumXX += dp.getX() * dp.getX();
  m_sumXY += dp.getX() * dp.getY();
  m_sumY += dp.getY();
  m_sumYY += dp.getY() * dp.getY();
}

double Interpolation::slope() const
{
  return (m_sumXY - (m_sumX * m_sumY / m_count)) /
    (m_sumXX - (m_sumX * m_sumX / m_count));
}

double Interpolation::intercept() const
{
  return (m_sumY / m_count) - slope() * (m_sumX / m_count);
}


double Interpolation::interpolate(double X) const
{
  return intercept() + slope() * X;
}


double Interpolation::correlate() const
{
  return m_sumXY / sqrt(m_sumXX * m_sumYY);
}

Answer 3

Are you limiting the number of polynomial coefficients (i.e. fitting to a max power of x in your polynomial)?

If not, then you don't need a "best fit" algorithm - you can always fit N data points EXACTLY to a polynomial of N coefficients.

Just use matrices to solve N simultaneous equations for N unknowns (the N coefficients of the polynomial).

If you are limiting to a max number of coefficients, what is your max?

Following your comments and edit:

What you want is a low-pass filter to filter out noise, not fit a polynomial to the noise.

Answer 4

Why not use a ring buffer of some fixed size (say, the last 1000 points) and do a standard QR decomposition-based least squares fit to the buffered data? Once the buffer fills, each time you get a new point you replace the oldest and re-fit. That way you have a bounded working set that still has some data locality, without all the challenges of live stream (memoryless) processing.

Answer 5

Given the nature of your data:

the points may lie anywhere on the X axis between 0.0 and 1.0, but the Y values will always be either 1.0 or 0.0.

Then you don't need even a single pass, as these two lines will pass exactly through every point:

• X = [0.0 ... 1.0], Y = 0.0
• X = [0.0 ... 1.0], Y = 1.0

Two short line segments, unit length, and every point falls on one line or the other.

Admittedly, an algorithm to find a good curve fit for arbitrary points in a single pass is interesting, but (based on your question), that's not what you need.

Answer 6

You need the solution to an overdetermined linear system. The popular methods are Normal Equations (not usually recommended), QR factorization, and singular value decomposition (SVD). Wikipedia has decent explanations, Trefethen and Bau is very good. Your options:

1. Out-of-core implementation via the normal equations. This requires the product A'A where A has many more rows than columns (so the result is very small). The matrix A is completely defined by the sample locations so you don't have to store it, thus computing A'A is reasonably cheap (very cheap if you don't need to hit memory for the node locations). Once A'A is computed, you get the solution in one pass through your input data, but the method can be unstable.

2. Implement an out-of-core QR factorization. Classical Gram-Schmidt will be fastest, but you have to be careful about stability.

3. Do it in-core with distributed memory (if you have the hardware available). Libraries like PLAPACK and SCALAPACK can do this, the performance should be much better than 1. The parallel scalability is not fantastic, but will be fine if it's a problem size that you would even think about doing in serial.

4. Use iterative methods to compute an SVD. Depending on the spectral properties of your system (maybe after preconditioning) this could converge very fast and does not require storage for the matrix (which in your case has 5-10 columns each of which are the size of your input data. A good library for this is SLEPc, you only have to find a the product of the Vandermonde matrix with a vector (so you only need to store the sample locations). This is very scalable in parallel.

Answer 7

Assuming that you don't know which point should belong to which curve, something like a Hough Transform might provide what you need.

The Hough Transform is a technique that allows you to identify structure within a data set. One use is for computer vision, where it allows easy identification of lines and borders within the field of sight.

Advantages for this situation:

• Each point need be considered only once
• You don't need to keep a data structure for each candidate line, just one (complex, multi-dimensional) structure
• Processing of each line is simple
• You can stop at any point and output a set of good matches
• You never discard any data, so it's not reliant on any accidental locality of references
• You can trade off between accuracy and memory requirements
• Isn't limited to exact matches, but will highlight partial matches too.

An approach

To find cubic fits, you'd construct a 4-dimensional Hough space, into which you'd project each of your data-points. Hotspots within Hough space would give you the parameters for the cubic through those points.

Answer 8

I believe I found the answer to my own question based on a modified version of this code. For those interested, my Java code is here.