15
$\begingroup$

Are there any packages to do piecewise linear regression, which can detect the multiple knots automatically? Thanks. When I use the strucchange package. I could not detect the change points. I have no idea how it detects the change points. From the plots, I could see there are several points I want it could help me to pick them out. Could anyone give an example here?

$\endgroup$
5
  • 1
    $\begingroup$ This appears to be the same question as stats.stackexchange.com/questions/5700/…. If it differs in any substantial way, please let us know by editing your question to reflect the differences; otherwise, we will close it as a duplicate. $\endgroup$ – whuber Nov 16 '11 at 4:43
  • 1
    $\begingroup$ I have edited the question. $\endgroup$ – Honglang Wang Nov 16 '11 at 5:18
  • 1
    $\begingroup$ I think you can do this as a non-linear optimisation problem. Just write the equation of the function to be fitted, with the coefficients and the knot locations as parameters. $\endgroup$ – mark999 Nov 16 '11 at 9:04
  • 1
    $\begingroup$ I think the segmented package is what you are looking for. $\endgroup$ – AlefSin Nov 24 '11 at 17:00
  • 1
    $\begingroup$ I had an identical problem, solved it with R's segmented package: stackoverflow.com/a/18715116/857416 $\endgroup$ – a different ben Apr 15 '14 at 0:34
9
$\begingroup$

Would MARS be applicable? R has the package earth that implements it.

$\endgroup$
9
$\begingroup$

In general, it's a bit odd to want to fit something as piece-wise linear. However, if you really wish to do so, then the MARS algorithm is the most direct. It will build up a function one knot at a time; and then usually prunes back the number of knots to combat over-fitting ala decision trees. You can access the MARS algotithm in R via earth or mda. In general, it's fit with GCV which is not so far removed from the other information criterion (AIC, BIC etc.)

MARS won't really give you an "optimal" fit since the knots are grown one at a time. It really would be rather difficult to fit a truly "optimal" number of knots since the possible permutations of knot placements would quickly explode.

Generally, this is why people turn towards smoothing splines. Most smoothing splines are cubic just so you can fool a human eye into missing the discontinuities. It would be quite possible to do a linear smoothing spline however. The big advantage of smoothing splines are their single parameter to optimize. That allows you to quickly reach a truly "optimal" solution without having to search through gobs of permutations. However, if you really want to seek inflection points, and you have enough data to do so, then something like MARS would probably be your best bet.

Here's some example code for penalized linear smoothing splines in R:

require(mgcv);data(iris);
gam.test <- gam(Sepal.Length ~ s(Petal.Width,k=6,bs='ps',m=0),data=iris)
summary(gam.test);plot(gam.test);

The actual knots chosen would not necessarily correlate with any true inflection points however.

$\endgroup$
3
$\begingroup$

I have programmed this from scratch once a few years ago, and I have a Matlab file for doing piece-wise linear regression on my computer. About 1 to 4 breakpoints is computationally possible for about 20 measurements points or so. 5 or 7 break points starts to be really too much.

The pure mathematical approach as I see it is to try all possible combinations as suggested by user mbq in the question linked to in the comment below your question.

Since the fitted lines are all consecutive and adjacent (no overlaps) the combinatorics will follow Pascals triangle. If there were overlaps between used data points by the line segments I believe that the combinatorics would follow Stirling numbers of the second kind instead.

The best solution in my mind is to choose the combination of fitted lines that has the lowest standard deviation of the R^2 correlation values of the fitted lines. I will try to explain with an example. Keep in mind though that asking how many break points one should find in the data, is similar to asking the question "How long is the coast of Britain?" as in one of Benoit Mandelbrots (a mathematician) papers about fractals. And there is a trade-off between number of break points and regression depth.

Now to the example.

Suppose we have the perfect data $y$ as a function of $x$ ($x$ and $y$ are integers):

$$\begin{array}{|c|c|c|c|c|c|} \hline &x &y &R^2 line 1 &R^2 line 2 &sum of R^2 values &standard deviation of R^2 \\ \hline &1 &1 &1,000 &0,0400 &1,0400 &0,6788 \\ \hline &2 &2 &1,000 &0,0118 &1,0118 &0,6987 \\ \hline &3 &3 &1,000 &0,0004 &1,0004 &0,7067 \\ \hline &4 &4 &1,000 &0,0031 &1,0031 &0,7048 \\ \hline &5 &5 &1,000 &0,0135 &1,0135 &0,6974 \\ \hline &6 &6 &1,000 &0,0238 &1,0238 &0,6902 \\ \hline &7 &7 &1,000 &0,0277 &1,0277 &0,6874 \\ \hline &8 &8 &1,000 &0,0222 &1,0222 &0,6913 \\ \hline &9 &9 &1,000 &0,0093 &1,0093 &0,7004 \\ \hline &10 &10 &1,000 &-1,978 &1,000 &0,7071 \\ \hline &11 &9 &0,9709 &0,0271 &0,9980 &0,6673 \\ \hline &12 &8 &0,8951 &0,1139 &1,0090 &0,5523 \\ \hline &13 &7 &0,7734 &0,2558 &1,0292 &0,3659 \\ \hline &14 &6 &0,6134 &0,4321 &1,0455 &0,1281 \\ \hline &15 &5 &0,4321 &0,6134 &1,0455 &0,1282 \\ \hline &16 &4 &0,2558 &0,7733 &1,0291 &0,3659 \\ \hline &17 &3 &0,1139 &0,8951 &1,0090 &0,5523 \\ \hline &18 &2 &0,0272 &0,9708 &0,9980 &0,6672 \\ \hline &19 &1 &0 &1,000 &1,000 &0,7071 \\ \hline &20 &2 &0,0094 &1,000 &1,0094 &0,7004 \\ \hline &21 &3 &0,0222 &1,000 &1,0222 &0,6914 \\ \hline &22 &4 &0,0278 &1,000 &1,0278 &0,6874 \\ \hline &23 &5 &0,0239 &1,000 &1,0239 &0,6902 \\ \hline &24 &6 &0,0136 &1,000 &1,0136 &0,6974 \\ \hline &25 &7 &0,0032 &1,000 &1,0032 &0,7048 \\ \hline &26 &8 &0,0004 &1,000 &1,0004 &0,7068 \\ \hline &27 &9 &0,0118 &1,000 &1,0118 &0,6987 \\ \hline &28 &10 &0,04 &1,000 &1,04 &0,6788 \\ \hline \end{array}$$

These y values have the graph:

idealized data

Which clearly has two break points. For the sake of argument we will calculate the R^2 correlation values (with the Excel cell formulas (European dot-comma style)):

=INDEX(LINEST(B1:$B$1;A1:$A$1;TRUE;TRUE);3;1)
=INDEX(LINEST(B1:$B$28;A1:$A$28;TRUE;TRUE);3;1)

for all possible non-overlapping combinations of two fitted lines. All the possible pairs of R^2 values have the graph:

R^2 values

The question is which pair of R^2 values should we choose, and how do we generalize to multiple break points as asked in the title? One choice is to pick the combination for which the sum of the R-square correlation is the highest. Plotting this we get the upper blue curve below:

sum of R squared and standard deviation of R squared

The blue curve, the sum of the R-squared values, is the highest in the middle. This is more clearly visible from the table with the value $1,0455$ as the highest value. However it is my opinion that the minimum of the red curve is more accurate. That is, the minimum of the standard deviation of the R^2 values of the fitted regression lines should be the best choice.

Piece wise linear regression - Matlab - multiple break points

$\endgroup$
1
$\begingroup$

There is a pretty nice algorithm described in Tomé and Miranda (1984).

The proposed methodology uses a least-squares approach to compute the best continuous set of straight lines that fit a given time series, subject to a number of constraints on the minimum distance between breakpoints and on the minimum trend change at each breakpoint.

The code and a GUI are available in both Fortran and IDL from their website: http://www.dfisica.ubi.pt/~artome/linearstep.html

$\endgroup$
0
$\begingroup$

... first of all you must to do it by iterations, and under some informative criterion, like AIC AICc BIC Cp; because you can get an "ideal" fit, if number of knots K = number od data points N, ok. ... first put K = 0; estimate L = K + 1 regressions, calculate AICc, for instance; then assume minimal number of data points at a separate segment, say L = 3 or L = 4, ok ... put K = 1; start from L-th data as the first knot, calculate SS or MLE, ... and step by step the next data point as a knot, SS or MLE, up to the last knot at the N - L data; choose the arrangement with the best fit (SS or MLE) calculate AICc ... ... put K = 2; ... use all previous regressions (that is their SS or MLE), but step by step divide a single segment into all possible parts ... choose the arrangement with the best fit (SS or MLE) calculate AICc ... if the last AICc occurs greater then the previous one: stop the iterations ! This is an optimal solution under AICc criterion, ok

$\endgroup$
1
  • $\begingroup$ AIC, BIC can't be used because they penalised for extra parameters, which is clearly not the case here. $\endgroup$ – SmallChess Nov 5 '15 at 8:12
0
$\begingroup$

I once came across a program called Joinpoint. On their website they say it fits a joinpoint model where "several different lines are connected together at the 'joinpoints'". And further: "The user supplies the minimum and maximum number of joinpoints. The program starts with the minimum number of joinpoint (e.g. 0 joinpoints, which is a straight line) and tests whether more joinpoints are statistically significant and must be added to the model (up to that maximum number)."

The NCI uses it for trend modelling of cancer rates, maybe it fits your needs as well.

$\endgroup$
0
$\begingroup$

In order to fit to data a piecewise function :

enter image description here

where $a_1 , a_2 , p_1 , q_1, p_2 , q_2 , p_3 , q_3$ are unknown parameters to be approximately computed, there is a very simple method (not iterative, no initial guess, easy to code in any math computer language). The theory given page 29 in paper : https://fr.scribd.com/document/380941024/Regression-par-morceaux-Piecewise-Regression-pdf and from page 30 :

enter image description here

For example, with the exact data provided by Mats Granvik the result is :

enter image description here

Without scattered data, this example is not very signifiant. Other examples with scattered data are shown in the referenced paper.

$\endgroup$
0
$\begingroup$

You can use the mcp package if you know the number of change points to infer. It gives you great modeling flexibility and a lot of information about the change points and regression parameters, but at the cost of speed.

The mcp website contains many applied examples, e.g.,

library(mcp)

# Define the model
model = list(
  response ~ 1,  # plateau (int_1)
  ~ 0 + time,    # joined slope (time_2) at cp_1
  ~ 1 + time     # disjoined slope (int_3, time_3) at cp_2
)

# Fit it. The `ex_demo` dataset is included in mcp
fit = mcp(model, data = ex_demo)

Then you can visualize:

plot(fit)

enter image description here

Or summarise:

summary(fit)

Family: gaussian(link = 'identity')
Iterations: 9000 from 3 chains.
Segments:
  1: response ~ 1
  2: response ~ 1 ~ 0 + time
  3: response ~ 1 ~ 1 + time

Population-level parameters:
    name match  sim  mean lower  upper Rhat n.eff
    cp_1    OK 30.0 30.27 23.19 38.760    1   384
    cp_2    OK 70.0 69.78 69.27 70.238    1  5792
   int_1    OK 10.0 10.26  8.82 11.768    1  1480
   int_3    OK  0.0  0.44 -2.49  3.428    1   810
 sigma_1    OK  4.0  4.01  3.43  4.591    1  3852
  time_2    OK  0.5  0.53  0.40  0.662    1   437
  time_3    OK -0.2 -0.22 -0.38 -0.035    1   834

Disclaimer: I am the developer of mcp.

$\endgroup$
1
  • $\begingroup$ The use of "detect" in the question indicates the number--and even the existence--of changepoints are not known beforehand. $\endgroup$ – whuber Jan 10 '20 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.