# Piecewise linear with constraints with the segmented function on R

I want to make a piecewise linear regression in R. I have a large dataset with 3 segments where I want the first and third segment to be without slope, i.e. parallel to x-axis and I also want the regression to be continuous. I have a small example to make piecewise regression with 2 breakpoints and slope1 =0

y <- c(4.5,4.3,2.57,4.40,4.52,1.39,4.15,3.55,2.49,4.27,4.42,4.10,2.21,2.9,1.42,1.5,1.45,1.7,4.6,3.8,1.9)
x<- c(320,419,650,340,400,800,300,570,720,480,425,460,675,600,850,920,975,1022,450,520,780)
plot(x, y, col="black",pch=16)

# slope = 0 before the first breakpoint
a<-lm(y~x)
g<- segmented(a, seg.Z = ~ x, psi=c(400,800))
g
fit.glm<-update(a,.~. -x)
fit.seg1 <- segmented.lm(fit.glm, seg.Z = ~x, psi=c(441.8,817.4))
points(x,fitted(fit.seg1),col=2)


or slope3 = 0

# slope = 0 after the second breakpoint
o<-lm(y~1)
xx<- -x
o2<-segmented(o,seg.Z=~xx,psi=c(-817.4,-441.8))
slope(o2)
points(x,fitted(o2),col=3)


but I can not find how to do slope1=0 and slope 3=0. Do you know if it's possible?

• Do you know the $x$ values at which the breaks occur, or should these be estimated? Jul 4 '19 at 8:43
• If this case, the breaks are estimated with this line : g<- segmented(a, seg.Z = ~ x, psi=c(400,800)) So in this case, breaks occur for x = 441.8 and x= 8.17.4 Jul 4 '19 at 8:54

If you know the $$x$$ values at which the breaks occur (or are happy with estimating them separately), then let's call them $$x_1=441.8$$ and $$x_2=817.4$$. In this case, you want to find parameters $$a$$ and $$b$$ such that the function

$$f(x) = \begin{cases} ax_1+b, & x

approximates your observed $$y$$ well. Note how this is constant outside the interval $$[x_1,x_2]$$, linear within the interval, and continuous.

I assume that you want a least squares solution.

However, this is nothing else than a straightforward linear regression of $$y$$ on a modified predictor

$$x' := \begin{cases} x_1, & x

In R, we can calculate $$x'$$ using the pmin() and pmax() functions. Here is how it works:

predictor <- pmin(817.4,pmax(441.8,x))
model <- lm(y~predictor)

predictor_fit <- seq(min(x),max(x),by=.01)
plot(x, y, col="black",pch=16)
lines(predictor_fit,predict(model,newdata=data.frame(predictor=pmin(817.4,pmax(441.8,predictor_fit)))))


• It works! Thanks! Jul 4 '19 at 9:54

The R package mcp was made exactly for these informed by-segment scenarios. Putting your data in a data.frame, you can do this:

model = list(
y ~ 1,  # Intercept (plateau)
~ 0 + x,  # Joined slope
~ 0  # Joined plateau
)

library(mcp)
fit = mcp(model, data = df)
plot(fit)


The distributions on the x-axis are the change point posteriors, and they are quite well defined for this model and dataset. Use summary(fit) and plot_pars(fit) to see individual parameter estimates.

A very easy method (no guessed initial value, no iterative calculus) is given in the paper : https://fr.scribd.com/document/380941024/Regression-par-morceaux-Piecewise-Regression-pdf

The case of piecewise linear function, first and third segments horizontal, is treated pp.17-18.

With the given data the result is :

PIECEWISE REGRESSION for three not horizontal segments :

This case is treated pp.29-30 in the paper referenced above. The result is :

NOTE :

The first part of the procedure is the calculus of $$a_1$$ and $$a_2$$ which determines the limits of the segments. Since the document referenced above is in French a translation of this part is added below.

Once having $$a_1$$ and $$a_2$$ a global linear regression for the parameters of the segments is classical.

FOR INFORMATION :

Since the piecewise function is a non-linear function (even made of linear segments) , the linearization of the global regression is based on an integral equation : $$y(x)=C_1\left(6\int x\,y\,dx-2x\int y\,dx-x^2y \right)+C_2\left(xy-2\int y\,dx \right)+C_3\:x+C_4$$ $$C_1=\frac{1}{a_1a_2}\quad;\quad C_2=\frac{a_1+a_2}{a_1a_2}$$ For more explanation see the referenced document.