R中的分段线性回归（segmented.lm）

Question

I appreciate any help to make segmented.lm (or any other function) find the obvious breakpoints in this example:

data = list(x=c(50,60,70,80,90) , y= c(703.786,705.857,708.153,711.056,709.257))
plot(data, type='b')
require(segmented)
model.lm = segmented(lm(y~x,data = data),seg.Z = ~x, psi = NA)

It returns with the following error:

Error in solve.default(crossprod(x1), crossprod(x1, y1)) : system is computationally singular: reciprocal condition number = 1.51417e-20

If I change K:

model.lm = segmented(lm(y~x,data = data),seg.Z = ~x, psi = NA, control = seg.control(K=1))

I get another error:

Error in segmented.lm(lm(y ~ x, data = data), seg.Z = ~x, psi = NA, control = seg.control(K = 1)) : only 1 datum in an interval: breakpoint(s) at the boundary or too close each other

Answer 1

A nice objective method to determine the break point is described in Crawley (2007: 427).

First, define a vector breaks for a range of potential break points:

breaks <- data$x[data$x >= 70 & data$x <= 90]

Then run a for loop for piecewise regressions for all potential break points and yank out the minimal residual standard error ( mse ) for each model from the summary output:

mse <- numeric(length(breaks))
for(i in 1:length(breaks)){
  piecewise <- lm(data$y ~ data$y*(data$x < breaks[i]) + data$y*(data$x >= breaks[i]))
  mse[i] <- summary(piecewise)[6]
}
mse <- as.numeric(mse)

Finally, identify the break point with the least mse :

breaks[which(mse==min(mse))] 

Hope this helps.