Model-based partitioning with "two-layer interaction" (segmented models) - r

Question

I am trying to build a model-based tree with a type of "two-layer interaction" where the models in the nodes of the tree are segmented again.
I am using the mob() function to this aim but I could not manage to make the argument for the fit function work with the lmtree() function.
In the following example a is function of b and the relationship between a and b depends on d and on b | d.
library("partykit")
set.seed(321)
b <- runif(200)
d <- sample(1:2, 200, replace = TRUE)
a <- jitter(ifelse(d == 1, 2 * b - 1, 4 * b - 1.2), amount = .1)
a[b < .5 & d == 1] <- jitter(rep(0, length(a[b < .5 & d == 1])))
a[b < .3 & d == 2] <- jitter(rep(0, length(a[b < .3 & d == 2])))
fit <- function(y, x, start = NULL, weights = NULL, offset = NULL, ..., estfun = FALSE, object = FALSE)
{
x <- x[, 2]
l <- lmtree(y ~ x | b)
return(l)
}
m <- mob(a ~ b | d, fit = fit) # not working
Of course with this simple example I could use lmtree(a ~ b | d + b) to find every interaction but is there a way to use as fit function of mob() a lmtree()?

Answer 1

No but yes ;-)
No, lmtree() cannot be used easily as a fitter for a mob().
The dimension of the inner tree (lmtree()) is not fixed, i.e., you may get a tree without any partition or with many subgroups, and this would be confusing for the outer tree (mob()).
Even if one worked around the dimension issue or fixed it by always forcing one break, one would need more work to set up the right coefficient vector, matrix of estimating functions, etc. This is also not straightforward because the convergence rate (and hence the inference) is different if breakpoints are given (e.g., for a binary factor) or have to be estimated (such as for your numeric variables b).
The way you set up your fit() function, the inner lmtree() does not know where to find b. All it has is a numeric vector y and a numeric matrix x but not the original data.
But yes, I think that all of these issues can be addressed if changing the view from fitting a "two-layer" tree to fitting a "segmented" model inside a tree. My impression is that you want to fit a model y ~ x (or a ~ b in your example) where a piecewise linear function is used with an additional breakpoint in x. If the piecewise linear function is supposed to be continuous in x, then the segmented package can be easily used. If not, then strucchange could be leveraged. Assuming you want the former (as you have simulated your data like this), I include a worked segmented example below (and also slightly modified your question to reflect this).
Changing the names and code a little bit, your data d has a segmented piecewise linear relationship of y ~ x with coefficients depending on a group variable g.
set.seed(321)
d <- data.frame(
x = runif(200),
g = factor(sample(1:2, 200, replace = TRUE))
)
d$y <- jitter(ifelse(d$g == "1",
pmax(0, 2 * d$x - 1),
pmax(0, 4 * d$x - 1.2)
), amount = 0.1)
Within every node of a tree I can then fit a model segmented(lm(y ~ x)) which comes with suitable extractors for coef(), logLik(), estfun() etc. Thus, the mobster function is simply:
segfit <- function(y, x, start = NULL, weights = NULL, offset = NULL, ...)
{
x <- as.numeric(x[, 2])
segmented::segmented(lm(y ~ x))
}
(Note: I haven't tried whether segmented() would also support lm() objects with weights and offset.)
With this we can obtain the full tree which simply splits in g in this basic example:
library("partykit")
segtree <- mob(y ~ x | g, data = d, fit = segfit)
plot(segtree, terminal_panel = node_bivplot, tnex = 2)
A hands-on introduction to segmented is available in: Muggeo VMR (2008). "segmented: An R Package to Fit Regression Models with Broken-Line Relationships." R News, 8(1), 20-25. https://CRAN.R-project.org/doc/Rnews/
For the underlying methodological background see: Muggeo VMR (2003). "Estimating Regression Models with Unknown Break-Points." Statistics in Medicine, 22(19), 3055-3071. doi:10.1002/sim.1545