问题
I'm looking to predict 'terms', especially ns splines, from an lmer model. I've replicated the problem with the mtcars dataset (technically poor example, but works to get the point across).
Here is what I'm trying to do with a linear model:
data(mtcars)
mtcarsmodel <- lm(wt ~ ns(drat,2) + hp + as.factor(gear), data= mtcars)
summary(mtcarsmodel)
coef(mtcarsmodel)
test <- predict(mtcarsmodel, type = "terms")
Perfect. However, there is no equivalent 'terms' option for lmer predict (unresolved issue here).
mtcarsmodellmer <- lmer(wt ~ ns(drat,2) + (hp|as.factor(gear)), data= mtcars)
summary(mtcarsmodellmer)
coef(mtcarsmodellmer)
ranef(mtcarsmodellmer)
Given there is no equivalent ‘predict, terms’ function, I was going to extract the fixed and random coefficients above and apply the coefficients to the mtcars data, but have no idea on how to extract an ns spline object from a model and 'predict' it to some new data. The same goes for a 'poly' transformed variable eg. poly(drat, 2) - extra kudos if you can get this as well.
回答1:
It is not difficult to do it yourself.
library(lme4)
library(splines)
X <- with(mtcars, ns(drat, 2)) ## design matrix for splines (without intercept)
## head(X)
# 1 2
#[1,] 0.5778474 -0.1560021
#[2,] 0.5778474 -0.1560021
#[3,] 0.5738625 -0.1792162
#[4,] 0.2334329 -0.1440232
#[5,] 0.2808520 -0.1704002
#[6,] 0.0000000 0.0000000
## str(X)
# ns [1:32, 1:2] 0.578 0.578 0.574 0.233 0.281 ...
# - attr(*, "dimnames")=List of 2
# ..$ : NULL
# ..$ : chr [1:2] "1" "2"
# - attr(*, "degree")= int 3
# - attr(*, "knots")= Named num 3.7
# ..- attr(*, "names")= chr "50%"
# - attr(*, "Boundary.knots")= num [1:2] 2.76 4.93
# - attr(*, "intercept")= logi FALSE
# - attr(*, "class")= chr [1:3] "ns" "basis" "matrix"
fit <- lmer(wt ~ X + (hp|gear), data= mtcars)
beta <- coef(fit)
#$gear
# hp (Intercept) X1 X2
#3 0.010614406 2.455403 -2.167337 -0.9246454
#4 0.014601363 2.455403 -2.167337 -0.9246454
#5 0.006342761 2.455403 -2.167337 -0.9246454
#
#attr(,"class")
#[1] "coef.mer"
If we want to predict the ns term, just do
## use `predict.ns`; read `?predict.ns`
x0 <- seq(1, 5, by = 0.2) ## example `newx`
Xp <- predict(X, newx = x0) ## prediction matrix
b <- with(beta$gear, c(X1[1], X2[1])) ## coefficients for spline
y <- Xp %*% b ## predicted mean
plot(x0, y, type = "l")
来源：https://stackoverflow.com/questions/41021658/r-extract-ns-spline-object-from-lmer-model-and-predict-on-new-data