# R: How to fit a GLMM in nlme

I want to compare lme4 and nlme packages for my data. But I'm confused by how to use syntax in nlme. I'm working with Mixed-Effects Models in S and S-Plus (Pinheiro, Bates 2000) and the current Version of the documentation Package 'nlme' (04/07/2018)

I tried to use groupedData() as well as nlsList() and SSlogis(), to fit my model.

For lme4 I can fit my models wihtout any trouble. Can you tell me, how to fit these models from lme4 in nlme?

My data has easy structure:

$$y$$ is a binary outcome in $$\{0,1\}$$, $$x$$ is continuous, $$group$$ is categorical grouping factor ($$N$$ different groups)

model1 <- glmer(y ~     (1 | group),       data = data , family = binomial(link="logit"))
model2 <- glmer(y ~ 1 + (x | group),       data = data , family = binomial(link="logit"))
model3 <- glmer(y ~     (x | group) group, data = data , family = binomial(link="logit"))
model4 <- glmer(y ~ x + (1 | group),       data = data , family = binomial(link="logit"))

• Your model2 seems nonsensical to me - you need to have x listed both in the fixed effects and random effects portion of that model. – Isabella Ghement Mar 22 '19 at 16:10
• When I lined up the code to make the structure easier to see, I notice there's an orphan group in model3. Can you clarify what that's supposed to be. (Also a word of warning: this Q seems borderline off topic to me.) – gung - Reinstate Monica May 21 '19 at 17:13

I was looking for a way to do what you're trying to accomplish and came across a couple of things. First was this question that was asked elsewhere on the website with this answer. The second was this paper, which did exactly what you were asking about. What they did was perform a logit transformation on the response variable, which could then be fitted into the lme function.

Full disclosure:

1. According to the answer I linked above, the method I'm about to share is apparently not recommended (though the comment by Aaron on the answer indicates that this is a very common way of doing it?)

2. There are very different results with the outputs I'm about to provide

Okay, onwards and outwards...

1. Create some simulated data

RNGkind('Mersenne-Twister')
set.seed(42)

x1 <- rnorm(1000)
x2 <- rnorm(1000)
x3 <- factor(rep(c('a', 'b', 'c'), length.out = 1000))
#x3 won't cause a lot of random noise, it's just for illustrative purposes.

beta0 <- 0.6
beta1 <- 0.9
beta2 <- 0.7
z <- beta0 + beta1*x1 + beta2*x2
pr <- 1/(1+exp(-z))
y <- rbinom(1000, 1, pr)
table(y) #Just check since I'm kiasu


2. Logit-transform the response variable and prepare our data frame

#####This is the part that isn't recommended#####
library(car)
logit.y <- logit(y)

df <- data.frame(y, logit.y, x1, x2, x3)


3. Run some models

library(nlme)
library(lme4)
library(glmmTMB)

test.lme <- lme(logit.y ~ x1 + x2, random = ~1|x3, data = df, method = 'ML') #We set this to maximum-likelihood as the default is REML (restricted maximum likelihood)
test.glmer <- glmer(y ~ x1 + x2 + (1|x3), data = df, family = binomial)
test.glmmTMB <- glmmTMB(y ~ x1 + x2 + (1|x3), data = df, family = binomial)


4. Look at the summaries

summary(test.lme)$tTable Value Std.Error DF t-value p-value (Intercept) 0.8431364 0.1020533 995 8.261726 4.546389e-16 x1 1.2517987 0.1018175 995 12.294535 1.920042e-32 x2 0.9074567 0.1035171 995 8.766248 7.863229e-18 summary(test.glmer)$coefficients
Estimate Std. Error   z value     Pr(>|z|)
(Intercept) 0.5834385 0.07423036  7.859836 3.846378e-15
x1          0.9044080 0.08409916 10.754067 5.670490e-27
x2          0.6545513 0.07965376  8.217456 2.078662e-16

summary(test.glmmTMB)$$coefficients$$cond
Estimate Std. Error   z value     Pr(>|z|)
(Intercept) 0.5834385 0.07423036  7.859836 3.846380e-15
x1          0.9044079 0.08409911 10.754072 5.670193e-27
x2          0.6545512 0.07965371  8.217460 2.078583e-16


The results from test.glmer and test.glmmTMB are quite similar, but the test.lme results (i.e. the one where we logit-transformed the response variable) is just a little bit different.

Also note that the nlme output uses a t-test, while glmer and glmmTMB use a z-test. More on the z-test used to test individual parameters can be found here.

5. Compare back-transformed estimates

logit.lme <- summary(test.lme)$$tTable[,1] logit.glmer <- summary(test.glmer)$$coefficients[,1]
logit.glmmTMB <- summary(test.glmmTMB)$$coefficients$$cond[,1]

prob.lme <- exp(logit.lme)/(1+exp(logit.lme))
prob.glmer <- exp(logit.glmer)/(1+exp(logit.glmer))
prob.glmmTMB <- exp(logit.glmmTMB)/(1+exp(logit.glmmTMB))
prob.brms <- apply(as.data.frame(test.brms\$fit), 2, mean)[1:3] #Example model not provided
names(prob.brms) <- names(prob.glmmTMB)

round(t(cbind(original = c('(Intercept)' = beta0, x1 = beta1, x2 = beta2),
lme = prob.lme, glmer = prob.glmer, glmmTMB = prob.glmmTMB, brms = prob.brms)), 2)
(Intercept)   x1   x2
original        0.60 0.90 0.70
lme             0.70 0.78 0.71
glmer           0.64 0.71 0.66
glmmTMB         0.64 0.71 0.66
brms            0.60 0.91 0.66

#Not incredible all round except for the brms method (which uses
#Bayesian anyway!), but nearly there. The point is that the results
#from the glmer, glmmTMB, and brms methods are more similiar to each
#other than the result from the lme method.


6. Let's look at some visualizations.

#Make some predicted data
#x1 axis
predicted.df.x1 <- data.frame(x1 = seq(min(df$$x1), max(df$$x1), length.out = 1000),
x2 = mean(df$$x2)) predicted.df.x1 <- as.data.frame(do.call(rbind, replicate(3, predicted.df.x1, simplify = F))) predicted.df.x1$$x3 <- rep(c('a', 'b', 'c'), each = 1000)

predict.lme.x1 <- logit.lme[1] + predicted.df.x1$$x1*logit.lme[2] + predicted.df.x1$$x2*logit.lme[3]
#Recall that we logit transformed our data, so we need to 'manually' get our predicted variables

predict.lme.x1 <- exp(predict.lme.x1)/(1+exp(predict.lme.x1))
predict.glmer.x1 <- predict(test.glmer, predicted.df.x1, type = 'response')
predict.glmmTMB.x1 <- predict(test.glmmTMB, predicted.df.x1, type = 'response')

#x2 axis
predicted.df.x2 <- data.frame(x2 = seq(min(df$$x2), max(df$$x2), length.out = 1000),
x1 = mean(df$$x1)) predicted.df.x2 <- as.data.frame(do.call(rbind, replicate(3, predicted.df.x2, simplify = F))) predicted.df.x2$$x3 <- rep(c('a', 'b', 'c'), each = 1000)

predict.lme.x2 <- logit.lme[1] + predicted.df.x2$$x1*logit.lme[2] + predicted.df.x2$$x2*logit.lme[3]
predict.lme.x2 <- exp(predict.lme.x2)/(1+exp(predict.lme.x2))
predict.glmer.x2 <- predict(test.glmer, predicted.df.x2, type = 'response')
predict.glmmTMB.x2 <- predict(test.glmmTMB, predicted.df.x2, type = 'response')

#Plot it
par(mfrow = c(2,3))
plot(df$$x1, (exp(predict(test.lme))/(1+exp(predict(test.lme)))), pch = 16, cex = 0.4, col = 'red', xlab = 'x1', ylab = 'Predicted Prob.', main = 'Logit-transormed') lines(predicted.df.x1$$x1[1:1000], predict.lme.x1[1:1000], col = 'black', lwd = 2)

plot(df$$x1, predict(test.glmer, type = 'response'), pch = 16, cex = 0.4, col = 'blue', xlab = 'x1', ylab = 'Predicted Prob.', main = 'glmer Function') lines(predicted.df.x1$$x1[1:1000], predict.glmer.x1[1:1000], col = 'black', lwd = 2)

plot(df$$x1, predict(test.glmmTMB, type = 'response'), pch = 16, cex = 0.4, col = 'darkgreen', xlab = 'x1', ylab = 'Predicted Prob.', main = 'glmerTMB function') lines(predicted.df.x1$$x1[1:1000], predict.glmmTMB.x1[1:1000], col = 'black', lwd = 2)

plot(df$$x2, (exp(predict(test.lme))/(1+exp(predict(test.lme)))), pch = 16, cex = 0.4, col = 'red', xlab = 'x2', ylab = 'Predicted Prob.', main = 'Logit-transormed') lines(predicted.df.x2$$x2[1:1000], predict.lme.x2[1:1000], col = 'black', lwd = 2)

plot(df$$x2, predict(test.glmer, type = 'response'), pch = 16, cex = 0.4, col = 'blue', xlab = 'x2', ylab = 'Predicted Prob.', main = 'glmer Function') lines(predicted.df.x2$$x2[1:1000], predict.glmer.x2[1:1000], col = 'black', lwd = 2)

plot(df$$x2, predict(test.glmmTMB, type = 'response'), pch = 16, cex = 0.4, col = 'darkgreen', xlab = 'x2', ylab = 'Predicted Prob.', main = 'glmerTMB function') lines(predicted.df.x2$$x2[1:1000], predict.glmmTMB.x2[1:1000], col = 'black', lwd = 2)


So the result we get from the logit-transformed response variable with the linear nlme is slightly different from outputs provided using the glmer and glmmTMB when specifying the binomial family. It also doesn't take too much to see that there is a difference between the methods when looking at the visuals.

Now the question is why is it different. I'm sure there is an answer, I just don't know what it is.

Some other things:

1. I'm not entirely sure if my syntax/calculations are correct in getting the logit-transformed response variable, so if someone is able to check and verify if I did it correctly, then maybe we'd be getting warmer.
2. I'm not going to profess to being an expert, I just happened to (vaguely) know how to code it, albeit with the help of the answer I highlighted earlier in this post. Would someone be able to shed some light on the differences?
3. The only reason I'd imagine anyone would want to use the lme method is to fit an autoregressive correlation structure of some kind to account for temporal/spatial autocorrelation, much like was done in the paper I mentioned earlier. Ben Bolker does provide a succinct brain dump (unlike this answer) on which packages can be used for fitting mixed models with temporally autocorrelative data, though I haven't tried any of them myself (I'm afraid I'm too intimidated to even try, lets I break R again).
• @IsabellaGhement I haven't got enough reputation to make a comment on your post, but would you be able to look through this answer and share your thoughts? I've seen your answers to a lot of questions on your website and I think you're better prepared to provide comment. – Lalochezia May 21 '19 at 17:51
• Maybe @BenBolker could share his insights, too, as he's provided a lot of really good insight into questions pertaining to GLMMs in general. Some of the links I provided were to his GLMM FAQ page anyway. – Lalochezia May 21 '19 at 17:53
• thank your for that detailed answer! I'll certainly make use of that! – Toby May 23 '19 at 10:53
• @Lalochezia: Somehow I missed that you tagged me to your nice answer. Better to join the party late than never, as they say. Applying the logit-transform to your response variable and then analyzing it with lme() means that you are modelling the expected value of the logit transformed response variable given the predictors and random effects. On the other side, with glmer() or other alternatives, you are modelling the logit transformed expected value of the response variable given the predictors and random effects. These two are entirely different quantities. – Isabella Ghement Dec 8 '20 at 16:27
• The logit transformation is a non-linear transformation, so the expected value of the logit transformed response variable given all else in the model is NOT the same as the logit transformed expected value of the response. What we really want to model is the latter, because we can use the inverse logit transformation to our advantage to learn more about the expected value of the response variable as a function of predictors and random effects. With the former, interpretability is a big concern. – Isabella Ghement Dec 8 '20 at 16:31

I don't think nlme can be used to fit a mixed effects logistic regression model. However, you have plenty of other options available for this task via packages such as the ones listed below, whose use is illustrated for your model 4.

install.packages("GLMMadaptive")

model4 <- mixed_model(fixed = y ~ x, random = ~ 1 | group,
data = data,


glmmTMB

install.packages("glmmTMB")

library(glmmTMB)

model4 <- glmmTMB(y ~ x + (1 | group),
data = data,


MASS

install.packages("MASS")

library(MASS)

model4 <- glmmPQL(fixed =  y ~ x,
random = ~ 1 | group,
data = data,


brms

install.packages("brms")

library(brms)

model4 <- brm(y ~ x + (1|group),
data = data,