Stata allows for fixed effects and random effects specification of the logistic regression through the xtlogit fe and xtlogit re commands accordingly. I was wondering what are the equivalent commands for these specifications in R.
The only similar specification I am aware of is the mixed effects logistic regression
mymixedlogit <- glmer(y ~ x1 + x2 + x3 + (1 | x4), data = d, family = binomial)
but I am not sure whether this maps to any of the aforementioned commands.
The glmer command is used to quickly fit logistic regression models with varying intercepts and varying slopes (or, equivalently, a mixed model with fixed and random effects).
To fit a varying intercept multilevel logistic regression model in R (that is, a random effects logistic regression model), you can run the following using the in-built "mtcars" data set:
data(mtcars)
head(mtcars)
m <- glmer(mtcars$am ~ 1 + mtcars$wt + (1|mtcars$gear), family="binomial")
summary(m)
# and you can examine the fixed and random effects
fixef(m); ranef(m)
To fit a varying-intercept slope model in Stata, you of course use the xtlogit command (using the similar but not identical in-built "auto" data set in Stata):
sysuse auto
xtset gear_ratio
xtlogit foreign weight, re
I'll add that I find the entire reference to "fixed" versus "random" effects ambiguous, and I prefer to refer to the structure of the model itself (e.g., are the intercepts varying? which slopes are varying, if any? is the model nested in 2 levels or more? are the levels cross-classified or not?). For a similar view, see Andrew Gelman's thoughts on "fixed" versus "random" effects.
Update: Ben Bolker's excellent comment below points out that in R it's more informative when using predict commands to use the data=mtcars option instead of, say, the dollar notation:
data(mtcars)
m1 <- glmer(mtcars$am ~ 1 + mtcars$wt + (1|mtcars$gear), family="binomial")
m2 <- glmer(am ~ 1 + wt + (1|gear), family="binomial", data=mtcars)
p1 <- predict(m1); p2 <- predict(m2)
names(p1) # not that informative...
names(p2) # very informative!
Related
I am wondering how to easily find the variation within the fixed effects used in a feols regression. In the below code, the fixed effects are type and CO2. How do I obtain the variance within each of the fixed effects (type and CO2)?
library(fixest)
library(car)
library(pander)
##Using the built-in CO2 data frame, run regression
i<- feols(conc ~ uptake + Treatment | Type, CO2, vcov = "hetero")
summary(i)
I am trying to fit a nonlinear mixed effects model with a categorical variable genotype that has around 200 levels.
So this is the linear version of the model.
mlinear <- lmer(WUE ~ moisture * genotype + (1|pot), data = d8)
Now I'm trying to make the same model, but with a logistic function instead of linear
mlogistic <- nlme(WUE ~ SSlogis(moisture, Asym, xmid, scal), data = d8, fixed = Asym + xmid + scal ~ 1, random = Asym + xmid + scal~1|pot)
Problem is, now I don't know how to incorporate genotype into this nonlinear model. Asym, xmid, and scal parameters should be able to vary between each genotype. Anyone know how to do this?
It looks like you’re using lme4::lmer for your linear model and nlme::nlme for the logistic? If you use lme4 for both, you should be able to keep the same model specification, and use lme4::glmer with family = binomial for the logistic model. The whole idea behind a GLM with a link function is that you shouldn’t have to do anything different with your predictor vs a linear model, as the link function takes care of that.
library(lme4)
mlinear <- lmer(WUE ~ moisture * genotype + (1|pot), data = d8)
mlogistic <- glmer(WUE ~ moisture * genotype + (1|pot), family = binomial, data = d8)
All that being said, how is WUE measured? You probably want to use either a logistic model (if binary) or linear (if continuous), not both.
Just to add to the answer by #zephryl, who has explained how you can do this, I would like to focus on:
Since my WUE responses are almost alway curved, I'm trying to fit a logistic function instead of a linear one.
Fitting a logistic model does not really make sense here. The distribution of WUE is not relevant. It is the conditional distribution that matters, and we typically assess this by inspecting the residuals. If the residuals show a nonlinear pattern then there are several ways to model this including
transformations
nonlinear terms
splines
I want to use an ordinal logistic regression (my response variable is ordinal) that works with 2 random variables and for quantitative predictor variable with interaction (my formula is: ordinal_variable~ quantitative_variable:habitat + (1|community) + (1|species).
I was analyzing my data with clmm (see the script below) and got the results I expected, however I noticed that clmm was designed to be used when response and predictor variables are factorial.
I then tried the polmer (see the script below), but I did not get any answers.
Would someone have any suggestions on how to analyze this data?
library(ordinal)
model1 <- clmm(as.factor(ordinal_variable)~
quantitative_variable:habitat + (1|community) + (1|species),
data=baseline)
summary(model1)
library(MPDiR)
library(lme4)
model2 <- polmer(as.factor(ordinal_variable) ~
quantitative_variable:habitat + (community - 1 | Obs) +
(species - 1 | Obs), data=baseline)
summary(model2)
I run a Cox Regression with two categorical variables (x1 and x2) and their interaction. I need to know the significance of the overall effect of x1, x2 and of the interaction.
The overall effect of the interaction:
I know how do find out the overall effect of the interaction using anova():
library(survival)
fit_x1_x2 <- coxph(Surv(time, death) ~ x1 + x2 , data= df)
fit_full <- coxph(Surv(time, death) ~ x1 + x2 + x1:x2, data= df)
anova(fit_x1_x2, fit_full)
But how are we supposed to use anova() to find out the overall effect of x1 or x2? What I tried is this:
The overall effect of x1
fit_x2_ia <- coxph(Surv(time, death) ~ x2 + x1:x2, data= df)
fit_full <- coxph(Surv(time, death) ~ x1 + x2 + x1:x2, data= df)
anova(fit_x2_ia, fit_full)
The overall effect of x2
fit_x1_ia <- coxph(Surv(time, death) ~ x1 + x1:x2, data= df)
fit_full <- coxph(Surv(time, death) ~ x1 + x2 + x1:x2, data= df)
anova(fit_x1_ia, fit_full)
I am not sure whether this is how we are supposed to use anova(). The fact that the output shows degree of freedom is zero makes me sceptical. I am even more puzzled that both times, for the overall effect of x1 and x2, the test is significant, although the log likelihood values of the models are the same and the Chi value is zero.
Here is the data I used
set.seed(1) # make it reproducible
df <- data.frame(x1= rnorm(1000), x2= rnorm(1000)) # generate data
df$death <- rbinom(1000,1, 1/(1+exp(-(1 + 2 * df$x1 + 3 * df$x2 + df$x1 * df$x2)))) # dead or not
library(tidyverse) # for cut_number() function
df$x1 <- cut_number(df$x1, 4); df$x2 <- cut_number(df$x2, 4) # make predictors to groups
df$time <- rnorm(1000); df$time[df$time<0] <- -df$time[df$time<0] # add survival times
The two models you have constructed for "overall effect" do really not appear to satisfy the statistical property of being hierarchical, i.e properly nested. Specifically, if you look at the actual models that get constructed with that code you should see that they are actually the same model with different labels for the two-way crossed effects. In both cases you have 15 estimated coefficients (hence zero degrees of freedom difference) and you will not that the x1 parameter in the full model has the same coefficient as the x2[-3.2532,-0.6843):x1[-0.6973,-0.0347) parameter in the "reduced" model looking for an x1-effect, namely 0.19729. The crossing operator is basically filling in all the missing cells for the main effects with interaction results.
There really is little value in looking at interaction models without all of the main effects if you want to stay within the bounds of generally accepted statistical practice.
If you type:
fit_full
... you should get a summary of the model that has p-values for x1 levels, x2 levels,and the interaction levels. Because you chose to categorize these by four arbitrary cutpoints each you will end up with a total of 15 parameter estimates. If instead you made no cuts and modeled the linear effects and the linear-by-linear interaction, you could get three p-values directly. I'm guessing there was suspicion that the effects were not linear and if so I thought a cubic spline model might be more parsimonious and distort the biological reality less than discretization into 4 disjoint levels. If you thought the effects might be non-linear but ordinal, there is an ordinal version of factor classed variables, but the results are generally confusion to the uninitiated.
The answer from 42- is informative but after reading it I still did not know how to determine the three p values or if this is possible at all. Thus I talked to the professor of biostatistic of my university. His answer was quite simple and I share it in case others have similar questions.
In this design it is not possible to determine the three p values for overall effect of x1, x2 and their interaction. If we want to know the p values of the three overall effects, we need to keep the continuous variables as they are. But breaking up the variables into groups answers a different question, hence we can not test the hypothesis of the overall effects no matter which statisstical model we use.
I am currently testing whether I should include certain random effects in my lmer model or not. I use the anova function for that. My procedure so far is to fit the model with a function call to lmer() with REML=TRUE (the default option). Then I call anova() on the two models where one of them does include the random effect to be tested for and the other one doees not. However, it is well known that the anova() function refits the model with ML but in the new version of anova() you can prevent anova() from doing so by setting the option refit=FALSE. In order to test for random effects should I set refit=FALSE in my call to anova() or not? (If I do set refit=FALSE the p-values tend to be lower. Are the p-values anti-conservative when I set refit=FALSE?)
Method 1:
mod0_reml <- lmer(x ~ y + z + (1 | w), data=dat)
mod1_reml <- lmer(x ~ y + z + (y | w), data=dat)
anova(mod0_reml, mod1_reml)
This will result in anova() refitting the models with ML instead of REML. (Newer versions of the anova() function will also output an info about this.)
Method 2:
mod0_reml <- lmer(x ~ y + z + (1 | w), data=dat)
mod1_reml <- lmer(x ~ y + z + (y | w), data=dat)
anova(mod0_reml, mod1_reml, refit=FALSE)
This will result in anova() performing its calculations on the original models, i.e. with REML=TRUE.
Which of the two methods is correct in order to test whether I should include a random effect or not?
Thanks for any help
In general I would say that it would be appropriate to use refit=FALSE in this case, but let's go ahead and try a simulation experiment.
First fit a model without a random slope to the sleepstudy data set, then simulate data from this model:
library(lme4)
mod0 <- lmer(Reaction ~ Days + (1|Subject), data=sleepstudy)
## also fit the full model for later use
mod1 <- lmer(Reaction ~ Days + (Days|Subject), data=sleepstudy)
set.seed(101)
simdat <- simulate(mod0,1000)
Now refit the null data with the full and the reduced model, and save the distribution of p-values generated by anova() with and without refit=FALSE. This is essentially a parametric bootstrap test of the null hypothesis; we want to see if it has the appropriate characteristics (i.e., uniform distribution of p-values).
sumfun <- function(x) {
m0 <- refit(mod0,x)
m1 <- refit(mod1,x)
a_refit <- suppressMessages(anova(m0,m1)["m1","Pr(>Chisq)"])
a_no_refit <- anova(m0,m1,refit=FALSE)["m1","Pr(>Chisq)"]
c(refit=a_refit,no_refit=a_no_refit)
}
I like plyr::laply for its convenience, although you could just as easily use a for loop or one of the other *apply approaches.
library(plyr)
pdist <- laply(simdat,sumfun,.progress="text")
library(ggplot2); theme_set(theme_bw())
library(reshape2)
ggplot(melt(pdist),aes(x=value,fill=Var2))+
geom_histogram(aes(y=..density..),
alpha=0.5,position="identity",binwidth=0.02)+
geom_hline(yintercept=1,lty=2)
ggsave("nullhist.png",height=4,width=5)
Type I error rate for alpha=0.05:
colMeans(pdist<0.05)
## refit no_refit
## 0.021 0.026
You can see that in this case the two procedures give practically the same answer and both procedures are strongly conservative, for well-known reasons having to do with the fact that the null value of the hypothesis test is on the boundary of its feasible space. For the specific case of testing a single simple random effect, halving the p-value gives an appropriate answer (see Pinheiro and Bates 2000 and others); this actually appears to give reasonable answers here, although it is not really justified because here we are dropping two random-effects parameters (the random effect of slope and the correlation between the slope and intercept random effects):
colMeans(pdist/2<0.05)
## refit no_refit
## 0.051 0.055
Other points:
You might be able to do a similar exercise with the PBmodcomp function from the pbkrtest package.
The RLRsim package is designed precisely for fast randomization (parameteric bootstrap) tests of null hypotheses about random effects terms, but doesn't appear to work in this slightly more complex situation
see the relevant GLMM faq section for similar information, including arguments for why you might not want to test the significance of random effects at all ...
for extra credit you could redo the parametric bootstrap runs using the deviance (-2 log likelihood) differences rather than the p-values as output and check whether the results conformed to a mixture between a chi^2_0 (point mass at 0) and a chi^2_n distribution (where n is probably 2, but I wouldn't be sure for this geometry)