What is the equivalent command in R for stphtest in Stata?
If there is not a 1-to-1 command, how do you test the assumption of proportional hazards in a Cox Proportional-Hazards Regression model?
Just to expand Glen's answer:
I know how different a mindset R poses for Stata users.
First run the coxph model:
model.coxph <- coxph(Surv(t1, t2, event) ~ var1 + var2, data = data)
> summary(model.coxph0)
coef exp(coef) se(coef) z Pr(>|z|)
var1 1.665e-01 1.181e+00 1.605e-01 1.038 0.29948
var2 -1.358e-03 9.986e-01 6.852e-04 -1.982 0.04746 *
Than run cox.zph on the model:
(viol.cox <- cox.zph(model.coxph0))
rho chisq p
var1 0.0486 1.095 0.295
var2 -0.0250 0.258 0.611
GLOBAL NA 1.462 0.691
Each p value ought to be > 0.05 so that the assumptions are not violated.
You can also plot the scaled schoenfeld residuals with a simple plot on the cox.zph:
> plot(viol.cox)
See ?survival:::cox.zph:
Description
Test the proportional hazards assumption for a Cox regression model fit (coxph).
Related
I have data from a longitudinal study and calculated the regression using the lme4::lmer function. After that I calculated the contrasts for these data but I am having difficulty interpreting my results, as they were unexpected. I think I might have made a mistake in the code. Unfortunately I couldn't replicate my results with an example, but I will post both the failed example and my actual results below.
My results:
library(lme4)
library(lmerTest)
library(emmeans)
#regression
regmemory <- lmer(memory ~ as.factor(QuartileConsumption)*Age+
(1 + Age | ID) + sex + education +
HealthScore, CognitionData)
#results
summary(regmemory)
#Fixed effects:
# Estimate Std. Error df t value Pr(>|t|)
#(Intercept) -7.981e-01 9.803e-02 1.785e+04 -8.142 4.15e-16 ***
#as.factor(QuartileConsumption)2 -8.723e-02 1.045e-01 2.217e+04 -0.835 0.40376
#as.factor(QuartileConsumption)3 5.069e-03 1.036e-01 2.226e+04 0.049 0.96097
#as.factor(QuartileConsumption)4 -2.431e-02 1.030e-01 2.213e+04 -0.236 0.81337
#Age -1.709e-02 1.343e-03 1.989e+04 -12.721 < 2e-16 ***
#sex 3.247e-01 1.520e-02 1.023e+04 21.355 < 2e-16 ***
#education 2.979e-01 1.093e-02 1.061e+04 27.266 < 2e-16 ***
#HealthScore -1.098e-06 5.687e-07 1.021e+04 -1.931 0.05352 .
#as.factor(QuartileConsumption)2:Age 1.101e-03 1.842e-03 1.951e+04 0.598 0.55006
#as.factor(QuartileConsumption)3:Age 4.113e-05 1.845e-03 1.935e+04 0.022 0.98221
#as.factor(QuartileConsumption)4:Age 1.519e-03 1.851e-03 1.989e+04 0.821 0.41174
#contrasts
emmeans(regmemory, poly ~ QuartileConsumption * Age)$contrast
#$contrasts
# contrast estimate SE df z.ratio p.value
# linear 0.2165 0.0660 Inf 3.280 0.0010
# quadratic 0.0791 0.0289 Inf 2.733 0.0063
# cubic -0.0364 0.0642 Inf -0.567 0.5709
The interaction terms in the regression results are not significant, but the linear contrast is. Shouldn't the p-value for the contrast be non-significant?
Below is the code I wrote to try to recreate these results, but failed:
library(dplyr)
library(lme4)
library(lmerTest)
library(emmeans)
data("sleepstudy")
#create quartile column
sleepstudy$Quartile <- sample(1:4, size = nrow(sleepstudy), replace = T)
#regression
model1 <- lmer(Reaction ~ Days * as.factor(Quartile) + (1 + Days | Subject), data = sleepstudy)
#results
summary(model1)
#Fixed effects:
# Estimate Std. Error df t value Pr(>|t|)
#(Intercept) 258.1519 9.6513 54.5194 26.748 < 2e-16 ***
#Days 9.8606 2.0019 43.8516 4.926 1.24e-05 ***
#as.factor(Quartile)2 -11.5897 11.3420 154.1400 -1.022 0.308
#as.factor(Quartile)3 -5.0381 11.2064 155.3822 -0.450 0.654
#as.factor(Quartile)4 -10.7821 10.8798 154.0820 -0.991 0.323
#Days:as.factor(Quartile)2 0.5676 2.1010 152.1491 0.270 0.787
#Days:as.factor(Quartile)3 0.2833 2.0660 155.5669 0.137 0.891
#Days:as.factor(Quartile)4 1.8639 2.1293 153.1315 0.875 0.383
#contrast
emmeans(model1, poly ~ Quartile*Days)$contrast
#contrast estimate SE df t.ratio p.value
# linear -1.91 18.78 149 -0.102 0.9191
# quadratic 10.40 8.48 152 1.227 0.2215
# cubic -18.21 18.94 150 -0.961 0.3379
In this example, the p-value for the linear contrast is non-significant just as the interactions from the regression. Did I do something wrong, or these results are to be expected?
Look at the emmeans() call for the original model:
emmeans(regmemory, poly ~ QuartileConsumption * Age)
This requests that we obtain marginal means for combinations of QuartileConsumption and Age, and obtain polynomial contrasts from those results. It appears that Age is a quantitative variable, so in computing the marginal means, we just use the mean value of Age (see documentation for ref_grid() and vignette("basics", "emmeans")). So the marginal means display, which wasn't shown in the OP, will be in this general form:
QuartileConsumption Age emmean
------------------------------------
1 <mean> <est1>
2 <mean> <est2>
3 <mean> <est3>
4 <mean> <est4>
... and the contrasts shown will be the linear, quadratic, and cubic trends of those four estimates, in the order shown.
Note that these marginal means have nothing to do with the interaction effect; they are just predictions from the model for the four levels of QuartileConsumption at the mean Age (and mean education, mean health score), averaged over the two sexes, if I understand the data structure correctly. So essentially the polynomial contrasts estimate polynomial trends of the 4-level factor at the mean age. And note in particular that age is held constant, so we certainly are not looking at any effects of Age.
I am guessing what you want to be doing to examine the interaction is to assess how the age trend varies over the four levels of that factor. If that is the case, one useful thing to do would be something like
slopes <- emtrends(regmemory, ~ QuartileConsumption, var = "age")
slopes # display the estimated slope at each level
pairs(slopes) # pairwise comparisons of these slopes
See vignette("interactions", "emmeans") and the section on interactions with covariates.
Sorry that this error has been discussed before, each answer on stackoverflow seems specific to the data
I'm attempting to run the following negative binomial model in lme4:
Model5.binomial<-glmer.nb(countvariable ~ waves + var1 + dummycodedvar2 + dummycodedvar3 + (1|record_id), data=datadfomit)
However, I receive the following error when attempting to run the model:
Error in f_refitNB(lastfit, theta = exp(t), control = control) :pwrssUpdate did not converge in (maxit) iterations
I first ran the model with only 3 predictor variables (waves, var1, dummycodedvar2) and got the same error. But centering the predictors fixed this problem and the model ran fine.
Now with 4 variables (all centered) I expected the model to run smoothly, but receive the error again.
Since every answer on this site seems to point towards a problem in the data, data that replicates the problem can be found here:
https://file.io/3vtX9RwMJ6LF
Your response variable has a lot of zeros:
I would suggest fitting a model that takes account of this, such as a zero-inflated model. The GLMMadaptive package can fit zero-inflated negative binomial mixed effects models:
## library(GLMMadaptive)
## mixed_model(countvariable ~ waves + var1 + dummycodedvar2 + dummycodedvar3, ## random = ~ 1 | record_id, data = data,
## family = zi.negative.binomial(),
## zi_fixed = ~ var1,
## zi_random = ~ 1 | record_id) %>% summary()
Random effects covariance matrix:
StdDev Corr
(Intercept) 0.8029
zi_(Intercept) 1.0607 -0.7287
Fixed effects:
Estimate Std.Err z-value p-value
(Intercept) 1.4923 0.1892 7.8870 < 1e-04
waves -0.0091 0.0366 -0.2492 0.803222
var1 0.2102 0.0950 2.2130 0.026898
dummycodedvar2 -0.6956 0.1702 -4.0870 < 1e-04
dummycodedvar3 -0.1746 0.1523 -1.1468 0.251451
Zero-part coefficients:
Estimate Std.Err z-value p-value
(Intercept) 1.8726 0.1284 14.5856 < 1e-04
var1 -0.3451 0.1041 -3.3139 0.00091993
log(dispersion) parameter:
Estimate Std.Err
0.4942 0.2859
Integration:
method: adaptive Gauss-Hermite quadrature rule
quadrature points: 11
Optimization:
method: hybrid EM and quasi-Newton
converged: TRUE
I am running a mixed effects model using the coxme() function in R. The model analyzes the event of product success of firms in different countries.
Fixed effects are for example GDP, population, technology and cultural variables. Random effects are the different countries.
I know that with coxph() it is possible to test for proportional hazard using the cox.zph() command.
My question: How can I check for proportional hazard with coxme()?
Fixed effects in a random-effects coxme model can be checked for proportional hazards (PH) with the same cox.zph() function used for standard coxph() models. According to the manual, the fit argument for cox.zph() is "the result of fitting a Cox regression model, using the coxph or coxme functions."
Random effects "are not checked for proportional hazards, rather they are treated as a fixed offset in model."
An example, borrowed from this Cross-Validated question:
> library(survival)
> library(coxme)
> df <- stanford2
> df$cid <- round(df$id / 10) + 1 ## generates some clusters
> fit <- coxme(Surv(time, status) ~ age + t5 + (1 | cid),data=df)
> fit
Cox mixed-effects model fit by maximum likelihood
Data: df
events, n = 102, 157 (27 observations deleted due to missingness)
Iterations= 2 12
NULL Integrated Fitted
Log-likelihood -451.0944 -446.8618 -446.8261
Chisq df p AIC BIC
Integrated loglik 8.47 3.00 0.037317 2.47 -5.41
Penalized loglik 8.54 2.04 0.014582 4.46 -0.88
Model: Surv(time, status) ~ age + t5 + (1 | cid)
Fixed coefficients
coef exp(coef) se(coef) z p
age 0.02960206 1.030045 0.01135724 2.61 0.0091
t5 0.17056610 1.185976 0.18330590 0.93 0.3500
Random effects
Group Variable Std Dev Variance
cid Intercept 0.0199835996 0.0003993443
> cox.zph(fit)
chisq df p
age 0.831 3.00 0.84
t5 2.062 2.04 0.36
GLOBAL 2.767 5.04 0.74
This was done with survival_3.1-11 and coxme_2.2-16.
Folks,
I'm trying to print a {texreg} table of lmer() {nlme} and lme() {lme4} models including variances. The variances however, differ significantly between the two model (several orders of magnitude). It seems that the lme() variances are the square root of the lmer() ones. Which ones are correct?
library(plm)
library(lme4)
library(nlme)
library(texreg)
data("Grunfeld", package="plm")
reML0 <-lmer(inv ~ value + capital + (1|firm), data=Grunfeld)
reML1 <- lme(inv ~ value + capital, data=Grunfeld, random=~1|firm)
screenreg(list(reML0, reML1), digits=3, include.variance=TRUE)
========================================================
Model 1 Model 2
--------------------------------------------------------
(Intercept) -57.864 * -57.864
(29.378) (29.378)
value 0.110 *** 0.110 ***
(0.011) (0.011)
capital 0.308 *** 0.308 ***
(0.017) (0.017)
--------------------------------------------------------
AIC 2205.851 2205.851
Num. obs. 200 200
Num. groups: firm 10
Variance: firm.(Intercept) 7366.992
Variance: Residual 2781.426
Num. groups 10
sigma 52.739
sigma. RE 85.831
========================================================
*** p < 0.001, ** p < 0.01, * p < 0.05
It seems pretty clear that it's reporting variance for model 1 and sigma for model 2. Variance is sigma squared, so the answer to your question is "both"
R version 3.1.0 (2014-04-10)
lmer package version 1.1-6
lmerTest package version 2.0-6
I am currently working with lmer and lmerTest for my analysis.
Every time I add an effect to the random structure, I get the following error when running summary():
#Fitting a mixed model:
TRT5ToVerb.lmer3 = lmer(TRT5ToVerb ~ Group + Condition + (1+Condition|Participant) + (1|Trial), data=AllData, REML=FALSE, na.action=na.omit)
summary(TRT5ToVerb.lmer3)
Error in `colnames<-`(`*tmp*`, value = c("Estimate", "Std. Error", "df", : length of 'dimnames' [2] not equal to array extent
If I leave the structure like this:
TRT5ToVerb.lmer2 = lmer(TRT5ToVerb ~ Group + Condition + (1|Participant) + (1|Trial), data=AllData, REML=FALSE, na.action=na.omit)
there is no error run summary(TRT5ToVerb.lmer2), returning AIC, BIC, logLik deviance, estimates of the random effects, estimates of the fixed effects and their corresponding p-values, etc., etc.
So, apparently something happens when I run lmerTest, despite the fact that the object TRT5ToVerb.lmer3 is there. The only difference between both is the random structure: (1+Condition|Participant) vs. (1|Participant)
Some characteristics of my data:
Both Condition and Group are categorical variables: Condition
comprises 3 levels, and Group 2
The dependent variable (TRT5ToVerb) is continuous: it corresponds to
reading time in terms of ms
This a repeated measures experiment, with 48 observations per
participant (participants=28)
I read this threat, but I cannot see a clear solution. Will it be that I have to transform my dataframe to long format?
And if so, then how do I work with that in lmer?
I hope it is not that.
Thanks!
Disclaimer: I am neither an expert in R, nor in statistics, so please, have some patience.
(Should be a comment, but too long/code formatting etc.)
This fake example seems to work fine with lmerTest 2.0-6 and a development version of lme4 (1.1-8; but I wouldn't expect there to be any relevant differences from 1.1-6 for this example ...)
AllData <- expand.grid(Condition=factor(1:3),Group=factor(1:2),
Participant=1:28,Trial=1:8)
form <- TRT5ToVerb ~ Group + Condition + (1+Condition|Participant) + (1|Trial)
library(lme4)
set.seed(101)
AllData$TRT5ToVerb <- simulate(form[-2],
newdata=AllData,
family=gaussian,
newparam=list(theta=rep(1,7),sigma=1,beta=rep(0,4)))[[1]]
library(lmerTest)
lmer3 <- lmer(form,data=AllData,REML=FALSE)
summary(lmer3)
Produces:
Linear mixed model fit by maximum likelihood ['merModLmerTest']
Formula: TRT5ToVerb ~ Group + Condition + (1 + Condition | Participant) +
(1 | Trial)
Data: AllData
AIC BIC logLik deviance df.resid
4073.6 4136.0 -2024.8 4049.6 1332
Scaled residuals:
Min 1Q Median 3Q Max
-2.97773 -0.65923 0.02319 0.66454 2.98854
Random effects:
Groups Name Variance Std.Dev. Corr
Participant (Intercept) 0.8546 0.9245
Condition2 1.3596 1.1660 0.58
Condition3 3.3558 1.8319 0.44 0.82
Trial (Intercept) 0.9978 0.9989
Residual 0.9662 0.9829
Number of obs: 1344, groups: Participant, 28; Trial, 8
Fixed effects:
Estimate Std. Error df t value Pr(>|t|)
(Intercept) 0.49867 0.39764 12.40000 1.254 0.233
Group2 0.03002 0.05362 1252.90000 0.560 0.576
Condition2 -0.03777 0.22994 28.00000 -0.164 0.871
Condition3 -0.27796 0.35237 28.00000 -0.789 0.437
Correlation of Fixed Effects:
(Intr) Group2 Cndtn2
Group2 -0.067
Condition2 0.220 0.000
Condition3 0.172 0.000 0.794