Cox model, coxph(), control treatment with no event, seed germination - r
I am performing survival analysis and I´m not sure if I am doing it correctly. My dataset is a result of a seed germination experiment. The main variable of interest is the "treat" one (categorical with 3 levels). In my script I am trying to figure out if there is a difference in between treatments, which one is the best, and at what extent, by comparing the PH coeff percentages. Could anyone help me with some of the problems that I'm dealing with?
1) Do I need to declare my variables as.factor() to use them? Or integer is interpreted equally?
2) If proportionality of hazards assumption (PH) is violated, what should I do with my data to proceed to a cox model building? I've intensely researched but haven't been able to understand the programming to add covariate*time interaction
or stratification to my model.
3) How to include frailty terms to cox model and detect random effect (e.g. plate in which seeds were germinated, categorical variable with 4 levels, representing repetition).
4) I also wasn't able to interpret the print(summary(cox.fra)).*
*see below
See below my two whole scripts with comments.
SCRIPT 1
rd01 <- read.table("sa_kb01.txt", header = T) # raw dataset, seed
survival
rd01
str(rd01)
rd01$begin <- as.factor(rd01$begin) # integers to factors
rd01$spp <- as.factor(rd01$spp)
rd01$cit <- as.factor(rd01$cit)
rd01$treat <- as.factor(rd01$treat)
rd01$plate <- as.factor(rd01$plate)
str(rd01)
summary(rd01)
names(rd01) # headers
### Survival analysis
# install.packages("survival")
library(survival)
library (survminer)
?survfit
?survfit.formula
?survfit.coxph
?ggsurvplot
## Fit Kaplan-Meier survivor function
km.fit <- survfit(Surv(day, status) ~ treat, data= rd01, type="kaplan-meier")
km.fit
print(summary(km.fit))
plot(km.fit, conf.int= T, fun = "event", mark.time = c(140), pch = c("S", "W", "A"), col = c("darkred","darkblue","darkgreen"), lty = c("solid","dotted","longdash"),lwd = 1.5, xlab = "time [days]", ylab = "germination probability [%]")
print(summary(km.fit))
## Comparison of Survivor Functions
# Log-rank tests
?survdiff
# Log-rank or Mantel-Haenszel test in "rho = 0" OR
# Peto & Peto modification of the Gehan-Wilcoxon test in "rho = 1"
# ... Assess all groups for heterogeneity
lrmh.123 <- survdiff(Surv(day,status) ~ treat, data= rd01, rho= 0)
print(lrmh.123) # If p<0.05 there are difference between all groups!
# ... Comparing groups pairwise
lrmh.120 <- survdiff(Surv(day,status) ~ treat, data= rd01, subset= {treat!=3}, rho= 0)
lrmh.103 <- survdiff(Surv(day,status) ~ treat, data= rd01, subset= {treat!=2}, rho= 0)
lrmh.023 <- survdiff(Surv(day,status) ~ treat, data= rd01, subset= {treat!=1}, rho= 0)
print(lrmh.120)
print(lrmh.103)
print(lrmh.023) # If p<0.05 there are difference pairwised groups!
## Checking Proportional Hazard (PH) assumption
# Define function mlogmlog() to calculate -log(-log(S(t)))
mlogmlog <- function(y){-log(-log(y))}
# Use estimated Kaplan-Meier survivor functions
km.fit
# ... to plot -log(-log(S(t))) versus log(t)
plot(km.fit, fun= mlogmlog, log="x", mark.time= c(140), pch = c("S", "W", "A"), col = c("darkred","darkblue","darkgreen"), lty = c("solid","dotted","longdash"), lwd = 1.5, xlab="time [days]", ylab= "-log(-log(S(t)))") # If lines do not cross, PH assumption is plausible!
# Interpretarion: http://www.sthda.com/english/wiki/cox-model-assumptions#testing-proportional-hazards-assumption
## Checking for multicollinearity
# install.packages("HH")
library(HH)
# Fit a generalized linear model predicting days from treatment
?glm
mc.glm <- glm(day ~ treat, data=rd01)
print(mc.glm) # doesn't need interpretation, only used to create object to VIF function
# Check for multicollinearity among covariates throught variance inflation factor (VIF)
?vif
mc.vif <- vif(mc.glm)
print(mc.vif) # VIF can determine what proportion of the variation in each covariate
# is explained by the other covariates:
# VIF > 10, serious multicollinearity; VIF = 5, evidence of multicollinearity;
# VIF < 1, no evidence of multicollinearity
## Adding covariates to the Cox model
# Create a Cox model
cox.mod <- coxph(Surv(day, status) ~ treat, data= rd01)
print(summary(cox.mod))
# Interpretation: http://www.sthda.com/english/wiki/cox-proportional-hazards-model
# Double check for PH assumption now with Cox model built
dc.ph <- cox.zph(cox.mod)
dc.ph
ggcoxzph(dc.ph) # if global and individual p-vale > 0.05, PH assumption is plausible!
## Including random effects
?frailty
# Adding plate variable as frailty term
cox.fra <- coxph(Surv(day, status) ~ treat + frailty(plate), data= rd01)
print(summary(cox.fra)) # if global and individual p-vale < 0.05,
# maintain frailty term while adding covariates 1 at a time in cox model!`
SCRIPT 2 - the same, but different dataset, control treat1 with no event!
rd01 <- read.table("sa_hal01.txt", header = T) # raw dataset, seed survival
rd01
str(rd01)
rd01$begin <- as.factor(rd01$begin) # integers to factors
rd01$spp <- as.factor(rd01$spp)
rd01$cit <- as.factor(rd01$cit)
rd01$treat <- as.factor(rd01$treat)
rd01$plate <- as.factor(rd01$plate)
str(rd01)
summary(rd01)
names(rd01) # headers
### Survival analysis
# install.packages("survival")
library(survival)
library (survminer)
?survfit
?survfit.formula
?survfit.coxph
?ggsurvplot
## Fit Kaplan-Meier survivor function
km.fit <- survfit(Surv(day, status) ~ treat, data= rd01, type="kaplan-meier")
km.fit
print(summary(km.fit))
plot(km.fit, conf.int= T, fun = "event", mark.time = c(140), pch = c("S", "W", "A"), col = c("darkred","darkblue","darkgreen"), lty = c("solid","dotted","longdash"),lwd = 1.5, xlab = "time [days]", ylab = "germination probability [%]")
print(summary(km.fit))
## Comparison of Survivor Functions
# Log-rank tests
?survdiff
# Log-rank or Mantel-Haenszel test in "rho = 0" OR
# Peto & Peto modification of the Gehan-Wilcoxon test in "rho = 1"
# ... Assess all groups for heterogeneity
lrmh.123 <- survdiff(Surv(day,status) ~ treat, data= rd01, rho= 0)
print(lrmh.123) # If p<0.05 there are difference between all groups!
# ... Comparing groups pairwise
lrmh.120 <- survdiff(Surv(day,status) ~ treat, data= rd01, subset= {treat!=3}, rho= 0)
lrmh.103 <- survdiff(Surv(day,status) ~ treat, data= rd01, subset= {treat!=2}, rho= 0)
lrmh.023 <- survdiff(Surv(day,status) ~ treat, data= rd01, subset= {treat!=1}, rho= 0)
print(lrmh.120)
print(lrmh.103)
print(lrmh.023) # If p<0.05 there are difference pairwised groups!
## Checking Proportional Hazard (PH) assumption
# Define function mlogmlog() to calculate -log(-log(S(t)))
mlogmlog <- function(y){-log(-log(y))}
# Use estimated Kaplan-Meier survivor functions
km.fit
# ... to plot -log(-log(S(t))) versus log(t)
plot(km.fit, fun= mlogmlog, log="x", mark.time= c(140), pch = c("S", "W", "A"), col = c("darkred","darkblue","darkgreen"), lty = c("solid","dotted","longdash"), lwd = 1.5, xlab="time [days]", ylab= "- log(-log(S(t)))") # If lines do not cross, PH assumption is plausible!
# Interpretarion: http://www.sthda.com/english/wiki/cox-model- assumptions#testing-proportional-hazards-assumption
## Checking for multicollinearity
# install.packages("HH")
library(HH)
# Fit a generalized linear model predicting days from treatment
?glm
mc.glm <- glm(day ~ treat, data=rd01)
print(mc.glm) # doesn't need interpretation, only used to create object to VIF function
# Check for multicollinearity among covariates throught variance inflation factor (VIF)
?vif
mc.vif <- vif(mc.glm)
print(mc.vif) # VIF can determine what proportion of the variation in each covariate
# is explained by the other covariates:
# VIF > 10, serious multicollinearity; VIF = 5, evidence of multicollinearity;
# VIF < 1, no evidence of multicollinearity
## Adding covariates to the Cox model
# Create a Cox model
cox.mod <- coxph(Surv(day, status) ~ treat, data= rd01)
print(summary(cox.mod))
# Interpretation: http://www.sthda.com/english/wiki/cox-proportional-hazards-model
# Double check for PH assumption now with Cox model built
dc.ph <- cox.zph(cox.mod)
dc.ph
ggcoxzph(dc.ph) # if global and individual p-vale > 0.05, PH assumption is plausible!
## Including random effects
?frailty
# Adding plate variable as frailty term
cox.fra <- coxph(Surv(day, status) ~ treat + frailty(plate), data= rd01)
print(summary(cox.fra)) # if global and individual p-vale < 0.05,
# maintain frailty term while adding covariates 1 at a time in cox model!
There seems to be a statistically significant difference and treat3 differs from the other groups in both scripts. In script 1 PH is violated and I don´t now what to do. Apart from that, Cox model in script 1 seems to work fine and the interpretation of hazard ratios are OK, but in script 2, no idea how to interpret or solve that (there was no event in control treat1).
1) Do I need to declare my variables as.factor() to use them? Or integer is interpreted equally?
I think in your case as.factor is correct. You can use integers if you have continuous numeric variables - for example if you would have time seeds have been stored before the experiment, you could use as.numeric for time variable.
2) If PH is violated, what should I do with my data to proceed to a cox model building? I've intensely researched but haven't been able to understand the programming to add covariate x time interaction or stratification to my model.
Cox regression, aka Cox proportional hazards model, is based on the assumption of proportional hazards. If that assumption is violated, you won´t get reliable results. You probably could try some data transformations to see if it would help. Or if it is violated in some subexperiment/group, you could just leave it out.
Related
Other than burn-in increase and priors, how can I help my multiple change point (mcp package in R) models converge?
I would like to identify changepoints in my data and an associated error term for their estimate. The mcp package seems to do a good job (visually) of identifying changepoints in my data, but the model parameters generally have rhat values >1.1. From my understanding, I cannot trust any Bayesian parameter estimates unless all rhat values in the model are =< 1.1. Aside from increasing my burn-in period with the adapt argument or using priors (see note below), how else can I improve these models? Alternatively, can I force mcp to fit a 'best' two and three segment model and return those parameter estimates with error? Ideally I would be able to provide changepoint estimates with an error term associated with each estimate, but packages like segmented and struccchange generally fail to identify changepoints in my data. The code looks like this: set.seed(42) x <- c(227,227,228,228,228,228,228,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,230,232,233,233,233,233,233,233,233,233,236,236,236,236,237,237,237,238,238,238,238,238,238,238,238,239,239,239,239,239,239,243,244,244,244,244,244,244,244,244,244,245,245,245,246,246,246,246,247,250,250,250,250,251,251,251,251,251,251,251,251,253,253,253,257,257,260,260,260,260,260,260,260,264,264,264,265,265,265,265,265,265,265,265,265,265,265,265,265,265,265,265,265,265,267,267,267,267,267,267,267,267,267,267,267,267,267,271,271,271,271,271,271,273,273,273,273,273,273,273,273,273,273,273,273,273,273,273,273,273,273,273,274,274,274,274,274,274,274) y <- c(8.43,6.9,8.93,7.33,7.28,7.24,6.62,8.36,8.17,8.07,8.07,7.63,7.6,7.54,7.37,7.31,7.21,7.03,6.93,6.88,6.82,6.78,6.7,6.5,8.35,10.97,7.48,7.46,7.28,7.17,6.72,6.68,6.08,7.42,7.14,6.92,6.68,7.49,7.28,6.67,9.4,7.54,7.04,6.89,6.88,6.52,6.45,6.39,8.48,8.04,7.52,7.35,6.9,6.57,6.86,7.46,7.39,7.16,7.08,6.83,6.83,6.7,6.54,6.47,9.75,7.38,5.96,10.49,8.32,7.22,7.05,8.55,10.34,8.23,7.9,7.31,8.18,7.8,7.31,7.18,7.17,7.13,7.02,6.84,10.62,10.09,9.26,10.8,10.37,10.9,10.52,10.23,9.28,9.18,8.85,8.81,11.03,8.84,6.29,11.36,10.91,10.87,10.4,10.17,9.61,9.5,9.36,9.17,9.13,8.88,8.73,8.55,8.37,8.33,8.25,7.82,6.9,9.77,9.53,9.39,9.1,8.93,8.68,8.64,8.47,8.41,8.38,8.28,8.18,7.74,10.67,10.64,10.54,10.36,10.35,7.03,9.51,9.37,9.24,9.22,9.18,8.96,8.95,8.94,8.89,8.82,8.79,8.72,8.35,8.22,8.13,8.07,7.91,7.85,7.79,8.82,8.59,8.44,8.42,8.37,8.06,7.34) df <- data.frame(x, y) #Writing the formula for a three-segment line three_segment_model = list( y ~ x, # intercept + slope y ~ 1 ~ 0 + x, #segment 2, specifying a changepoint and joined slope y ~ 1 ~ 0 + x #segment 3, specifying a changepoint and joined slope ) #Writing the formula for a two-segment line two_segment_model = list( y ~ x, # intercept + slope y ~ 1 ~ 0 + x #segment 2, specifying a changepoint and joined slope ) #Disjointing the slopes of two segments test_segment_model = list( y ~ x, # intercept + slope y ~ 1 ~ 1 + x #segment 2, specifying a changepoint and disjoined slope ) #Disjointing the slopes of the three expected segments test_three_segment_model = list( y ~ x, # intercept + slope y ~ 1 ~ 1 + x, #segment 2, specifying a changepoint and disjoined slope y ~ 1 ~ 1 + x #segment 3, specifying a changepoint and disjoined slope ) These are the models I've tried, but none achieve consistently tolerable rhat values. The code I use to test them is as follows: #An example run, although each of these models fails to provide suitable rhat values fit1 <- mcp(three_segment_model, df, chains = 4, iter = 10000, cores = 3) plot(fit1) fit1_summary <- data.frame(summary(fit1)) Given the literature, I have also tried priors (though admittedly I cannot make heads or tails of the truncate and distribution code with mcp), which do not seem to reflect the patterns poorly identified with this modelling approach.
If your issue is primarily a practical one, upping the number of iterations and chains reveals that the posterior is quite reproducible across chains for the "problematic" parameters: fit1 <- mcp(three_segment_model, df, chains = 6, iter = 50000, cores = 6) plot_pars(fit1, c("cp_1", "cp_2", "x_1")) And you have good rhat values for the non-three-segment models; all pointing to the three-segment model being unidentifiable with this data. Without knowing the process, I did try some fairly informative priors: prior = list( x_1 = 0, # fixed horizontal! x_2 = "dnorm(0, 1) T(0, )" # Positive slope ) but it did not improve rhat values for the change point parameters.
Is there a difference between gamma hurdle (two-part) models and zero-inflated gamma models?
I have semicontinuous data (many exact zeros and continuous positive outcomes) that I am trying to model. I have largely learned about modeling data with substantial zero mass from Zuur and Ieno's Beginner's Guide to Zero-Inflated Models in R, which makes a distinction between zero-inflated gamma models and what they call "zero-altered" gamma models, which they describe as hurdle models that combine a binomial component for the zeros and a gamma component for the positive continuous outcome. I have been exploring the use of the ziGamma option in the glmmTMB package and comparing the resulting coefficients to a hurdle model that I built following the instructions in Zuur's book (pages 128-129), and they do not coincide. I'm having trouble understanding why not, as I know that the gamma distribution cannot take on the value of zero, so I suppose every zero-inflated gamma model is technically a hurdle model. Can anyone illuminate this for me? See more comments about the models below the code. library(tidyverse) library(boot) library(glmmTMB) library(parameters) ### DATA id <- rep(1:75000) age <- sample(18:88, 75000, replace = TRUE) gender <- sample(0:1, 75000, replace = TRUE) cost <- c(rep(0, 30000), rgamma(n = 37500, shape = 5000, rate = 1), sample(1:1000000, 7500, replace = TRUE)) disease <- sample(0:1, 75000, replace = TRUE) time <- sample(30:3287, 75000, replace = TRUE) df <- data.frame(cbind(id, disease, age, gender, cost, time)) # create binary variable for non-zero costs df <- df %>% mutate(cost_binary = ifelse(cost > 0, 1, 0)) ### HURDLE MODEL (MY VERSION) # gamma component hurdle_gamma <- glm(cost ~ disease + gender + age + offset(log(time)), data = subset(df, cost > 0), family = Gamma(link = "log")) model_parameters(hurdle_gamma, exponentiate = T) # binomial component hurdle_binomial <- glm(cost_binary ~ disease + gender + age + time, data = df, family = "binomial") model_parameters(hurdle_binomial, exponentiate = T) # predicted probability of use df$prob_use <- predict(hurdle_binomial, type = "response") # predicted mean cost for people with any cost df_bin <- subset(df, cost_binary == 1) df_bin$cost_gamma <- predict(hurdle_gamma, type = "response") # combine data frames df2 <- left_join(df, select(df_bin, c(id, cost_gamma)), by = "id") # replace NA with 0 df2$cost_gamma <- ifelse(is.na(df2$cost_gamma), 0, df2$cost_gamma) # calculate predicted cost for everyone df2 <- df2 %>% mutate(cost_pred = prob_use * cost_gamma) # mean predicted cost mean(df2$cost_pred) ### glmmTMB with ziGamma zigamma_model <- glmmTMB(cost ~ disease + gender + age + offset(log(time)), family = ziGamma(link = "log"), ziformula = ~ disease + gender + age + time, data = df) model_parameters(zigamma_model, exponentiate = T) df <- df %>% predict(zigamma_model, new data = df, type = "response") # doesn't work # "no applicable method for "predict" applied to an object of class "data.frame" The coefficients from the gamma component of my hurdle model and the fixed effects components of the zigamma model are the same, but the SEs are different, which in my actual data has substantial implications for the significance of my predictor of interest. The coefficients on the zero-inflated model are different, and I also noticed that the z values in the binomial component are the negative inverse of those in my binomial model. I assume this has to do with my binomial model modeling the probability of presence (1 is a success) and glmmTMB presumably modeling the probability of absence (0 is a success)? In sum, can anyone point out what I am doing wrong with the glmmTMB ziGamma model?
The glmmTMB package can do this: glmmTMB(formula, family=ziGamma(link="log"), ziformula=~1, data= ...) ought to do it. Maybe something in VGAM as well? To answer the questions about coefficients and standard errors: the change in sign of the binomial coefficients is exactly what you suspected (the difference between estimating the probability of 0 [glmmTMB] vs the probability of not-zero [your/Zuur's code]) The standard errors on the binomial part of the model are close but not identical: using broom.mixed::tidy, round(1-abs(tidy(hurdle_g,component="zi")$statistic)/ abs(tidy(hurdle_binomial)$statistic),3) ## [1] 0.057 0.001 0.000 0.000 0.295 6% for the intercept, up to 30% for the effect of age ... the nearly twofold difference in the standard errors of the conditional (cost>0) component is definitely puzzling me; it holds up if we simply implement the Gamma/log-link in glmmTMB vs glm. It's hard to know how to check which is right/what the gold standard should be for this case. I might distrust Wald p-values in this case and try to get p-values with the likelihood ratio test instead (via drop1). In this case the model is badly misspecified (i.e. the cost is uniformly distributed, nothing like Gamma); I wonder if that could be making things harder/worse?
Getting estimated means after multiple imputation using the mitml, nlme & geepack R packages
I'm running multilevel multiple imputation through the package mitml (using the panimpute() function) and am fitting linear mixed models and marginal models through the packages nlme and geepack and the mitml:with() function. I can get the estimates, p-values etc for those through the testEstimates() function but I'm also looking to get estimated means across my model predictors. I've tried the emmeans package, which I normally use for getting estimated means when running nlme & geepack without multiple imputation but doing so emmeans tell me "Can't handle an object of class “mitml.result”". I'm wondering is there a way to get pooled estimated means from the multiple imputation analyses I've run? The data frames I'm analyzing are longitudinal/repeated measures and in long format. In the linear mixed model I want to get the estimated means for a 2x2 interaction effect and in the marginal model I'm trying to get estimated means for the 6 levels of 'time' variable. The outcome in all models is continuous. Here's my code # mixed model fml <- Dep + time ~ 1 + (1|id) imp <- panImpute(data=Data, formula=fml, n.burn=50000, n.iter=5000, m=100, group = "treatment") summary(imp) plot(imp, trace="all") implist <- mitmlComplete(imp, "all", force.list = TRUE) fit <- with(implist, lme(Dep ~ time*treatment, random = ~ 1|id, method = "ML", na.action = na.exclude, control = list(opt = "optim"))) testEstimates(fit, var.comp = TRUE) confint.mitml.testEstimates(testEstimates(fit, var.comp = TRUE)) # marginal model fml <- Dep + time ~ 1 + (1|id) imp <- panImpute(data=Data, formula=fml, n.burn=50000, n.iter=5000, m=100) summary(imp) plot(imp, trace="all") implist <- mitmlComplete(imp, "all", force.list = TRUE) fit <- with(implist, geeglm(Dep ~ time, id = id, corstr ="unstructured")) testEstimates(fit, var.comp = TRUE) confint.mitml.testEstimates(testEstimates(fit, var.comp = TRUE))
is there a way to get pooled estimated means from the multiple imputation analyses I've run? This is not a reprex without Data, so I can't verify this works for you. But emmeans provides support for mira-class (lists of) models in the mice package. So if you fit your model in with() using the mids rather than mitml.list class object, then you can use that to obtain marginal means of your outcome (and any contrasts or pairwise comparisons afterward). Using example data found here, which uncomfortably loads an external workspace: con <- url("https://www.gerkovink.com/mimp/popular.RData") load(con) ## imputation library(mice) ini <- mice(popNCR, maxit = 0) meth <- ini$meth meth[c(3, 5, 6, 7)] <- "norm" pred <- ini$pred pred[, "pupil"] <- 0 imp <- mice(popNCR, meth = meth, pred = pred, print = FALSE) ## analysis library(lme4) # fit multilevel model mod <- with(imp, lmer(popular ~ sex + (1|class))) library(emmeans) # obtain pooled estimates of means (em <- emmeans(mod, specs = ~ sex) ) pairs(em) # test comparison
How to check for overdispersion in a GAM with negative binomial distribution?
I fit a Generalized Additive Model in the Negative Binomial family using gam from the mgcv package. I have a data frame containing my dependent variable y, an independent variable x, a factor fac and a random variable ran. I fit the following model gam1 <- gam(y ~ fac + s(x) + s(ran, bs = 're'), data = dt, family = "nb" I have read in Negative Binomial Regression book that it is still possible for the model to be overdisperesed. I have found code to check for overdispersion in glm but I am failing to find it for a gam. I have also encountered suggestions to just check the QQ plot and standardised residuals vs. predicted residuals, but I can not decide from my plots if the data is still overdisperesed. Therefore, I am looking for an equation that would solve my problem.
A good way to check how well the model compares with the observed data (and hence check for overdispersion in the data relative to the conditional distribution implied by the model) is via a rootogram. I have a blog post showing how to do this for glm() models using the countreg package, but this works for GAMs too. The salient parts of the post applied to a GAM version of the model are: library("coenocliner") library('mgcv') ## parameters for simulating set.seed(1) locs <- runif(100, min = 1, max = 10) # environmental locations A0 <- 90 # maximal abundance mu <- 3 # position on gradient of optima alpha <- 1.5 # parameter of beta response gamma <- 4 # parameter of beta response r <- 6 # range on gradient species is present pars <- list(m = mu, r = r, alpha = alpha, gamma = gamma, A0 = A0) nb.alpha <- 1.5 # overdispersion parameter 1/theta zprobs <- 0.3 # prob(y == 0) in binomial model ## simulate some negative binomial data from this response model nb <- coenocline(locs, responseModel = "beta", params = pars, countModel = "negbin", countParams = list(alpha = nb.alpha)) df <- setNames(cbind.data.frame(locs, nb), c("x", "yNegBin")) OK, so we have a sample of data drawn from a negative binomial sampling distribution and we will now fit two models to these data: A Poisson GAM m_pois <- gam(yNegBin ~ s(x), data = df, family = poisson()) A negative binomial GAM m_nb <- gam(yNegBin ~ s(x), data = df, family = nb()) The countreg package is not yet on CRAN but it can be installed from R-Forge: install.packages("countreg", repos="http://R-Forge.R-project.org") Then load the packages and plot the rootograms: library("countreg") library("ggplot2") root_pois <- rootogram(m_pois, style = "hanging", plot = FALSE) root_nb <- rootogram(m_nb, style = "hanging", plot = FALSE) Now plot the rootograms for each model: autoplot(root_pois) autoplot(root_nb) This is what we get (after plotting both using cowplot::plot_grid() to arrange the two rootograms on the same plot) We can see that the negative binomial model does a bit better here than the Poisson GAM for these data — the bottom of the bars are closer to zero throughout the range of the observed counts. The countreg package has details on how you can add an uncertain band around the zero line as a form of goodness of fit test. You can also compute the Pearson estimate for the dispersion parameter using the Pearson residuals of each model: r$> sum(residuals(m_pois, type = "pearson")^2) / df.residual(m_pois) [1] 28.61546 r$> sum(residuals(m_nb, type = "pearson")^2) / df.residual(m_nb) [1] 0.5918471 In both cases, these should be 1; we see substantial overdispersion in the Poisson GAM, and some under-dispersion in the Negative Binomial GAM.
Predicting relative risk with predict.coxph, simPH and the formula
There is a great post about the interpretation of the predict.coxph() output. However, I keep getting different results comparing the output from predict.coxph, simPH and the formula for relative risk. Since my hypothesis includes a quadratic effect, I am going to include a polynomial with power 2 in my example. I use the example from this post. data("lung") Predicting relative risk with predict() # Defining the quadratic predictor lung$meal.cal_q <- lung$meal.cal^2 # conduct a cox regression with the predictor meal.cal, its quadratic version and some covariates. cox_mod <- coxph(Surv(time, status) ~ ph.karno + pat.karno + meal.cal + meal.cal_q, data = lung) # a vector of fitted values to predict for meal.cal_new <- seq(min(lung$meal.cal, na.rm= TRUE), max(lung$meal.cal, na.rm= TRUE), by= 1) # a vector of fitted values to predict for, the quadratic effect meal.cal_q_new <- meal.cal_new^2 # the length of the vector with the values to predict for n <- length(meal.cal_new) # a dataframe with all the values to predict for lung_new <- data.frame(ph.karno= rep(mean(lung$ph.karno, na.rm= TRUE), n), pat.karno= rep(mean(lung$pat.karno, na.rm= TRUE), n), meal.cal= meal.cal_new, meal.cal_q = meal.cal_q_new) # predict the relative risk lung_new$rel_risk <- predict(cox_mod, lung_new, type= "risk") Predicting the relative risk with the formula (see the post mentioned above) # Defining the quadratic predictor lung$meal.cal_q <- lung$meal.cal^2 # run a cox regression with the predictor meal.cal, its quadratic version and some covariates. cox_mod <- coxph(Surv(time, status) ~ ph.karno + pat.karno + meal.cal + meal.cal_q, data = lung) # a vector of fitted values to predict for meal.cal_new <- seq(min(lung$meal.cal, na.rm= TRUE), max(lung$meal.cal, na.rm= TRUE), by= 1) # a vector of fitted values to predict for, the quadratic effect meal.cal_q_new <- meal.cal_new^2 # length of the vector to predict for n <- length(meal.cal_new) # A dataframe with the values to make the prediction for lung_new2 <- data.frame( ph.karno= rep(mean(lung$ph.karno, na.rm= TRUE), n), pat.karno= rep(mean(lung$pat.karno, na.rm= TRUE), n), meal.cal= meal.cal_new, meal.cal_q = meal.cal_q_new) # A dataframe with the values to compare the prediction with lung_new_mean <- data.frame( ph.karno= rep(mean(lung$ph.karno, na.rm= TRUE), n), pat.karno= rep(mean(lung$pat.karno, na.rm= TRUE), n), meal.cal= rep(mean(lung$meal.cal, na.rm= TRUE), n), meal.cal_q = rep(mean(lung$meal.cal_q, na.rm= TRUE), n)) # extract the coefficients coefCPH <- coef(cox_mod) # make the prediction for the values of interest cox_risk <- exp(coefCPH["ph.karno"] * lung_new2[ , "ph.karno"] + coefCPH["pat.karno"] * lung_new2[ , "pat.karno"] + coefCPH["meal.cal"] * lung_new2[ , "meal.cal"] + coefCPH["meal.cal_q"] * lung_new2[ , "meal.cal_q"]) # make the predictions for the values to compare with cox_risk_mean <- exp(coefCPH["ph.karno"] * lung_new_mean[ , "ph.karno"] + coefCPH["pat.karno"] * lung_new_mean[ , "pat.karno"] + coefCPH["meal.cal"] * lung_new_mean[ , "meal.cal"] + coefCPH["meal.cal_q"] * lung_new_mean[ , "meal.cal_q"]) # calculate the relative risk lung_new2$rel_risk <- unlist(cox_risk)/ unlist(cox_risk_mean) Now the plot with the predicted relative risk using predict() and using the formula: ggplot(lung_new, aes(meal.cal, rel_risk)) + geom_smooth() + geom_smooth(data= lung_new2, col= "red") The plot shows that the predictions are different. I do not understand why this is the case, although the mentioned post shows that the predict function and the formula should give the same result. Because of this confusion I tried to solve the problem with the simPH package. Here is what I did: # Defining the quadratic predictor lung$meal.cal_q <- lung$meal.cal^2 # run a cox regression with the predictor, its quadratic version and some covariates. cox_mod <- coxph(Surv(time, status) ~ ph.karno + pat.karno + meal.cal + meal.cal_q, data = lung) # a vector of fitted values to predict for meal.cal_new <- seq(min(lung$meal.cal, na.rm= TRUE), max(lung$meal.cal, na.rm= TRUE), by= 1) # length of the vector to predict for n <- length(meal.cal_new) # A vector with the values to compare the prediction with meal.cal_new_mean <- rep(mean(lung$meal.cal, na.rm= TRUE), n) # running 100 simulations per predictor value with coxsimPoly Sim <- coxsimPoly(obj= cox_mod, b = "meal.cal", pow = 2, qi = "Relative Hazard", Xj = meal.cal_new, Xl = meal.cal_new_mean, ci = .95, nsim = 100, extremesDrop = FALSE) # plot the result simGG(Sim) This gives an empty plot with the warning Warning messages: 1: In min(obj$sims[, x]) : no non-missing arguments to min; returning Inf 2: In max(obj$sims[, x]) : no non-missing arguments to max; returning -Inf And the Sim$simsobject appears indeed to be empty. My questions are: Why do the results from predict() and the use of the formula differ? Why does the simPH package not calculate the relative risk? Which method should I choose? My hypothesis is a quadratic effect in a cox regression and I need a plot for this predictor with its relative risk (compared to the predictor being at its mean value), just like in the example.
Quick answer to the simPH issue: the polynomial terms need to be specified in the coxph call using the I function, e.g.: cox_mod <- coxph(Surv(time, status) ~ ph.karno + pat.karno + meal.cal + I(meal.cal^2), data = lung) (The error handling in your use case is pretty poor.) Using this modification (and 1000 simulations) with your code above should return something like: Differences between simPH and predict My guess to the differences is that simPH doesn't create confidence intervals around the transformed point estimates like predict. It draws simulations from the multivariate normal distribution specified by the fitted model, then shows the central 50% and 95% of this simulated distribution. The central line is just the median of the sims. It is explicitly a different logic from predict. For very non-monotonic quantities of interest, like this one, predict point estimates give highly substantively misleading results compared to simPH. There is little evidence for such a form based on 4 observations.