isFitCoxPH | R Documentation |
The isFitCoxPH A simple check if object inherits either "coxph" or "crr" class indicating that it is a survival function.
isFitCoxPH(fit) isFitLogit(fit)
fit |
Regression object |
boolean
Returns TRUE
if the object is of that type
otherwise it returns FALSE
.
# simulated data to use set.seed(10) ds <- data.frame( ftime = rexp(200), fstatus = sample(0:1, 200, replace = TRUE), x1 = runif(200), x2 = runif(200), x3 = runif(200) ) library(survival) library(rms) dd <- datadist(ds) options(datadist = "dd") s <- Surv(ds$ftime, ds$fstatus == 1) fit <- cph(s ~ x1 + x2 + x3, data = ds) if (isFitCoxPH(fit)) { print("Correct, the cph is of cox PH hazard type") } fit <- coxph(s ~ x1 + x2 + x3, data = ds) if (isFitCoxPH(fit)) { print("Correct, the coxph is of cox PH hazard type") } library(cmprsk) set.seed(10) ftime <- rexp(200) fstatus <- sample(0:2, 200, replace = TRUE) cov <- matrix(runif(600), nrow = 200) dimnames(cov)[[2]] <- c("x1", "x2", "x3") fit <- crr(ftime, fstatus, cov) if (isFitCoxPH(fit)) { print(paste( "Correct, the competing risk regression is", "considered a type of cox regression", "since it has a Hazard Ratio" )) } # ** Borrowed code from the lrm example ** # Fit a logistic model containing predictors age, blood.pressure, sex # and cholesterol, with age fitted with a smooth 5-knot restricted cubic # spline function and a different shape of the age relationship for males # and females. n <- 1000 # define sample size set.seed(17) # so can reproduce the results age <- rnorm(n, 50, 10) blood.pressure <- rnorm(n, 120, 15) cholesterol <- rnorm(n, 200, 25) sex <- factor(sample(c("female", "male"), n, TRUE)) label(age) <- "Age" # label is in Hmisc label(cholesterol) <- "Total Cholesterol" label(blood.pressure) <- "Systolic Blood Pressure" label(sex) <- "Sex" units(cholesterol) <- "mg/dl" # uses units.default in Hmisc units(blood.pressure) <- "mmHg" # To use prop. odds model, avoid using a huge number of intercepts by # grouping cholesterol into 40-tiles # Specify population model for log odds that Y = 1 L <- .4 * (sex == "male") + .045 * (age - 50) + (log(cholesterol - 10) - 5.2) * (-2 * (sex == "female") + 2 * (sex == "male")) # Simulate binary y to have Prob(y = 1) = 1/[1+exp(-L)] y <- ifelse(runif(n) < plogis(L), 1, 0) cholesterol[1:3] <- NA # 3 missings, at random ddist <- datadist(age, blood.pressure, cholesterol, sex) options(datadist = "ddist") fit_lrm <- lrm(y ~ blood.pressure + sex * (age + rcs(cholesterol, 4)), x = TRUE, y = TRUE ) if (isFitLogit(fit_lrm) == TRUE) { print("Correct, the lrm is a logistic regression") } fit_lm <- lm(blood.pressure ~ sex) if (isFitLogit(fit_lm) == FALSE) { print("Correct, the lm is not a logistic regression") } fit_glm_logit <- glm(y ~ blood.pressure + sex * (age + rcs(cholesterol, 4)), family = binomial() ) if (isFitLogit(fit_glm_logit) == TRUE) { print("Correct, the glm with a family of binomial is a logistic regression") } fit_glm <- glm(blood.pressure ~ sex) if (isFitLogit(fit_glm) == FALSE) { print("Correct, the glm without logit as a family is not a logistic regression") }
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