244 lines
5.1 KiB
Text
244 lines
5.1 KiB
Text
---
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title: foo
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execute:
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cache: true
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freeze: auto
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number-sections: true
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---
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```{r}
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library(tidyverse)
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library(survival)
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# library(gtsummary)
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```
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```{r}
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dat <- read.csv("./unos.txt", sep = "\t")
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head(dat)
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```
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```{r}
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names(dat) <- c("hla.match", "age.donor", "age.rec", "cold.isc", "death",
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"year", "sex", "tx.type", "follow.up")
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```
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# Exercise
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## Exercise 1
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> Illustrate in a table the characteristics of the population (age, sex, race,
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donor, . . . ).
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```{r}
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g <- ggplot(dat)
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```
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```{r}
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g + geom_point(aes(x = follow.up, y = death))
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g + geom_density(aes(x = follow.up))
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g + geom_density(aes(x = age.donor))
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g + geom_bar(aes(x = age.rec))
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g + geom_boxplot(aes(x = age.rec, y = age.donor, group = age.donor))
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# dat$age.1 |> table()
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```
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## Exercise 2
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Plot the Kaplan-Meier overall survival curve for pediatric kid-
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ney transplant recipients for the first 12 years after transplantation.
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```{r}
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km <- survfit(Surv(follow.up, death) ~ 1, data = dat[dat$follow.up <= 12, ])
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plot(km)
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```
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## Exercise 3
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We are going to compare mortality rates (hazard functions)
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between children whose transplanted kidney was provided by a living donor
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(in general a family member) and those whose source was recently deceased
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(variable donor type: `txtype`). Use the life table method to calculate the death
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rates for the first 5 years for each group (take in the first year intervals of 4
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months and then look at each year) and show the results in a table. Estimate
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the hazard ratio in each time interval as the ratio between the mortality rates
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in the two groups. What do you notice?
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```{r}
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dat.5 <- dat[dat$follow.up <= 5, ]
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head(dat.5)
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classify_time_interval = function(fu) {
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if (fu <= 1/3) {
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return(1/3)
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} else if (fu <= 2/3) {
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return(2/3)
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} else if (fu <= 1) {
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return(1)
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}
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ceiling(fu)
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}
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dat.5$fu.interval <- sapply(dat.5$follow.up, classify_time_interval)
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table(dat.5$fu.interval)
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head(dat.5)
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```
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```{r}
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dat.5.life <- dat.5 |>
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group_by(fu.interval) |>
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summarize(
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n.censored = sum(death == 0),
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n.event = sum(death),
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n.at.risk = nrow(dat),
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)
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for (i in 2:nrow(dat.5.life)) {
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j <- i - 1
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n.censored.pre <- dat.5.life$n.censored[j]
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n.event.pre <- dat.5.life$n.event[j]
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n.at.risk.pre <- dat.5.life$n.at.risk[j]
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n.at.risk <- n.at.risk.pre - n.event.pre - n.censored.pre
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dat.5.life$n.at.risk[i] <- n.at.risk
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}
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print(nrow(dat))
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dat.5.life
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```
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```{r}
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dat.5.life <- dat.5.life |>
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mutate(
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hazard.rate = n.event / n.at.risk
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)
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```
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---
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```{r}
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get_life_table = function(dat) {
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dat <- dat |>
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group_by(fu.interval) |>
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summarize(
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n.censored = sum(death == 0),
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n.event = sum(death),
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n.at.risk = nrow(dat),
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)
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for (i in 2:nrow(dat)) {
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j <- i - 1
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n.censored.pre <- dat$n.censored[j]
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n.event.pre <- dat$n.event[j]
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n.at.risk.pre <- dat$n.at.risk[j]
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n.at.risk <- n.at.risk.pre - n.event.pre - n.censored.pre
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dat$n.at.risk[i] <- n.at.risk
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}
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dat <- dat |>
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mutate(
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hazard.rate = n.event / n.at.risk
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)
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return(dat)
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}
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```
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```{r}
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dat.5.tx0 = dat.5[dat.5$tx.type == 0, ]
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dat.5.tx1 = dat.5[dat.5$tx.type == 1, ]
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```
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```{r}
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tx0.life <- get_life_table(dat.5.tx0)
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tx0.life
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```
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```{r}
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tx1.life <- get_life_table(dat.5.tx1)
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tx1.life
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```
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```{r}
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tx1.life$hazard.rate / tx0.life$hazard.rate
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```
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```{r}
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hazard.df <- data.frame(
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fu.interval = tx1.life$fu.interval,
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hazard.rate.0 = tx0.life$hazard.rate,
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hazard.rate.1 = tx1.life$hazard.rate,
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hazard.ratio = tx1.life$hazard.rate / tx0.life$hazard.rate
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)
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ggplot(hazard.df, aes(x = fu.interval)) +
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geom_line(aes(y = hazard.rate.0), color = "blue") +
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geom_line(aes(y = hazard.rate.1), color = "orange")
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ggplot(hazard.df, aes(x = fu.interval)) +
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geom_line(aes(y = hazard.ratio))
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```
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## Exercise 4
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Show a plot with Kaplan-Meier survival curves for the two donor types.
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```{r}
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km.tx <- survfit(Surv(follow.up, death) ~ tx.type, data = dat[dat$follow.up <= 12, ])
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plot(km.tx, col = c("blue", "orange"))
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legend(legend = c("cadaveric", "living"), "bottomleft", lwd = 2, col = c("blue", "orange"))
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```
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## Exercise 5
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Fit a univariate Cox model with predictor donor type. Report
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the hazard ratio and 95% confidence interval and interpret the result obtained.
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```{r}
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cox <- coxph(Surv(follow.up, death) ~ tx.type, data = dat)
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summary(cox)
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```
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```{r}
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1.90539
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```
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```{r}
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1.90539 + 1.96 * exp(0.09558)
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(2.298 - 1.58) / 2 |> exp() / 2
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```
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Exercise 6 — Research shows that an important determinant of mortality
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after kidney transplant is the age of the recipient. Fit a Cox model with age
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as predictor and estimate the hazard ratio and its confidence interval. Consider
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age first as continuous variable and then divide into categories.
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Exercise 7 — Fit a multivariate Cox model by using other predictors and
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describe your results.
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Exercise 8 — Estimate the survival function for specific covariate patterns.
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Based on the previous results choose the best predictors.
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Exercise 9 — Check the proportional hazards assumption. You may use the
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function cox.zph. Discuss the result and possible implications.
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Exercise 10 — Plot the Schoenfeld residuals and comment.
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