709 lines
16 KiB
Text
709 lines
16 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(dplyr)
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library(ggplot2)
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library(survival)
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library(emmeans)
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library(foreign)
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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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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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Complete case
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```{r}
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nrow(dat)
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```
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```{r}
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sapply(names(dat), function(col) {
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is.na(dat[, col]) |> sum()
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})
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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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hist(dat$age.donor)
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hist(dat$age.rec)
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hist(dat$cold.isc)
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hist(dat$year)
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hist(dat$follow.up)
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unique(dat$hla.match)
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unique(dat$year)
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```
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Change the order!!
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```{r}
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dat.table1 <- dat |>
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mutate(
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sex = factor(sex, levels = c(0,1), labels = c("Female","Male")),
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tx.type = factor(tx.type, levels = c(0,1), labels = c("Cadaveric","Living")),
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hla.match = factor(hla.match),
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year = factor(year)
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)
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```
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```{r}
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table1 <- dat.table1 |>
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select(hla.match, age.donor, age.rec, cold.isc, year, sex, tx.type) |>
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tbl_summary(
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statistic = list(
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all_continuous() ~ "{median} ({p25}, {p75})"
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),
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label = list(
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hla.match ~ "HLA matches, n(%)",
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age.donor ~ "Donor age, median (IQR)",
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age.rec ~ "Recipient age, median (IQR)",
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cold.isc ~ "Cold ischemic time (hours), median (IQR), ",
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year ~ "Year of transplant",
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sex ~ "Sex, n(%)",
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tx.type ~ "Transplant type, n(%)"
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),
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missing = "ifany"
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) |>
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modify_footnote(all_stat_cols() ~ NA)
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table1
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```
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```{r}
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table.split <- dat.table1 |>
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select(hla.match, age.donor, age.rec, cold.isc, year, sex, tx.type) |>
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tbl_summary(
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by = tx.type,
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statistic = list(
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all_continuous() ~ "{median} ({p25}, {p75})"
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),
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label = list(
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hla.match ~ "HLA matches, n(%)",
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age.donor ~ "Donor age, median (IQR)",
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age.rec ~ "Recipient age, median (IQR)",
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cold.isc ~ "Cold ischemic time (hours), median (IQR), ",
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year ~ "Year of transplant",
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sex ~ "Sex, n(%)"
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),
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missing = "ifany"
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) |>
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add_overall() |>
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modify_footnote(all_stat_cols() ~ NA)
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table.split
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```
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Calculate median follow-up (reverse Kaplan-Meier method), with 95% CI
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```{r}
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rev_km <- survfit(Surv(follow.up, death == 0) ~ 1, data = dat)
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summary(rev_km)$table
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median_followup <- summary(rev_km)$table["median"]
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confidence_interval <- c(summary(rev_km)$table["0.95LCL"], summary(rev_km)$table["0.95UCL"])
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median_followup
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confidence_interval
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```
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```{r}
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g <- ggplot(dat)
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```
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```{r}
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g + geom_point(aes(x = follow.up, y = death))
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g + geom_density(aes(x = follow.up))
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g + geom_density(aes(x = age.donor))
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g + geom_bar(aes(x = age.rec))
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g + geom_boxplot(aes(x = age.rec, y = age.donor, group = age.donor))
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# dat$age.1 |> table()
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```
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```{r}
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g + geom_boxplot(aes(x = year, y = follow.up, group = year))
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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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!!!!!!!! just set xlim to 12, do not remove those individuals !!!!!!!
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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$n.at.risk[i] <- n.at.risk
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}
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print(nrow(dat))
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dat.5.life
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```
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n.removed.acc
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```{r}
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dat.5.life <- dat.5.life |>
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mutate(
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hazard.rate = n.event / n.at.risk
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)
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```
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---
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```{r}
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get_life_table = function(dat.sub, dat) {
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dat.sub <- dat.sub |>
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group_by(fu.interval) |>
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summarize(
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n.censored = sum(death == 0),
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n.event = sum(death),
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n.at.risk = nrow(dat),
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)
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for (i in 2:nrow(dat.sub)) {
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j <- i - 1
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n.censored.pre <- dat.sub$n.censored[j]
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n.event.pre <- dat.sub$n.event[j]
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n.at.risk.pre <- dat.sub$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.sub$n.at.risk[i] <- n.at.risk
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}
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dat.sub <- dat.sub |>
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mutate(
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hazard.rate = n.event / n.at.risk
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)
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return(dat.sub)
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}
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```
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```{r}
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dat.5.tx0 = dat.5[dat.5$tx.type == 0, ]
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dat.5.tx1 = dat.5[dat.5$tx.type == 1, ]
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```
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```{r}
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tx0.life <- get_life_table(dat.5.tx0, dat[dat$tx.type == 0, ])
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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, dat[dat$tx.type == 1, ])
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tx1.life
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```
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```{r}
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tx1.life$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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---
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General Solution:
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```{r}
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get.life.table <- function(dat, time.intervals) {
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n.pop <- nrow(dat)
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dat |>
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recode.dat(time.intervals) |>
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group_by(fu.interval) |>
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summarize(
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n.censored = sum(.data$death == 0),
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n.event = sum(.data$death),
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) |>
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ungroup() |>
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calculate.hazard(n.pop)
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}
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get.life.table.by.groups <- function(dat, time.intervals, grps) {
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grps |>
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lapply(function(grp) {
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dat |>
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get.life.table.by.group(time.intervals, grp) |>
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mutate(
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grp.name = grp,
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grp.value = pick(1)[[1]]
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) |>
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select(-1)
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}) |>
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bind_rows()
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}
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get.life.table.by.group <- function(dat, time.intervals, grp) {
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dat |>
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recode.dat(time.intervals) |>
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group_by(fu.interval, .data[[grp]]) |>
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summarize(
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n.censored = sum(.data$death == 0),
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n.event = sum(.data$death),
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.groups = "keep"
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) |>
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ungroup(fu.interval) |>
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group_modify(function(df.sub, grp) {
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grp.name <- names(grp)
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grp.value <- grp[[1]]
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n.pop <- (dat[[grp.name]] == grp.value) |> sum()
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calculate.hazard(df.sub, n.pop)
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}) |>
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ungroup()
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}
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calculate.hazard <- function(life.table, n.pop) {
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n.removed <- life.table$n.event + life.table$n.censored
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n.removed.accum <- c(0, cumsum(n.removed)[-length(n.removed)])
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life.table |>
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mutate(
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n.at.risk = n.pop - n.removed.accum,
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# TODO: how to account for censored? How do we adjust for uneven interval?
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hazard.rate = n.event / n.at.risk
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)
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}
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recode.dat <- function(dat, time.intervals) {
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df <- dat[dat$follow.up <= sum(time.intervals), ]
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time.points <- cumsum(time.intervals)
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df$fu.interval <- sapply(df$follow.up, function(time) {
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time.points[time <= time.points][1]
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})
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df
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}
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```
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```{r}
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time.intervals <- c(1/3, 1/3, 1/3, 1, 1, 1, 1)
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grps <- c("tx.type", "sex", "hla.match", "cold.isc")
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get.life.table.by.groups(dat, time.intervals, grps)
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```
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```{r}
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grps <- c("tx.type", "sex", "hla.match", "cold.isc")
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grps <- c("tx.type")
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plot.hazard.rate.by.groups <- function(dat, time.intervals, grps) {
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dat |>
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get.life.table.by.groups(time.intervals, grps) |>
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ggplot(aes(x = fu.interval, y = hazard.rate,
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color = paste(grp.name, grp.value, sep = " = "))) +
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geom_point() +
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geom_line() +
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labs(
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color = "Group"
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) +
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theme(
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legend.position = "bottom"
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)
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}
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```
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```{r}
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time.intervals <- c(rep(1/3, 3), rep(1, 4))
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plot.hazard.rate.by.groups(dat, time.intervals, c("tx.type"))
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```
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```{r}
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time.intervals <- rep(1, 5)
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plot.hazard.rate.by.groups(dat, time.intervals, c("tx.type"))
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```
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```{r}
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time.intervals <- rep(1/3, 15)
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plot.hazard.rate.by.groups(dat, time.intervals, c("tx.type"))
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```
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```{r}
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time.intervals <- c(rep(1/3, 3*3), rep(1, 2))
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plot.hazard.rate.by.groups(dat, time.intervals, c("tx.type"))
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```
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```{r}
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time.intervals <- c(rep(1/4, 4), rep(1, 4))
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plot.hazard.rate.by.groups(dat, time.intervals, c("tx.type"))
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```
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```{r}
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time.intervals <- c(rep(1/4, 4*2), rep(1, 3))
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plot.hazard.rate.by.groups(dat, time.intervals, c("tx.type"))
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```
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```{r}
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time.intervals <- c(rep(1/4, 4*2), rep(1/3, 3*3), rep(1, 6))
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plot.hazard.rate.by.groups(dat, time.intervals, c("tx.type"))
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```
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---
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Compare
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```{r}
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time.intervals <- c(1/3, 1/3, 1/3, 1, 1, 1, 1)
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get.life.table.by.groups(dat, time.intervals, grps)
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```
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```{r}
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hazard.df
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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.tx <- coxph(Surv(follow.up, death) ~ tx.type, data = dat)
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summary(cox.tx)
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```
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```{r}
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log.hazard.ratio.hat <- cox.tx$coefficient
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hazard.ratio.hat <- log.hazard.ratio.hat |> exp()
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```
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```{r}
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log.hazard.ratio.se <- summary(cox.tx)$coefficients[3]
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log.hazard.ratio.hat.lower.bound <- log.hazard.ratio.hat - qnorm(0.975) * log.hazard.ratio.se
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log.hazard.ratio.hat.upper.bound <- log.hazard.ratio.hat + qnorm(0.975) * log.hazard.ratio.se
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c(log.hazard.ratio.hat.lower.bound, log.hazard.ratio.hat.upper.bound) |> exp()
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```
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::: {.callout-note}
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Patients receiving kidney transplants from living donors (`tx.type = 1`) has
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1.91 higher hazard rate than those receiving from cadaveric donors
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(`tx.type = 0`).
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Since 1 is not in the 95% confidence interval, the result is statistically
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significant.
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:::
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# Exercise 6
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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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## Continuous
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```{r}
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cox.age.rec <- coxph(Surv(follow.up, death) ~ age.rec, data = dat)
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cox.age.rec |> summary()
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```
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## Categorical (Case 1)
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```{r}
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recode.age <- function(dat, ages) {
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df$age.group <- sapply(df$follow.up, function(time) {
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time.points[time <= time.points][1]
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})
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df
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}
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```
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```{r}
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dat.age.rec <- dat[!is.na(dat$age.rec), ]
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class.age <- function(age) {
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if (age < 2) {
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return("0,1")
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} else if (age < 6) {
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return("2,3,4,5")
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} else if (age < 11) {
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return("6,7,8,9,10")
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} else if (age < 19) {
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return("11-18")
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}
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}
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```
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```{r}
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dat.age.rec$age.group <- sapply(dat.age.rec$age.rec, class.age) |>
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factor(levels = c(
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"0,1",
|
|
"2,3,4,5",
|
|
"6,7,8,9,10",
|
|
"11-18"
|
|
))
|
|
|
|
|
|
cox.age.rec <- coxph(Surv(follow.up, death) ~ age.group, data = dat.age.rec)
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|
cox.age.rec |> summary()
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|
```
|
|
|
|
```{r}
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|
emmeans(cox.age.rec, pairwise ~ age.group, type = "response")
|
|
```
|
|
|
|
|
|
# Exercise 7
|
|
|
|
Fit a multivariate Cox model by using other predictors and
|
|
describe your results.
|
|
|
|
|
|
```{r}
|
|
sapply(names(dat), function(col) {
|
|
is.na(dat[, col]) |> sum()
|
|
})
|
|
```
|
|
|
|
```{r}
|
|
names(dat)
|
|
```
|
|
|
|
```{r}
|
|
Waldtest <- function(cox.fit, i.betas){
|
|
q <- length(i.betas)
|
|
coefs <- cox.fit$coefficients
|
|
# print(cox.fit$var)
|
|
var <- cox.fit$var[i.betas, i.betas]
|
|
Wald <- coefs[i.betas]%*%solve(var)%*%coefs[i.betas]
|
|
p.value <- 1-pchisq(as.numeric(Wald), df = q)
|
|
return(cbind(Wald, p.value))
|
|
}
|
|
```
|
|
|
|
|
|
```{r}
|
|
fit.cox.model <- function(surv, i.betas) {
|
|
fit.tx <- coxph(surv, data = dat)
|
|
Waldtest(fit.tx, i.betas)
|
|
}
|
|
```
|
|
|
|
Start with `tx.type`
|
|
|
|
```{r}
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec
|
|
fit.cox.model(surv, 1:2) # NOTE: Why is this 1???
|
|
```
|
|
|
|
Iteration 1: `tx.type + age + ?`
|
|
|
|
```{r}
|
|
names(dat)
|
|
```
|
|
|
|
```{r}
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + hla.match
|
|
fit.cox.model(surv, 3)
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + age.donor
|
|
fit.cox.model(surv, 3)
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + cold.isc
|
|
fit.cox.model(surv, 3)
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + year
|
|
fit.cox.model(surv, 3)
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + sex
|
|
fit.cox.model(surv, 3)
|
|
```
|
|
|
|
Add `hla.match`
|
|
|
|
Iteration 2: `tx.type + age.rec + hla.match + ?`
|
|
|
|
```{r}
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + hla.match + age.donor
|
|
fit.cox.model(surv, 4)
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + hla.match + cold.isc
|
|
fit.cox.model(surv, 4)
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + hla.match + year
|
|
fit.cox.model(surv, 4)
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + hla.match + sex
|
|
fit.cox.model(surv, 4)
|
|
```
|
|
|
|
Add `age.donor`
|
|
|
|
Iteration 2: `tx.type + age.rec + hla.match + age.donor + ?`
|
|
|
|
```{r}
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + hla.match + age.donor +
|
|
cold.isc
|
|
fit.cox.model(surv, 5)
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + hla.match + age.donor +
|
|
year
|
|
fit.cox.model(surv, 5)
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + hla.match + age.donor +
|
|
sex
|
|
fit.cox.model(surv, 5)
|
|
```
|
|
|
|
Add `year`
|
|
|
|
```{r}
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + hla.match + age.donor +
|
|
year + cold.isc
|
|
fit.cox.model(surv, 6)
|
|
surv <- Surv(follow.up, death) ~ tx.type + age.rec + hla.match + age.donor +
|
|
year + sex
|
|
fit.cox.model(surv, 6)
|
|
```
|
|
|
|
```{r}
|
|
|
|
```
|
|
|
|
|
|
Next, we test for each remaining variables
|
|
|
|
# Exercise 8
|
|
|
|
Estimate the survival function for specific covariate patterns.
|
|
Based on the previous results choose the best predictors.
|
|
|
|
Exercise 9 — Check the proportional hazards assumption. You may use the
|
|
function cox.zph. Discuss the result and possible implications.
|
|
|
|
H0: PH-assumption holds
|
|
H1: PH-assumption doesn't hold
|
|
|
|
```{r}
|
|
# different cox model!!
|
|
cox.final <- coxph(Surv(follow.up, death) ~ tx.type + age.rec + hla.match + age.donor, data = dat)
|
|
cox.zph(cox.final)
|
|
```
|
|
|
|
|
|
Exercise 10 — Plot the Schoenfeld residuals and comment.
|
|
A non-random pattern or slope in a plot of scaled residuals against time indicates a violation, suggesting the covariate's impact changes, while a horizontal line around zero supports the assumption.
|
|
|
|
```{r}
|
|
plot(cox.zph(cox.final))
|
|
```
|
|
|
|
|
|
|
|
NOTES:
|
|
left censoring: event happened before time zero -> not applicable
|
|
left truncation: not applicable??
|
|
right censoring: we saw this in the follow up times, administrative censoring high
|
|
|
|
! No left truncation, since our target population is patients who received a
|
|
kidney transplant. So we don't care about patients who died before receiving
|
|
a transplant.
|
|
|
|
! ADD: we used a Cox model but we already saw that the PH assumption doesn't hold !
|
|
|
|
? remove year from prediction model, since it is not a patient characteristic and may not be stable for future predictions. It mainly captures changes in clinical practice rather than individual risk.
|
|
|
|
|