给定因变量列表或向量，返回生存 (survfit) 和 kaplan-meier (ggsurvplot) 的函数

Question

给定 R 中具有不同列的数据框可以作为因变量，我尝试创建一个函数来接收数据框“df”、具有因变量“vars”的列表或向量、时间变量“time”状态变量“status”使用“survfit”返回生存结果，并使用 ggsurvplot 返回卡普兰-迈耶曲线。

目的是避免过多复制和粘贴代码。

以下数据为例：

library(ggplot2)
library(survival)
library("dplyr")

df <- lung %>%
  transmute(time,
            status,  # censoring status 1=censored, 2=dead
            Age = age,
            Sex = factor(sex, labels = c("Male", "Female")),
            ECOG = factor(lung$ph.ecog),
            `Meal Cal` = as.numeric(meal.cal))

# help(lung)

# Turn status into (0=censored, 1=dead)
df$status <- ifelse(df$status == 2, 1, 0)

我当然可以进行这样的生存分析：

fit <- survfit(Surv(time, status) ~ ECOG, data = df)

ggsurvplot(fit,
           pval = TRUE, pval.coord = c(750, 0.3), 
           conf.int = FALSE, 
           surv.median.line = "hv", 
           legend = c(0.8, 0.6), 
           legend.title = "",
           risk.table = "absolute", 
           risk.table.y.text = FALSE,  
           xlab = "Time (days)", ylab = "Survival", 
           palette="jco",
           title="Overall Survival", font.title = c(16, "bold", "black"), 
)

但是，如果我想对“性”做同样的事情，我就必须再次复制并粘贴所有内容。因此，我想在 R 中创建一个函数，它将数据框“df”、因变量“vars”列表、时间变量“time”和状态变量“status”作为输入，并返回生存结果使用“survfit”和使用“ggsurvplot”的 Kaplan-Meier 曲线，如下所示：

vars <- c("ECOG", "Sex")

surv_plot_func <- function(df, vars, time, status) {
  results_list <- lapply(vars, function(var, time, status) {
    
    # Fit a survival model
    fit <- survfit(Surv(as.numeric(df[[time]]), as.logical(df[[status]])) ~ as.factor(df[[var]]), data = df)
    
    # Plot the Kaplan-Meier curve using ggsurvplot
    ggsurv <- ggsurvplot(fit, pval = TRUE, conf.int = TRUE,
                         risk.table = TRUE, legend.title = "",
                         surv.median.line = "hv", xlab = "Time", ylab = "Survival Probability")
    
    # Return the fit and ggsurv as a list
    list(fit = fit, ggsurv = ggsurv)
  })
  
  # Return the list of results
  results_list
}

res_list <- surv_plot_func(df, vars, "time", "status")

然而，这并没有成功。有什么想法吗？

Answer 1

下面的代码对我有用。
正如我在评论中提到的，我发现错误是由于

ggsurvplot()

造成的，并且此函数无法读取

form

中的

lapply()

。

所以我尝试用

form

全局制作

<<-

，终于成功了。

library(survival)
library(survminer)
library(dplyr)

df <- lung %>%
  transmute(time,
            status,  # censoring status 1=censored, 2=dead
            Age = age,
            Sex = factor(sex, labels = c("Male", "Female")),
            ECOG = factor(lung$ph.ecog),
            `Meal Cal` = as.numeric(meal.cal))

vars <- c("ECOG", "Sex")

surv_plot_func <- function(df, vars, time, status) {

  results_list <- lapply(vars, \(x){
    # # Creating a formula as a string
    form <<- paste0("Surv(", time, ", ", status,") ~ ",x)
    fit <- survfit(as.formula(form), data=df)
    
    # # Plot the Kaplan-Meier curve using ggsurvplot
    ggsurv <- ggsurvplot(fit, pval = TRUE, conf.int = TRUE,
                         risk.table = TRUE, legend.title = "",
                         surv.median.line = "hv", xlab = "Time", ylab = "Survival Probability")

    # Return the fit and ggsurv as a list
    list(fit = fit, ggsurv = ggsurv)
  })
  
  # Return the list of results
  return(results_list)
}

res_list <- surv_plot_func(df, vars, "time", "status")
res_list[[1]]$ggsurv

res_list[[2]]$ggsurv

^{创建于 2023-04-12，使用 reprex v2.0.2}

Answer 2

如果您遇到如下错误：

• Error in as.formula(form) : object 'form' not found Called from: as.formula(form)

• Error: object of type 'symbol' is not subsettable

例如，可能是Shiny应用程序的问题引起的，您可以使用

eval(parse(...))

df <- lung %>%
  transmute(time,
            status,  # censoring status 1=censored, 2=dead
            Age = age,
            Sex = factor(sex, labels = c("Male", "Female")),
            ECOG = factor(lung$ph.ecog),
            `Meal Cal` = as.numeric(meal.cal))
            
categorical_variable <- "Sex"
fit_command <- paste0("survfit(Surv(time, status) ~ ", categorical_variable, ", data = df, conf.int = conf.int.survfit)")
fit <- eval(parse(text = fit_command))
ggsurvplot(fit, data = df)