根据Survreg预测绘制生存曲线 [英] Plotting a survival curve from a survreg prediction

问题描述

我在生存分析方面还比较陌生，并且已经使用了一些标准的电信用户流失数据示例，下面的示例称为电信":

I'm relatively new to survival analysis and have been used some standard telco churn data example with a sample below called 'telco':

telco <- read.csv(text = "State,Account_Length,Area_Code,Intl_Plan,Day_Mins,Day_Calls,Day_Charge,Eve_Mins,Eve_Calls,Eve_Charge,Night_Mins,Night_Calls,Night_Charge,Intl_Mins,Intl_Calls,Intl_Charge,CustServ_Calls,Churn
IN,65,415,no,129.1,137,21.95,228.5,83,19.42,208.8,111,9.4,12.7,6,3.43,4,TRUE
RI,74,415,no,187.7,127,31.91,163.4,148,13.89,196,94,8.82,9.1,5,2.46,0,FALSE
IA,168,408,no,128.8,96,21.9,104.9,71,8.92,141.1,128,6.35,11.2,2,3.02,1,FALSE
MT,95,510,no,156.6,88,26.62,247.6,75,21.05,192.3,115,8.65,12.3,5,3.32,3,FALSE
IA,62,415,no,120.7,70,20.52,307.2,76,26.11,203,99,9.14,13.1,6,3.54,4,FALSE
NY,161,415,no,332.9,67,56.59,317.8,97,27.01,160.6,128,7.23,5.4,9,1.46,4,TRUE")

我已经跑步:

library(survival)

dependentvars = Surv(telco$Account_Length, telco$Churn)

telcosurvreg = survreg(dependentvars ~ -Churn -Account_Length, dist="gaussian",data=telco)

telcopred = predict(telcosurvreg, newdata=telco, type="quantile", p=.5)

...获得每个客户的预期寿命.

...to get the predicted lifetime of each customer.

我正在努力的是如何可视化生存曲线.是否有办法(最好在ggplot2中)从我拥有的数据中执行此操作?

What I'm struggling with is how to visualise a survival curve for this. Is there a way (preferably in ggplot2) to do this from the data I have?

推荐答案

此处为 base R版本，用于绘制预测的生存曲线.我已经更改了公式，因此每一行的曲线都不同

Here is a base R version that plots the predicted survival curves. I have changed the formula so the curves differ for each row

> # change setup so we have one covariate
> telcosurvreg = survreg(
+   Surv(Account_Length, Churn) ~ Eve_Charge, dist = "gaussian", data = telco)
> telcosurvreg # has more than an intercept 
Call:
survreg(formula = Surv(Account_Length, Churn) ~ Eve_Charge, data = telco, 
    dist = "gaussian")

Coefficients:
(Intercept)  Eve_Charge 
 227.274695   -3.586121 

Scale= 56.9418 

Loglik(model)= -12.1   Loglik(intercept only)= -12.4
    Chisq= 0.54 on 1 degrees of freedom, p= 0.46 
n= 6 
> 
> # find linear predictors
> vals <- predict(telcosurvreg, newdata = telco, type = "lp")
> 
> # use the survreg.distributions object. See ?survreg.distributions
> x_grid <- 1:400
> sur_curves <- sapply(
+   vals, function(x) 
+     survreg.distributions[[telcosurvreg$dist]]$density(
+       (x - x_grid) / telcosurvreg$scale)[, 1])
> 
> # plot with base R
> matplot(x_grid, sur_curves, type = "l", lty = 1)

这是结果

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