根据survreg预测绘制生存曲线

我对生存分析比较陌生，曾被一些标准的电信公司客户流失数据示例使用，下面的示例名为“电信公司”：

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")

…以获得每个客户的预测寿命

---

我正在努力解决的问题是如何想象这一点的生存曲线。有没有一种方法（最好是在ggplot2中）可以从我掌握的数据中实现这一点？

这里有一个

base

R版本，它绘制了预测生存曲线。我更改了

公式

，因此每行的曲线不同

> # 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)

#更改设置，使我们有一个协变量
>telcosurvreg=survreg(
+Surv（账户长度，客户流失）~Eve\u费用，dist=“gaussian”，data=telco）
>telcosurvreg#不仅仅是一个截获
电话：
survreg（公式=Surv（账户长度、客户流失）~Eve费用，数据=电信公司，
dist=“高斯”）
系数：
（截获）Eve_冲锋
227.274695   -3.586121 
比例=56.9418
Loglik（型号）=-12.1 Loglik（仅限截距）=-12.4
在1个自由度上，Chisq=0.54，p=0.46
n=6
> 
>#寻找线性预测值
>瓦尔斯
>#使用survreg.distributions对象。见？survreg
>x_栅格曲线
>#以R为底的绘图
>matplot（x网格，表面曲线，type=“l”，lty=1）

结果如下

---

可能会对您有所帮助。该帖子犯的错误与此问题相同。不要在survreg函数之外构建Surv对象。

> # 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)