如何绘制survreg生成的生存曲线(R的包生存)? [英] How to plot the survival curve generated by survreg (package survival of R)?

问题描述

我正在尝试将 Weibull 模型拟合并绘制到生存数据中.数据只有一个协变量，即队列，运行时间为 2006 年到 2010 年.那么，对于绘制 2010 年队列生存曲线的以下两行代码有什么想法吗?

I’m trying to fit and plot a Weibull model to a survival data. The data has just one covariate, cohort, which runs from 2006 to 2010. So, any ideas on what to add to the two lines of code that follows to plot the survival curve of the cohort of 2010?

library(survival)
s <- Surv(subSetCdm$dur,subSetCdm$event)
sWei <- survreg(s ~ cohort,dist='weibull',data=subSetCdm)

使用 Cox PH 模型完成相同的操作非常简单，如下几行.问题是 survfit() 不接受 survreg 类型的对象.

Accomplishing the same with the Cox PH model is rather straightforward, with the following lines. The problem is that survfit() doesn’t accept objects of type survreg.

sCox <- coxph(s ~ cohort,data=subSetCdm)
cohort <- factor(c(2010),levels=2006:2010)
sfCox <- survfit(sCox,newdata=data.frame(cohort))
plot(sfCox,col='green')

使用数据肺(来自生存包)，这就是我想要完成的.

Using the data lung (from the survival package), here is what I'm trying to accomplish.

#create a Surv object
s <- with(lung,Surv(time,status))

#plot kaplan-meier estimate, per sex
fKM <- survfit(s ~ sex,data=lung)
plot(fKM)

#plot Cox PH survival curves, per sex
sCox <- coxph(s ~ as.factor(sex),data=lung)
lines(survfit(sCox,newdata=data.frame(sex=1)),col='green')
lines(survfit(sCox,newdata=data.frame(sex=2)),col='green')

#plot weibull survival curves, per sex, DOES NOT RUN
sWei <- survreg(s ~ as.factor(sex),dist='weibull',data=lung)
lines(survfit(sWei,newdata=data.frame(sex=1)),col='red')
lines(survfit(sWei,newdata=data.frame(sex=2)),col='red')

推荐答案

希望这会有所帮助，我没有犯一些误导性的错误:

Hope this helps and I haven't made some misleading mistake:

从上面复制:

    #create a Surv object
    s <- with(lung,Surv(time,status))

    #plot kaplan-meier estimate, per sex
    fKM <- survfit(s ~ sex,data=lung)
    plot(fKM)

    #plot Cox PH survival curves, per sex
    sCox <- coxph(s ~ as.factor(sex),data=lung)
    lines(survfit(sCox,newdata=data.frame(sex=1)),col='green')
    lines(survfit(sCox,newdata=data.frame(sex=2)),col='green')

对于 Weibull，请使用 predict，这是 Vincent 的评论:

for Weibull, use predict, re the comment from Vincent:

    #plot weibull survival curves, per sex,
    sWei <- survreg(s ~ as.factor(sex),dist='weibull',data=lung)

    lines(predict(sWei, newdata=list(sex=1),type="quantile",p=seq(.01,.99,by=.01)),seq(.99,.01,by=-.01),col="red")
    lines(predict(sWei, newdata=list(sex=2),type="quantile",p=seq(.01,.99,by=.01)),seq(.99,.01,by=-.01),col="red")

这里的技巧是颠倒绘制与预测的分位数顺序.可能有更好的方法来做到这一点，但它在这里有效.祝你好运！

The trick here was reversing the quantile orders for plotting vs predicting. There is likely a better way to do this, but it works here. Good luck!

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