如何预测R中的生存概率?

问题描述

我将名为 veteran 的数据存储在 R 中.我创建了一个生存模型，现在希望预测生存概率.例如，具有 80 karno 值、10diagtime、age 65 和 prior=10 和 trt = 2 寿命超过 100 天?

I have data called veteran stored in R. I created a survival model and now wish to predict survival probability predictions. For example, what is the probability that a patient with 80 karno value, 10diagtime, age 65 and prior=10 and trt = 2 lives longer than 100 days?

在这种情况下，设计矩阵是 x = (1,0,1,0,80,10,65,10,2)

In this case the design matrix is x = (1,0,1,0,80,10,65,10,2)

这是我的代码:

library(survival)
attach(veteran)
weibull <- survreg(Surv(time,status)~celltype + karno+diagtime+age+prior+trt ,dist="w")

这是输出:

知道如何预测生存概率吗?

Any idea how to predict the survival probabilities?

推荐答案

您可以使用 predict.survreg 生成单个案例的预测生存时间(您将向 newdata 传递值)不同的分位数:

You can get predict.survreg to produce predicted times of survival for individual cases (to which you will pass values to newdata) with varying quantiles:

 casedat <- list(celltype="smallcell", karno =80, diagtime=10, age= 65 , prior=10 , trt = 2)
 predict(weibull, newdata=casedat,  type="quantile", p=(1:98)/100)
 [1]   1.996036   3.815924   5.585873   7.330350   9.060716  10.783617
 [7]  12.503458  14.223414  15.945909  17.672884  19.405946  21.146470
[13]  22.895661  24.654597  26.424264  28.205575  29.999388  31.806521
[19]  33.627761  35.463874  37.315609  39.183706  41.068901  42.971927
[25]  44.893525  46.834438  48.795420  50.777240  52.780679  54.806537
[31]  56.855637  58.928822  61.026962  63.150956  65.301733  67.480255
[37]  69.687524  71.924578  74.192502  76.492423  78.825521  81.193029
[43]  83.596238  86.036503  88.515246  91.033959  93.594216  96.197674
[49]  98.846083 **101.541291** 104.285254 107.080043 109.927857 112.831032
[55] 115.792052 118.813566 121.898401 125.049578 128.270334 131.564138
[61] 134.934720 138.386096 141.922598 145.548909 149.270101 153.091684
[67] 157.019655 161.060555 165.221547 169.510488 173.936025 178.507710
[73] 183.236126 188.133044 193.211610 198.486566 203.974520 209.694281
[79] 215.667262 221.917991 228.474741 235.370342 242.643219 250.338740
[85] 258.511005 267.225246 276.561118 286.617303 297.518110 309.423232
[91] 322.542621 337.160149 353.673075 372.662027 395.025122 422.263020
[97] 457.180183 506.048094
#asterisks added

然后您可以找出哪个比指定的时间长，并且它看起来在第 50 个百分位左右，就像人们在家庭作业问题中所期望的那样.

You can then figure out which one is greater than the specified time and it looks to be around the 50th percentile, just as one might expect from a homework question.

png(); plot(x=predict(weibull, newdata=casedat,  type="quantile",
             p=(1:98)/100),  y=(1:98)/100 , type="l")
dev.off()

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