Question

我需要手动计算Cox PH模型的线性预测因子。

我可以获取连续变量和二进制变量，以匹配predict.coxph（指定“ lp”）的输出，但是我似乎无法弄清楚如何为两个以上级别的分类变量进行计算。

我的目标是在我自己的数据中评估已发布模型的校准-我只有系数，因此需要能够手工完成。

这篇关于stackoverflow的文章描述了如何计算连续变量...

（Coxph predictions don't match the coefficients）

任何建议将不胜感激！谢谢

R示例...

URL   <- "http://socserv.mcmaster.ca/jfox/Books/Companion/data/Rossi.txt"
Rossi <- read.table(URL, header=TRUE)                  
summary(Rossi[,c("week", "arrest", "fin")])
 #      week           arrest        fin     
 # Min.   : 1.00   Min.   :0.0000   no :216  
 # 1st Qu.:50.00   1st Qu.:0.0000   yes:216  
 # Median :52.00   Median :0.0000            
 # Mean   :45.85   Mean   :0.2639            
 # 3rd Qu.:52.00   3rd Qu.:1.0000            
 # Max.   :52.00   Max.   :1.0000 

library(survival)     

#for binary variable
fitCPH <- coxph(Surv(week, arrest) ~ fin, data=Rossi)    #Cox-PH model
(coefCPH <- coef(fitCPH))                               # estimated coefficients
#   finyes 
#-0.3690692 

head(predict(fitCPH,type="lp"))
#[1]  0.1845346  0.1845346  0.1845346 -0.1845346  0.1845346  0.1845346

head(((as.numeric(Rossi$fin) - 1) - mean(as.numeric(Rossi$fin) - 1))    * coef(fitCPH))
#[1]  0.1845346  0.1845346  0.1845346 -0.1845346  0.1845346  0.1845346

#for categorical variable
set.seed(170981)
Rossi$categorical.example <-    as.factor(sample(1:3,nrow(Rossi),replace = TRUE))
summary(Rossi[,c("week", "arrest", "categorical.example")])
 #      week           arrest       categorical.example
 # Min.   : 1.00   Min.   :0.0000   1:156              
 # 1st Qu.:50.00   1st Qu.:0.0000   2:138              
 # Median :52.00   Median :0.0000   3:138              
 # Mean   :45.85   Mean   :0.2639                      
 # 3rd Qu.:52.00   3rd Qu.:1.0000                      
 # Max.   :52.00   Max.   :1.0000               

fitCPH2 <- coxph(Surv(week, arrest) ~ categorical.example, data=Rossi)    #Cox-PH model
(coefCPH2 <- coef(fitCPH2))  
#categorical.example2 categorical.example3 
#           -0.181790            -0.103019 

head(predict(fitCPH2,type="lp"))
#[1]  0.09098066 -0.01203832 -0.01203832  0.09098066 -0.09080938 -0.09080938

#How to calculate manually??

Answer 1

如果有人感兴趣，我想我已经知道了...

#How to calculate manually??
dv1 <- as.numeric(Rossi$categorical.example)-1    #make 0,1,2 rather than 1,2,3
dv1[dv1==2] <- 0

dv2 <- as.numeric(Rossi$categorical.example)-2    #make -1,0,1 rather than 1,2,3
dv2[dv2==-1] <- 0

meandv1  <- mean(dv1)   
meandv2  <- mean(dv2) 

head(((dv1-meandv1)*coefCPH2 [1])+((dv2-meandv2)*coefCPH2 [2]))
#[1]  0.09098066 -0.01203832 -0.01203832  0.09098066 -0.09080938 -0.09080938

all(round(predict(fitCPH2,type="lp"),5)==round(((dv1-meandv1)*coefCPH2 [1])+((dv2-meandv2)*coefCPH2 [2]),5))
#[1] TRUE

仅来自predict.coxph的线性预测因子

1 个答案: