绘制Kaplan-Meier进行Cox回归 [英] Plot Kaplan-Meier for Cox regression

问题描述

我有一个Cox比例风险模型，该模型使用R中的以下代码来预测死亡率。添加协变量A，B和C只是为了避免混淆（即年龄，性别，种族），但我们对预测变量X确实很感兴趣。X是连续变量。

  cox.model<-coxph（生存（时间，死亡）〜A + B + C + X，数据= df）
  

现在，我很难为此绘制Kaplan-Meier曲线。我一直在寻找如何创建这个数字的方法，但是运气不高。我不确定是否可以为Cox模型绘制Kaplan-Meier？ Kaplan-Meier是否针对我的协变量进行调整还是不需要它们？

我在下面尝试过，但被告知这是不对的。

 图（survfit（cox.model），xlab =时间（年），ylab =生存概率）$我还试图绘制一个数字，显示死亡的累积危害。我不知道自己是否做对了，因为我尝试了几种不同的方法并获得了不同的结果。理想情况下，我想绘制两条线，一条线显示X的第75个百分位数的死亡风险，另一条线显示X的第25个百分位数。我该怎么做？
 
 
 
我可以列出我尝试过的所有其他内容，但我不想让任何人感到困惑！
 
 
 
非常感谢。
解决方案
以下是从 
 
 
 
我们得到了一个图（置信区间为95％）存活率。对于累积危害率，您可以执行以下操作
 ＃plot（survfit（mod.allison）$ cumhaz）
  
但这不会给出置信区间。但是，不用担心！我们知道H（t）= -ln（S（t）），我们有S（t）的置信区间。我们所需要做的就是
  sfit<-survfit（mod.allison）
 cumhaz.upper<- -log（sfit $ upper）
 cumhaz.lower<---log（sfit $ lower）
 cumhaz<-sfit $ cumhaz＃与-log（sfit $ surv）
  
然后只需绘制这些
 图（cumhaz，xlab =未来几周，ylab =累积危害，
 ylim = c（min（cumhaz.lower），max（cumhaz.upper）））
线（cumhaz .lower）
行（cumhaz.upper）
  
  
 
 
 
您将要使用 survfit（...，conf.int = 0.50）获取乐队75％和25％，而不是97.5％和2.5％。
 
I have a Cox proportional hazards model set up using the following code in R that predicts mortality. Covariates A, B and C are added simply to avoid confounding (i.e. age, sex, race) but we are really interested in the predictor X. X is a continuous variable.
cox.model <- coxph(Surv(time, dead) ~ A + B + C + X, data = df)
Now, I'm having troubles plotting a Kaplan-Meier curve for this. I've been searching on how to create this figure but I haven't had much luck. I'm not sure if plotting a Kaplan-Meier for a Cox model is possible? Does the Kaplan-Meier adjust for my covariates or does it not need them?

What I did try is below, but I've been told this isn't right.
plot(survfit(cox.model), xlab = 'Time (years)', ylab = 'Survival Probabilities')
I also tried to plot a figure that shows cumulative hazard of mortality. I don't know if I'm doing it right since I've tried it a few different ways and get different results. Ideally, I would like to plot two lines, one that shows the risk of mortality for the 75th percentile of X and one that shows the 25th percentile of X. How can I do this?

I could list everything else I've tried, but I don't want to confuse anyone!

Many thanks.
解决方案
Here is an example taken from this paper.
url <- "http://socserv.mcmaster.ca/jfox/Books/Companion/data/Rossi.txt"
Rossi <- read.table(url, header=TRUE)
Rossi[1:5, 1:10]

#   week arrest fin age  race wexp         mar paro prio educ
# 1   20      1  no  27 black   no not married  yes    3    3
# 2   17      1  no  18 black   no not married  yes    8    4
# 3   25      1  no  19 other  yes not married  yes   13    3
# 4   52      0 yes  23 black  yes     married  yes    1    5
# 5   52      0  no  19 other  yes not married  yes    3    3

mod.allison <- coxph(Surv(week, arrest) ~ 
                        fin + age + race + wexp + mar + paro + prio,
                        data=Rossi)
mod.allison

# Call:
# coxph(formula = Surv(week, arrest) ~ fin + age + race + wexp + 
#    mar + paro + prio, data = Rossi)
#
#
#                   coef exp(coef) se(coef)      z      p
# finyes         -0.3794     0.684   0.1914 -1.983 0.0470
# age            -0.0574     0.944   0.0220 -2.611 0.0090
# raceother      -0.3139     0.731   0.3080 -1.019 0.3100 
# wexpyes        -0.1498     0.861   0.2122 -0.706 0.4800
# marnot married  0.4337     1.543   0.3819  1.136 0.2600
# paroyes        -0.0849     0.919   0.1958 -0.434 0.6600
# prio            0.0915     1.096   0.0286  3.194 0.0014
#
# Likelihood ratio test=33.3  on 7 df, p=2.36e-05  n= 432, number of events= 114    
Note that the model uses fin, age, race, wexp, mar, paro, prio to predict arrest. As mentioned in this document the survfit() function uses the Kaplan-Meier estimate for the survival rate.
plot(survfit(mod.allison), ylim=c(0.7, 1), xlab="Weeks",
     ylab="Proportion Not Rearrested")


We get a plot (with a 95% confidence interval) for the survival rate. For the cumulative hazard rate you can do
# plot(survfit(mod.allison)$cumhaz)
but this doesn't give confidence intervals. However, no worries! We know that H(t) = -ln(S(t)) and we have confidence intervals for S(t). All we need to do is
sfit <- survfit(mod.allison)
cumhaz.upper <- -log(sfit$upper)
cumhaz.lower <- -log(sfit$lower)
cumhaz <- sfit$cumhaz # same as -log(sfit$surv)
Then just plot these
plot(cumhaz, xlab="weeks ahead", ylab="cumulative hazard",
     ylim=c(min(cumhaz.lower), max(cumhaz.upper)))
lines(cumhaz.lower)
lines(cumhaz.upper)


You'll want to use survfit(..., conf.int=0.50) to get bands for 75% and 25% instead of 97.5% and 2.5%.

                        
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