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在 RI 中,可以使用生存包中的 coxph 来拟合生存时间(timeevent)和自变量的 Cox 比例风险模型stat,但我不知道如何控制可能的影响,比如说,pol除了将其包含在模型作为另一个自变量。以下是两个模型,一个有控制变量,一个没有控制变量。

> colnames(first.data)
 [1] "state"           "year"            "event"           "time"          "contag_per"      "one_neigh_dis"  
 [7] "pol"         "demveto"         "dem_cont"        "tr.ma"           "stat"
> coxph = coxph(Surv(time, event) ~ stat, data = first.data, method = "breslow")
> summary(coxph)
Call:
coxph(formula = Surv(time, event) ~ stat, data = first.data, 
method = "breslow")

n= 48, number of events= 45 
(2 observations deleted due to missingness)

                  coef  exp(coef)   se(coef)      z Pr(>|z|)    
 stat -4.429e+01  5.835e-20  1.318e+01 -3.361 0.000777 ***
 ---
 Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

      exp(coef) exp(-coef) lower .95 upper .95
 stat 5.835e-20  1.714e+19 3.542e-31 9.612e-09

 Concordance= 0.708  (se = 0.059 )
 Rsquare= 0.239   (max possible= 0.997 )
 Likelihood ratio test= 13.12  on 1 df,   p=0.0002923
 Wald test            = 11.3  on 1 df,   p=0.000777
 Score (logrank) test = 11.77  on 1 df,   p=0.0006022

具有独立 ( stat) 和控制变量 ( pol)的第二个模型

 > coxph = coxph(Surv(time, event) ~ stat + pol, data = first.data, method = "breslow")
 > summary(coxph)
Call:
coxph(formula = Surv(time, event) ~ stat + pol, 
    data = first.data, method = "breslow")

  n= 47, number of events= 44 
   (3 observations deleted due to missingness)

                      coef  exp(coef)   se(coef)      z Pr(>|z|)   
stat -4.121e+01  1.264e-18  1.341e+01 -3.072  0.00212 **
pol          3.778e+00  4.372e+01  1.745e+00  2.165  0.03042 * 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

                exp(coef) exp(-coef) lower .95 upper .95
stat 1.264e-18  7.911e+17 4.830e-30 3.308e-07
pol         4.372e+01  2.287e-02 1.429e+00 1.337e+03

Concordance= 0.746  (se = 0.059 )
Rsquare= 0.318   (max possible= 0.997 )
Likelihood ratio test= 17.99  on 2 df,   p=0.0001242
Wald test            = 14.29  on 2 df,   p=0.0007899
Score (logrank) test = 15.27  on 2 df,   p=0.0004833

使用 R 生存包将 cox 比例风险模型与控制变量拟合还有哪些其他想法?

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