Question

我在flexsurvreg中使用逆威布尔分布拟合了生存模型：

 if (require("actuar")){
  invweibull <- list(name="invweibull",
                     pars=c("shape","scale"),
                     location="scale",
                     transforms=c(log, log),
                     inv.transforms=c(exp, exp),
                     inits=function(t){ c(1, median(t)) })
  invweibull <- flexsurvreg(formula = kpnsurv~iaas, data = kpnrs2,
                            dist=invweibull)
  invweibull
}

我得到了以下输出：

    Call:
flexsurvreg(formula = kpnsurv ~ iaas, data = kpnrs2, dist = invweibull)

Estimates: 
 data.     mean.       est       L95%      U95%      se        exp(est)  L95%      U95%    
shape        NA     0.4870    0.4002    0.5927    0.0488        NA        NA        NA
scale        NA    62.6297   36.6327  107.0758   17.1371        NA        NA        NA
iaas     0.4470    -0.6764   -1.2138   -0.1391    0.2742    0.5084    0.2971    0.8701

N = 302,  Events: 54,  Censored: 248
Total time at risk: 4279
Log-likelihood = -286.7507, df = 3
AIC = 579.5015

如何获得协变量估计值的p值（在本例中为iaas）？谢谢你的帮助。

Answer 1

以防万一这仍然对任何人有用，这对我有用。首先从模型中提取系数信息矩阵：

invweibull.res <- invweibull$res

然后将估计的系数除以它们的标准误差，以计算具有渐近标准正态分布的Wald统计量：

invweibull.wald <- invweibull.res[,1]/invweibull.res[,4]

最后，获取p值：

invweibull.p <- 2*pnorm(-abs(invweibull.wald))

flexsurvreg中估计值的p值

1 个答案: