好的，所以我将使用 DWin 描述的 R 帮助在此处发布答案。使用rweibullR 中的函数给出了 Weibull 分布的通常形式，其累积函数为：

$$F(x)=1-\exp(-\left ( \frac{x}{b}\right )^a)$$

所以我们将形状参数表示rweibull为$a$和比例参数rweibullby$b$.

现在的问题是 的输出survreg给出的形状和比例参数与 的形状和比例参数不同rweibull。让我们将形状参数表示survreg为$a_s$和比例参数survregby$b_s$.

然后，?survreg我们有：

survreg 的比例 = 1/(rweibull 形状)

survreg 的截距 = log(rweibull scale)

所以这给了我们：

$$a=\frac{1}{b_s}\quad \mbox{and} \quad b=\exp(a_s)$$

所以如果我们假设我们运行这个survreg函数$n$协变量，那么输出将是：

$\alpha_0,\ldots, \alpha_{n-1}$，协变量的系数和一些尺度参数$k$. 以标准形式给出的 Weibull 模型由下式给出：

$$F(x)=1-\exp\left (- \left (\frac{x}{\exp(\alpha_0+\alpha_1+\ldots +\alpha_{n-1})} \right ) ^{\frac{1}{k}}\right )$$

为了进一步清楚，我想添加一个带有代码示例的答案。

我们基本上追求的是获取survreg输出模型并从中导出生存函数。为了避免常见的符号混淆，我将继续展示执行此操作的代码：

fit <- survreg(Surv(time,status) ~ age, data=stanford2) # this is the survreg output model
survreg.lp <- predict(fit, type = "lp")
survreg.scale <- fit$scale

# this is the survival function!
S_t <- function(t, survreg.scale, survreg.lp){
  shape <- 1/survreg.scale 
  scale <- exp(survreg.lp)
  ans <- 1 - pweibull(t, shape = shape, scale = scale)
}

正如vkehayas R 的pweibull参数化所提到的：

$$F(x) = 1-exp(-\left(\frac{x}{b}\right)^a$$

其中a是威布尔分布形状，b是尺度。

然后我们得到a = 1/fit$scale了b = exp(predict(fit, type = "lp"))

我们可以在下面验证导出的生存函数

# next let's verify it's correct:
fit <- survreg(Surv(time,status) ~ age, data=stanford2) # this is the survreg output model

# this is the survival function!
S_t <- function(t, survreg.scale, survreg.lp){
  shape <- 1/survreg.scale 
  scale <- exp(survreg.lp)
  ans <- 1 - pweibull(t, shape = shape, scale = scale)
}

new_dat <- data.frame(age = c(0, seq(min(stanford2$age), max(stanford2$age), length.out = 10)))

pct <- seq(0.01, 0.99, 0.01)

surv_curves <- sapply(pct,
                      function(x) predict(fit, type = "quantile", p = 1 - x,
                                          newdata = new_dat))

matplot(y = pct, t(surv_curves), type = "l")

# you can vary the below subject_i variable to see it works for all of them
subject_i <- 1
single_curve <- surv_curves[subject_i, ]
plot(single_curve, pct, type = "l") # this is we know to be true
  
times <- round(seq(1, max(single_curve), length.out = 100))
lp <- predict(fit, newdata = new_dat, type = "lp")[subject_i]
surv <- sapply(times, function(t) S_t(t, survreg.scale = fit$scale, survreg.lp = lp))
lines(times, surv, col = "red", lty = 2) # this is the new S_t function
# They match!

所以，总结一下：

a = 1/fit$scale和b = exp(predict(fit, type = "lp"))

希望这可以帮助。我知道在弄清楚这一点之前我拉了几根头发。

帮助页面?Weibull说：

具有形状参数 a 和尺度参数 b 的 Weibull 分布的密度为

f(x) = (a/b) (x/b)^(a-1) exp(- (x/b)^a)

然后帮助页面?survreg说：

# There are multiple ways to parameterize a Weibull distribution. The survreg 
# function embeds it in a general location-scale family, which is a 
# different parameterization than the rweibull function, and often leads
# to confusion.
#   survreg's scale  =    1/(rweibull shape)
#   survreg's intercept = log(rweibull scale)

并且在示例部分中详细说明了这是如何处理的 ?survreg.distributions