[英]Plot survival and hazard function of survreg using curve()
I have the following survreg model: 我有以下幸存模型:
Call:
survreg(formula = Surv(time = (ev.time), event = ev) ~ age,
data = my.data, dist = "weib")
Value Std. Error z p
(Intercept) 4.0961 0.5566 7.36 1.86e-13
age 0.0388 0.0133 2.91 3.60e-03
Log(scale) 0.1421 0.1208 1.18 2.39e-01
Scale= 1.15
Weibull distribution
I would like to plot the hazard function and the survival function based on the above estimates. 我想根据上述估计绘制危险函数和生存函数。
I don't want to use predict()
or pweibull()
(as presented here Parametric Survival or here SO question . 我不想使用
predict()
或pweibull()
(如此处所示的Parametric Survival或这里的SO问题 。
I would like to use the curve()
function. 我想使用
curve()
函数。 Any ideas how I can accomplish this? 我有什么想法可以做到这一点? It seems the Weibull function of the survreg uses other definitions of scale and shape than the usual (and different that for example rweibull).
似乎幸存者的Weibull函数使用了比平常更多的尺度和形状定义(并且与例如rweibull不同)。
UPDATE: I guess what I really require it to express hazard / survival as a function of the estimates Intercept
, age (+ other potential covariates)
, Scale
without using any ready made *weilbull
function. 更新:我想我真的需要它来表达危害/存活的估计的函数
Intercept
, age (+ other potential covariates)
, Scale
不使用任何现成的*weilbull
功能。
The first link you provided actually has a clear explanation on the theory of how this works, along with a lovely example. 您提供的第一个链接实际上对其工作原理有一个清晰的解释,以及一个可爱的例子。 (Thank you for this, it is a nice resource I will use in my own work.)
(谢谢你,这是我将在我自己的工作中使用的一个很好的资源。)
To use the curve
function, you will need to pass some function as an argument. 要使用
curve
函数,您需要将一些函数作为参数传递。 It is true that the *weibull
family of functions use a different parameterization for the Weibull than survreg
, but it can be easily transformed, as explained your first link. 确实,
*weibull
weibull函数系列对Weibull使用不同于survreg
参数survreg
,但它可以很容易地转换,如第一个链接所述。 Also, from the documentation in survreg
: 另外,从
survreg
的文档:
There are multiple ways to parameterize a Weibull distribution.
有多种方法可以参数化Weibull分布。 The survreg function imbeds it in a general location-scale familiy, which is a different parameterization than the rweibull function, and often leads to confusion.
幸存函数将其嵌入到一般位置规模的家族中,这是与rweibull函数不同的参数化,并且经常导致混淆。
survreg's scale = 1/(rweibull shape) survreg's intercept = log(rweibull scale)
Here is an implementation of that simple transformation: 以下是该简单转换的实现:
# The parameters
intercept<-4.0961
scale<-1.15
par(mfrow=c(1,2),mar=c(5.1,5.1,4.1,2.1)) # Make room for the hat.
# S(t), the survival function
curve(pweibull(x, scale=exp(intercept), shape=1/scale, lower.tail=FALSE),
from=0, to=100, col='red', lwd=2, ylab=expression(hat(S)(t)), xlab='t',bty='n',ylim=c(0,1))
# h(t), the hazard function
curve(dweibull(x, scale=exp(intercept), shape=1/scale)
/pweibull(x, scale=exp(intercept), shape=1/scale, lower.tail=FALSE),
from=0, to=100, col='blue', lwd=2, ylab=expression(hat(h)(t)), xlab='t',bty='n')
par(mfrow=c(1,1),mar=c(5.1,4.1,4.1,2.1))
I understand that you mentioned in your answer that you did not want to use the pweibull
function, but I am guessing that you did not want to use it because it uses a different parameterization. 我知道你在答案中提到你不想使用
pweibull
函数,但我猜你不想使用它,因为它使用了不同的参数化。 Otherwise, you could simply write your own version of pweibull
that uses that survreg
's parameterization: 否则,您可以简单地编写自己的
pweibull
版本,该版本使用了survreg
的参数化:
my.weibull.surv<-function(x,intercept,scale) pweibull(x,scale=exp(intercept),shape=1/scale,lower.tail=FALSE)
my.weibull.haz<-function(x,intercept,scale) dweibull(x, scale=exp(intercept), shape=1/scale) / pweibull(x,scale=exp(intercept),shape=1/scale,lower.tail=FALSE)
curve(my.weibull.surv(x,intercept,scale),1,100,lwd=2,col='red',ylim=c(0,1),bty='n')
curve(my.weibull.haz(x,intercept,scale),1,100,lwd=2,col='blue',bty='n')
As I mentioned in the comments, I don't know why you would do this (unless this is homework), but you could handcode pweibull
and dweibull
if you like: 正如我在评论中提到的,我不知道为什么你会这样做(除非这是家庭作业),但你可以手动
pweibull
和dweibull
如果你喜欢:
my.dweibull <- function(x,shape,scale) (shape/scale) * (x/scale)^(shape-1) * exp(- (x/scale)^shape)
my.pweibull <- function(x,shape,scale) exp(- (x/scale)^shape)
Those definitions come straight out of ?dweibull
. 这些定义直接来自
?dweibull
。 Now just wrap those, slower, untested functions instead of pweibull
and dweibull
directly. 现在直接包装那些较慢的,未经测试的函数而不是
pweibull
和dweibull
。
Your parameters are: 你的参数是:
scale=exp(Intercept+beta*x)
in your example and lets say for age=40 在您的示例中,
scale=exp(Intercept+beta*x)
,让我们说年龄= 40
scale=283.7
your shape parameter is the reciprocal of the scale that the model outputs 您的形状参数是模型输出的比例的倒数
shape=1/1.15
Then the hazard is: 然后危险是:
curve((shape/scale)*(x/scale)^(shape-1), from=0,to=12,ylab=expression(hat(h)(t)), col="darkblue",xlab="t", lwd=5)
The cumulative hazard function is: 累积危险函数是:
curve((x/scale)^(shape), from=0,to=12,ylab=expression(hat(F)(t)), col="darkgreen",xlab="t", lwd=5)
The Survival function is 1-the cumulative hazard function, so: 生存函数是1-累积危险函数,因此:
curve(1-((x/scale)^(shape)), from=0,to=12,ylab=expression(hat(S)(t)), col="darkred",xlab="t", lwd=5, ylim=c(0,1))
Also check out the eha
package, and the function hweibull
and Hweibull
还可以查看
eha
包,以及hweibull
和Hweibull
功能
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