我想找到一个能够确定危险率(死亡率)存活率的函数的最佳参数。
eh <- function(kappa,lambda,time,delay){
FoM <- -log(1-(1-exp(-kappa*(time+delay)))*(exp(-lambda*time)))
return(FoM)
}
survival <- function(k, l, t, d){
theSurvs <- array(1, c(1, length(t)))
S = array(1)
Qs <- array(0)
for(j in 1:length(t)){
S[1] = 1
for(i in 1:(t[j]+1)){
theEHs <- eh(k, l, i-1, d)
theQs <- hazards2Qs(theEHs)
S[i+1] <- S[i]-S[i]*theQs
}
theSurvs[j] <- S[i]
}
return(theSurvs)
}
hazards2Qs <- function(hazards){
Qs<- 1-exp(-hazards)
return(Qs)
}
timeX = 0 # time cycles of arbitrary length
kappaX = 0.001
lambdaX = 0.05
delayX = 50
theYears <- floor(runif(20)*100)
theYears
EHs <- eh(kappaX, lambdaX, theYears, delayX)
theS <- survival(kappaX, lambdaX, theYears, delayX)
theS
theData <- data.frame(theYears, t(theS))
theData
survival(kappaX, lambdaX, theYears, delayX)
mod <- nls(~ survival(kappa, lambda, theYears, delay), start=list(lambda=0.0015, kappa=0.0013, delay=48), data=theData, trace=1)
当我运行它时,我收到此错误:
3.647752 : 0.0015 0.0013 48.0000
Error in qr.qty(QR, resid) :
'qr' and 'y' must have the same number of rows
回溯():
4: stop("'qr' and 'y' must have the same number of rows")
3: qr.qty(QR, resid)
2: (function ()
{
if (npar == 0)
return(0)
rr <- qr.qty(QR, resid)
sqrt(sum(rr[1L:npar]^2)/sum(rr[-(1L:npar)]^2))
})()
> dim(theS)
[1] 1 20
> dim(theYears)
NULL
> dim(theData)
[1] 20 2
> typeof(theData)
[1] "list"
> typeof(theS)
[1] "double"
> typeof(theYears)
[1] "double"
我一直在挣扎一天而没有深究这一点。有什么想法吗?
答案 0 :(得分:1)
在花了很长时间试图弄清楚自己发生了什么之后,我认为问题在于你如何塑造你的survival
功能结果。我不确定你为什么要尝试返回数组,但是你应该在这里返回一个向量。所以只需将第一行改为
#OLD: theSurvs <- array(1, c(1, length(t)))
theSurvs <- rep.int(1, length(t)
这意味着您还必须更改
#OLD: theData <- data.frame(theYears, t(theS))
theData <- data.frame(theYears, theS)
通过返回这样的形状对象,它干扰了使用qr()
的渐变计算。尽可能远离单维数组,只需使用简单的向量。
现在,正如所说,解决这个问题似乎导致了另一个问题。似乎当它试图采用数值导数时,它会遇到NaN/Inf
值。您可以在return(theSurvs)
行上方添加此代码,以查看发生这种情况时调用的参数
if(any(!is.finite(theSurvs))) {
dput(c(k,l,d))
dput(t)
print(theSurvs)
}