我是R编程的新手,我没有得到一个解决我使用nls函数时出现的错误的方法。 我尝试将ecdf中的数据(值被提取并保存在y中)与这个函数模型拟合,有四个参数:
fitsim <- nls(y ~ exp(-(((a-Abfluss)/(c*(Abfluss-b)))^d)),
start = list( a=max(Abfluss), b=min(Abfluss),
c=3, d=1))
当我启动nls功能时,会发生以下错误:
Error in numericDeriv(form[[3L]], names(ind), env) :
Fehlender Wert oder etwas Unendliches durch das Modell erzeugt
这意味着存在缺失值,通过模型生成具有无穷大的某个值。
我的向量Abfluss和y都有相同的长度。目的是获得参数估计。
也许问题是,该模型只能在这种条件下工作:
c> 0,d> 0,b <= Abfluss&lt; = a。
我已经尝试了na.rm=True
命令。然后出现另一个错误:
Error in model.frame.default(formula = ~y + Abfluss, na.rm = TRUE) :
Variablenlängen sind unterschiedlich (gefunden für '(na.rm)')
这意味着,变量的长度是不同的。
我感谢各种帮助和建议。
为了更好地理解,我将整个代码附加到整个数据中:
time<-c(1851:2013)
Abfluss<- c(4853,4214,5803,3430,4645,4485,3100,4797,4030,3590,5396,9864,3683,4485,4064,3420,5396,
4895,3931,4238,3790,3520,4263,5474,3790,4700,5109,4525,4007,6340,4993,6903,8160,3600,3480,3540,
3540,4565,3333,7764,
4755,7940,3112,3169,4435,5365,9422,3150,10500,4512,3790,4618,6126,3769,3704,
5938,5669,4552,5458,5854,4867,6057,4783,5753,5736,4618,6091,5820,5007,7984, 4435,
4645,7465,5820,5988,6022,4300,6062,3302,4877,4586,5275,4410,3174,4966,4939,4638,
5541,5760,6495,5435,4952,4912,6092,5182,5820,5129,6436,6648,3063,5550,5160,4400,
9600,6400,6380,6300,6180,6899,4360,5550,4580,3894,5277,7520,6780,5100,5430,4550,
6620,4050,4560,5290,6610,8560,4943,6940,4744,6650,5700,7440,6200,4597,3697,7300,
4644,5456,6302,3741,5398,9500,6296,5279,5923,6412,6559,6559,5891,5737,5010,5790,
10300,4150,4870,6740,7560,8010,5120,8170,7430, 7330,5900, 11150)
#EV4-Distribution
dEV4 <- function(x, a, b, c,d) {
m<-exp(-(((a-Abfluss)/(c*(Abfluss-b)))^d))
return(m)
}
#Simulation example
Sim<-dEV4(Abfluss,a=max(Abfluss),b=min(Abfluss), c=3, d=1)
dEV4cdf<-cbind(Abfluss,Sim)
#Empirical cdf
p = ecdf(Abfluss)
y<- p(Abfluss) #Extracting of cumulated probabilities
m<-cbind(Abfluss,y)
#plot EV4 and ecdf
plot(dEV4cdf, type="p",main="EV4")
plot(ecdf(Abfluss), add=T)
#Fitting EV4 nls
fitsim <- nls(y ~ exp(-(((a-Abfluss)/(c*(Abfluss-b)))^d)),
start = list( a=max(Abfluss), b=min(Abfluss),
c=3, d=1), na.rm=TRUE)
答案 0 :(得分:2)
不要使用可行区域边界上的起始值,而是在nlmrt中尝试nlxb
(可以使用相同的参数,但data =
不可选):
library(nlmrt)
fitsim <- nlxb(y ~ exp(-(((a - Abfluss) / (c * (Abfluss - b))) ^ d)),
data = data.frame(y, Abfluss),
start = list(a = 2 * max(Abfluss), b = min(Abfluss) / 2, c = 3, d = 1))
plot(y ~ Abfluss, pch = 20)
o <- order(Abfluss)
fit <- y - fitsim$resid
lines(fit[o] ~ Abfluss[o], col = "red")
,并提供:
nlmrt class object: x
residual sumsquares = 0.02908 on 163 observations
after 5001 Jacobian and 6060 function evaluations
name coeff SE tstat pval gradient JSingval
a 20047.7 NA NA NA 1.119e-07 3251
b -1175384 NA NA NA 1.432e-09 0.1775
c 0.0129414 NA NA NA -0.1296 5.808e-06
d 12.146 NA NA NA -2.097e-06 6.798e-11