我不是 R 方面的专家，我正在尝试将热死亡细菌数据拟合到威 bool 模型中:
log._CFU.ml=c(9.6,9.2,9,8.5,9,7.5,6.8,5.8)
time_min=c(0,10,15,20,25,30,35,45)
ecoli_52=data.frame(log._CFU.ml,time_min)
weibull=function(x,a,b){-(1/2.303)*((x/a)^b)} #first i defined the function
然后，我应用了nls:
mod.ecoli_52=nls(ecoli_52$log._CFU.ml~weibull(ecoli_52$time_min,a,b),
ecoli_52,start=list(a=1,b=1))
之后，我发现了错误
Error in numericDeriv(form[[3L]], names(ind), env) :
Missing value or an infinity produced when evaluating the model
我确信这个错误与 start 中的 a 和 b 值有关。是否存在估计这些值的方法？
最佳答案
如果N是每毫升菌落形成单位的数量，N0是时间 0 时的数字，那么您可能想要模型 log(N/N0) ~ ...而不是log(N) ~ ...自log(N/N0) = log(N) - log(N0)以来我们有:
log_ratio <- with(ecoli_52, log._CFU.ml - log._CFU.ml[1])
fm <- nls(log_ratio ~ weibull(time_min, a, b), ecoli_52, start = list(a = 1, b = 1))
plot(log_ratio ~ time_min, ecoli_52)
lines(fitted(fm) ~ time_min, ecoli_52, col = "red")
给予:
> fm
Nonlinear regression model
model: log_ratio ~ weibull(time_min, a, b)
data: ecoli_52
a b
13.314 1.804
residual sum-of-squares: 0.7894
Number of iterations to convergence: 7
Achieved convergence tolerance: 6.363e-07
关于r - 使用 nls 拟合威 bool 模型，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/34082396/