df <- data.frame(
time = c(0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17),
var = c(12.69,16.35,20.29,25.08,30.81,38.75,45,49.16,55.15,62.852,68.63,76.64,82.47,85.68,89.14,91.86,95.28,98.17)
)
logisticmodel <- nls(var ~ SSlogis(time, phi1, phi2, phi3), data = df)
summary(logisticmodel)
coef(logisticmodel)
#predict(logisticmodel, data.frame(time = 18))
R给出的输出如下:
phi1 phi2 phi3
105.737368 7.432555 3.852865
但是the website给了我们:
我知道有些语言有不同的输出。正常,但我想知道您的想法是什么?
提前感谢。
问题在于,不同的模型都适用。
具有自启动功能nls
的SSlogis
适合该型号(请参见help('SSlogis')
)
Asym/(1+exp((xmid-input)/scal))
或使用您的符号,
phi1/(1 + exp((phi2 - input)/phi3))
笔和纸表明,以下转换在网页中给出了结果。
fit <- nls(var ~ SSlogis(time, phi1, phi2, phi3), data = df)
kappa <- coef(fit)[1]
alpha <- exp(coef(fit)[2]/coef(fit)[3])
beta <- 1/coef(fit)[3]
c(kappa = unname(kappa), alpha = unname(alpha), beta = unname(beta))
# kappa alpha beta
#105.7373679 6.8832991 0.2595471
因此,为了使其自动化,编写一个简单的函数。
transf <- function(x){
kappa <- coef(x)[1]
alpha <- exp(coef(x)[2]/coef(x)[3])
beta <- 1/coef(x)[3]
c(kappa = unname(kappa), alpha = unname(alpha), beta = unname(beta))
}
transf(fit)