R nls: fitting a curve to data

别等时光非礼了梦想. 提交于 2019-12-11 00:02:06

问题


I'm having trouble finding the right curve to fit to my data. If someone more knowledgeable than me has an idea/solution for a better fitting curve I would be really grateful.

Data: The aim is to predict x from y

dat <- data.frame(x = c(15,25,50,100,150,200,300,400,500,700,850,1000,1500),
                  y = c(43,45.16,47.41,53.74,59.66,65.19,76.4,86.12,92.97,
                        103.15,106.34,108.21,113) ) 

This is how far I've come:

model <- nls(x ~ a * exp( (log(2) / b ) * y),
             data = dat, start = list(a = 1, b = 15 ), trace = T)

Which is not a great fit:

dat$pred <- predict(model, list(y = dat$y))
plot( dat$y, dat$x, type = 'o', lty = 2)
points( dat$y, dat$pred, type = 'o', col = 'red')

Thanks, F


回答1:


Predicting x from y a 5th degree polynomial is not so parsimonius but does seem to fit:

fm <- lm(x ~ poly(y, 5), dat)
plot(x ~ y, dat)
lines(fitted(fm) ~ y, dat)

(continued after plot)

You could also consider the UCRS.5b model of the drc package:

library(drc)
fm <- drm(x ~ y, data = dat, fct = UCRS.5b())
plot(fm)

Note: Originally, I assumed you wanted to predict y from x and had written the answer below.

A cubic looks pretty good:

plot(y ~ x, dat)
fm <- lm(y ~ poly(x, 3), dat)
lines(fitted(fm) ~ x, dat)

(continued after plot)

A 4 parameter logistic also looks good:

library(drc)
fm <- drm(y ~ x, data = dat, fct = LL.4())
plot(fm)



来源:https://stackoverflow.com/questions/33478848/r-nls-fitting-a-curve-to-data

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