Hi Rui,
Looks like a bug:
###if(names(model)[1] = "(Intercept)")
Should this be:
if(names(model)[1] == "(Intercept)")
A.K.
----- Original Message -----
From: Rui Barradas <ruipbarradas at sapo.pt>
To: Mike Rennie <mikerennie.r at gmail.com>
Cc: r-help Mailing List <r-help at r-project.org>
Sent: Friday, March 1, 2013 3:18 PM
Subject: Re: [R] solving x in a polynomial function
Hello,
Try the following.
a <- 1:10
b <- c(1, 2, 2.5, 3, 3.5, 4, 6, 7, 7.5, 8)
dat <- data.frame(a = a, b = b)? # for lm(), it's better to use a df
po.lm <- lm(a~b+I(b^2)+I(b^3)+I(b^4), data = dat); summary(po.lm)
realroots <- function(model, b){
??? is.zero <- function(x, tol = .Machine$double.eps^0.5) abs(x) < tol
??? if(names(model)[1] = "(Intercept)")
??? ??? r <- polyroot(c(coef(model)[1] - b, coef(model)[-1]))
??? else
??? ??? r <- polyroot(c(-b, coef(model)))
??? Re(r[is.zero(Im(r))])
}
r <- realroots(po.lm, 5.5)
predict(po.lm, newdata = data.frame(b = r))? # confirm
Hope this helps,
Rui Barradas
Em 01-03-2013 18:47, Mike Rennie escreveu:
Hi there,
Does anyone know how I solve for x from a given y in a polynomial
function? Here's some example code:
##example file
a<-1:10
b<-c(1,2,2.5,3,3.5,4,6,7,7.5,8)
po.lm<-lm(a~b+I(b^2)+I(b^3)+I(b^4)); summary(po.lm)
(please ignore that the model is severely overfit- that's not the point).
Let's say I want to solve for the value b where a = 5.5.
Any thoughts? I did come across the polynom package, but I don't think
that does it- I suspect the answer is simpler than I am making it out
to be. Any help would be welcome.