Re: [R] Find the prediction or the fitted values for an lm model
See in-line below. On 11/28/13 20:50, jpm miao wrote: Hi, I would like to fit my data with a 4th order polynomial. Now I have only 5 data point, I should have a polynomial that exactly pass the five point Then I would like to compute the "fitted" or "predict" value with a relatively large x dataset. How can I do it? BTW, I thought the model "prodfn" should pass by (0,0), but I just wonder why the const is unequal to zero Because poly() produces orthonormalized polynomials, Look at poly(x1,4). It is not much like cbind(x1,x1^2,x1^3,x1^4), is it? cheers, Rolf Turner x1<-c(0,3,4,5,8) y1<-c(0,1,4,7,8) prodfn<-lm(y1 ~ poly(x1, 4)) x<-seq(0,8,0.01) temp<-predict(prodfn,data.frame(x=x)) # This line does not work.. prodfn Call: lm(formula = y1 ~ poly(x1, 4)) Coefficients: (Intercept) poly(x1, 4)1 poly(x1, 4)2 poly(x1, 4)3 poly(x1, 4)4 4.000e+00 6.517e+00-4.918e-16-2.744e+00-8.882e-16 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Find the prediction or the fitted values for an lm model
On Thu, 28 Nov 2013, jpm miao wrote: Hi, I would like to fit my data with a 4th order polynomial. Now I have only 5 data point, I should have a polynomial that exactly pass the five point Then I would like to compute the "fitted" or "predict" value with a relatively large x dataset. How can I do it? BTW, I thought the model "prodfn" should pass by (0,0), but I just wonder why the const is unequal to zero x1<-c(0,3,4,5,8) y1<-c(0,1,4,7,8) prodfn<-lm(y1 ~ poly(x1, 4)) x<-seq(0,8,0.01) temp<-predict(prodfn,data.frame(x=x)) # This line does not work.. You need to call the variable x1 because that is the name you used in the original data: plot(x, predict(prodfn,data.frame(x1=x)), type = "l") points(x1, y1) prodfn Call: lm(formula = y1 ~ poly(x1, 4)) Coefficients: (Intercept) poly(x1, 4)1 poly(x1, 4)2 poly(x1, 4)3 poly(x1, 4)4 4.000e+00 6.517e+00-4.918e-16-2.744e+00-8.882e-16 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Find the prediction or the fitted values for an lm model
Hi, I would like to fit my data with a 4th order polynomial. Now I have only 5 data point, I should have a polynomial that exactly pass the five point Then I would like to compute the "fitted" or "predict" value with a relatively large x dataset. How can I do it? BTW, I thought the model "prodfn" should pass by (0,0), but I just wonder why the const is unequal to zero x1<-c(0,3,4,5,8) y1<-c(0,1,4,7,8) prodfn<-lm(y1 ~ poly(x1, 4)) x<-seq(0,8,0.01) temp<-predict(prodfn,data.frame(x=x)) # This line does not work.. > prodfn Call: lm(formula = y1 ~ poly(x1, 4)) Coefficients: (Intercept) poly(x1, 4)1 poly(x1, 4)2 poly(x1, 4)3 poly(x1, 4)4 4.000e+00 6.517e+00-4.918e-16-2.744e+00-8.882e-16 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
