similar to: products and polynomials in formulae

This very likely falls in the category of an unexpected result due to user ignorance. I generated the following data: time <- 0:10 set.seed(4302007) y <- 268 + -9*time + .4*(time^2) + rnorm(11, 0, .1) I then fit models using both orthogonal and raw polynomials: fit1 <- lm(y ~ poly(time, 2)) fit2 <- lm(y ~ poly(time, degree=2, raw=TRUE)) > predict(fit1, data.frame(time =

Hi Folks! Im using the function poly to generate orthogonal polynomials, but Id like to see the actual polynomials so that I could convert it to a polynomial in my original variable. Is that possible and if so how do I do it? /E

Dear All, I have found in the poly help this sentence: The orthogonal polynomial is summarized by the coefficients, which can be used to evaluate it via the three-term recursion given in Kennedy & Gentle (1980, pp. 343–4), and used in the predict part of the code. My question: which type of orthogonal polynomials are used by this function? Hrmite, legendre.. TIA Giovanni [[alternative HTML

How can ordinary polynomial coefficients be calculated from an orthogonal polynomial fit? I'm trying to do something like find a,b,c,d from lm(billions ~ a+b*decade+c*decade^2+d*decade^3) but that gives: "Error in eval(expr, envir, enclos) : Object "a" not found" > decade <- c(1950, 1960, 1970, 1980, 1990) > billions <- c(3.5, 5, 7.5, 13, 40) > #

Hello everyone, I hope this is a fair enough question, but I don’t have access to a copy of Bates and Pinheiro. It is probably quite obvious but the answer might be of general interest. If I fit a fixed effect with an added quadratic term and then do it as an orthogonal polynomial using maximum likelihood I get the expected result- they have the same logLik.

Looking to the wonderful statistical advice that this group can offer. In behavioral science applications of stats, we are often introduced to coefficients for orthogonal polynomials that are nice integers. For instance, Kirk's experimental design book presents the following coefficients for p=4: Linear -3 -1 1 3 Quadratic 1 -1 -1 1 Cubic -1 3 -3 1 In R orthogonal

Dear All, Need some help in polynomials transformation to get the coefficients. I have tried "poly.transform" as applied in S-plus but it does not work. Thanks in advanced for any helps. Regards. Abd. Rahman Kassim (PhD) Head Forest Ecology Branch Forest Management & Ecology Program Forestry and Conservation Division Forest Research Institute Malaysia Kepong 52109 Selangor,

Folks, I'm doing fine with using orthogonal polynomials in a regression context: # We will deal with noisy data from the d.g.p. y = sin(x) + e x <- seq(0, 3.141592654, length.out=20) y <- sin(x) + 0.1*rnorm(10) d <- lm(y ~ poly(x, 4)) plot(x, y, type="l"); lines(x, d$fitted.values, col="blue") # Fits great! all.equal(as.numeric(d$coefficients[1] + m

As usual, the R newsgroup set me straight (thanks to Douglas Bates, Robert Balshaw and Albyn Jones). There is really no difference between using orthogonal polynomials of the form: Linear -3 -1 1 3 Quadratic 1 -1 -1 1 Cubic -1 3 -3 1 Versus > poly(c(1:4),3) 1 2 3 [1,] -0.6708204 0.5 -0.2236068 [2,] -0.2236068 -0.5 0.6708204 [3,] 0.2236068

When doing polynomial regression I believe it is a good idea to use the poly function to generate orthogonal polynomials. When doing this in Splus there is a handy function (transform.poly I think) to convert the coefficients produced by regression with the poly function back to the original scale. Has somebody written something similar for R ? Robert

What does the warning message "1: Singular precision matrix in level -1, block 1" mean? I get this warning 50+ times when I try to fit the following model lme( response ~ covariateA + poly(covariateB,3), ~poly(covariateB,3)|group ) It's not a small dataset - a set of up to 20 blood pressure readings on just over 2000 people, and I don't get the error message when I try to fit

Full_Name: Renaud Lancelot Version: Version 2.3.0 (2006-04-24) OS: MS Windows XP Pro SP2 Submission from: (NULL) (82.239.219.108) I think there is a bug in predict.lme, when a polynomial generated by poly() is used as an explanatory variable, and a new data.frame is used for predictions. I guess this is related to * not * using, for predictions, the coefs used in constructing the orthogonal

Hi Everyone, I'm continuing to run into trouble with polyfit. I'm using the fitting function of the form; fit <- lm(y ~ poly(x,degree,raw=TRUE)) and I have found that in some cases a polynomial of certain degree can't be fit, the coefficient won't be calculated, because of a singularity. If I use orthogonal polynomials I can fit a polynomial of any degree, but I don't get

Many thanks for your very useful comments and suggestions. Renaud 2006/5/30, Prof Brian Ripley <ripley at stats.ox.ac.uk>: > On Tue, 30 May 2006, Prof Brian Ripley wrote: > > > This is not really a bug. See > > > > http://developer.r-project.org/model-fitting-functions.txt > > > > for how this is handled in other packages. All model-fitting in R used =

Dear Keith, >>>>> <Keith.Jewell at campdenbri.co.uk> >>>>> on Thu, 16 Jul 2015 08:58:11 +0000 writes: > Dear R Core Team, > Last week I made a post to the R-help mailing list > ?predict.poly for multivariate data? > <https://stat.ethz.ch/pipermail/r-help/2015-July/430311.html> > but it has had no responses so I?m

Dear all, Can any one tell me how can i perform Orthogonal Polynomial Regression parameter estimation in R? -------------------------------------------- Here is an "Orthogonal Polynomial" Regression problem collected from Draper, Smith(1981), page 269. Note that only value of alpha0 (intercept term) and signs of each estimate match with the result obtained from coef(orth.fit). What

R users, I was a bit surprised to find that when I attempted to add a polynomial term to a linear model using either lm or glm as could be done in S resulted in a fit without that term included and without warning(!!), e.g. > lm(response ~ x + x^2, data). As far as I can gather, there is no poly() yet in R, and if lm/glm do not allow functions of variables as their formula arguements, is

By accident, I left out the lines defining n1 and n2. Here it is as a function. Peter B. hotelling <- function(d1,d2){ k <- ncol(d1) n1 <- nrow(d1) n2 <- nrow(d2) xbar1 <- apply(d1,2,mean) xbar2 <- apply(d2,2,mean) dbar <- xbar2-xbar1 v <- ((n1-1)*var(d1)+(n2-1)*var(d2))/(n1+n2-2) t2 <- n1*n2*dbar%*%solve(v)%*%dbar/(n1+n2) f <-

Hi, I am curious about how to interpret the results of a polynomial regression-- using poly(raw=TRUE) vs. poly(raw=FALSE). set.seed(123456) x <- rnorm(100) y <- jitter(1*x + 2*x^2 + 3*x^3 , 250) plot(y ~ x) l.poly <- lm(y ~ poly(x, 3)) l.poly.raw <- lm(y ~ poly(x, 3, raw=TRUE)) s <- seq(-3, 3, by=0.1) lines(s, predict(l.poly, data.frame(x=s)), col=1) lines(s,

Dear all, I am using poly() in lm() in the following form. 1> DelsDPWOS.lm3 <- lm(DelsPDWOS[,1] ~ poly(DelsPDWOS[,4],3)) 2> DelsDPWOS.I.lm3 <- lm(DelsPDWOS[,1] ~ poly(I(DelsPDWOS[,4]),3)) 3> DelsDPWOS.2.lm3 <- lm(DelsPDWOS[,1]~DelsPDWOS[,4]+I(DelsPDWOS[,4]^2)+I(DelsPDWOS[,4]^3)) 1 and 2 lead to identical but wrong results. 3 is correct. Surprisingly (to me) the residuals