similar to: Find the prediction or the fitted values for an lm model

Suppose you have a vector of data in x and response values in y. How do you plot together both the points (x,y) and the curve that results from the fitted model, if the model is not y ~ x, but a higher order polynomial, e.g. y~poly(x,2)? (In other words, abline doesn't work for this case.) Thanks, --Paul -- Paul Lynch Aquilent, Inc. National Library of Medicine (Contractor)

Platform: WIN2000 Version of R: 1.6.2 I'm interested in plotting fitted values in a trellis xyplot. I believe the following should work; however, I only get the points (not the fitted lines). library(lattice) trellis.device(bg="white") xyplot(MULTDV~TIME|SUBNUM,data=TEMP, panel=function(x,y){ panel.xyplot(x,y) lines(x,fitted(lm(y~poly(x,1),na.action=na.omit)))

Hi R-community, I have the code as follows,i Fitted model as follows lbeer<-log(beer_monthly) t<-seq(1956,1995.2,length=length(beer_monthly)) #beer_monthly contains 400+ entries t2=t^2 beer_fit_parabola=lm(lbeer~t+t2) Below is not working for me. Please help me in preparing the new data set for the below prediction

I am fitting a regression model with a bs term and then making predictions based on the model. According to some info on the internet at http://www.stat.auckland.ac.nz/~yee/smartpred/DummiesGuide.txt there are some problems with using predict.lm when you have a model with terms such as bs,ns,or poly. However when I used one of the examples they said would illustrate the problems I get virtually

I have a question about some strange results I get when using lmer to build a logistic mixed effects model. I have a data set of about 30k points, and I'm trying to do backwards selection to reduce the number of fixed effects in my model. I've got 3 crossed random effects and about 20 or so fixed effects. At a certain point, I get a model (m17) where the fixed effects are like this

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

This smells like a bug to me. The error is triggered by the line: variables <- eval(predvars, data, env) inside model.frame.default(). At that point, na.action has not been applied, so poly() ended being called on data that still contains missing values. The qr() that issued the error is for generating the orthogonal basis when evaluating poly(), not for fitting the linear model itself.

This smells like a bug to me. The error is triggered by the line: variables <- eval(predvars, data, env) inside model.frame.default(). At that point, na.action has not been applied, so poly() ended being called on data that still contains missing values. The qr() that issued the error is for generating the orthogonal basis when evaluating poly(), not for fitting the linear model itself.

Often in practical situations a predictor has missing values, so that poly crashes. For instance: > x<-1:10 > y<- x - 3 * x^2 + rnorm(10)/3 > x[3]<-NA > lm( y ~ poly(x,2) ) Error in poly(x, 2) : missing values are not allowed in 'poly' > > lm( y ~ poly(x,2) , subset=!is.na(x)) # This does not help?!? Error in poly(x, 2) : missing values are not allowed in

The columns of the model matrix are all orthogonal. So the problem lies with poly(), not with lm(). > x = rep(1:5,3) y = rnorm(15) z <- model.matrix(lm(y ~ poly(x, 12))) x = rep(1:5,3) > y = rnorm(15) > z <- model.matrix(lm(y ~ poly(x, 12))) > round(crossprod(z),15) (Intercept) poly(x, 12)1 poly(x, 12)2 poly(x, 12)3 poly(x, 12)4 (Intercept)

Dear R experts, Excuse me if my question will be stupid... I'd like to fit data with x^2 polynomial: d <- read.table(file = "Oleg.dat", head = TRUE) d X T 3720.00 4.113 3715.00 4.123 3710.00 4.132 ... out <- lm(T ~ poly(X, 4), data = d) out Call: lm(formula = T ~ poly(X, 2), data = d) Coefficients: (Intercept) poly(X, 2)1 poly(X, 2)2

Hi, I can not comprehend the linear fitting results of polynoms. For example, given the following data (representing y = x^2): > x <- 1:3 > y <- c(1, 4, 9) performing a linear fit > f <- lm(y ~ poly(x, 2)) gives weird coefficients: > coefficients(f) (Intercept) poly(x, 2)1 poly(x, 2)2 4.6666667 5.6568542 0.8164966 However the fitted() result makes sense: >

Full_Name: Russell Lenth Version: 2.6.2 OS: Windows XP Pro Submission from: (NULL) (128.255.132.36) The poly() function allows a higher-degree polynomial than it should, when raw=FALSE. For example, consider 5 distinct 'x' values, each repeated twice. we can fit a polynomial of degree 8: ===== R> x = rep(1:5, 2) R> y = rnorm(10) R> lm(y ~ poly(x, 8)) Call: lm(formula = y ~

Hello, I'm trying to fit some points with a 8-degrees polynom (result of lm is stored in pfit). In most of the case, it is ok but for some others, some coefficients are "NA". I don't really understand the meaning of these "NA". And the problem is that I can't perform a derivation (pderiv<-as.function((deriv(polynomial(pfit$coefficients))))) on pfit due to the

Here's my little discussion example for a quadratic regression: http://pj.freefaculty.org/R/WorkingExamples/regression-quadratic-1.R Students press me to know the benefits of poly() over the more obvious regression formulas. I think I understand the theory on why poly() should be more numerically stable, but I'm having trouble writing down an example that proves the benefit of this. I

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) > #

> On Jul 13, 2017, at 7:43 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > Below. > > -- Bert > Bert Gunter > > > > On Thu, Jul 13, 2017 at 3:07 AM, Luigi Biagini <luigi.biagini at gmail.com> wrote: >> I have two ideas about it. >> >> 1- >> i) Entering variables in quadratic form is done with the command I >>

?? If I haven't misunderstood, they are completely different! 1) NIR must be a matrix, or poly(NIR,...) will fail. 2) Due to the previously identified bug in poly, degree must be explicitly given as poly(NIR, degree =2,raw = TRUE). Now consider the following example: > df <-matrix(runif(60),ncol=3) > y <- runif(20) > mdl1 <-lm(y~df*I(df^2)) > mdl2

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

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,