similar to: Improvements (?) in stats::poly and stats::polym.

Suppose, 'fr' is data.frame with columns 'Y', 'A' and 'B'. 'A' has levels 'Aa' 'Ab' and 'Ac', and 'B' has levels 'Ba', 'Bb', 'Bc' and 'Bd'. 'Y' columns are numbers. I tried the following three sets of commands. I understand that A*B is equivalent to A+B+A:B. However, A:B in A+B+A:B is

Hi, I think he's talking about how much a statistical estimator is influenced by extreme datapoints, e.g. https://en.m.wikipedia.org/wiki/Robust_statistics#Breakdown_point Olivier -- Olivier Crouzet Assistant Professor @LLING UMR6310 - Universit? de Nantes / CNRS Guest Scientist @UMCG - University Medical Center Groningen / RijksUniversiteit Groningen > Le 19 avr. 2018 ? 11:00,

Here is an example given from ?yags library(methods) data(stackloss) Y1 <- yags(stack.loss~Air.Flow,id=1:21, data=stackloss) How can I access parts of the output. I tried: > str(Y1) Formal class 'yagsResult' [package "yags"] with 25 slots ..@ coefficients : num [1:2] -44.13 1.02 ..@ coefnames : chr(0) > Y1$coefnames Error in Y1$coefnames : $

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 r-help, Is there any available description of the components of lqs objects found in the package "lqs"? > names(slts) [1] "crit" "sing" "coefficients" "bestone" [5] "fitted.values" "residuals" "scale" "terms" [9] "call"

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

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

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

DDear developeRs, about a year ago, Alex Stolpovsky posted an issue with predict.lm on a fit generated using poly with the raw=TRUE option and too few new data (slightly modified reproducible example below). Alex did not get any reply. I have just stumbled on the same problem, and I think that this is a bug of function poly, which arises from the check whether the polynomial degree is

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

Hello r-help, I'm fitting a model with lm() and using the orthogonal polynomials from poly() as my basis: dat <- read.csv("ConsolidatedData.csv", header=TRUE) attach(dat) nrows <- 1925 Rad <- poly(Radius, 2) ntheta <- 14 Theta <- poly(T.Angle..deg., ntheta) nbeta <- 4 Beta <- poly(B.Beta..deg., nbeta) model.1 <- lm( Measurement ~ Block + Rad + Theta + Beta

can you recommend a good manual for R that starts with a data set and gives demonstrations on what can be done using R? I downloadedR Langauage definition and An introduction to R but haven't found them overly useful. I'd really like to be able to follow some tutorials using a dataset or many datasets. The datasets I have available on R are Data sets in package 'datasets':

Hi, in coxph, coxph(Surv(time, status)~x,data=alm). How to encode x? For example, if x has two groups, the treatment group and control group. if I encode them as 1 and 2. Is the HR results treatment/control? What if I encode them as 0 and 1. I am confused with how survival package works for HR. What should I do to get a treatment/control HR? Thank you. -- View this message in context:

Hi Bert, Ok, to your initial point, the key nuance is that if 'x' is a vector, you can leave the 'degree' argument unnamed, however, if 'x' is a matrix, you cannot. That aspect of the behavior does not seem to change if poly() is called stand alone or, as suggested in ?poly, within a formula to be parsed. Working on tracing through the code using debug(), the error is

How do I add to a trellis plot the best fit line from a robust fit? I can use panel.lm to add a least squares fit, but there is no panel.rlm function. -- Alan S Barnett <asb at mail.nih.gov> NIMH/CBDB

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

Hi all, I want to fit data called "metal" with a polynominal function as dP ~ a.0 + a.1 * U0 + a.2 * U0^2 + a.3 * U0^3 + a.4 * U0^4 The data set includes, the independant variable U0 and the dependant variable dP. I've seen that the combination of lm() and poly() can do that instead of using the nls() function. But I don't get how to interpret the results from the linear

Full_Name: Jens Keienburg Version: 2.3.0 OS: Windows XP Submission from: (NULL) (193.174.53.122) I used the function lm() to calculate the coefficients of a polynome. If I used the function poly(t,2) to denote a polynome of form 1 + x + x^2, the coefficients are wrong. I appended an excerpt below: > t=1:100 > p=-20 - 10 * t + 2 * t^2 > p [1] -28 -32 -32 -28 -20 -8 8

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

Full_Name: David Clifford Version: Version 1.5.1 (2002-06-17) OS: Red Hat 7.3 Submission from: (NULL) (128.135.149.55) For n values above 100 there appears to be a bug in contr.poly(n). The contrast matrix should have rank n-1. Running the code below gives output (ie errors) at n=98, 100 and every value greater than 102. for(n in 2:150) { K <- contr.poly(n) rnk <-