similar to: lsfit residuals

I'm trying to fit a simple linear regression of just Y ~ X, but both X and Y are noisy. Thus instead of fitting a standard linear model minimizing vertical residuals, I would like to minimize orthogonal/perpendicular residuals. I have tried searching the R-packages, but have not found anything that seems suitable. I'm not sure what these types of residuals are typically called

Dear R-devel, It seems that lsfit incorrectly reports coefficients when the input matrix 'x' is rank-deficient, see the example below: ## here values of 'b' and 'c' are incorrectly swapped > x <- cbind(a=rnorm(100), b=0, c=rnorm(100)); y <- rnorm(100); lsfit(x, y)$coef Intercept a b c -0.0227787 0.1042860 -0.1729261 0.0000000 Warning

Hello, I am working on a project to create some scatter plots. I have syntax for 26 plots, and 22 of them display as they should. But here, for example, is a sample of the command syntax I am using: good <- complete.cases(affect1,adh1scr) plot (jitter(affect1,2.0),jitter(adh1scr,1.0),xlim=c(1,35),ylim=c(1,35),pch=1 6, main='Adherence Score by Affectiveness Level - Visit 1',

Hi, in R-3.0 I get the following error when calling lsfit with more observations than variables, which seems to come from an error in the formatting of the error message (note that this was not happening in 2.15.3): > nobs <- 5; nvar <- 6; lsfit(matrix(runif(nobs*nvar), ncol=nvar), runif(nobs), intercept=FALSE) Error in sprintf(ngettext(nry, "%d response", "%d

ls.print produces error that I don't seem to be able to trace. Output of the commands as follows: (hyeung is a 24x2 matrix of data) ------------------------------------------------- > summary(hyeung) x.1 x.2 Min. : 28.0 Min. : 10.0 1st Qu.: 72.0 1st Qu.: 87.5 Median : 86.5 Median : 92.5 Mean : 81.0 Mean : 82.5 3rd Qu.: 97.0 3rd Qu.:100.0 Max.

ls.print produces error that I don't seem to be able to trace. Output of the commands as follows: (hyeung is a 24x2 matrix of data) ------------------------------------------------- > summary(hyeung) x.1 x.2 Min. : 28.0 Min. : 10.0 1st Qu.: 72.0 1st Qu.: 87.5 Median : 86.5 Median : 92.5 Mean : 81.0 Mean : 82.5 3rd Qu.: 97.0 3rd Qu.:100.0 Max.

I suspect that the line in the added variable plot (library car) is a SLR of the residuals, but I can't seem to find this written anywhere. Can someone confirm this? Thanks. Tom [[alternative HTML version deleted]]

Dear list, I would like to know how the function residuals.lm calculates the partial residuals from an lm object with more than one predictor variable. In other words what is residuals.lm(...,type="partial") doing behind the scenes? According to the help file for residuals.lm (?residuals.lm), "The partial residuals are a matrix with each column formed by omitting a term from

Hello, I have a customized pairs () fonction as follows that displays correctely my data. ------------------------------------------------------------------------ panel.cor1 <- function (x, y, digits=2, prefix="") { usr <- par("usr"); on.exit(par(usr)) par(usr = c(0, 1, 0, 1)) r <- cor(x, y,use="pairwise.complete.obs",

Dear All, I have seldom needed to use loops in R, but now I need to code a loop with a stride different from one. In the R manual I downloaded I have the example: > xc <- split(x, ind) > yc <- split(y, ind) > for (i in 1:length(yc)) { plot(xc[[i]], yc[[i]]); abline(lsfit(xc[[i]], yc[[i]])) } but in my case I'd like to add a condition so that i varies by 4 from one go

To the purpose of fitting a 2nd order polynomial (a + b*x + c*x^2) to the chunk of signal falling in a 17 consecutive samples window I wrote the following very crude script. Since I have no previous experience of using Least Square Fit with R I would appreciate your supervision and suggestion. I guess the returned coefficients of the oolynomial are: a = -1.3191398 b = 0.1233055 c = 0.9297401

This isn't a question about R, but I'm hoping someone will be willing to help. I've been looking at calibration plots in multiple regression (plotting observed response Y on the vertical axis versus predicted response [Y hat] on the horizontal axis). According to Frank Harrell's "Regression Modeling Strategies" book (pp. 61-63), when making such a plot on new data

Hello there This is the first time I have used r-help message board so I hope I have got the right address. I am trying to check the residuals of a GLMM model(run using the package lme4). I have been able to check the residiuals of REMLs in lme4 using the following: m1<-lmer(vTotal~Week+fCollar+ (1|fCat), collars) res<-resid(m1) plot(res) qqnorm(res) library(MASS) par(mfrow=c(2,3))

Dear R-users, I want to plot residuals vs fitted for multiple groups with ggplot2. I try this code, but unsuccessful. library("plyr") models<-dlply(dat1,"d",function(df) mod<-lm(y~x,data=df) ggplot(models,aes(.fitted,.resid), color=factor(d))+ geom_hline(yintercept=0,col="white",size=2)+ geom_point()+ geom_smooth(se=F) -- --- Catalin-Constantin ROIBU

Not really r-specific: Z = (b1 - b2) / SQRT ( SEb1^2 + SEb2^2) -------Original Message------- From: Gijsbert Stoet <stoet at volition.wustl.edu> Sent: 07/20/03 09:51 PM To: r-help at stat.math.ethz.ch Subject: [R] how to test whether two slopes are sign. different? > > Hi, suppose I do want to test whether the slopes (e.g. determined with lsfit) of two different population are

Hello experts, as newcomer in pca, i have a question, concerning the princomp algorithm. With a dataset "r" containing 18 "input" parameters and 1 "output" parameter r[19], i got with the following fit ls <- lsfit(r[1:18],r[19]); lsdiag <- ls.diag(ls); lsdiag$std.dev a prediction error of: [1] 8.879561 what is quite reasonable. If i take only two

Dear All, Sorry to bother you again. I have a model: coxfita=coxph(Surv(rem.Remtime/365,rem.Rcens)~all.sex,data=nearma) and I'm trying to do a plot of Schoenfeld residuals using the code: plot(cox.zph(coxfita)) abline(h=0,lty=3) The error message I get is: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In sqrt(x$var[i, i] * seval) : NaNs

Hi all, I divide the graphical device into 4 ---mfrow=c(2,2)--- to plot 3 boxplots graph. Thus I get an empty space for the fourth graph. I d'like to place a legend in this space. legend(xxxx) creates a legend in the last graph (the 3rd) I'd like a legend in the empty space. How may I do it? #graph parameter par( fin=c(6,6), #dim graph mai=c(1,1,0.5,0.5) #marges graph

This is not a solution but your lsfit attempt #Error in lsfit(A, b) : only 3 cases, but 4 variables lsfit(A,b) gave that error because lsfit adds a column of 1 to its first argument unless you use intercept=FALSE. Then it will give you an answer (but I think it converts your sparse matrix into a dense one before doing any linear algebra). Bill Dunlap TIBCO Software wdunlap tibco.com On

Hello, First, sorry for my poor english, I will try to be understood. It's the first time I try this "r-help mailing list" and I hope it will be a success. I am working on heteroscedasticity analysis. I would like to get the "Box-Ljung" and the "Lagrange multipliers" test. I found the first one in the library "ts", but I can't find the second one.