similar to: princomp

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

I'm trying to extract the points above and below a particular lsfit. I can only get the residuals from the original fit though. x = runif(100, 0, 10) plot(x) abline(lsfit(1:100, test)) abline(lsfit(1:100, test + sd(test))) #I want the points above THIS line. Is there a way to use the coefficients from the fit to do this? Thanks for any help.

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

Hi experts, 1. I am working with large datasets (80000 rows, 20 columns) with R 1.3.0 und linux; so i try to save memory and remove unused objects. But even if i remove all objects with rm() and force the garbage collector to work with gc(), the process monitor utiliity kpm does not show a change of a single byte in memory consumption. So what is wrong. 2. I order to scale a second dataset

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.

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

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, 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 All, I have millions of regression lines to fit. So I am looking for the most efficient approach in R. Details: I have a large desing matrix X. The dimension is n by p. Each time when fitting the model, select rows from this matrix X and form a new design matrix, called X_current. There is another binary matrix M, with dim m by n, and each row is a 1*n vector. It helps to determin X_current.

I have a situation in R where I would like to find any x (if one exists) that solves the linear system of equations Ax = b, where A is square, sparse, and singular, and b is a vector. Here is some code that mimics my issue with a relatively simple A and b, along with three other methods of solving this system that I found online, two of which give me an error and one of which succeeds on the

Dear all, I’m not sure if it is O.K. to ask this question here. But where can I find the code for the function ‘C_Cdqrls’ which is called by the R function ‘lsfit‘. Thank you all. Sorry for being naïve if so. -------------------- Ziqiang Zhao 2013-04-23 [[alternative HTML version deleted]]

Hi folks, Wish y'all a Happy New Year 2010! I need some help with the following: Say I have lots of data sets, on which I have to apply a certain function on the same set of columns in each of the data set. Let's take, for ex, the typical data set is: df1 <- as.data.frame(cbind(rnorm(10),rnorm(10))) names(df1)[1] <- "A" names(df1)[2] <- "B" There are many

The function crossval (in library bootstrap) works well for the first degree polynomial model, but in the case of the second degree model I got an error message (see below). I would be very greatfull if somebody could give some advices for the following: > library(bootstrap) > > x<-c(22,23.4,24.9,28.5,29.8,31.6,34.2,36.4,37.7,39) >

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

Hello, I'm using fArma package to estimate the value of Hurst exponent using R/S method. However, for a certain set of data I get H ~ 1.8. How do I interpret this? Following are the output that I get for this set: > mean(data[,2]) [1] 400.5433 > sd(data[,2]) [1] 1139.786 > > rsFit(data[,2], levels = 64) Title: Hurst Exponent from R/S Method Call: rsFit(x = data[, 2], levels

dear professor: thank you for your help,witn your help i develop the nomogram successfully. after that i want to do the internal validation to the model.i ues the bootpred to do it,and then i encounter problem again,just like that.(´íÎóÓÚerror to :complete.cases(x, y, wt) : ²»ÊÇËùÓеIJÎÊý¶¼Ò»Ñù³¤(the length of the augment was different)) i hope you tell me where is the mistake,and maybe i have

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",

Hi there, I was tryting to use "Rcmd SHLIB --help" on Windows RGui. However it always gives me sytax error message. I wonder whether you have any suggestion. Thanks in advance! Best, Yunda -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help",

> On 20 Apr 2016, at 13:22, A A via R-help <r-help at r-project.org> wrote: > > > > > I have a situation in R where I would like to find any x (if one exists) that solves the linear system of equations Ax = b, where A is square, sparse, and singular, and b is a vector. Here is some code that mimics my issue with a relatively simple A and b, along with three other