similar to: non-linear fitting (nls) and confidence limits

for unweighted fits using `nls' I compute confidence intervals for the fitted model function by using: #------------------- se.fit <- sqrt(apply(rr$m$gradient(), 1, function(x) sum(vcov(rr)*outer(x,x)))) luconf <- yfit + outer(se.fit, qnorm(c(probex, 1 - probex))) #------------------- where `rr' contains an `nls' object, `x' is the independent variable vector, `yfit'

dear list members, I apologize in advance for posting a second time, but probably after one week chances are, the first try went down the sink.. my question concerns computation of confidence intervals in nonlinear fits with `nls' when weigthing the fit. the seemingly correct procedure does not work as expected, as is detailed in my original post below. any remarks appreciated. greetings

Dear all, I found really difficult with the time series questions, please help me with this monthly airline series! I have run the following r code, and there is an error appeared at the end. The data files was enclosed in the email. I'm sorry the errors message appeared in chinese, but it says "plot.xy(xy.coords(x, y), type = type, ...) : errors in argument has more than 3

Greetings everyone, I have the following problem (illustrating R-code at bottom of mail): Given a training sample with binary outcomes (-1/+1), I train a linear Support Vector Machine to separate them. Afterwards, I compute the weight vector w in the usual way, and obtain the fitted values as w'x + b > 0 ==> yfitted = 1, otherwise -1. However, upon verifying with the

I am trying to do a MLE fit of the weibull to some data, which I attach. fitweibull<-function() { rt<-scan("r/rt/data2/triam1.dat") rt<-sort(rt) plot(rt,ppoints(rt)) a<-9 b<-.27 fn<-function(p) -sum( log(dweibull(rt,p[1],p[2])) ) cat("starting -log like=",fn(c(a,b)),"\n") out<-nlm(fn,p=c(a,b), hessian=TRUE)

Dear useRs, I'm trying to simply fill in the area under a curve using RGL. Here' the set up: x <- c(0.75,75.75,150.75,225.75,300.75,375.75,450.75,525.75,600.75,675.75, 0.5,50.5,100.5,150.5,200.5,250.5,300.5,350.5,400.5,450.5, 0.25,25.25,50.25,75.25,100.25,125.25,150.25,175.25,200.25,225.25) y <- c(0.05,4.91,9.78,14.64,19.51,24.38,29.24,34.11,38.97,43.84,

Hello, I'm new to this and am trying to teach myself some R by plotting biological data. The growth curve in question is supposed to be fitted to the Verhulst equation, which may be transcribed as follows: f(x)=a/(1+((a-0.008)/0.008)*exp(-(b*x))) - for a known population density (0.008) at t(0). I am trying to rework the example from "An Introduction to R" (p. 72) for my case and

Hi, I'm trying to generate a distance matrix between sample pairs (example below). I'm not very familiar with the loop command which I expect I will need for this. The example below demosntrates what I'd like to get out of the data - essentially, to calculate the proportion of positions where two samples differ. Any help much appreciated! Also, any notes on how the functions work

Dear all, Here I come with another stupid question. Suppose I want to use nls() to fit a series of data (here modelled by generated points), then plot the points and the fitting curve. I figured out some way of doing it: x <- runif(1:20, 0, 10) y <- 0.1*x^2 - rep(3, length(x)) + rnorm(length(x), sd = 0.5) yfit <- nls(y ~ a*x^2 + b*x + c, start = list(a = 1, b = 1, c = 1),

Hi all, I defined the following ############################# myhist=function(x){ hist(x,xlab="",main="") h=hist(x) xfit=seq(min(x),max(x),length=100) yfit=dnorm(xfit,mean(x),sd=sd(x)) yfit=yfit*diff(h$mids[1:2])*length(x) lines(xfit, yfit, col="blue", lwd=2) } ############################# individually, it worked fine however, if I used par(mfrow=c(2,2))

I am using R on a Windows XP professional platform. The following code is part of a bigger one CODE press=function(y,x){ library(qpcR) models.press=numeric(0) cat("\n") dep=y print(dep) indep=log(x) print(indep) yfit=dep-PRESS(lm(dep~indep))[[2]] cat("\n yfit\n") print(yfit) yfit.orig=yfit presid=y-yfit.orig press=sum(presid^2)

Currently, I am doing it this way. x <- mtcars$mpg h<-hist(x, breaks=10, col="red", xlab="Miles Per Gallon", main="Histogram with Normal Curve") xfit<-seq(min(x),max(x),length=40) yfit<-dnorm(xfit,mean=mean(x),sd=sd(x)) yfit <- yfit*diff(h$mids[1:2])*length(x) lines(xfit, yfit, col="blue", lwd=2) But since, ggplot2 has more appealing

Dear R developers: I am a PHD candidate student in the school of public health of Peking University and my major is genetic epidemiology. I am learning the FAM-MDR algorithm, which is used to detect the gene-gene and gene-environment interactions in the data of pedigree. The codes were written by Tom Cattaert of the University of Liege. The algorithms and the sample datasets are available at

Here are some patches to expand some of the examples in R-intro.texi. -- Brian Gough Network Theory Ltd, Publishing the R Reference Manuals --- http://www.network-theory.co.uk/R/ --- R-intro.texi~ Tue Aug 24 11:21:37 2004 +++ R-intro.texi Tue Aug 24 11:21:37 2004 @@ -6288,6 +6288,21 @@ use @example +> help(package = "@var{name}") +@end example + +A complete list of the

Dear lattice wizards, I am trying to figure out how to plot predicted values in xyplot, where the intercept, but not the slope, varies among conditioning factor levels. I am sure it involves the groups, but I have been unsuccessful in my search in Pinhiero and Bate, in the help files, or in the archive, or in my attempts on my own. My example follows: FACT is a factor with levels a,b,c

Hi, I want to get a histogram with the legend for my data. I drew a normal density curve and kernel density curve in the histogram, and I also label mean and median in the X axis. From the code, I got two legend: One shows "Normal Density" and "Kernel Density" and their corresponding lines, the other shows "Mean = value" and "Median = value" and their

Hi people, I´m trying to development a simple routine to run many Arima models result from some parâmeters combination. My data test have one year and daily level. A part of routine is: for ( d in 0:1 ) { for ( p in 0:3 ) { for ( q in 0:3 ) { for ( sd in 0:1 ) { for ( sp in 0:3 ) { for ( sq in 0:3 ) {

Hello I followed the example in page 59, chapter 11 of the 'Introduction to R' manual. I entered my own x,y data. I used the least squares. My function has 5 parameters: p[1], p[2], p[3], p[4], p[5]. I plotted the x-y data. Then I used lines(spline(xfit,yfit)) to overlay best curves on the data while changing the parameters. My question is how do I calculate the residual sum of squares.