similar to: Substituting the extracted coefficients into the formula, exctracted from the result of nls()

Hello everybody, I am sure this is a beginners' problem which is being asked recurrently every few months, but nevertheless I wasn't able to find the answer searching through the r-help list. So here is my problem: I would like to plot a set of points (y vs. x), a (linear) regression line through them, and on the second graph underneath the first one I would want to plot residuals vs. x.

Hello, Fitting a piecewise smooth curve to a set of points (and a piecewise linear function in particular) seems to be a recurring question on this list. Nevertheless, I was not able to find an answer to a question that bothers me. Suppose I have the following data set, and would want to fit it with a piecewise smooth curve, In this model data, one curve is valid for up to 3 and another one for

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

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 ) {

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

Hello everybody, Is it possible to coax legend() into displaying more than one simbol per line in legend? I have a graph like the one attached to this mail; I would like to reorganize the legend in such a way that the duplicate text would be omitted, i.e., the first line would read <square> <triangledown> "increasing frequency" and the second one would read <circle>

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.

Dear all, when working on a project with embedded R, I found out that R-2.0.0 causes a problem where older versions worked OK. Rf_findVar(...) causes the following error when used to find a generic function (such as print): fun = Rf_findVar(...); R_tryEval(fun, ...); Error in function (object, ...) : Invalid generic function in usemethod Alternatively, using Rf_findFun(...) seems OK in this

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

Dear R people: I would like to superimpose a normal curve on a histogram. I've seen this example in a book, somewhere. I know that you draw the hist, get the mean and sd of the data set, but then I'm stuck. Could you help, please? Thanks! Erin hodgess at uhddx01.dt.uh.edu -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

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

. I am looking for a histogram or box plot with the adding normal distribution curve I think that must be possible, but I am not able to find out how to do. Regards Knut

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

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

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 All, I am trying to do some simple mixture analyses. For instance, I have a sample of n observations and I suspect they come from two different exponential distributions with parameters rate1 and rate2, respectively. So, I want to estimate rate1, rate2, and the proportions of both kinds of individuals in the sample. I had a look at the packages mda and mclust, but they do not seem to do this