search for: yfitted

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

...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 'predict' method, the outcomes do not match up as they should. I've already tried to find information concerning this issue on the R-help board, but to no avail. Can any of you point me in the right direction? Signed, Johan Van Kerc...

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

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

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.

...ssible to compute a R2, eg the ratio of variation explained by a given model. My model is : bivmod<-spgm(logIKA~NBLITRE0+NBLITRE1,data=mydatap,listw=comsKnn.nbW,spatial.error=TRUE) I know that we can calculate the R^2 as the variance of the fitted values from the reduced form of the model (Yfitted) over the variance of y (here logIKA). Since I am using the option lag = FALSE; the fitted value for an error model are Yfitted = X\beta. If I well understood : *1) We can get X* (NT * k matrix of observations on the non-stochastic regressors. with N: spatial units , T: time unit et k : number...

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 list members, my question concerns computation of confidence intervals in nonlinear fits with `nls' when weigthing the fit. the seemingly correct procedure does not work as (I) expected. I'm posting this here since: (A) the problem might suggest a modification to the `m' component in the return argument of `nls' (making this post formally OK for this list) and (B) I got no

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

# Dear R list, # # I'm needing help with lattice, regression and respective lines. # My data is below: bra = gl(2, 24, label = c('c', 's')) em = rep(gl(3, 8, label = c('po', 'pov', 'ce')), 2) tem = rep(c(0, 0, 30, 30, 60, 60, 90, 90), 6) tem2 = tem^2 r = rep(1:2, 24) y = c(40.58, 44.85, 32.55, 35.68, 64.86, 51.95, 42.52, 52.21,