search for: yfit

...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), trace = TRUE) plot(x, y) curve(coef(yfit)[1]*x^2 + coef(yfit)[2]*x + coef(yfit)[3], 0, 10, add = TRUE) However, this code is rather fortranesque, and most likely there exists a much more elegant way in...

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)) each time i run myhist it produces TWO plots, one without the 'lines',...

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) cat("\n") cat("PRESS =",press,"\n") } On the command R window I define - Ignored: > x=c(12,24,13,11,23,10,9,17,11,...

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 graphics, I wonder how can it be done. -G.V.

...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 K...

...ls 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 ) { Yfit=arima(Yst[,2],order=c(p,d,q),seasonal=list(order=c(sp,sd,sq),period=7),include.mean=TRUE,xreg=DU0) }}}}}} Until the step 187 it´s run normally, but in the step 187 return a error and stop the program. > Yfit=arima(Yst[,2],order=c(1,0,1),seasonal=list(order=c(2,1,2),period=...

...ws the right predictions if the relation is the same for all factor levels, but I can't figure out how to have the same slopes but different intercepts. # Function to draw predictions in xyplot panel.predfinal <- function(mod, x, y) { xfit <- seq(min(x), max(x), length=21) yfit <- predict(mod, newdata=data.frame(COV=xfit)) llines(xfit,yfit,lty=1) } xyplot(Y ~ COV | FACT, panel=function(x,y,groups,subscripts){ panel.xyplot(x,y) panel.predfinal(mod,x,y) } I would be very grateful for pointers (books, chapters, pages, archiv...

...line and point types.My code is as following: hist(CDR3,xlab="",ylab="",main=NULL, xlim=c(min(CDR3),max(CDR3)),freq=FALSE,col="lightgrey") # draw the overlaying normal curve and kernel curve, draw mean and median on X-axis xfit<-seq(min(CDR3),max(CDR3),length=100) yfit<-dnorm(xfit,mean=mean(CDR3),sd=sd(CDR3)) lines(xfit, yfit, col="red",lty=1,lwd=1) lines(density(CDR3,from=min(CDR3),to=max(CDR3)),col="blue",lty=1,lwd=1) mean(CDR3) median(CDR3) points(x=mean(CDR3),y=0,pch=19,col="red",cex=1) points(x=median(CDR3),y=0,pch=19,col=&q...

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

...llelic(k) } SNPS = as.data.frame(SNPS) SNPS[SNPS==-1] = NA # GRAMMAR calculations # here one has to include main effect and/or covariate adjustments in the polygenic model statement pkin = kinship(pedigree[,2],pedigree[,3],pedigree[,4]) maineff1 = as.factor(SNPS[,1]) maineff2 = as.factor(SNPS[,2]) Yfit = polygenic(trait~maineff1+maineff2,pkin,simulation.GenABEL,trait.type="gaussian") resi = Yfit$pgresidualY When I calculated the GRAMMAR and ran the command, "*Yfit = polygenic(trait~maineff1+maineff2,pkin,simulation.GenABEL,trait.type="gaussian") *", the software...

. 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

...[,2] +[1,] 11.94725 -7661.319 +[2,] -7661.31875 8039421.153 + +$code +[1] 3 + +$iterations +[1] 26 +@end example + + To obtain the approximate standard errors (SE) of the estimates we do: @example @@ -5032,7 +5057,7 @@ @example > plot(x, y) > xfit <- seq(.02, 1.1, .05) -> yfit <- 212.68384222 * xfit/(0.06412146 + xfit) +> yfit <- out$estimate[1] * xfit/(out$estimate[2] + xfit) > lines(spline(xfit, yfit)) @end example --- R-intro.texi~ Tue Aug 24 11:21:38 2004 +++ R-intro.texi Tue Aug 24 11:21:38 2004 @@ -3307,8 +3307,8 @@ @R{} is an expression language i...

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. In the example they have the following: df <- data.frame( x=x, y=y) fit <- nls(y ~SSmicmen(s, Vm, K), df) fit In the second line how would I input my functio...

...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 : num...

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' the corresponding model prediction (`fitted(rr)'), `se.fit' the corresponding standard error and `luconf' the lower and...

...ny remarks appreciated. greetings joerg original post: -------------- 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' the corresponding model prediction (`fitted(rr)'), `se.fit' the corresponding standard error and `luconf' the lower and...

...("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) xfit<-seq(min(rt),max(rt),(max(rt)-min(rt))/100) yfit<-pweibull(xfit,out$estimate[1], out$estimate[2]) lines(xfit,yfit,lty=2) yfit2<-pweibull(xfit,a, b) lines(xfit,yfit2) list(out=out) } I got the starting values a=9, b=.27 from fitting the Weibull CDF by eye to a quantile plot of the data. The final values fitted by nlm() are a= 4.8299357, b=...

...uot; motivation for posting here, hoping for some clarifications...): for _unweighted_ fits using `nls' one can compute confidence intervals for the fitted model function via #------------------- se.fit <- sqrt(apply(res$m$gradient(), 1, function(x) sum(vcov(res)*outer(x,x)))) luconf <- yfit + outer(se.fit, qnorm(c(probex, 1 - probex))) #------------------- where `res' contains an `nls' object, `x' is the independent variable vector, `yfit' the corresponding model prediction (`fitted(res)'), `se.fit' the corresponding standard error and `luconf' the lower a...

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,