similar to: bootstrap help

Displaying 20 results from an estimated 30000 matches similar to: "bootstrap help"

2010 Nov 08
3
how do i plot this "hist"?
Hi all, I have the following data in abc.dat ======================= 50 0 1 0 0 55 1 14 0 1 60 7 86 0 3 65 22 324 2 3 70 58 1035 1 7 75 30 2568 0 34 80 9 2936 15 162 85 27 2169 46 365 90 80 1439 212 432 95 236 1670 521 281 100 332 827 709 172 105 156 311 556
2010 Nov 28
4
how to divide each column in a matrix by its colSums?
Hi, I have a matrix, say m=matrix(c( 983,679,134, 383,416,84, 2892,2625,570 ),nrow=3 ) i can find its row/col sum by rowSums(m) colSums(m) How do I divide each row/column by its rowSum/colSums and still return in the matrix form? (i.e. the new rowSums/colSums =1) Thanks. Casper -- View this message in context:
2010 Nov 20
3
how to add frequencies to barplot
Hi, I have count data x2=rep(c(0:3),c(13,80,60,27)) x2 0 1 2 3 13 80 60 27 I want to graph to be ploted as barplot(table(x2),density=4) how do I add relative frequency to it, like in hist(x2,labels=T) above the 'bar's Thanks. casper -- View this message in context: http://r.789695.n4.nabble.com/how-to-add-frequencies-to-barplot-tp3051923p3051923.html Sent from the R help
2002 Jan 21
2
a Bootstrap understanding problem
I tried to reproduce a result from a former colleague which he got with S-plus bootstrap method. I don't have S-plus at hand. In R, there are 2 packages related to bootstrap method, bootstrap and boot. The former has a function called 'bootstrap' but this does not seem to conform either to the function used in S-plus nor to that described in MASS, 3d ed., p.144. The latter seems to be
2010 Mar 07
3
barplot with factors problem
http://www.harding.edu/fmccown/R/#autosdatafile http://www.harding.edu/fmccown/R/#autosdatafile I am tring to get a barchat by factors, following the example in that link above. =========================== x=c(145,40,40,120,180, 140,155,90,160,95, 195,150,205,110,160, 45,40,195,65,145, 195,230,115,235,225, 120,55,50,80,45 ) y2=c( rep(as.character(1),5), rep(as.character(2),5),
2010 Dec 18
3
use of 'apply' for 'hist'
Hi all, ########################################## dof=c(1,2,4,8,16,32) Q5=matrix(rt(100,dof),100,6,T,dimnames=list(NULL,dof)) par(mfrow=c(2,6)) apply(Q5,2,hist) myf=function(x){ qqnorm(x);qqline(x) } apply(Q5,2,myf) ########################################## These looks ok. However, I would like to achieve more. Apart from using a loop, is there are fast way to 'add' the titles to be
2010 Apr 13
3
writing function ( 'plot' and 'if') problem
=========================== myf=function(ds=1){ x=rnorm(10) y=rnorm(10) { #start of if if (ds==1) { list(x,y) } else (ds==2) { plot(x,y) } } # end of if } # end of function =========================== Hi All, the problem i am having here is, that I want to be able to control the display, lf ds=1, i want to just have a list, but it seem to always plot... Thanks. casper -- View this
2004 Jul 18
2
bootstrap and nls
Hi, I am trying to bootstrap the difference between each parameters among two non linear regression (distributed loss model) as following: # data.frame > Raies[1:10,] Tps SolA Solb 1 0 32.97 35.92 2 0 32.01 31.35 3 1 21.73 22.03 4 1 23.73 18.53 5 2 19.68 18.28 6 2 18.56 16.79 7 3 18.79 15.61 8 3 17.60 13.43 9 4 14.83 12.76 10 4 17.33 14.91 etc... # non
2004 Apr 09
1
bootstrap function coefficients
Dear R community, Please, can you help me with a problem concerning bootstrap. The data table called «RMika», contained times (Tps) and corresponding concentration of a chemical in a soil (SolA). I would like to get, by bootstraping, 10 estimations of the parameters C0 and k from the function: SolA = C0*exp(-k*Tps). # First, I fit the data and all is OK >
2010 Nov 16
3
plot linear model problem
Hi all, Say I fit a linear model, and saved it as 'test.lm' Then if I use plot(test.lm) it gives me 4 graphs How do I ask for a 'subset' of it?? say just want the 1st graph, the residual vs fitted values, or the 1,3,4th graph? I think I can use plot(test.lm[c(1,3,4)]) before, but now, it's not working... Every time, it goes to the end, the only thing I can click is
2009 Dec 07
5
confint for glm (general linear model)
Hi, I have a glm gives summary as follows, Estimate Std. Error z value Pr(>|z|) (Intercept) -2.03693352 1.449574526 -1.405194 0.159963578 A 0.01093048 0.006446256 1.695633 0.089955471 N 0.41060119 0.224860819 1.826024 0.067846690 S -0.20651005 0.067698863 -3.050421 0.002285206 then I use confint(k.glm)
2010 Nov 10
1
par mfrow in "function" problem
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$mids[1:2])*length(x) lines(xfit, yfit, col="blue", lwd=2) } ############################# individually, it worked fine however, if I used par(mfrow=c(2,2))
2012 Mar 23
2
show and produce PDF file with pdf() and dev.off( ) in function
Hi all, I know how to use pdf() and dev.off() to produce and save a graph. However, when I put them in a function say myplot(x=1:20){ pdf("xplot.pdf") plot(x) dev.off() } the function work. But is there a way show the graph in R as well as saving it to the workspace? Thanks. casper ----- ################################################### PhD candidate in Statistics School
2010 Apr 10
1
writing function (plot problem)
Hi, ======================= x=rnorm(20) y=rnorm(20) t=lm(y~x) plot(t) ======================= you will get "click or hit enter to next page..." how do I write a function to archieve this ? say plot(x,y) then pause, wait plot(y,x) Thanks! casper -- View this message in context: http://n4.nabble.com/writing-function-plot-problem-tp1835723p1835723.html Sent from the R help mailing
2010 Dec 08
1
how to add these "axis" label?
Hi All, How do I add these axis labels? ############################################### p=seq(0,1,length.out=500) p=p[-c(1,length(p))] g1=log(p/(1-p)) g2=qnorm(p) g3=log(-log(1-p)) g4=-log(-log(p)) plot(p,g1, 'n',ylim=c(-5,5),las=1, bty='n', xaxt='n',yaxt='n', xlab="",ylab="" ) lines(p,g1,lty=1,col=1) lines(p,g2,lty=1,col=2)
2012 Mar 23
1
Vectorize (scalar) function
Hi all, myint=function(mu,sigma){ integrate(function(x) dnorm(x,mu,sigma)/(1+exp(-x)),-Inf,Inf)$value } x=seq(0,50,length=3000) x=x[-1] plot(x,myint(4,x)) # not working yet I think I have to 'Vectorize' it somehow? What's a scalar function? and a primitive function? Thanks. casper ----- ################################################### PhD candidate in Statistics School
2012 Apr 03
1
Create Model Object (setClass?setMethod?)
Hi all, I have a self written likelihood as a model and functions to optimize and get fitted values, confidence intervals ect. I wonder if there is a way to define a 'class', or a 'model' (or a certain object)? so that I can use 'summary' to produce a summary like it does for a lm object. Also, it should be able to use 'predict' and 'plot' and other
2012 Mar 23
3
R numerical integration
Hi all, Is there any other packages to do numerical integration other than the default 'integrate'? Basically, I am integrating: integrate(function(x) dnorm(x,mu,sigma)/(1+exp(-x)),-Inf,Inf)$value The integration is ok provided sigma is >0. However, when mu=-1.645074 and sigma=17535.26 It stopped working. On the other hand, Maple gives me a value of 0.5005299403. It is an
2010 Feb 03
1
legend help
i=1 for(rate in c(2,4) ){ for(shape in c(1,3,5) ){ curve(dgamma(x,rate,shape),xlim=c(0,3),ylab="",col=i,lty=i,add=T) i=i+1 } } How can I add some legend to represent these lines? i.e. the legend is displayed as col=1 lty=1 lambda=2 theta=1 col=2 lty=2 lambda=2 theta=3 col=3 lty=3 lambda=2 theta=5 col=4 lty=4 lambda=4 theta=1 col=5 lty=5 lambda=4 theta=3 col=6 lty=6 lambda=4 theta=5
2010 Dec 21
3
how to see what's wrong with a self written function?
Hi all, I am writing a simple function to implement regularfalsi (secant) method. ################################################### regulafalsi=function(f,x0,x1){ x=c() x[1]=x1 i=1 while ( f(x[i])!=0 ) { i=i+1 if (i==2) { x[2]=x[1]-f(x[1])*(x[1]-x0)/(f(x[1])-f(x0)) } else { x[i]=x[i-1]-f(x[i-1])*(x[i-1]-x[i-2])/(f(x[i-1])-f(x[i-2])) } } x[i] }