similar to: Generate Numbers

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible Url : https://stat.ethz.ch/pipermail/r-help/attachments/20080422/9bb74193/attachment.pl

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible URL : <https://stat.ethz.ch/pipermail/r-help/attachments/20100501/c1302c2e/attachment.pl>

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible URL : <https://stat.ethz.ch/pipermail/r-help/attachments/20101008/0bc9c0cc/attachment.pl>

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible URL : <https://stat.ethz.ch/pipermail/r-help/attachments/20111202/b4d64bba/attachment.pl>

Hello, I have a question regarding competing risk regression using cmprsk package (function crr()). I am using R2.9.1. How can I do to assess the effect of qualitative predictor (gg) with more than two categories (a,b,c) categorie c is the reference category. See above results, gg is considered like a ordered predictor ! Thank you for your help Jan > # simulated data to test > set.seed(10)

Dear all, I am trying to use the crr function in the cmprsk package version 2.2 to analyse 198 observations.I have receive the error in solve.default. Can anyone give me some insights into where the problem is? Thanks here is my script : cov=cbind(x1,x2) z<-crr(ftime,fstatus,cov)) and data file: x1 x2 fstatus ftime 0 .02 1 263 0 .03 1 113 0 .03 1 523

Greetings R users, My goal is to generate quartile groups of each variable in my data set. I would like each experiment to have its designated group added as a subsequent column. I can accomplish this individually with the following code: brks <- with(data_variables, cut2(var2, g=4)) #I don't want the actual numbers, I need a numbered group data$test1=factor(brks,

Hi I need some help getting results from multiple linear models into a dataframe. Let me explain the problem. I have a dataframe with ejection fraction results measured over a number of quartiles and grouped by base_study. My dataframe (800 different base_studies) looks like > afvtprelvefs basestudy quartile ef ef_std entropy CBP0908020 1 21.6 0.53 3.27

Hey all... I am trying to connect to a Epson FX-85 24-pin printer on my file/print/samba server. I am running SuSE 9.1/Samba 3.0.2a with CUPS as the print spool. When I try to connect to the printer from W2K land, I get this error: "Permission denied" I am at a loss as to where to look (sorry I dont have my smb.conf file posted. I am away from home for a couple of days). ALl else

OK, I have a very simple setup problem. I have a server and three printers which I want to share. These printers should need no password. This does not work However when I login and map a network drive I can print just fin How do I bypass the need for a password or network login on these printers. Thanks I am running Fedora 3 core and samba Thanks Chip Here is the printers section of my

I've just started using R and am still a neophyte, but I found the following curious result. I'm using the current version of R (2.5.1 (2007-06-27) ). Why are the results for the third quartile different in the output from the summary and fivenum commands? For the following data set 457 514 530 530 538 560 687 745 745 778 786 790 792

Hi,everyone i need to calculate quartile values of a variable grouped by the other variable . same as in aggregate function(only median,mean or functions is possible-i think so) Could you please help me to achieve the same for other quartile values(5,10,25,75,90) as for median using aggregate. Thanks in advance. data : zip price 60000 567000 60001 478654 60004 485647 60001

Hi every one, I am trying to generate a normally distributed random variable with the following descriptive statistics, min=1, max=99, variance=125, mean=38.32, 1st quartile=38, median=40, 3rd quartile=40, skewness=-0.274. I know the "rnorm" will allow me to simulate random numbers with mean 38.32 and Sd=11.18(sqrt(125)). But I need to have the above mentioned descriptive

Hi! My name is Martin and I have a problem concerning the boxplot function in R. I want my boxes to be limited by the 1st and 3rd quartile and NOT the 'hinges' values that are the default setting in R. Do anyone knows if there is any command that I could do to change this default setting? Sincerely Martin Olofsson

I have data that looks like lake,loglength,logweight 1,2.369215857,1.929418926 1,2.426511261,2.230448921 1,2.434568904,2.298853076 1,2.437750563,2.298853076 1,2.442479769,2.230448921 1,2.445604203,2.356025857 ... 102,2.722633923,3.310268367 102,2.781755375,3.502153893 102,2.836324116,3.683407299 102,2.802773725,3.583312152 102,2.790285164,3.546419267 102,2.806179974,3.599118565

> On Oct 13, 2017, at 2:51 AM, PIKAL Petr <petr.pikal at precheza.cz> wrote: > > Hi > > You expect us to solve your problem but you ignore advice already recieved. > > Your data are unreadable, use dput(yourdata) instead. see ?dput > >> test<-read.table("clipboard", heade=T) > Error in scan(file = file, what = what, sep = sep, quote = quote,

Why am I getting a wrong result for quartiles? here is my code: > cbiomass = c(910, 1058, 929, 1103, 1056, 1022, 1255, 1121, 1111, 1192, > 1074, 1415) > summary(cbiomass) > IQR(cbiomass) The result R gives me is: For the summary > Min. 1st Qu. Median Mean 3rd Qu. Max. 910 1048 1088 1104 1139 1415 For IQR > 91.25 ********* The true Q1 is 1039

Dear list, How can I add the mean and standard deviation to each of the boxplots using the example provided in the boxplot function? boxplot(len ~ dose, data = ToothGrowth, boxwex = 0.25, at = 1:3 - 0.2, subset = supp == "VC", col = "yellow", main = "Guinea Pigs' Tooth Growth", xlab = "Vitamin C dose mg",

Hello All, Using the standard "summary" function in 'R', I ran across some odd behavior that I cannot understand. Easy to reproduce: Typing: summary(c(6,207936)) Yields:: Min. *1st Qu. Median Mean 3rd Qu. Max.* 6 *51990 104000 104000 156000 207900* None of these values are correct except for the minimum. If I perform "quantile(c(6,

I have the following code. > f <- cph(formula = Surv(TimeToDeath, Dead == "Yes") ~1,data=single.dat, x=T, y=T, surv=T) > for(i in c('A', 'B', 'C', 'D', 'E', 'F')){ > f <-update(f,as.formula(paste('Surv(TimeToDeath, Dead == "Yes")~',i,sep=''))) > print(summary(f, paste(i,"=1st