similar to: distribution functions and lists

Hello everyone, I have programmed a function and, among other things, it has to compute the differential of a function. I have been using the function grad() and I have programmed some simulated studies. I have testes with 72,576 different combinations of the initial values and now I got an error in only 4 situations. The error comes from the function grad() and I don't understand why. This

Dear all, I'm trying to fit a curve to some 1 year failure-time data, so that I can extrapolate and predict failure rates up to 3 years. The data is in the general form: Treatment Time Status Treatment A 28 0 Treatment B 28 0 Treatment B 28 0 Treatment A 28

Hi, I have a Excel file with three spreadsheets: PlanA, PlanB and PlanC. I'm trying to read the three spreadsheets and then adding them together. But, when I try read the PlanA there is an error message: rm(list=ls()) setwd('C:/Test/Dados/Teste') require(RODBC) Arquivo = odbcConnectExcel('T070206_1347.xls') (Geral = sqlFetch(Arquivo, 'PlanA')) (Lactacao =

Hello, I am working with the nls() function and inserting a formula into it that uses the pweibull function. However the pweibull function is annoyingly producing NaNs, which nls() refuses to handle. I have put a sample of the code below. Is there a way to prevent these NaNs from interfering, for example a method to catch them? I get the following error when I try to run the code: res.nls <-

In rw1050, I found that > pweibull(3:10, 2) [1] 0.9998766 0.9999999 1.0000000 1.0000000 NaN NaN [7] NaN NaN Warning message: NaNs produced in: pweibull(q, shape, scale, lower.tail, log.p) more surprisingly, > pweibull(3:10, 2.1) [1] 0.9999566 1.0000000 1.0000000 -Inf NaN NaN [7] NaN NaN Warning message: NaNs produced in: pweibull(q,

Dear R gurus, I have the following code, but I still not know how to estimate and extract confidence intervals (95%CI) from resampling. Thanks! ~Adriana #data penta<-c(770,729,640,486,450,410,400,340,306,283,278,260,253,242,240,229,201,198,190,186,180,170,168,151,150,148,147,125,117,110,107,104,85,83,80,74,70,66,54,46,45,43,40,38,10) x<-log(penta+1) plot(ecdf(x),

Please allow me to add just a little more about this: nothing wrong with pweibull(), namely, the two cases I reported: pweibull(3:10, 2) and pweibull(3:10, 2.1), in rw1041 and earlier version. I wonder this might just due to the change from rw1041 to rw1050, however, I can't find anything relevant (seems to me) in the News or Readme. Thanks Sundar for the suggestion of using 1 -

Dear all - I have followed the thread the reply to which was lead by Thomas Lumley about using pweibull to generate fitted survival curves for survreg models. http://tolstoy.newcastle.edu.au/R/help/04/11/7766.html Using the lung data set, data(lung) lung.wbs <- survreg( Surv(time, status)~ 1, data=lung, dist='weibull') curve(pweibull(x, scale=exp(coef(lung.wbs)),

OK this is bound to be something silly as I'm completely new to R - having started using it yesterday. However I am already warming to its lack of 'proper' GUI... I like being able to rerun a command by editing one parameter easily... try and do that in a Excel Chart Wizzard! I eventually want to use it to analyse some chemotherapy response / survival data. That data will not be

Hi, I'm trying to install Fedora Core 3 on an AMD 64-based machine using PXE (via DHCP and TFTP). I've managed to get the machine to load the kernel over the network and start booting, but instead of loading the root fs from initrd.img (that I appended as a config via pxelinux.cfg), it loads the pre-existing rootfs on my harddisk. At the top of dmesg is the message: "so old

I have an R function (about 1000 lines long) that takes more than 20 times as long to run under R Windows 1.9.0 and 1.8.1 than it does under 1.7.1. Profile results indicate that the $<-.data.frame operation is the culprit, but I don't understand exactly what that is (assignment of data frame elements to another variable?), or why it's only a problem under 1.8.1 and 1.9.0. Any advice?

Had to changed the Subject header because the mailing list software thought the previous one ("so old bootloader" ...) was suspicious. Just a backgrounder and a little more info I've gathered: Trying to install Fedora Core 3 via PXE using DHCP-TFTP. Kernel loads but initrd isn't used for rootfs. The machine's an AMD Athlon 64-based box with 512MB physical memory (one DDR

Hi I am trying to determine the mean of a Weibull function that has been fit to a data set, adjusted for a categorical covariate , gender (0=male,1=female). Here is my code: library(survival) survdata<-read.csv("data.csv") ##Fit Weibull model to data WeiModel<-survreg(Surv(survdata$Time,survdata$Status)~survdata$gender) summary(WeiModel) P<-pweibull(n,

Dear all, I have a question regarding performing test if the data fits chi-squared distribution. For example, using ks.test() I found in the examples how to fit it to gamma or weibull x<-rnorm(100) ks.test(x, "pweibull", shape=2,scale=1) for the gamma, pgamma can be used But I cannot find the value of this second parameter for the chi-squared distribution. Maybe someone

Dear R helpers, I need to use KS and AD test for Generalized Pareto and Generalized extreme value. E.g. if I need to use KS for Weibull, I have teh syntax ks.test(x.wei,"pweibull", shape=2,scale=1) Similarly, for AD I use ad.test(x, distr.fun, ...) My problem is fir given data, I have estimated the parameters of GPD and GEV using lmom. But I am not able to find out the distribution

Dear Sir/madam, I'm getting a problem with a R-code which calculate Fisher Information Matrix for Hybrid Censored Weibull Distribution. My problem is that: when I take weibull(scale=1,shape=2) { i.e shape>1} I got my desired result but when I take weibull(scale=1,shape=0.5) { i.e shape<1} it gives error : Error in integrate(int2, lower = 0, upper = t) : the integral is probably

Hello all, I write a decision-support column for a computing magazine, Intelligent Enterprise. In my next column, I'll be revisiting a topic I wrote on 3 years ago, open-source analytical software. R is perhaps the most successful open-source analytical package. I'd like to hear from users, especially those who chose R for some other reason than that it's free, about what

Hi folks, I tried to retain the Weibull distribution using the `nls' function and proceeding along the lines of the example provided in the `SSweibull' help (at least I thought so): t <- (1:200)/100 v <- pweibull(t, shape=3, scale=1) df <- data.frame(Time=t, Value=v) Asym <- 1.0; Drop <- 1.0; lrc <- 0; pwr <- 1 df.estimate <- nls(Value ~ SSweibull(Time,

Full_Name: M Welinder Version: 1.4 OS: (src) Submission from: (NULL) (192.5.35.38) It seems to me that pweibull can be improved in the lower_tail=TRUE and log_p=FALSE case by using expm1. Something like -expm1(-pow(x / scale, shape)), I think. -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-devel mailing list -- Read

(1) ...need to speed up a monte-carlo sampling...any suggestions about how I can get R to use all 8 cores of a mac pro would be most useful and very appreciated... (2) spent the last few hours trying to get pnmath to compile under os- x 10.5.4... using gcc version 4.2.1 (Apple Inc. build 5553) as downloaded from CRAN, xcode 3.0... ...xcode 3.1 installed over top of above after