similar to: survreg and pweibull

Dear R users, Please can you help me with a relatively straightforward problem that I am struggling with? I am simply trying to plot a baseline survivor and hazard function for a simple data set of lung cancer survival where `futime' is follow up time in months and status is 1=dead and 0=alive. Using the survival package: lung.wbs <- survreg( Surv(futime, status)~ 1, data=lung,

Hello Everyone, I am trying to excess the inbuit .Fortran and .C codes of R. Can any one help me in that. For example in kmeans clustering the algorithms are written in .Fortran I want to access them and see the .Fortran syntax of the codes. Can any one help me how can I do that? Thanx, Nitin Kumar On Thu, Nov 27, 2008 at 12:00 PM, <r-help-request@r-project.org> wrote: > Send R-help

Hello 1/ 'priors' is a table looking like: "W123" "T678" "S789" 23 42 11 12 35 9 etc 2/ WBS <- labels(priors) gives me a result of class list and length 1 looking like: "W123" "T678" "S789" I want to read W123 into X[1] as W, T687 into X[2] as T and S789 into X[3] as S using

Calum had a long question about drawing survival curves after fitting a Weibull model, using pweibull, which I have not reproduced. It is easier to get survival curves using the predict function. Here is a simple example: > library(survival) > tfit <- survreg(Surv(time, status) ~ factor(ph.ecog), data=lung) > table(lung$ph.ecog) 0 1 2 3 <NA> 63 113 50 1

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 -

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 all, What is the cumulative distribution (with parameterization) used within survreg with respect to the log-logistic distribution? That is, how are the parameters linked to the survivor function? Best regards, Mario [[alternative HTML version deleted]]

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

Hi, I'm sure this should be simple but I can't figure it out! I want to get the median survival calculated by the survfit function and use the value rather than just be able to print it. Something like this: library(survival) data(lung) lung.byPS = survfit(Surv (time, status) ~ ph.ecog, data=lung) # lung.byPS Call: survfit(formula = Surv(time, status) ~ ph.ecog, data = lung) 1

Hi, Is it normal to get intercept in the list of covariates in the output of survreg function with standard error, z, p.value etc? Does it mean that intercept was fitted with the covariates? Does Value column represent coefficients or some thing else? Regards, ------------------------------------------------- tmp = survreg(Surv(futime, fustat) ~ ecog.ps + rx, ovarian,

I found an issue with the data() command this evening when working on the survival package. 1. I have a lot of data sets in the package, almost all used in at least one vignette, help file, or test.? As a space saving measure, I have bundled many of them together, i.e., the file data/cancer.rda contains 19 data sets, many of them small. The resulting file (using xz compression) is quite a bit

Hi everybody, this is a real dummy thing. I sorted a matrix based on a given column, and what I get is right, until it comes to columns of negative and positive values; than, "order" orders everything from max to min in the negative values, and then AGAIN from max to min in the positive values!!! Why isn't everything order from max to min, and that's it? Thank you!!! Attached

Hi, I've lost my mind on it... I have to scatterplot two vectors, grouped by a third variable, with two different dimensions according to whether each cell line in the plot is sensitive or resistant to a given drug, and with a different color for each of 9 tissues of origin. Here's what I've done:

I had trouble with some tests for the survival suite last night that I cannot explain. Framework: Ubuntu Linux, R2.11. For testing survival I have a separate directory and Makefile. I pull everything into the local .RData, no packages, library, or namespace. (It's easier to add test modifications to a routine in a chain of calls). A test of survreg + psline would fail because

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

Reading "Modern Applied Statististics with S" and trying the corresponding examples both in the book and in ../lib/R/library/MASS/script, I'm now trying chapter 4 plotting bars with the following code on a linux box with R 1.8.1: ---------------------- library(MASS) library(lattice) options(echo=T, width=65, digits=5) lung.deaths <- aggregate(ts.union(mdeaths, fdeaths), 1)

I was looking at how the model.frame method for lm works and comparing it to my own for coxph. The big difference is that I try to retain xlevels and predvars information for a new model frame, and lm does not. I use a call to model.frame in predict.coxph, which is why I went that route, but never noted the difference till now (preparing for my course in Nashville). Could someone shed light

I have a sample of lung capacities from a population measured against height. I need to know the 95% CI of the lung capacity of a person of average height. I have fitted a regression line. How do I get a minimum and maximum values of the 95% CI? My thinking was that this has something to do with covariance, but how? My other thinking was that I could derive the 0.975 (sqrt 0.95) CI for the

Hello, I have a maybe trivial question, but I simply don't understand well enought how to work with text/strings: I have a rather compelx data structure, a big list with several sub-lists/dataframes and for certain calculations (which I do in loops), I only need a certain group of sub-lists/dataframes, which I want to specify with a name vector and paste together the object name and pass it