search for: lungs

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,

I?m trying to set up proportional hazard model that is stratified with respect to covariate 1 and has an interaction between covariate 1 and another variable, covariate 2. Both variables are categorical. In the following, I try to illustrate the two problems that I?ve encountered, using the lung dataset. The first problem is the warning: To me, it seems that there are too many dummies

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)),

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

First, here is your message as it appears on R-help. On 10/14/2012 05:00 AM, r-help-request@r-project.org wrote: > I?m trying to set up proportional hazard model that is stratified with > respect to covariate 1 and has an interaction between covariate 1 and > another variable, covariate 2. Both variables are categorical. In the > following, I try to illustrate the two problems that

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'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

On 23/10/2020 9:25 p.m., Therneau, Terry M., Ph.D. via R-devel wrote: > 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

All, I was having trouble trying to create a new class of data and pass it on to splom (in the lattice library). I mentioned this to Martin Morgan after a talk he gave. Following is not so much a question, but rather an answer from Morgan that might be useful to others. Here is the edited part of an email conversation with him: On Thursday, August 26, 2010 1:36 PM, Martin Morgan wrote:

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:

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

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 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

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 just configured a server to use the CUPS printing system and have all my printers shared out by Samba. I made all the necessary changes to the smb.conf file but don't want to restart samba in the middle of the day since it is a high traffic server and don't want to break connections. Is there a way to just have samba load all these printers without restarting the service????

In the past few months we have seen a couple of our production servers crash that are running samba-3.0.14a-1. What seems to happen is the smbd pid's seem to grow and grow until an out of memory error occurs and the smbd process just hangs. A simple restart does not even clear out the pid's. I end up rebooting to just clear it all out and start fresh. It probably has happened at

Hello R helpers: *My first message didn't pass trough filter so here it's again* I would like to obtain probability of an event for one single patient as a function of time (from survfit.coxph) object, as I want to find what is the probability of an event say at 1 month and what is the probability of an event at 80 months and compare. So I tried the following but it fails miserably. I

Hello: I would like to obtain probability of an event for one single patient as a function of time (from survfit.coxph) object, as I want to find what is the probability of an event say at 1 month and what is the probability of an event at 80 months and compare. So I tried the following but it fails miserably. I looked at some old posts but could not figure out the solution. Here's what I did

http://wiki.rubyonrails.com/rails/show/HomePage Now sure if anyone has noticed yet, but the above page has been hacked.

Hi, I am new to model selection by coefficient shrinkage method such as lasso. And I became particularly interested in variable selection in Cox regression by lasso. I became aware of the coxpath() in R package glmpath does lasso on Cox model. I have tried the sample script on the help page of coxpath(), but I have difficult time understanding the output. Therefore, I would greatly appreciate if