similar to: cox model

I'm a young physician and i work on the breast cancer. I made a cox model and i want to do a test of heterogeneity. Do you know how i can do this with R. I'm sorry for this question wich seems to be easy for you. I read the manual and the help and i don't find it. Thanks for your help Bertrand Billemont

Hello, i'm a french student of medical oncology and i'm working on breast cancer. I have a variable with the histologic type of tumor wich is between 1 and 5. I use as.factor function to make some variable with level between 1 and 5. When i put it in the cox model i have only the level between 2 and 5. The level 1 doesn't appear. I think i have to change the number of level but i

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:

The warnings from make.packages.html() on the Apple Mac OS X platform can be dealt with as follows: ------------------------------------------------ (1) make.packages.html() uses the function tempdir() and attempts to create a temporary directory in the default location /tmp/ which fails due to the /tmp directory architecture on the Mac. I set up a .Renviron file in my user account

Hi useRs, A recent coding infelicity along these lines yielded a corrupt data frame. foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4] > bar X1 X2 X3 X4 NewCol 1 1 4 7 10 <NA> 2 2 5 8 11

Hi useRs, Perhaps I am having a senior moment? I have a nested variable situation to model, toy example: > df <- data.frame(A = factor(c("a", "a", "x", "x"), levels = c("x", "a")), + B = factor(c("b", "x", "x", "x"), levels = c("x", "b"))) > >

Hi i'm a french medical student, i have some data that i import from excel. My colomn of the datafram are the localisations of metastasis. If there is a metatsasis there is the symbol "_". i want to exclude the row without metastasis wich represent the NA data. so, i wrote this mela is the data fram mela1=ifelse(mela[,c(11:12,14:21,23,24)]=="_",1,0) # selection of the

Dear professor, I am currently using Design package and the cph formula for assessing multivariable analysis. I am tryng to get the C-index for my survival model based on Dxy coefficient. I am confused since there is a negative value. Do I need to used the absolute Dxy ? index.orig training test optimism index.corrected n Dxy -0.341357727 -0.344002740

Hi all, I have been having problems generating png files on a Mac running OS X 10.6.8. Here's a simple example > png("foo.png", type = "cairo"); plot(1:10); dev.off(); null device 1 libpng warning: Application built with libpng-1.2.26 but running with 1.5.2 The resultant file is of size 0 Kb. Is this the proper place to report this issue? Or should

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 all, I believe this is a bug in the model.matrix function. I'd like a second opinion before filing a bug report. If I have a nested covariate B with multiple values for just one level of A, I can not get a non-singular design matrix out of model.matrix > df <- data.frame(A = factor(c("a", "a", "x", "x"), levels = c("x",

Running R2.4.0 on Apple Mac OS X 10.4.8, in Emacs ESS mode, and also R.app. In an attempt to learn a bit more about a particular method (geneNames in package affy) I invoked getMethods("geneNames") which produced geneNames methods, but not the one in affy (output below). I had to know the signature (AffyBatch) in order to find the method > getMethod("geneNames",

On 7/28/2009 6:30 PM, smckinney at bccrc.ca wrote: > Hi all, > > When I install a new package, and don't yet know any function names, > I have to play the "poor man's game" to get to the standard > help system "Index" page for the package: You could complain to the package author or maintainer, who should have created a help alias with the package name

Hi all, When I install a new package, and don't yet know any function names, I have to play the "poor man's game" to get to the standard help system "Index" page for the package: Poor Man's Game -=20 Load new package;=20 issue search() command; find position (say N) of loaded package;=20 issue objects(pos =3D N) command; get name of a random function (san

Running R.app on Mac OS X 10.4 > version _ platform powerpc-apple-darwin8.6.0 arch powerpc os darwin8.6.0 system powerpc, darwin8.6.0 status major 2 minor 3.1 year 2006

Running R.app on Mac OS X 10.4 > version _ platform powerpc-apple-darwin8.6.0 arch powerpc os darwin8.6.0 system powerpc, darwin8.6.0 status major 2 minor 3.1 year 2006

Hi all, Can anyone explain why the following use of the subset() function produces a different outcome than the use of the "[" extractor? The subset() function as used in density(subset(mydf, ht >= 150.0 & wt <= 150.0, select = c(age))) appears to me from documentation to be equivalent to density(mydf[mydf$ht >= 150.0 & mydf$wt <= 150.0, "age"])

Hi R users I need to estimate, with kaplan Meier methodology, a Quality adjusted survival analysis. It is possible doing this at R? Thanks in advance. Best Regards Pedro Mota Veiga -- View this message in context: http://r.789695.n4.nabble.com/How-can-I-doing-Quality-adjusted-survival-analysis-in-R-tp4295868p4295868.html Sent from the R help mailing list archive at Nabble.com.

Full_Name: Steven McKinney Version: 2.9.0 OS: Mac OS X 10.5.6 Submission from: (NULL) (142.103.207.10) A corrupt data frame can be constructed as follows: foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4]

> -----Original Message----- > From: Duncan Murdoch [mailto:murdoch at stats.uwo.ca] > Sent: Tuesday, August 04, 2009 8:03 AM > To: Steven McKinney > Cc: r-devel at stat.math.ethz.ch; R-bugs at r-project.org > Subject: Re: [Rd] Wishlist: Navigate to "Index" page of help when no > topic specified (PR#13860) >=20 > On 7/28/2009 6:30 PM, smckinney at bccrc.ca