similar to: survfit

Displaying 20 results from an estimated 400 matches similar to: "survfit"

2010 Dec 14
0
Urgent help requested using survfit(individual=T):
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
2009 Jan 16
2
Predictions with GAM
Dear, I am trying to get a prediction of my GAM on a response type. So that I eventually get plots with the correct values on my ylab. I have been able to get some of my GAM's working with the example shown below: * model1<-gam(nsdall ~ s(jdaylitr2), data=datansd) newd1 <- data.frame(jdaylitr2=(244:304)) pred1 <- predict.gam(model1,newd1,type="response")* The problem I am
2006 Jan 17
2
help with parsing multiple coxph() results
Dear All: I have a question on using coxph for multiple genes: I have written code to loop through all 22283 genes in the Hgu-133A and apply coxph on survival data. However, I don't know how to work with the result for each gene: survtest<-coxph(Surv(pcc.primary.stg.3.cox[,'fup_interval'],pcc.primary.stg.
2007 May 16
2
log rank test p value
How can I get the Log - Rank p value to be output? The chi square value can be output, so I was thinking if I can also have the degrees of freedom output I could generate the p value, but can't see how to find df either. > (survtest <- survdiff(Surv(time, cens) ~ group, data = surv,rho=0)) Call: survdiff(formula = Surv(time, cens) ~ group, data = surv, rho = 0) N Observed
2006 Sep 15
3
Crashes and tests failures again with 0.10.4
In the beginning 0.10.4 looked promising, but now that my index has grown to > 100 MB I''m getting segfaults on some searches again: >> Post.find_by_contents(''rubyforum'') # ok >> Post.find_by_contents(''ruby-forum'') /usr/local/lib/ruby/gems/1.8/gems/ferret-0.10.4/lib/ferret/index.rb:351: [BUG] Segmentation fault ruby 1.8.4 (2005-12-24)
2006 Jan 17
0
help with coxph() for multiple genes
Dear All: I have a question on using coxph for multiple genes: I have written code to loop through all 22283 genes in the Hgu-133A and apply coxph on survival data. However, I don't know how to work with the result for each gene: survtest<-coxph(Surv(pcc.primary.stg.3.cox[,'fup_interval'],pcc.primary.stg.
2004 Aug 09
2
Approaches to using RUnit
Having used JUnit and PyUnit, I was pleased to see the release of the RUnit package on CRAN. I'm wondering if there are any RUnit users out there that would be willing to share some tips on how they organize their code to work with RUnit. Specifically, I'm wondering about the best way to load/import/source the functions to be tested. I would like to end up with a script, testall or some
2006 Sep 13
2
recursive methods for concatenating sets of files
Hello, I would like to read sets of files within a folder, perhaps using recursive methods. Right now, I rename the files before import. It would be even better to do this without renaming files, without providing explicit filenames, perhaps by importing files based on chronology, and translating each filename into a header? Please excuse my ignorance, and help cure my clunky programming
2009 Feb 25
3
survival::survfit,plot.survfit
I am confused when trying the function survfit. my question is: what does the survival curve given by plot.survfit mean? is it the survival curve with different covariates at different points? or just the baseline survival curve? for example, I run the following code and get the survival curve #### library(survival) fit<-coxph(Surv(futime,fustat)~resid.ds+rx+ecog.ps,data=ovarian)
2008 Aug 20
1
read.csv : double quoted numbers
Hello; I am new user of R; so pardon me. I am reading a .txt file that has around 50+ numeric columns with '\t' as separator. I am using read.csv function along with colClasses but that fails to recognize double quoted numeric values. (My numeric values are something like "1,001.23"; "1,008,000.456".) Basically read.csv fails with - "scan() expected 'a
2018 Dec 01
20
[Bug 1305] New: Rules in second chain same hook ignored if first chain has policy drop
https://bugzilla.netfilter.org/show_bug.cgi?id=1305 Bug ID: 1305 Summary: Rules in second chain same hook ignored if first chain has policy drop Product: nftables Version: unspecified Hardware: x86_64 OS: Debian GNU/Linux Status: NEW Severity: normal Priority: P5
2012 Oct 16
2
R Kaplan-Meier plotting quirks?
Hello. I apologize in advance for the VERY lengthy e-mail. I endeavor to include enough detail. I have a question about survival curves I have been battling off and on for a few months. No one local seems to be able to help, so I turn here. The issue seems to either be how R calculates Kaplan-Meier Plots, or something with the underlying statistic itself that I am misunderstanding. Basically,
2012 Oct 18
1
Kaplan-Meier plotting quirks
Better would be to use interval censored data. Create your data set so that you have (time1, time2) pairs, each of which describes the interval of time over which the tag was lost. So an animal first captured at time 10 sans tag would be (0,10); with tag at 5 and without at 20 would be (5,20), and last seen with tag at 30 would be (30, NA). Then survit(Surv(time1, time2,
2008 Mar 02
0
coxpath() in package glmpath
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
2009 Sep 16
2
Teasing out logrank differences *between* groups using survdiff or something else?
R Folk: Please forgive what I'm sure is a fairly na?ve question; I hope it's clear. A colleague and I have been doing a really simple one-off survival analysis, but this is an area with which we are not very familiar, we just happen to have gathered some data that needs this type of analysis. We've done quite a bit of reading, but answers escape us, even though the question below
2002 Apr 04
0
Basle/ Allerød: Survival Analysis in S-PLUS with Terry Therneau
SURVIVAL ANALYSIS IN S-PLUS by Dr Terry Therneau 14/15 May Aller?d, Denmark 16/17 May Basel, Switzerland Dr. Terry Therneau has worked in medical research statistics for over 15 years. He has written several papers on the use of residuals in the Cox model, and is the author of the survival routines found in S-PLUS, as well as the SAS routines COXREG and SURVTEST. Note: Due to his
2006 Aug 15
8
Ferret 0.10 series
Hi David, >> Otherwise I''m fully committed to getting 0.10 out. >> When it is out, I''d recommend getting it into your >> development app as soon as you can. Any idea of timeframe to a 0.10 alpha release? Kind Regards Neville
2011 Jul 08
1
survConcordance with 'counting' type Surv()
Dear Prof. Therneau I was impressed to discover that the 'survConcordance' now handles Surv() objects in counting format (example below to clarify what I mean). This is not documented in the help page for the function. I am very curious to see how a c-index is estimated in this case, using just the linear predictors. It was my impression that with left truncation the ordering of
2009 Sep 08
1
Obtaining value of median survival for survfit function to use in calculation
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
2011 Jan 24
1
How to measure/rank ?variable importance when using rpart?
--- included message ---- Thus, my question is: *What common measures exists for ranking/measuring variable importance of participating variables in a CART model? And how can this be computed using R (for example, when using the rpart package)* ---end ---- Consider the following printout from rpart summary(rpart(time ~ age + ph.ecog + pat.karno, data=lung)) Node number 1: 228 observations,