similar to: Survival Analysis Questions

Hi, May be this helps: ?set.seed(24) ?mydata4<- as.data.frame(matrix(sample(1:100,10*38,replace=TRUE),ncol=38)) ?dim(mydata4) #[1] 10 38 ?library(matrixStats) res<-do.call(cbind,lapply(1:400, function(i) {permutation<-sample(mydata4); (rowMeans(permutation[,1:27])-rowMeans(permutation[,28:38]))/(rowSds(permutation[,1:27])+rowSds(permutation[,28:38]))} )) ?dim(res) #[1]? 10 400 A.K.

Hello, I have questions regarding penalized Cox regression using survival package (functions coxph() and ridge()). I am using R 2.8.0 on Ubuntu Linux and survival package version 2.35-4. Question 1. Consider the following example from help(ridge): > fit1 <- coxph(Surv(futime, fustat) ~ rx + ridge(age, ecog.ps, theta=1), ovarian) As I understand, this builds a model in which `rx' is

I have a gene expression matrix n (genes) X p (cases), where n = 8000 and p = 100. I want to fit each gene as univariate in a coxph model, i.e., fitting 8000 models. I do something like this: res <- apply(data, 1, coxph.func) which takes about 4 min, not bad. But I need to do large numbers of permutations of the data (permuting the columns), for example, 2000, which would take 5 days. I would

Dear R users, There is a error message when I run the following code. It is used to load microarray data and use the top 1000 genes for training svm to classify test set . > library(e1071) Loading required package: class > f=read.table("F:\\lab\\ microarray analysis\\VEH LPS\\exprs.txt",

For any given pre-specified gene or short list of genes, yes the Cox model works fine. Two important caveats: 1. Remeber the rule of thumb for a Cox model of 20 events per variable (not n=20). Many microarray studies will have very marginal sample size. 2. If you are looking at many genes then a completely different strategy is required. There is a large and growing literature; I like Newton

Hello, all R-users! I am working on fitting a survival analysis model using the coxph function for Cox proportional hazards regression model. Data look like usual: ========================== group block death censor Group1 1 4 1 Group1 1 12 1 ... Group2 30 4 1 Group2 30 4 1 ... Group3 57 16

Hi, I am performing Cox proportional hazards regression on a microarray dataset with 15000 genes. The p values generated from the Cox regression (based on normal distribution of large sample theory) showed only 2 genes have a p value less than 0.05. However, when I did a permutation on the dataset to obtained permutated p values, and it turned out about 750 genes had a permutated p value less than

Hi all, I try to generate sets of strategies that contain probability distributions for a defined number of elements, e.g. imagine an animal that can produce 5 different types of offspring and I want to figure out which percentage of each type it should produce in order to maximize its fitness. In order to do so, I need to calculate the fitness for all potential strategies. As an example, if I

Hi there, I would like to find a more efficient way of permuting the rows and columns of a symmetrical matrix that represents ecological or actual distances between objects in space. The permutation is of the type used in a Mantel test. Specifically, the permutation has to accomplish something like this: Original matrix addresses: a11 a12 a13 a21 a22 a23 a31 a32 a33 Example

The latest version of the survival package has two important additions. In prior code the call coxph(Surv(time, status) ~ age + strata(inst), data=lung) could fail if a version of either Surv() or strata() existed elsewhere on the search path; the wrong function could be picked up. Second, a model with survival::strata(inst) in the formula would not do what users expect. These

I have what I *think* should be a simple problem in R, and hope someone might be able to help me. I'm working with cancer survival data, and would like to calculate adjusted survival figures based on the age of the patient and the tumour classification. A friendly statistician told me I should use Cox proportional hazards to do this, and I've made some progress with using the

[code]>require("survival") > coxph(Surv(futime,fustat)~age + strata(rx),ovarian) Call: coxph(formula = Surv(futime, fustat) ~ age + strata(rx), data = ovarian) coef exp(coef) se(coef) z p age 0.137 1.15 0.0474 2.9 0.0038 Likelihood ratio test=12.7 on 1 df, p=0.000368 n= 26, number of events= 12 > coxph(Surv(futime,fustat)~age, ovarian, subset=rx==1)

I wonder if anyone know why survest (a function in Design package) and standard survfit.coxph (survival) returned different confidence intervals on survival probability estimation (say 5 year). I am trying to estimate the 5-year survival probability on a continuous predictor (e.g. Age in this case). Here is what I did based on an example in "help cph". The 95% confidence intervals

Dear all, I am using glmnet (Coxnet) for building a Cox Model and to make actual prediction, i.e. to estimate the survival function S(t,Xn) for a new subject Xn. If I am not mistaken, glmnet (coxnet) returns beta, beta*X and exp(beta*X), which on its own cannot generate S(t,Xn). We miss baseline survival function So(t). Below is my code which takes beta coefficients from glmnet and creates coxph

Dear all, I am confused with the output of survfit.coxph. Someone said that the survival given by summary(survfit.coxph) is the baseline survival S_0, but some said that is the survival S=S_0^exp{beta*x}. Which one is correct? By the way, if I use "newdata=" in the survfit, does that mean the survival is estimated by the value of covariates in the new data frame? Thank you very much!

Hello, When I try to to obtain the expected risk for a new dataset using coxph in the survival package I get an error. Using the example from ?coxph: > test1 <- list(time= c(4, 3,1,1,2,2,3),+ status=c(1,NA,1,0,1,1,0),+ x= c(0, 2,1,1,1,0,0),+ sex= c(0, 0,0,0,1,1,1))> cox<-coxph( Surv(time, status) ~ x + strata(sex), test1)

Hi, I'm running a tutorial ("Meta-analyses of data from two (or more) microarray data sets"), which use wgcna package. I have an error in the function modulePreservation (it is below). I'm using R2.13 Can you help me? Do you know, what is happens? Thanks Raquel multiExpr = list(A = list(data=t(badea)),B = list(data=t(mayo))) # two independent datasets (dim = 13447 x 36) mp =

Dear all, I would have some questions on the coxph function for survival analysis, which I use with frailty terms. My model is: mdcox<-coxph(Surv(time,censor)~ gender + age + frailty(area, dist='gauss'), data) I have a very large proportion of censored observations. - If I understand correctly, the function mdcox$frail will return the random effect estimated for each group on the

It seems to me that summary for ridge coxph() prints summary but returns NULL. It is not a big issue because one can calculate statistics directly from a coxph.object. However, for some reason the score test is not calculated for ridge coxph(), i.e score nor rscore components are not included in the coxph object when ridge is specified. Please find the code below. I use 2.9.2 R with 2.35-4 version

Dear all, I am confused with the behaviour of survfit with newdata option. I am using the latest version R-2-15-0. In the simple example below I am building a coxph model on 90 patients and trying to predict 10 patients. Unfortunately the survival curve at the end is for 90 patients. Could somebody please from the survival package confirm that this behaviour is as expected or not - because I