Displaying 20 results from an estimated 200 matches similar to: "coxpath() in package glmpath"
2010 Mar 23
0
glmpath and coxpath variables
Hi,
I am analyzing a set of variables in order to create a survival model for a
set of patients. I have checked the reference manual for glm path and
coxpath in order to achieve it. However I have a doubt about the class of
the covariates I can use with the last mentioned package.
In the example, the package loads a list called "lung.data". This object has
a matrix with the covariate
2006 Mar 02
0
glmpath (new version 0.91)
We have uploaded to CRAN a new version of glmpath, a package
which fits the L1 regularization path for generalized linear models.
The revision includes:
- coxpath, a function for fitting the L1-regularization path for the Cox
ph model;
- bootstrap functions for analyzing sparse solutions;
- the ability to mix in L2 regularization along with L1 (elasticnet).
We have also completed a report that
2006 Mar 02
0
glmpath (new version 0.91)
We have uploaded to CRAN a new version of glmpath, a package
which fits the L1 regularization path for generalized linear models.
The revision includes:
- coxpath, a function for fitting the L1-regularization path for the Cox
ph model;
- bootstrap functions for analyzing sparse solutions;
- the ability to mix in L2 regularization along with L1 (elasticnet).
We have also completed a report that
2012 Jul 26
0
Using pspline in bic.surv of BMA package
Hi,
I'm trying to using pspline in bic.surv{BMA}.
#############################
library(BMA)
library(survival)
data(veteran)
test.bic.surv<- bic.surv(Surv(time,status) ~ karno+pspline(age,df=3)+diagtime+prior, data = veteran, factor.type = TRUE)
summary(test.bic.surv, conditional=FALSE, digits=2)
#############################
The results are:
2010 Jun 04
0
glmpath crossvalidation
Hi all,
I'm relatively new to using R, and have been trying to fit an L1
regularization path using coxpath from the glmpath library.
I'm interested in using a cross validation framework, where I crossvalidate
on a training set to select the lambda that achieves the lowest error, then
use that value of lambda on the entire training set, before applying to a
test set. This seems to entail
2008 Feb 22
0
R CMD check for glmpath on Windows (PR#10823)
The problem first appeared in R 2.6.1 and is still there in R 2.6.2
On Windows running R CMD check command for glmpath package fails. The reason
seems to be that when R is running the examples file (glmpath-Ex.R), it skips
about 50 lines and as a result gives a syntax error.
I'm working with a modified version of the CRAN glmpath 0.94. My version
happens to give a more clear example of a
2010 Dec 14
1
survfit
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
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
2008 Sep 29
0
nomogram function (design library)
Dear colleagues,
I hope someone can help me with my problem.
I have fitted a cox model with the following syntax:
# cox01def <-cph(Surv(TEVENT,EVENT) ~ ifelse(AGE>50, (AGE-50)^2,0) +
BMI +
# HDL+DIABETES +HISTCAR2 + log(CREAT)+
as.factor(ALBUMIN)+STENOSIS+IMT,data # = XC, x=T, y=T, surv=T) *1
Furthermore I have estimated my beta's also with a Lasso method -
Coxpath ( from
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
2009 Aug 21
1
LASSO: glmpath and cv.glmpath
Hi,
perhaps you can help me to find out, how to find the best Lambda in a
LASSO-model.
I have a feature selection problem with 150 proteins potentially
predicting Cancer or Noncancer. With a lasso model
fit.glm <- glmpath(x=as.matrix(X), y=target, family="binomial")
(target is 0, 1 <- Cancer non cancer, X the proteins, numerical in
expression), I get following path (PICTURE
2013 May 02
0
Questions regarding use of predict() with glmpath
I'm trying to do LASSO in R with the package glmpath. However, I'm not sure
if I am using the accompanying prediction function *predict.glmpath()*
correctly.
Suppose I fit some regularized binomial regression model like so:
library(glmpath);load(heart.data);attach(heart.data);
fit <- glmpath(x, y, family=binomial)
Then I can use predict.glmpath() to estimate the value of the
2005 Nov 28
0
glmpath: L1 regularization path for glms
We have uploaded to CRAN the first version of glmpath,
which fits the L1 regularization path for generalized linear models.
The lars package fits the entire piecewise-linear L1 regularization
path for
the lasso. The coefficient paths for L1 regularized glms, however,
are not piecewise linear.
glmpath uses convex optimization - in particular predictor-corrector
methods-
to fit the
2005 Nov 28
0
glmpath: L1 regularization path for glms
We have uploaded to CRAN the first version of glmpath,
which fits the L1 regularization path for generalized linear models.
The lars package fits the entire piecewise-linear L1 regularization
path for
the lasso. The coefficient paths for L1 regularized glms, however,
are not piecewise linear.
glmpath uses convex optimization - in particular predictor-corrector
methods-
to fit the
2009 May 19
0
error glmpath()
Hi R-users!
I am trying to learn how to use the glmpath package. I have a dataframe like this
> dim(data)
[1] 605 109
and selected the following
> response <- data[,1]
> features<-as.matrix(data[,3:109])
> mymodel <- glmpath(features,response, family = binomial)
Error in if (lambda <= min.lambda) { :
missing value where TRUE/FALSE expected
Reading the glmpath pdf, I
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
2007 Sep 23
0
glmpath: how to choose best lambda
Hi all,
I am using glampath package for L1 regularized logistic regression. I have
read the article " L1 regularization path algorithm for GLM" by park and
Hastie (2006). One thing I can't understand that how to find best lambda for
my prediction. I want to use that lambda for the prediction not the entire
set.
thanks.
--
View this message in context:
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,
2010 Apr 06
1
glmpath in R
Hi Claire,
I'm replying and CC-ing to the R-help list to get more eyes on your
question since others will likely have more/better advice, and perhaps
someone else in the future will have a similar question, and might
find this thread handy.
I've removed your specific research aim since that might be private
information, but you can include that later if others find it
necessary to know
2002 Aug 28
0
user defined function in rpart
Hi,
I am trying to use the rpart library with my own set of functions on a
survival object. I get an immeadiate segmentation fault when i try
calling rpart with my list of functions. I get the same problem with the
logrank example from Therneau,s S-rpart library though their anova
example works. Should I report this as a bug, as even if my functions
are structured improperly, that should lead to