Displaying 20 results from an estimated 600 matches similar to: "turning labels into a vector"
2008 Nov 26
1
survreg and pweibull
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)),
2004 Nov 23
6
Weibull survival regression
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,
2005 Jun 11
0
Xend won''t start: attribute VIRQ_DOM_EXC (fwd)
I''ll step back to Xen stable, because of my network probs with Xen
unstable. So I configured and compiled Xen 2.0.6, but I came to the
following error after starting Xend:
Exception starting xend: ''module'' object has no attribute ''VIRQ_DOM_EXC''
Traceback (most recent call last):
File "/usr/lib/python/xen/xend/server/SrvDaemon.py", line 302,
2005 Mar 01
3
Anova with Scheffe Tests
Hi R-people,
I am wanting to run Factorial ANOVA followed by Scheffe tests on some spatial subjective data. I'm comparing X-Y independent coordinates against x-y dependent coordinates. There are only four independent spatial coordinates that form a square.
I am wondering whether I am doing the right thing, because there doesn't seem to be a simple way of doing this. I have attempted to
2005 Mar 22
2
LME correlation structures: user defined
Let me modify my question about user-defined covariance structures for LME models: Can somebody tell me how I can see the code for the definition of the correlation structures that come with the NLME package. Specifically I like to see the code for the functions coef, corMatrix, and intialize for any of the pre-defined correlation structures, and use this as a template to define a new correlation
2009 Jun 03
4
Excel Export in a beauty way
Hallo all,
I`ve read a lot of things in this forum about an Excel export via R. It is
no problem to export my data frames via write.table or write.xls (xls or
csv), but some things are not very convenient for me: I always have to
adjust the column with to see all the numbers or the text and there is no
frame between the cells. And I missing the possibility to make some headers
bold or coloured.
2010 Aug 17
2
how to selection model by BIC
Hi All:
the package "MuMIn" can be used to select the model based on AIC or AICc.
The code is as follows:
data(Cement)
lm1 <- lm(y ~ ., data = Cement)
dd <- dredge(lm1,rank="AIC")
print(dd)
If I want to select the model by BIC, what code do I need to use? And when
to select the best model based on AIC, what the differences between the
function "dredge" in
2005 Apr 11
1
extracting correlations from nlme
Hi,
I would like to know how (if) I can extract some of the information from
the summary of my nlme.
at present, I get a summary looking something like this:
> summary(fit.nlme)
Nonlinear mixed-effects model fit by maximum likelihood
Model: MLKYLD ~ W4(DIM, logA, B, C)
Data: ADHIS.x0
AIC BIC logLik
265314 265401.6 -132647
Random effects:
Formula: list(logA ~ 1 , B ~
2005 Apr 29
2
Iterative process for reading in text files
Hello
Instead of reading in group1.txt I want to read in groups1 for the first iteration of i, then groups2 for the second and so on. Obviously I can't use groups(i) but assume there is a way to do this.
group<-read.table("C:/Data/April 2005/group1.txt",header=T)
thanks in advance
Meredith
2005 May 04
1
Double hurdle model in R
I am interested in utilizing this so called "double hurdle" model
in my study. We can write the model in the following way:
if (z'a + u > 0 & x'b + e > 0) y = x'b + e, else y = 0
In the model, consumption y is the (left-) censored dependent variable. e
and u are the normally distributed error terms. z'a is the participation
equation and x'b is the
2005 May 27
1
logistic regression
Hi
I am working on corpora of automatically recognized utterances, looking
for features that predict error in the hypothesis the recognizer is
proposing.
I am using the glm functions to do logistic regression. I do this type
of thing:
* logistic.model = glm(formula = similarity ~., family = binomial,
data = data)
and end up with a model:
> summary(logistic.model)
Call:
2005 Jan 11
8
Calculate Mean of Column Vectors?
Hello,
I've got an array defined as y <- rnorm(3000), dim(y) <- c(3, 1000).
I'd like to produce a 1000-element vector z that is the mean of the
corresponding elements of y (like z[1,1] <- mean(y[1,1], y[2,1],
y[3,1])), but being new to R, I'm not sure how to do this for all
elements at once (or, at least, simply). Any help is appreciated.
Thanks,
Tom
2006 Mar 11
2
weird! QDA does not depend on priors?
Hi all,
If I run LDA on the same data (2-class classification) with default(no
priors specified in the lda function) vs. "prior=c(0.5, 0.5)", the results
are different.
The (0.5, 0.5) priors give better 1-classify-to-1 rate, and the proportional
priors(default, no priors specified in the lda function) give better
0-classify-to-0 rate, for both training and testing data sets.
However,
2011 Jul 14
1
Amelia_Multiple_Imputation_with_observational_priors_noms
I am fairly new at using R/programming in general so I apologize if I am
leaving crucial parts of the puzzle out, but here goes.
First and foremost this is the error I am receiving:
Error in muPriors[priors[, 1:2]] <- priors[, 3] :
NAs are not allowed in subscripted assignments
This occurs only when I am using observational priors and some number of
nominal variables, it does not
2010 May 16
1
predict.lda breaks when priors are specified
Dear R help,
What am I doing wrong here? when I don't specify the priors it works
just fine but when I specify the priors it breaks.? Does anyone know
why and how I can fix it?
----
> N=20000
> ncontrol=ncases=50
> X <- as.matrix(rnorm(N,0,1))
> eta <- -5.3 + X * 1.7
> p <- exp(eta)/(1+exp(eta))
> Y <- rbinom(N,1,p)
> controls <- sample(seq_len(N),
2006 Nov 14
2
Problem with file size
Hi everyone,
I have 2 environments (2 different R sessions) as described below:
Session 1:
Name of the environment: "CrlmmInfo"
Objects in the environment:
index1: logical index - length 238304
index2: logical index - length 238304
priors: list of 4 - (matrix 6x6, 2 vectors of length 6, vector of
length 2) - all num
params: list of 4:
centers [238304 x 3 x
2006 Nov 14
2
Problem with file size
Hi everyone,
I have 2 environments (2 different R sessions) as described below:
Session 1:
Name of the environment: "CrlmmInfo"
Objects in the environment:
index1: logical index - length 238304
index2: logical index - length 238304
priors: list of 4 - (matrix 6x6, 2 vectors of length 6, vector of
length 2) - all num
params: list of 4:
centers [238304 x 3 x
2008 Mar 28
4
rsync: writefd_unbuffered failed to write 4 bytes [sender]: Broken pipe
Hi,
i installed rsync 3.0 on a Sun Solaris 10 Sparc Server and tried to
Backup the Server via BackupPC
the complete rsync command from the backup server is this:
/usr/bin/ssh -q -x -l root IP /usr/bin/rsync --numeric-ids --perms
--owner --group -D --links --hard-links --times --block-size=2048
--recursive --checksum-seed=32761 --log-file=/var/log/rsync.log
--ignore-times . /
in the log on
2012 Sep 12
7
multinomial MCMCglmm
Dear all,
I would like to add mixed effects in a multinomial model and I am trying
to use MCMCglmm for that.
The main problem I face: my data set consits of a trapping data set,
where the observation at eah trap (1 or 0 for each species) have been
aggregated per traplines. Therefore we have a proportion of
presence/absence for each species per trapline.
ex:
ID_line mesh habitat Apsy Mygl
2009 Nov 12
1
multiple connections per imap/pop3 process in 2.0
When I set this in master.conf:
service imap {
service_count = 5
}
I see this error when two imap users log in:
Nov 11 16:54:16 server dovecot[5432]: imap-login: Login: user=<user1>, method=PLAIN, rip=10.100.0.84, lip=10.80.0.163, pid=5573
Nov 11 16:54:31 server dovecot[5432]: imap-login: Login: user=<user2>, method=PLAIN, rip=::1, lip=::1, secured, pid=5573
Nov 11