Displaying 20 results from an estimated 10000 matches similar to: "xlim in plot.survfit() [with a discussion on "..."] (PR#2173)"
2003 Jun 05
2
Fwd: Re: legend() with option adj=1
Is there a simpler way then the solution to the one that was posted here? I'm
not very proficient with legend, and I don't understand this solution. All
I have is two or more lines on one plot that I want to put a legend on and I
can't figure out how to do it from the examples. Can you give a very simple
example? It does not have to be fancy!! I have never worked with a
2003 Feb 24
2
"trace" argument in legend() (PR#2578)
Full_Name: Jerome Asselin
Version: 1.6.2
OS: RedHat Linux 7.2
Submission from: (NULL) (142.103.173.179)
Should be an easy fix...
Consider the examble below:
plot(0,0)
legend(0,0,c("Hello!","Hi!"),pch=1:2,lty=1:2,trace=T)
It gives the following trace:
> plot(0,0)
> legend(0,0,c("Hello!","Hi!"),pch=1:2,lty=1:2,trace=T)
xchar= 0.05178 ;
2003 May 20
1
legend() with option adj=1
Hi there,
I want to justify to right the text of my legend. Consider this short
reproducable example.
x <- 1:5
y1 <- 1/x
y2 <- 2/x
plot(rep(x,2),c(y1,y2),type="n",xlab="x",ylab="y")
lines(x,y1)
lines(x,y2,lty=2)
legend(5,2,c("1,000","1,000,000"),lty=1:2,xjust=1,yjust=1)
2003 Aug 07
2
segmentation fault: formula() with long variable names (PR#3680)
R version: 1.7.1
OS: Red Hat Linux 7.2
In this example, I would expect an error for the overly long variable
name. This is always reproducable for me.
> formula(paste("y~",paste(rep("x",50000),collapse="")))
Segmentation fault
Sincerely,
Jerome Asselin
--
Jerome Asselin (Jérôme), Statistical Analyst
British Columbia Centre for Excellence in HIV/AIDS
St.
2003 Jul 15
2
"na.action" parameter in princomp() (PR#3481)
Full_Name: Jerome Asselin
Version: 1.7.1
OS: Red Hat Linux 7.2
Submission from: (NULL) (24.77.125.119)
Setting the parameter na.action=na.omit should remove
incomplete records in princomp. However this does not
seem to work as expected. See example below.
Sincerely,
Jerome Asselin
data(USArrests)
princomp(USArrests, cor = TRUE) #THIS WORKS
USArrests[1,3] <- NA
princomp(USArrests, cor =
2003 Oct 24
2
Segmentation fault in .Call() (PR#4761)
Full_Name: Jerome Asselin
Version: 1.8.0
OS: RedHat Linux 7.2
Submission from: (NULL) (142.103.177.13)
I would not expect a segmentation fault; perhaps an error message.
> .Call("log")
Segmentation fault
This is always reproducable for me.
Sincerely,
Jerome Asselin
2003 Aug 08
1
covmat argument in princomp() (PR#3682)
R version: 1.7.1
OS: Red Hat Linux 7.2
When "covmat" is supplied in princomp(), the output value "center" is all
NA's, even though the input matrix was indeed centered. I haven't read
anything about this in the help file for princomp(). See code below for an
example: pc2$center is all NA's.
Jerome Asselin
x <- rnorm(6)
y <- rnorm(6)
X <- cbind(x,y)
2002 Nov 22
1
Segmentation fault using "survival" package (PR#2320)
Full_Name: Jerome Asselin
Version: 1.6.1
OS: RedHat Linux 7.2
Submission from: (NULL) (142.103.173.179)
Hello,
I get a segmentation fault when I run the following code. I wouldn't expect
meaningful results because my response variable contains only missing values.
However, I would expect something like a regular error (not a segmentation
fault).
library(survival)
data <-
2003 Mar 12
1
plot() with type="s" and lty=2 (PR#2630)
Full_Name: Jerome Asselin
Version: 1.6.2
OS: RedHat Linux 7.2
Submission from: (NULL) (142.103.173.179)
In the following example, the line type lty=2 does not show properly
across the entire line.
x <- c(seq(0,.5,.001),seq(.6,1,.1))
y <- rep(1,length(x))
plot(x,y,type="s",lty=2)
Sincerely,
Jerome Asselin
2003 Feb 27
2
interval-censored data in survreg()
I am trying to fit a lognormal distribution on interval-censored
data. Some of my intervals have a lower bound of zero.
Unfortunately, it seems like survreg() cannot deal with lower
bounds of zero, despite the fact that plnorm(0)==0 and
pnorm(-Inf)==0 are well defined. Below is a short example to
reproduce the problem.
Does anyone know why survreg() must behave that way?
Is there an alternate
2003 Aug 14
0
How to get the pseudo left inverse of a singular square m atrix?
I'm rusty, but not *that* rusty here, I hope.
If W (=Z*Z' in your case) is singular, it can not have inverse, which by
definition also mean that nothing multiply by it will produce the identity
matrix (for otherwise it would have an inverse and thus nonsingular).
The definition of a generalized inverse is something like: If A is a
non-null matrix, and G satisfy AGA = A, then G is called
2003 May 21
1
axis() default values for "lty", "lwd", and "col"
Hi,
I would like to recommend a minor modification in axis() which I believe
can simplify the making of plots for publications. I am trying to define
default values for par() in order to make labels bigger and lines thicker,
so that the resulting plots look good when resized for publication
purposes. I ran into the following problem...
axis() does not use par() values as default for
2003 May 07
0
frailty models in survreg() -- survival package (PR#2933)
I am confused on how the log-likelihood is calculated in a parametric
survival problem with frailty. I see a contradiction in the frailty() help
file vs. the source code of frailty.gamma(), frailty.gaussian() and
frailty.t().
The function frailty.gaussian() appears to calculate the penalty as the
negative log-density of independent Gaussian variables, as one would
expect:
>
2003 Jul 04
0
degrees of freedom in nlme() (PR#2384)
I would like to document my findings (with a potential FIX) regarding the
issue of calculation of the degrees of freedom with nlme().
The program given at the bottom of this email generates and fit 20 data
sets with a mixed-effects LINEAR model, but using the function nlme()
instead of lme(). In each case, the correct number of degrees of freedom
for the intercept parameter is 12. However, in
2003 Aug 30
3
fisher.test() gives wrong confidence interval (PR#4019)
The problem occurs when the sample odds ratio is Inf, such as in the
following example. Given the fact that both upper bounds of the two 95%
confidence intervals are Inf, I would have expected that the two lower
bounds be equal, but they aren't.
x <- matrix(c(9,4,0,2),2,2)
x
# [,1] [,2]
#[1,] 9 0
#[2,] 4 2
rbind("two.sided.95CI"=fisher.test(x)$conf.int,
2003 Aug 07
2
model.frame() call from inside a function (PR#3671)
R version: 1.7.1
OS: Red Hat Linux 7.2
Hi all,
The formula object in model.frame() is not retrieved properly when
model.frame() is called from within a function and the "subset" argument
is supplied.
foo <- function(formula,data,subset=NULL)
{
cat("\n*****Does formula[-3] == ~y ?**** TRUE *****\n")
print(formula[-3] == ~y)
cat("\n*****Result of model.frame()
2003 Jan 24
1
table() with option "exclude=NULL" (PR#2491)
Full_Name: Jerome Asselin
Version: 1.6.2
OS: redhat linux 7.2
Submission from: (NULL) (142.103.173.179)
Bug or feature? Hard to say...
But it sure would be nice if table()
would label the frequency of NA's as
it does for NaN's.
Regards,
Jerome
> table(c(2,NA,1,1,1),exclude=NULL)
1 2
3 1 1
> table(c(2,NA,1,1,1,NaN),exclude=NULL)
1 2 NaN
3 1 1 1
#For sake of
2003 Jan 24
2
hist() with option "sub" (PR#2492)
Full_Name: Jerome Asselin
Version: 1.6.2
OS: redhat linux 7.2
Submission from: (NULL) (142.103.173.179)
This is certainly not a big problem, but should there really
be a warning message when I run this?
> x <- c(1,1,2,2,2,2,3,3)
> hist(x,sub="Sub Title")
Warning messages:
1: parameter "sub" couldn't be set in high-level plot() function
2: parameter
2010 Jul 15
1
Standard Error for individual patient survival with survfit and summary.survfit
I am using the coxph, survfit and summary.survfit functions to calculate an estimate of predicted survival with confidence interval for future patients based on the survival distribution of an existing cohort of subjects. I am trying to understand the calculation and interpretation of the std.err and confidence intervals printed by the summary.survfit function.
Using the default confidence
2004 Jan 15
1
random effects with lme() -- comparison with lm()
Hi all,
In the (very simple) example below, I have defined a random effect for
the residuals in lme(). So the model is equivalent to a FIXED effect
model. Could someone explain to me why lme() still gives two standard
deviation estimates? I would expect lme() to return either:
a) an error or a warning for having an unidentifiable model;
b) only one standard deviation estimate.
Thank you for your