Hello
Thanks a lot Marc, for the suggestion to explore the issue a bit more
systematically
So I did and the conclusion is that with the old drc 1.4-2, I get a
SE=0.003, with the new drc 1.5-2, I get a SE=0.4
irrespective of the R version or the version of the packages drc depends
on
I hope
Thomas Zumbrunn wrote:
...
Thanks for your answers. This was almost what I was looking for, except that I
would need something for a 2-dimensional context (my question was not specific
enough).
Hi Thomas,
The thigmophobe.labels function just works out how to place each label
away from the
Hello all,
I'm trying to replicate a confirmatory factor analysis done in Amos. The
idea is to compare a one-factor and a two-factor model. I get the following
warning message when I run either model:
Could not compute QR decomposition of Hessian.
Optimization probably did not converge.
I
Dear R users,
I've used the function qr.fit.sfn to estimate a quantile regression on a
panel data set. Now I would like to compute an statistic to measure the
goodness of fit of this model.
Does someone know how could I do that? I could compute a pseudo R2 but in
order to do that I would need
Hi Hans,
I hope I can resolve your problems below (Marc, thank you very much for cc'ing
me on your
initial response!).
Have a look at the following R lines:
## Fitting the model using drm() (from the latest version)
m1- drm(response ~ dose, data = d, fct = LL.4())
summary(m1)
plot(m1)
##
I have a problem related to measuring likelihood between
-an observed presence absence dataset (containing 0 or 1)
-a predicted simulation matrix of the same dimensions (containing values from 0
to 1)
This must be a common problem but I am struggling to find the answer in the
literature.
Hi,
Perhaps you can try this,
seq.weave - function(froms, by, length, ... ){
c(
matrix(c(sapply(froms, seq, by=by, length = length/2, ...)),
nrow=length(froms), byrow=T)
)
}
seq.weave(c(2, 3), by=3, length=8)
seq.weave(c(2, 3, 4),
maxb wrote:
Can someone help me. I am very new to R. I am fitting a Cox model using Frank
Harrell's cph as I want to produce a Nomogram. This is what I have done:
Srv- Surv(time,cens)
f.cox- cph(Srv~ v1+v2+v3+v4, x=T, y=T, surv=T)
This is not reproducible for us. Where are the data?
What is
tsunhin wong wrote:
Dear R Users,
I have some dynamic selection rules that I want to pass around for my functions:
rules - paste(g$TrialList==1 g$Session==2)
I guess you do not want to paste() at all:
rules - g$TrialList==1 g$Session==2
Uwe Ligges
myfunction - function(rules) {
Michael wrote:
Could anybody point me to the latest status of the most user-friendly
debugger in R?
I always have been happy with ?debug, ?recover and friends cited there.
Although you also find additional software like the debug package ..
Best,
Uwe Ligges
How I wish I don't have to
Yes, thanks that's very useful
Apart from checking the fit with nls() as you suggested, I've also used Prism,
which gave the following results
Equation 1
Best-fit values
BOTTOM 10.96
TOP106.4
LOGEC50-5.897
HILLSLOPE 0.9501
EC50 1.2670e-006
Rowe, Brian Lee Yung (Portfolio Analytics) wrote:
To get the value I want, I am using the following code:
sort(as.vector(apply(array(c(2,3)), 1, seq, by=3,length.out=4)))
[1] 2 3 5 6 8 9 11 12
So two questions:
1. Is seq designed/intended to be used with a vector from argument,
Yes, you're right: taking logarithms is no longer needed!
Christian
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented,
Dear R-list,
the following code had been running well over the last months:
exam - matrix(rnorm(100,0,1), 10, 10)
gg - factor(c(rep(A, 5), rep(B, 5)))
mlmfit - lm(exam ~ 1); mlmfitG - lm(exam ~ gg)
result - anova(mlmfitG, mlmfit, X=~0, M=~1)
Until, all of a sudden the
Dear all,
I have seen a response from Duncan Murdoch which comes close to solving this
one, but I can't quite seem to tailor it to fit my needs!
I am trying to make just the title in my legend as bold font, with the legend
'items' in normal typeface.
I've tried setting par(font=2) external
Dear All,
I am running Debian testing on my box and I have R 2.9.0 installed from
the standard repositories.
I downloaded the package source from
http://cran.r-project.org/web/packages/Cairo/index.html
but when I try to install it on my system, this is what I get
$ sudo R CMD INSTALL
On 22 May 2009 at 13:42, Lorenzo Isella wrote:
| Dear All,
| I am running Debian testing on my box and I have R 2.9.0 installed from
| the standard repositories.
| I downloaded the package source from
| http://cran.r-project.org/web/packages/Cairo/index.html
| but when I try to install it on my
Dirk Eddelbuettel wrote:
On 22 May 2009 at 13:42, Lorenzo Isella wrote:
| Dear All,
| I am running Debian testing on my box and I have R 2.9.0 installed from
| the standard repositories.
| I downloaded the package source from
| http://cran.r-project.org/web/packages/Cairo/index.html
| but
Dear all ,
I am just wondering if there is a way to read inputs from adjacency matrix . I
am using igraph module. I want to directly load the input as adjacency instead
of reading as edgelist or pajek format. Can anypne help ?
Thanks ,
Pankaj Barah
Mathematical modelling Computational
Tim Clark mudiver1200 at yahoo.com writes:
I need some help in coming up with a function that will take two data sets,
determine if a value is missing in
one, find a value in the second that was taken at about the same time, and
substitute the second value in for
where the first should have
To all. I need your help. The big question is: How do I make an R library
with Fortran 95 files? You may assume that I am a pretty decent programmer
in both R and Fortran. I lay out a scenario and need your help!
I know how to make an ordinary R package and have dabbled with R + Fortran
95
Greetings,
is there are way to simply compute the score-statistics for a logistic
model generated with lrm?
For example, I want to compare the wald-statistics for a given model
against the score-statistics in order to find the relevant predictors.
attach(iris)
model = lrm
Dear Solomon,
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of S. Messing
Sent: May-22-09 1:27 AM
To: r-help@r-project.org
Subject: [R] Confirmatory factor analysis problems using sem package
(works
in Amos)
Hello all,
Further I appreciate your new function survmean(). At the moment it
seems to be intended as internal, and not documented in the help.
The computations done by print.survfit are now a part of the results returned
by
summary.survfit. See 'table' in the output list of ?summary.survfit.
maxb wrote:
Can someone help me. I am very new to R. I am fitting a Cox model using Frank
Harrell's cph as I want to produce a Nomogram. This is what I have done:
Srv- Surv(time,cens)
f.cox- cph(Srv~ v1+v2+v3+v4, x=T, y=T, surv=T)
As soon as I press enter, Windows XP crashes. If I remove
On Thu, May 21, 2009 at 8:28 PM, Gabor Grothendieck ggrothendi...@gmail.com
wrote:
It uses hours/minutes/seconds for values 1 day and uses days and
fractions
of a day otherwise.
Yes, my examples were documenting this idiosyncracy.
For values and operations that it has not considered it
On May 21, 2009, at 6:31 PM, Ted Harding wrote:
On 21-May-09 23:02:28, David Scott wrote:
Well most people deal with that problem by not using Acrobat to
read .pdf files. On linux you can use evince or xpdf. On windows
just use gsview32. Those readers don't lock the .pdf.
I am with Peter and
On 5/22/2009 8:18 AM, sjnovick wrote:
To all. I need your help. The big question is: How do I make an R library
with Fortran 95 files? You may assume that I am a pretty decent programmer
in both R and Fortran. I lay out a scenario and need your help!
I know how to make an ordinary R
Hi friends,
I have a query regarding na.omit function.Please ,someone help me.
I have a function
xyz_function-function(arguments)
{
some code
return(list(matrix=dataset))
}
xyz_function_returnvalue-xyz_function(passed argumentss)
*Case-I*
Dear All,
I am attempting to use forward and/or backward selection to determine
the best model for the variables I have. Unfortunately, because I am
dealing with patients and every patient is receiving treatment I need
to force the variable for treatment into the model. Is there a way to
do
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
We have no idea of the structure of 'dataset' that is being returned by your
function. This example works
You could create a subclass of times with its own print and format
methods that printed and formatted hours/minutes/seconds even if
greater than one day if that is the main item you need.
Regarding division you could contribute that to the chron package.
I've contributed a few missing items and
On 5/22/2009 9:58 AM, Laura Bonnett wrote:
Dear All,
I am attempting to use forward and/or backward selection to determine
the best model for the variables I have. Unfortunately, because I am
dealing with patients and every patient is receiving treatment I need
to force the variable for
Hi Jim,
xyz_function-function(arguments)
{
some code
pair_raw_data-changeMissingValuestoNA(pair_raw_data$matrix,pair_raw_data$dim,missing_values)
return(list(matrix=pair_raw_data))
}
Type of my dataset was a list.I tried checking that ,changing my dataset
to some other
Hi, i would to know, if someone have ever write the code to estimate the
parameter (mixing proportion, mean, a var/cov matrix) of a mixture of two
multivariate normal distribution. I wrote it and it works (it could find
mean and mixing proportion, if I fix the var/cov matrix), while if I fix
Greetings,
I checked the Indian diabetes data again and get one tree for the data with
reordered columns and another tree for the original data. I compared these
two trees, the split points for these two trees are exactly the same but the
fitted classes are not the same for some cases. And the
I am choosing a file like this:
#Bring up file selection box
fn-file.choose()
fp-file.path(fn,fsep='\\')
Unfortunately, the file path contains the short file name and extension as
well. I had hoped to get only the path so I could make my own long
filenames (for output graphs) by concatenation
So if I want to concatenate the output of multiple seq calls, there's no clear
way to to do this?
For background, I have a number of data.frames with the same structure in a
list. I want to 'collapse' the list into a single data.frame but only keeping
certain columns from each underlying
Hi all,
Could anybody point me to some existing code in R for fitting
Autoregressive Conditional Duration and Cox proportional hazard model
and with model selection and model specification tests?
Thank you!
__
R-help@r-project.org mailing list
Really I think if there is a Visual Studio strength debugger, our
collective time spent in developing R code will be greatly reduced.
2009/5/22 Uwe Ligges lig...@statistik.tu-dortmund.de:
Michael wrote:
Could anybody point me to the latest status of the most user-friendly
debugger in R?
Hi, i would to know, if someone have ever write the code to estimate the
parameter (mixing proportion, mean, a var/cov matrix) of a mixture of two
multivariate normal distribution. I wrote it and it works (it could find
mean and mixing proportion, if I fix the var/cov matrix), while if I fix
Content-Type: text/plain; charset=ISO-8859-15; format=flowed
Content-Transfer-Encoding: 8bit
Dear R Users,
I'm using pgfSweave() form R-Forge.
How can I disable prevention of overlapping tick labels in axis()?
If I use axis() with Sweave(Test1, driver = pgfSweaveDriver), R
treats the
On Fri, May 22, 2009 at 10:03 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Regarding division you could contribute that to the chron package.
I've contributed a few missing items and they were incorporated.
Good to know. Maybe I'll do that
Giving an error when it does not
Skotara wrote:
Dear R-list,
the following code had been running well over the last months:
exam - matrix(rnorm(100,0,1), 10, 10)
gg - factor(c(rep(A, 5), rep(B, 5)))
mlmfit - lm(exam ~ 1); mlmfitG - lm(exam ~ gg)
result - anova(mlmfitG, mlmfit, X=~0, M=~1)
On 5/22/2009 10:45 AM, am...@xs4all.nl wrote:
I am choosing a file like this:
#Bring up file selection box
fn-file.choose()
fp-file.path(fn,fsep='\\')
file.path() constructs a path from component parts, it doesn't extract
the path from a filename.
You want dirname(fn). You may also want
Try it this way:
# test list of data frames
L - list(anscombe[1:4], anscombe[5:8], anscombe[1:4], anscombe[5:8])
# get columns 2 and 3 from each component; cbind those together
do.call(cbind, lapply(L, [, 2:3))
On Fri, May 22, 2009 at 11:01 AM, Rowe, Brian Lee Yung (Portfolio
Analytics)
Are you looking for choose.dir() ?
Or basename() and dirname()?
Uwe Ligges
am...@xs4all.nl wrote:
I am choosing a file like this:
#Bring up file selection box
fn-file.choose()
fp-file.path(fn,fsep='\\')
Unfortunately, the file path contains the short file name and extension as
well. I had
On 5/22/2009 10:59 AM, Michael wrote:
Really I think if there is a Visual Studio strength debugger, our
collective time spent in developing R code will be greatly reduced.
If someone who knows how to write a debugger plugin for Eclipse wants to
help, we could have that fairly easily. All the
Dear Terry,
sorry that I did not see this change, and thank you for it. It is very useful.
Heinz
At 15:28 22.05.2009, Terry Therneau wrote:
Further I appreciate your new function survmean(). At the moment it
seems to be intended as internal, and not documented in the help.
The
##Is this what you mean ??
do.call(rbind,lapply(yourlist,[,,seq(from=1,by=3,length=4)))
## note the ,, that omits the row argument in the call to [
-- Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: r-help-boun...@r-project.org
Yet another way of doing it:
x - c(2,3)
x + rep(seq(0, by=3, length=4), each=length(x))
[1] 2 3 5 6 8 9 11 12
On Fri, May 22, 2009 at 11:01 AM, Rowe, Brian Lee Yung (Portfolio Analytics)
b_r...@ml.com wrote:
So if I want to concatenate the output of multiple seq calls, there's no
With levelplot I would like to combine fitted values and raw values in
one plot. The surface should be based on fitted values and on top of
those I would like distinguisable points based on the raw data with a
colorscheme the same as the surface.
# raw data
x1 = rep(1:10, times = 10)
x2 =
Brilliant! Thanks, Brian
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: Friday, May 22, 2009 11:20 AM
To: Rowe, Brian Lee Yung (Portfolio Analytics)
Cc: Peter Dalgaard; r-help@r-project.org
Subject: Re: [R] Behavior of seq with vector from
Try it
Alan wrote:
Hi,
How do you obtain the limits of the plotting region in a scatterplot3d
plot? `par('usr')' does not seem to give sensible values, and that
vector only has 4 elements (not the expected 6).
Well, I never designed anything to do that, but it is possible with the
following
Dear R-Users,
Is there a way to check within the following dummy function if the strip
argument is different from the default set in the function declaration?
FYI, this argument should be used downstream in a xyplot call of my
actual function.
dummyfunction - function(..., strip =
On Fri, May 22, 2009 at 11:01 AM, Stavros Macrakis
macra...@alum.mit.edu wrote:
On Fri, May 22, 2009 at 10:03 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Regarding division you could contribute that to the chron package.
I've contributed a few missing items and they were
Have you tried the maxLik package? If that is not adequate, you
can check CRAN Task View: Optimization and Mathematical Programming
(http://cran.fhcrc.org/web/views/Optimization.html). Or try the new
RSiteSearch.function in the RSiteSearch package. If none of these
adequate, please
A subset of my raw data looks like this:
--
Grip Technique Baseline.integrated Task
Stroke..direction.Engag Disen
PenDG PenUG PenDS
PenUS Duration
-
Hi,
I would like to have lineplot.CI and barplot.CI to actually plot
confidence intervals , instead of standard error.
I understand I have to use the ci.fun option, but I'm not quite sure
how.
Like this :
qt(0.975,df=n-1)*s/sqrt(n)
but how can I apply it to visualize the length of
Hi all,
I had some trouble in regrouping factor levels for a variable. After some
experiments, I have figured out how I can recode to modify the factor levels. I
would now like some help to understand why some methods work and others don't.
Here's my code :
rm(list=ls())
###some trials in
Laura: Part of the issue may depend on what you mean by goodness-of-ft.
If you are looking for some global measure like a pseudo R or AIC to
select among models, you ought to be able to make those calculations off
the objective function that was minimized as you recognized. If
qr.fit.sfn()
Error message is self-explanatory: there is an unused parameter
'na.rm=TRUE'. You are calling your function 'truecost' which only has a
single parameter 'time' and you are attempting to pass in 'na.rm=TRUE' which
it will not accept. You don't need it.
On Fri, May 22, 2009 at 12:36 PM, Thomas
Here are 2 functions, which.just.above and which.just.below,
which may help you. They will tell which element in a reference
dataset is the first just above (or just below) each element
in the main dataset (x). They return NA if there is no reference
element above (or below) an element of x.
I think this does what you want. It uses 'findInterval' to determine where
a possible match is:
myvscan-data.frame(c(1,NA,1.5),as.POSIXct(c(12:00:00,12:14:00,12:20:00),
format=%H:%M:%S))
# convert to numeric
names(myvscan)-c(Latitude,DateTime)
myvscan$tn - as.numeric(myvscan$DateTime) #
Duncan Murdoch wrote:
On 5/22/2009 10:59 AM, Michael wrote:
Really I think if there is a Visual Studio strength debugger, our
collective time spent in developing R code will be greatly reduced.
If someone who knows how to write a debugger plugin for Eclipse wants
to help, we could have that
Yuanyuan wrote:
Greetings,
I checked the Indian diabetes data again and get one tree for the data with
reordered columns and another tree for the original data. I compared these
two trees, the split points for these two trees are exactly the same but the
fitted classes are not the same for
On Fri, May 22, 2009 at 12:28 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
...The way this might appear in code is if someone wanted to calculate the
number of one hour intervals in 18 hours. One could write:
t18 - times(18:00:00)
t1 - times(1:00:00)
as.numeric(t18) /
Thanks Gregory. I see that that with
boxcox.lm() the optimal lambda is obtained and
plotted against log-likelihood.
library(MASS)
boxcox(Volume ~ log(Height) + log(Girth), data = trees,
lambda = seq(-0.25, 0.25, length = 10))
But has how can I see the fit of the same linear
model
On Fri, May 22, 2009 at 1:55 PM, Stavros Macrakis macra...@alum.mit.edu wrote:
On Fri, May 22, 2009 at 12:28 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
...The way this might appear in code is if someone wanted to calculate the
number of one hour intervals in 18 hours. One could
The first exaample on the lmer help page uses a formula
Reaction ~ Days + (Days|Subject). Here, Subject is the name of a
column in the data.frame sleepstudy, with levels 308, 309, ... .
Does this answer your question? If no, please provide commented,
minimal, self-contained,
Hello,
I run into a problem:
ftable(table(Fire, Standard, StoAll), col.vars=c(Fire,Standard))
Error in table(Fire, Standard, StoAll) : object 'Fire' not found
I do not understand that because when I read the table everything seems
correct.
You read them into a dataframe so you need to do
ftable(table(Stocking_all$Fire, Stocking_all$Standard, StoAll),
col.vars=c(Fire,Standard))
On Fri, May 22, 2009 at 2:33 PM, Gaertner, Stefanie
stefanie.gaert...@ales.ualberta.ca wrote:
Hello,
I run into a problem:
ftable(table(Fire,
str(time)
function (x, ...)
str(t_p1)
num [1:576] 190 180 190 200 210 200 220 190 230 230 ...
str(Baseline.integrated)
Factor w/ 2 levels Baseline,Integrated: 1 1 1 1 1 1 1 1 1 1 ...
str(Technique)
Factor w/ 2 levels Barrel,NonPrefHand: 1 1 1 1 1 1 1 1 1 1 ...
str(Grip)
Factor w/ 2 levels
Thank you very much, Gabor!
On Thu, May 21, 2009 at 9:37 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
If you are willing to go down one level and work at the grid level
then you can do it without modifying the panel function.
Below gg.ls$name lists the grid object names. Within that
Hi,
Romain Francois wrote:
Duncan Murdoch wrote:
On 5/22/2009 10:59 AM, Michael wrote:
Really I think if there is a Visual Studio strength debugger, our
collective time spent in developing R code will be greatly reduced.
If someone who knows how to write a debugger plugin for Eclipse wants
As a newbie I'm trying to figure out how much more than RKWard does is
wanted. The code turns colors as syntax is checked and errors are noted.
It seems like a reasonable IDE. Maybe someone is looking for the same in
windows?
John F Lindsey
803-790-5006 Home , 803-790-5008 Cell
919-439-9088
On Thu, 21 May 2009, Prew, Paul wrote:
Hello,
I'm trying to use the vcd package to analyze survey data. Expert judges
ranked possible features for product packaging. Seven features were
listed, and 19 judges split between 2 cities ranked them.
The following code (1) works, but the
Here is a test data set and code. I am including the data set after the
code and discussion to make reading easier. Apologies for the size of the
data set, but my problem occurs when there are a lot of Z
variables. Thanks for your time.
# Enter data below
# Sample code
library(lme4)
mb-
Take for example the following stripchart that's created:
b - 1:5
a - 11:15
e - 21:25
f - -11:-15
foo - rbind(b,a,e,f)
stripchart(foo ~ rownames(foo))
In this case, I would like the bottom part of the plot to be the f vector,
followed by the e vector, etc.
However, R reorders the matrix and
Dear Achim, thank you very much for the suggestions, they work well. Agreed
that an ordinal logistic regression seems a more powerful choice of analysis
method, and I'm using that also.
Thanks for providing the vcd package, it's proving quite helpful.
Regards, Paul
Paul Prew | Statistician
Hi all,
How do I evaluate how good is my MLE fit? Moreover, suppose I am
having 30 data points, and 50 points, how do I compare which one gives
better goodness-of-fit?
What are the common test procedures to assess the goodness of fit for MLE?
Thanks!
I miscommunicated: In every application I've seen with large numbers
of parameters to estimate, most of those parameters are specific
instances of different levels of a random effect. For example, a
colleague recently did a fixed effects analysis of a longitudinal
abundance survey of a large
Is there a way to code the mle() function in library stats4 such that it
switches optimizing methods midstream (i.e. BFGS to Newton and back to
BFGS, etc.)?
Thanks,
Stephen Collins, MPP | Analyst
Health Benefits | Aon Consulting
[[alternative HTML version deleted]]
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of spencerg
Sent: Friday, May 22, 2009 3:01 PM
To: Leigh Ann Starcevich
Cc: r-help@r-project.org
Subject: Re: [R] Naming a random effect in lmer
[ ... elided statistical advice
I'm an experienced programmer, but learning R is making me lose the little
hair I have left...
list(NULL)
[[1]]
NULL
length(list(NULL))
[1] 1
x - list()
x[[1]] - NULL
x
list()
length(x)
[1] 0
From the above experiment, it is clear that, although one can create a
one-element list
Continuing your example below,
x[4] - list(NULL)
I won't try to defend the semantics,which have been complained about before.
However, note that with lists, x[i] is the list which consists of one
member, the ith component of x, which is not the same as x[[i]], the ith
component, itself; so the
Hi Kynn: this oddity is discussed in Patrick Burn's document called The R
Inferno. I don't recall the fix so I'm not sure if below is the same as
what his book says to do but it seems to do what you want.
x - list()
x[[1]] - 2
x
length(x)
print(str(x))
x[2] -
Jim,
Thanks! I like the way you use indexing instead of the loops. However, the
find.Interval function does not give the right result. I have been playing
with it and it seems to give the closest number that is less than the one of
interest. In this case, the correct replacement should
Hi all,
Suppose I have 100 simultaneous time series, what's the best
statistical procedure to figure out a transformation of the data, and
see if we could squeeze most of the information into a few transformed
time series?
Thanks!
__
Here is a modification that should now find the closest:
myvscan-data.frame(c(1,NA,1.5),as.POSIXct(c(12:00:00,12:14:00,12:20:00),
+ format=%H:%M:%S))
# convert to numeric
names(myvscan)-c(Latitude,DateTime)
myvscan$tn - as.numeric(myvscan$DateTime) # numeric for findInterval
Thanks, that works great.
Andrew
On Fri, May 22, 2009 at 5:42 PM, William Dunlap wdun...@tibco.com wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Andrew Yee
Sent: Friday, May 22, 2009 2:16 PM
To:
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