On Mon, 2004-11-22 at 17:43, Barry Rowlingson wrote:
Deepayan Sarkar wrote:
Pretty much what 'append' does.
A shame then, that help.search(insert) doesn't find 'append'! I cant
think why anyone looking for a way of _inserting_ a value in the middle
of a vector would think of looking
Dear all,
I want to find out the mode for a data set, anyone knows how to do it in
R?
I tried the codes below, but it seems too long:
tt-table(data1) #get the frequency tables of data1
oo-order(tt);#get the order of frequencies
len-length(tt) #the length of the
Troels Ring wrote:
I hope you will forgive me this simple question on titration.
I'm trying to find very small values from the algorithm below, which I
believe is
correctly formatted, and the constants are also correct. When SID goes
over ATOT, fitted vales are
much too low compared to the
LONG Yu wrote:
Dear all,
I want to find out the mode for a data set, anyone knows how to do it in
R?
I tried the codes below, but it seems too long:
tt-table(data1) #get the frequency tables of data1
oo-order(tt);#get the order of frequencies
len-length(tt) #the
Dear friends - just to prevent further loss of time on the problem below -
I realized that it was really a root finding exercise and not a
minimization - uniroot solved the problem.
Sorry for the confusion.
Best wishes
Troels
At 00:06 11/23/04, you wrote:
I hope you will forgive me this simple
Hi Yu,
this has been recently discussed in the list, look at:
http://tolstoy.newcastle.edu.au/R/help/04/11/7054.html
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel:
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permitted sender hosts)
X-Virus-Scanned: by amavisd-new at stat.math.ethz.ch
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I'm using R since version 1.6, and I didn't redefine plot in my
workspace (I would never even dream of doing anything like that), I
didn't change any R-environment variable or option in the last 4 month.
In my example x is a vector of 30 real values ...
I installed the windows service pack2
giovanna jona lasinio wrote:
I'm using R since version 1.6, and I didn't redefine plot in my
workspace (I would never even dream of doing anything like that), I
didn't change any R-environment variable or option in the last 4 month.
In my example x is a vector of 30 real values ...
I installed
Hello,
In order to increase the performance of a script I'd like to sort very large
vectors containing repeated integer values.
I'm not interesting in having the values sorted, but only grouped.
I also need the equivalent of index.return from the standard sort function:
The problem I'm pointing out do not happen if I type commands directly
on the command window (as I wrote in the yesterday email) but it shows
when I source any ascii file containing R-command lines, even files I
used in several of my classes and always worked before.
I'm starting to suspect a
I do not belive there is a plot method for survreg object.
#You can plot survuval using the coeficents it returns:
curve(exp(-(exp(-lung.wbs$coef[1])*x)^(1/lung.wbs$scale)))
#Here is an exemple for lung data from library survival
library(survival)
data(lung)
lung.wbs - survreg( Surv(time,
PS: You can plot hazard in a similar way
Best regards,
Ale iberna
- Original Message -
From: Eric Lim [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, November 23, 2004 7:56 AM
Subject: [R] Weibull survival regression
Dear R users,
Please can you help me with a relatively
Hello,
I suspect that vglm would do the job for the logit link. Try :
data(pneumo)
pneumo$let = log(pneumo$exposure.time)
vglm(cbind(normal, mild, severe) ~ let, multinomial, pneumo)
vglm(cbind(normal, mild, severe) ~ let, acat, pneumo)
This is an adjacent category model
Hello,
The search engine in Netscape 7.1 does not seem to work for me -- it used to
work fine before. I do get the html help page but nothing happens when I enter
a keyword for search.
Java plug in and it is enabled -- so this is not the problem.
Any help would be appreciated.
Thanks Marwan
On 22 Nov 2004 at 16:43, [EMAIL PROTECTED] wrote:
I didn't know about rules, I found Email link in search engine as I
remember. Anyhow, thanks for answer. I did use Hitory, just I was
wondering if there is a command to use it in R and bring up everything
On Windows each line of saved
Hi Giovanna
I have encountered such problem some time ago (so it had to be R
1.8-1.9 approximately). It was connected with some complicated
plot statements and usually also with **non english** characters in
annotation texts. I did not found any traceable or repeatable
pattern and sometime it
Thanks, at least this crazy behavior is not my exclusive problem :-)
I think that if we understand how the OS operates when copying and
pasting we may find a solution...
-Messaggio originale-
Da: Petr Pikal [mailto:[EMAIL PROTECTED]
Inviato: marted, 23. novembre 2004 12:04
A: giovanna
On Tue, 23 Nov 2004, Petr Pikal wrote:
On 22 Nov 2004 at 16:43, [EMAIL PROTECTED] wrote:
I didn't know about rules, I found Email link in search engine as I
remember. Anyhow, thanks for answer. I did use Hitory, just I was
wondering if there is a command to use it in R and bring up everything
for
On Tue, 23 Nov 2004, Marwan Khawaja wrote:
The search engine in Netscape 7.1 does not seem to work for me -- it used to
work fine before.
Perhaps you need to undo what you changed, but we can't guess what that is.
I do get the html help page but nothing happens when I enter
a keyword for search.
Marc Mamin [EMAIL PROTECTED] writes:
Hello,
In order to increase the performance of a script I'd like to sort very large
vectors containing repeated integer values.
I'm not interesting in having the values sorted, but only grouped.
I also need the equivalent of index.return from the
Yhe question of finding the mode of a dataset has been discussed previously in
the list not only for categorical data :
search How can I get the mode in the search engine of the list.
Hope this helps.
Romain.
Selon Uwe Ligges [EMAIL PROTECTED]:
LONG Yu wrote:
Dear all,
I want to find
On Tue, Nov 23, 2004 at 06:56:56AM -, Eric Lim wrote:
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
Eric Lim wrote:
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
Zoltan,
On Tue, Nov 23, 2004 at 04:04:14AM -0800, Zoltan Nagy wrote:
Hi,
I wrote you this email because I have some problems
with
the R library's. I use R Version 2.0.0 and it cannot
load
some libraries, eg:
library();
MASS** No title available
(pre-2.0.0 install?)
Hi
The problem was with my Netscape 7.1.
Many thanks to Prof Ripley for pointing me to the right direction.
Best Marwan
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Prof
Brian Ripley
Sent: Tuesday, November 23, 2004 1:36 PM
To: Marwan
Hi Marc,
continuing on Prof. Dalgaard's proposal, you could use:
ix - unlist(split(seq(along=v), v), use.names=FALSE)
but even with this, `sort()' seems faster if you are interseted only
in grouping:
v - sample(1:25000, 5, TRUE)
##
system.time(ix - do.call(c,split(seq(along=v),v)),
On Tue, 23 Nov 2004, Peter Dalgaard wrote:
Marc Mamin [EMAIL PROTECTED] writes:
Hello,
In order to increase the performance of a script I'd like to sort very
large vectors containing repeated integer values.
I'm not interesting in having the values sorted, but only grouped.
I also need the
You haven't given much detail to go on. So I have to make some assumptions
(feel free to correct me if I am wrong):
1. You already know how to plot something in R from the R command line.
2. You know how to save your plotting commands in a text file (say
fancyplots.R).
If these are true, then
Hi All,
If you had run into the problem with ROracle on Linux: fetch()
returns zero rows or empty dataset, here is an easy safe
work-around for you to try out. It works for me, and very likely it
will work for you too. You must have root privilege to do this on your
machine.
Quick Instructions
is there a smart way of determining the number of pairwise present data
in a data matrix with missings (maybe as a by-product of some
statistical function?)
so far, i used several loops like:
for (column1 in 1:99) {
for (column2 in 2:100) {
for (row in 1:500) {
if
Try a `very long' vector as originally specified:
v - sample(1:25000, 1e6, TRUE)
system.time(ix - sort.list(v, method=radix), gcFirst=TRUE)
[1] 0.14 0.01 0.15 0.00 0.00
system.time(x - sort(v), gcFirst=TRUE)
[1] 0.42 0.02 0.44 0.00 0.00
system.time(x - sort(v, method=quick, index.return=TRUE),
Dear Duncan,
I don't think that there is an automatic, nearly costless way of providing
an effective solution to locating R resources. The problem seems to me to be
analogous to indexing a book. There's an excellent description of what that
process *should* look like in the Chicago Manual of
Andreas Wolf andreas.wolf at uni-jena.de writes:
:
: is there a smart way of determining the number of pairwise present data
: in a data matrix with missings (maybe as a by-product of some
: statistical function?)
:
: so far, i used several loops like:
:
: for (column1 in 1:99) {
: for
Suppose your matrix is called A (`matrix' is not a good name). Then
crossprod(!is.na(A)) is pretty efficient. Test:
A - matrix(1, 6, 3)
A[1,1] - A[3, 1] - A[2,2] - NA
A
[,1] [,2] [,3]
[1,] NA11
[2,]1 NA1
[3,] NA11
[4,]111
[5,]111
[6,]
Andreas Wolf wrote:
is there a smart way of determining the number of pairwise present data
in a data matrix with missings (maybe as a by-product of some
statistical function?)
so far, i used several loops like:
for (column1 in 1:99) {
for (column2 in 2:100) {
for (row in 1:500) {
if
Hi Andreas,
maybe something like this could do it:
mat - sample(0:3, 20*2, TRUE); dim(mat) - c(20,2)
mat[sample(1:20, 4),] - NA
mat
sum(rowMeans(mat)==mat[,1], na.rm=TRUE)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
I am looking for up to the millisecond resolution. Is there a package
that has that?
On Mon, 22 Nov 2004 21:48:20 + (UTC), Gabor Grothendieck
[EMAIL PROTECTED] wrote:
Yasser El-Zein abu3ammar at gmail.com writes:
From the document it is apparent to me that I need as.POSIXct (I have
Sorry my first reply was not relevant, I understood a different thing.
Dimitris
- Original Message -
From: Andreas Wolf [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, November 23, 2004 2:42 PM
Subject: [R] number of pairwise present data in matrix with missings
is there a smart
On Tue, 23 Nov 2004, Eric Lim wrote:
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
You might want to check out 'chron'. This stores the time as days and
fractions of a day.
If you take the current date,
as.numeric(chron(dates.=11/23/2004))
[1] 12745
you get the value above. If you change this to millisecond, you get
as.numeric(chron(dates.=11/23/2004)) * 86400 *
Having just finished an index I would like to second John's comments.
Even as an author, it is difficult to achieve some degree of
completeness and consistency.
Of course, maybe a real whizz at clustering could assemble something
very useful quite easily. All of us who have had the frustration
I want to create a character vector based on the table (shortened for
display) below:
Where there are ones in the matrix I want the column name to appear and
where there are zeros nothing, which would make the vector in this
shortened case:
combinations - (A B,A C,A E H,A F G,B C D,E G H,A C D
james == james holtman [EMAIL PROTECTED]
on Tue, 23 Nov 2004 10:39:04 -0500 writes:
james You might want to check out 'chron'. This stores the
james time as days and fractions of a day.
james If you take the current date,
as.numeric(chron(dates.=11/23/2004))
james
On Tue, 2004-11-23 at 17:40, roger koenker wrote:
Having just finished an index I would like to second John's comments.
Even as an author, it is difficult to achieve some degree of
completeness and consistency.
Of course, maybe a real whizz at clustering could assemble something
very
Hi,
Here is something that does the job (though I am sure other people will
find smarter solutions!):
samp=matrix(sample(0:1,size=100,replace=TRUE ,prob=c(0.8,0.2)),ncol=10)
colnames(samp)-LETTERS[1:10]
unlist(lapply(apply(samp,1,FUN=function(vec)
colnames(samp)[as.logical(vec)]
Dear all,
As our previous email did not get any response, we try again with a
reformulated question!
We are trying to do something which needs an efficient loop over a huge
array, possibly functions such as apply and related (tapply,
lapply...?), but can't really understand syntax and examples
Hi all,
As part of a research project we are creating a statistical software
tool and will be using R as the computational engine. I was wandering
whether we should also use R for plotting. R has a good plotting
flexibility and an extensive library of available plot types. However,
our
Hi
I want to extract a row from a data.frame but I want that object to
be a vector . After trying some different ways I end up always with a
data.frame or with the wrong vector. Any pointers?
x - data.frame(a = factor(c('a',2,'b')), b = c(4,5,6))
I want to get
a 4
I tried:
as.vector(x[1,])
a
MM == Martin Maechler [EMAIL PROTECTED]
on Tue, 23 Nov 2004 17:04:41 +0100 writes:
james == james holtman [EMAIL PROTECTED]
on Tue, 23 Nov 2004 10:39:04 -0500 writes:
james You might want to check out 'chron'. This stores the
james time as days and fractions of a day.
On Tue, 23 Nov 2004, Martin Maechler wrote:
james == james holtman [EMAIL PROTECTED]
on Tue, 23 Nov 2004 10:39:04 -0500 writes:
james You might want to check out 'chron'. This stores the
james time as days and fractions of a day.
james If you take the current date,
x - c(1,2,3.095,1,1,0,0,0,0,0,0,
2,2,1.687,1,0,1,0,0,0,0,0,
46,3,3.47,1,0,0,0,1,0,0,1,
47,3,1.563,1,0,0,0,0,1,1,0,
50,3,6.234,0,1,1,1,0,0,0,0,
148,4,3.663,0,0,1,0,1,0,1,1,
151,4,3.47,0,0,0,1,1,1,0,1,
177,5,5.411,1,0,1,1,1,0,0,1,
I think John has exactly the right image -- index to a book --
but I disagree with his conclusions.
I read somewhere that an index should not be done by the
author. It was probably written by someone who was bored
of indexing, but the logic was precisely because indices should
be about concepts.
Hello,
I would like to know if it possible to use locator() only on one region of
a graphic device.
I would like to fragment a graphic device into sub-regions (using layout or
split.screen or any other functions that I do not know) and then use
locator. but I want that the locator is only used
Here's another way:
dat - read.table(clipboard, header=TRUE)[,-(1:3)]
dat
A B C D E F G H
1 1 1 0 0 0 0 0 0
2 1 0 1 0 0 0 0 0
3 1 0 0 0 1 0 0 1
4 1 0 0 0 0 1 1 0
5 0 1 1 1 0 0 0 0
6 0 0 1 0 1 0 1 1
7 0 0 0 1 1 1 0 1
8 1 0 1 1 1 0 0 1
9 1 0 1 1 0 1 1 0
apply(dat 0, 1, function(idx)
There has to be a better (more readable) way, but this works...
set.seed(323)
foo.df - data.frame(A = round(runif(5)), B = round(runif(5)), C =
round(runif(5)))
foo.df
A B C
1 0 1 1
2 1 1 1
3 1 1 1
4 0 1 1
5 1 1 0
names.list - lapply( apply( foo.df, 1, function( x ) colnames(
foo.df )[
Dear R community, once again I request your generous help on an R issue
I cannot seem to figure out.
I am trying to do some analysis for multiple sites, in my example I am
using only two, but the same will done on many sites. It's just ONE
line, so I hope somebody can give a me a hand by
A data frame is a list, and a list is a vector. Once you understand that,
yoy may understand what you are seeing.
as.matrix(x)[1,] seems to be one of the easiest ways to get what you want
On Tue, 23 Nov 2004, Tiago R Magalhaes wrote:
Hi
I want to extract a row from a data.frame but I want that
Dear list members,
In re-running with GLMM() from the lme4 package a generalized-linear mixed
model that I had previously fit with glmmPQL() from MASS, I'm getting a
warning of a convergence failure, even when I set the method argument of
GLMM() to PQL:
bang.mod.1 - glmmPQL(contraception ~
I'll give it half a crack:
Steps a through c can be done via nested ifelse(), treating A and M as
vectors (as they really are). Step d is the hard one. I'd write a simple
Fortran code and use .Fortran() for that.
I don't see how any of the *apply() functions can help here, as your
operations
Hi Neela,
Just my ¢2 regarding the R vs. SAS issue. I started to use SAS since my
undergraduate studies in 1993, and I am now a migrating to R user, and I
hope not to ever go back to SAS.
My comments are:
- SAS is huge it takes over 1GB of space of my PC while R takes just
over 100 MB, this
Is this what you want?
as.matrix(x[1,])[1,]
a b
a 4
HTH,
Andy
From: Tiago R Magalhaes
Hi
I want to extract a row from a data.frame but I want that object to
be a vector . After trying some different ways I end up always with a
data.frame or with the wrong vector. Any pointers?
Michal Blazejczyk [EMAIL PROTECTED] 11/23/04 09:21AM
Hi all,
As part of a research project we are creating a statistical software
tool and will be using R as the computational engine. I was
wandering
whether we should also use R for plotting. R has a good plotting
flexibility and an
Please disregard my message I've seen the light now...
curve(predict(out, data.frame(x = x)), add = T)
Thanks,
Jose
--
Jose A. Hernandez
Department of Soil, Water, and Climate
University of Minnesota
1991 Upper Buford Circle
St. Paul, MN 55108
Ph. (612) 625-0445, Fax. (612) 625-2208
On Tue, 2004-11-23 at 16:27 +, Tiago R Magalhaes wrote:
Hi
I want to extract a row from a data.frame but I want that object to
be a vector . After trying some different ways I end up always with a
data.frame or with the wrong vector. Any pointers?
x - data.frame(a =
R/S Advanced Programming course in San Francisco on December 20-21st,
2004
Please check out the full description and Agenda of the 2-day class on
the website:
www.xlsolutions-corp.com/Radv.htm
And let us know if we should hold seats for you.
Ask for group discount!
Here's the outline:
Course
On Tue, 2004-11-23 at 12:28 -0500, Liaw, Andy wrote:
I'll give it half a crack:
Steps a through c can be done via nested ifelse(), treating A and M as
vectors (as they really are). Step d is the hard one. I'd write a simple
Fortran code and use .Fortran() for that.
I don't see how any
Hi
Mulholland, Tom wrote:
This raises the question of best practice. My answer was predicated
on the fact that Jin Li had been attempting to use grid.circle in the
first place without success. I rashly made the assumption that there
was already a move to try and use some of the more
I appreciate if anyone can help me,
I have a table as follow,
rate
DATE VALUE
1 1997-01-10 5.30
2 1997-01-17 5.30
3 1997-01-24 5.28
4 1997-01-31 5.30
5 1997-02-07 5.29
6 1997-02-14 5.26
7 1997-02-21 5.24
8 1997-02-28 5.26
9 1997-03-07 5.30
10 1997-03-14 5.30
R-community,
Assuming a file with 3 columns: site, nrate, yield. In SAS I can easily
run the analysis BY SITE using the following code:
*--*;
proc nlin method=MARQUARDT;
by farm;
parms b0=100 b1=0.33 b2=-0.00111 x0=120;
model yield = (b0 + b1 * nrate
By 'ignore', can we delete those from the list of data? I would then
assume that if you have a sequence of +0+0+ that you would want the last
+ for the increase of three.
If that is the case, then do a 'diff' and delete the entries that are 0.
Then create a new 'diff' and then use 'rle' to
Perhaps you would be interested in the nlsList() function in the nlme
package.
HTH,
Andy
From: Jose A. Hernandez
R-community,
Assuming a file with 3 columns: site, nrate, yield. In SAS I
can easily
run the analysis BY SITE using the following code:
From: Patrick Burns
I think John has exactly the right image -- index to a book --
but I disagree with his conclusions.
I read somewhere that an index should not be done by the
author. It was probably written by someone who was bored
of indexing, but the logic was precisely because
Patrick Burns wrote:
[]
No, I'm not volunteering to build the system.
Too bad! ;-)
Indeed, the idea to index tens of thousands of functions could not be
appealing to many of us! Why not to consider to test such ideas at the
package level? I mean, building a system that points out the
AFAICS, there are two errors:
1. curve() is probably not what you want: it plots functions, and that's not
what predict() gives you.
2. predict() needs to be given the correct data frame: The name of the
variable(s) need to match those used in the formula. The formula has `x' on
the righthand
On Tuesday 23 November 2004 11:14, John Fox wrote:
Dear list members,
In re-running with GLMM() from the lme4 package a generalized-linear
mixed model that I had previously fit with glmmPQL() from MASS, I'm
getting a warning of a convergence failure, even when I set the
method argument of
At 11/23/2004 11:45 AM Tuesday, Patrick Burns wrote:
...There could be an index builder that accepts a search phrase and
the function or package that is the successful answer to the search.
If this were open, then R users could contribute to the index who
don't feel qualified to submit code. It
A big thank you to everyone who responded...
Using Ale iberna solution with the lung dataset, I obtained:
plot(survfit(Surv(time,status),data=lung))
Error in Surv(time, status) : Time variable is not numeric
In addition: Warning message:
is.na() applied to non-(list or vector) in: is.na(time)
Dear Deepayan,
Thanks for the explanation (and for pointing out that the data set is also
in the lme4 package). I had tried control=lmeControl(tolerance=1e-4,
PQLmaxIt=100) and control=lmeControl(PQLmaxIt=1000), but gave up at that
point. I just tried lmeControl(tolerance=1e-4, PQLmaxIt=1000),
On Tue, 23 Nov 2004, Eric Lim wrote:
Using Thomas Lumley's solution with the lung dataset, I obtained:
curve(pweibull(x, scale=1/coef(lung.wbs), shape=1/lung.wbs$scale,
lower.tail=FALSE),from=0, to=max(lung$time))
That's because you read the part of the message where I got the
From: Liaw, Andy [EMAIL PROTECTED]
Date: Tue, 23 Nov 2004 12:28:48 -0500
I'll give it half a crack:
Steps a through c can be done via nested ifelse(), treating A and M as
vectors (as they really are). Step d is the hard one. I'd write a simple
Fortran code and use .Fortran() for that.
It has been a while since I looked at those two sources. My main impetus
has been the detail. I think that one of the aims of my project has been
to document the variants that might not be given a name. I have a
function called tufte158 (Tufte, The Visual Display of Quantitative
Information
I have a short simulation problem (well it's not *that* short) that
requires I create all the possible routes amongst a set of destinations. I
didn't seem to find anything regarding route generation on the website and
wanted to know if it's possible to generate routes using R (via
On Tue, 23 Nov 2004, Philippe Grosjean wrote:
Patrick Burns wrote:
[]
No, I'm not volunteering to build the system.
Too bad! ;-)
Indeed, the idea to index tens of thousands of functions could not be
appealing to many of us! Why not to consider to test such ideas at the
package
Yasser El-Zein abu3ammar at gmail.com writes:
:
: I am looking for up to the millisecond resolution. Is there a package
: that has that?
:
: On Mon, 22 Nov 2004 21:48:20 + (UTC), Gabor Grothendieck
: ggrothendieck at myway.com wrote:
: Yasser El-Zein abu3ammar at gmail.com writes:
:
:
Hello all,
As a general programming question I can't seem to figure out how to make
a list of lists in R.
As matrices won't work as they have to be rectangular.
I am sure that there is an easy solution but...
the specific situation is this:
- I have created a Tukey confidence interval table and
From: Karla Sartor
Hello all,
As a general programming question I can't seem to figure out
how to make
a list of lists in R.
As matrices won't work as they have to be rectangular.
I am sure that there is an easy solution but...
the specific situation is this:
- I have created a
Is the following more like what you want:
a=c(1,1,1,1,1) # generate the first list
b=c(2,2,2)# generate a second list
d=list(a,b)# make a list of a and b
(
e=c(d,a)
)
[[1]]
[1] 1 1 1 1 1
[[2]]
[1] 2 2 2
[[3]]
[1] 1
[[4]]
[1] 1
Inspired by the functions that Barry Rawlingson and Dave Forrest posted for
searching Rwiki and R-help archive, I've made up a function that does the
search on Prof. Baron's site (Thanks to Prof. Baron's help on setting up the
query string!):
RSiteSearch - function(string, restrict=Rhelp,
Hi.
I'd like to do a t-test to compare the Delta values of items with Crit=1
with Delta values of items with Crit=0. What is the t.test syntax?
It should produce a result like this below (I can't get in touch with the
person who originally did this for me)
Welch Two Sample t-test
data:
I was trying to install some more packages and ran into a problem
I hadn't seen before.
Version:
platform sparc-sun-solaris2.9
arch sparc
os solaris2.9
system sparc, solaris2.9
status
major2
Liaw, Andy andy_liaw at merck.com writes:
:
: Inspired by the functions that Barry Rawlingson and Dave Forrest posted for
: searching Rwiki and R-help archive, I've made up a function that does the
: search on Prof. Baron's site (Thanks to Prof. Baron's help on setting up the
: query string!):
What did this have to do with GLMM? I've changed the subject line.
On Wed, 24 Nov 2004, Richard A. O'Keefe wrote:
I was trying to install some more packages and ran into a problem
I hadn't seen before.
We've seen it for C, and test it for C in R CMD check. I think we should
check C++ and
Hi everyone,
I've been opening up R 2.0.1 from the dock and then starting the Mac OSX
X11 server.
After that I call
jpeg(test.jpg,plot(rnorm(10)))
and r pops up a quartz window with the appropriate graph in it and puts a
file called test.jpg into the HOME which is 0kb in size and is completely
On Wed, 24 Nov 2004, Thuan-Jin Kee wrote:
I've been opening up R 2.0.1 from the dock and then starting the Mac OSX
X11 server.
After that I call
jpeg(test.jpg,plot(rnorm(10)))
Please do read the help page: the syntax is
jpeg(test.jpg)
plot(rnorm(10))
dev.off()
and r pops up a quartz window with
Off hand, the costs of GPL'd software are not hidden at all. R for instance
demands that a would be user sit down and learn the language. This in turn
pushes a user into learning more about statistics than the simple overview
that Stat 1 presents a student.
In contrast, any program that
Good morning,
I have to apply the Ks test with the the t distribution.
I know I have to write ks.test(data_name,distribution_name, parameters..)
but I don't know what is the name fot t distribution and which parameters
to introduce? may be mean=0 and freedom degrees in my case?
Thank you for
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