Dear all,
I would need a function which convert small letter into capital letter (at
least the first letter of a character variable).
Does such a function exist in R ?
Thanks by advance
Jessica
__
R-help@stat.math.ethz.ch mailing list
Try ?toupper
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: Tuesday, May 15, 2007 12:11 PM
To: R-help@stat.math.ethz.ch
Subject: [R] conversion into capital letter
Dear all,
I would need a function which convert small letter
[EMAIL PROTECTED] napsal dne 15.05.2007 08:41:27:
Dear all,
I would need a function which convert small letter into capital letter
(at
least the first letter of a character variable).
Does such a function exist in R ?
?toupper, ?tolower
Regards
Petr
Thanks by advance
Jessica
help(toupper)
b
On May 15, 2007, at 2:41 AM, [EMAIL PROTECTED] wrote:
Dear all,
I would need a function which convert small letter into capital
letter (at
least the first letter of a character variable).
Does such a function exist in R ?
Thanks by advance
Jessica
--
Benilton
On Tue, 2007-05-15 at 08:41 +0200, [EMAIL PROTECTED] wrote:
Dear all,
I would need a function which convert small letter into capital letter (at
least the first letter of a character variable).
Does such a function exist in R ?
Thanks by advance
Jessica
See ?toupper and take note
Dear R users,
I'm having some problems trying to create a routine for a bootstrap
resampling issue. Suppose I've got a dataset like this:
Header inr inf .weeks ...
insideaboveunder
12800012.75 2.5 ..0
Dear peRsons,
I have a Sweave document which demonstrates the usage of step()
function. With current R version 2.5.0 the step() function was changed
so that the heading of trace=TRUE output for each model is printed using
command message():
if (trace)
message(\nStep: AIC=,
hello,
I don't understand what's happen just before the textConnection function runs
good but now it doesn't run
Line[1]
[1] if C325=. then C743=(C152/C103)*100| else C743=(C152/C325)*100
textConnection(Line[1])
Erreur dans textConnection(Line[1]) : toutes les connexions sont utilisées
why R
It means what it says. You do need to close() connections, as there is a
finite number available. (The number depends on your unstated version of
R, but is at least 50.)
On Tue, 15 May 2007, elyakhlifi mustapha wrote:
hello,
I don't understand what's happen just before the textConnection
Dear all,
I would like to know if there is any R package that uses a sliding window
approach to assess statistical significance out of my data. My data is composed
of DNA sequences (of variable length) that are mapped to a genome with a
determined score of alignment.
So, I want to see if I
hello,
can you help me I need to seperate words and symbol in a mathematics formula as
follow
C744=(C627*C177)/100
How could I do please?
_
[[alternative HTML version deleted]]
elyakhlifi mustapha wrote:
hello,
can you help me I need to seperate words and symbol in a mathematics
formula as follow
C744=(C627*C177)/100
How could I do please?
If you need to simply split a character vector, use strsplit.
This and previous your posts suggest you need to
Dear R users,
I'm having some problems trying to create a routine for a bootstrap
resampling issue. Suppose I've got a dataset like this:
Header inr weeks .
12800012.47 0 ...
12800011.48 1 ...
1280001
Vladimir Eremeev wrote:
Dear R experts,
I have a Matlab code which I am translating to R in order to examine and
enhance it.
First of all, I need to reproduce in R the results which were already
obtained in Matlab (to make sure that everything is correct).
There are some matrix
Vladimir Eremeev wrote:
Dear R experts,
I have a Matlab code which I am translating to R in order to examine and
enhance it.
First of all, I need to reproduce in R the results which were already
obtained in Matlab (to make sure that everything is correct).
There are some matrix
2007/5/14, Andy Fugard [EMAIL PROTECTED]:
Hello all,
There's a new mailing list for researchers in psychology who are
learning and using R. New users of R are especially welcome.
Hi List
Does anyone know if there is a special list for researchers in medicine using R?
--
A. Goralczyk,
1. Use this gsub:
txt - C744=(C627*C177)/100
gsub(\\b|([^[:alnum:]]), \\1 , txt)
and then strsplit or scan as in prior response.
2. If your text consists of valid R expressions then we can use the
R parse function can traverse the tree as shown:
txt - C744=(C627*C177)/100
e - parse(text =
the following seems a bit better:
set.seed(1)
nc - 30
nr - 25000
x - matrix(rnorm(nc*nr), ncol = nc)
g - matrix(sample(1:3, nr*nc, rep = TRUE), ncol = nc)
#
trimmedMeanByGroup1 - function(y, grp, trim=.05)
tapply(y, factor(grp, levels=1:3), mean, trim=trim)
Hi,
I'm finding that readMM() cannot read a file written with writeMM().
Example:
library(Matrix)
a = Matrix(c(1,0,3,0,0,5), 10, 10)
a = as(a, CsparseMatrix)
writeMM(a, kk.mm)
b = readMM(kk.mm)
Error in validObject(.Object) : invalid class dgTMatrix object: all row
indices must be between 0
I am using R to make two-way ANOVA on a number of variables using
g - aov(var ~ fact1*fact2)
where var is a matrix containing the variables.
However the outcome seem to be dependent on the order of fact1 and fact2
(i.e. fact2*fact1) gives a slightly (factor of 1.5) different result.
Any ideas
Dear R people,
I do not have much knowledge about linear algebra but currently I need
to understand what the function qr.qty is actually doing. The
documentation states that it calculates t(Q) %*% y via a previously
performed QR matrix decomposition.
In order to do that, I tried following
Anders Malmendal wrote:
I am using R to make two-way ANOVA on a number of variables using
g - aov(var ~ fact1*fact2)
where var is a matrix containing the variables.
However the outcome seem to be dependent on the order of fact1 and fact2
(i.e. fact2*fact1) gives a slightly (factor of
You need complete = TRUE. See ?qr.Q
m - matrix(c(1,0,0,0,1,0,0,0,1,0,0,1), ncol = 3)
y - matrix(c(1,2,3,4,2,3,4,5,1,1,1,1,2,2,2,2), nrow = 4)
t(qr.Q(qr(m), complete = TRUE)) %*% y
[,1] [,2] [,3] [,4]
[1,] -1 -2 -1 -2
[2,] -4 -5 -1 -2
[3,]3412
[4,] -2
You've missed the complete=TRUE argument (and also crossprod) as in
crossprod(qr.Q(qr(m), TRUE),
matrix(c(1,2,3,4,2,3,4,5,1,1,1,1,2,2,2,2),nrow=4))
On Tue, 15 May 2007, Sebastian Bauer wrote:
Dear R people,
I do not have much knowledge about linear algebra but currently I need
Hi,
If it was me I would have done this
- First reshape the data frame to get some thing like
header measure1 measure3 measure3
12800012.471.482.23 ...
Since you have same number of measure for all subject. The you define you
statistic with the data
hello,
I have an argument of a the list a like this
a[[18]]
[1] C744=(C627*C177)/100
and I wanna seperate the character and the mathematics symbol to use it like a
formula
and why when I used the strsplit function i obtain as follow
strsplit(a[[18]], '\\W')
[[1]]
[1] C744 C627 C177
All,
Happy to say that the problem could be solved. The key idea was from
Patrick Burns (Convert the data-frame to a matrix!). As written
earlier, the steps were to first get a object (call it ad) containing
the non-missing entries at each row. Then run a sum over each row,
selecting only
Hi - I'm having a problem trying to use the function GLMM() from lme4. Here
is what happens:
library(lme4)
Loading required package: Matrix
Loading required package: lattice
f1 - GLMM(success~yearF, data=quality, random=~1|bandnumb,
family=binomial, method=PQL)
Error: couldn't find
Sending in plain text, as the html version doesn't seem to go through..
Best,
-Tir
From: Patnaik, Tirthankar [GWM-CIR]
Sent: Tuesday, May 15, 2007 2:55 PM
To: '[EMAIL PROTECTED]'
Cc: r-help@stat.math.ethz.ch; [EMAIL PROTECTED]
Subject: RE: [R] Conditional
On 5/15/07, elyakhlifi mustapha [EMAIL PROTECTED] wrote:
hello,
I have an argument of a the list a like this
a[[18]]
[1] C744=(C627*C177)/100
and I wanna seperate the character and the mathematics symbol to use it like
a formula
and why when I used the strsplit function i obtain as
the lme4 function you want is probably lmer()
type
?lmer
btw. your R is very old we are at 2.5.0 ...
Stefan
Amelie LESCROEL wrote:
Hi - I'm having a problem trying to use the function GLMM() from lme4. Here
is what happens:
library(lme4)
Loading required package: Matrix
Hi.
I'm having trouble testing for existence of an object inside a function.
Suppose I have a function:
f-function(x){
...
}
and I call it with argument y:
f(y)
I'd like to check inside the function whether argument y exists. Is this
possible, or do I have to either check outside the
Hi,
I am very new to R. I am trying to perform an Anova Test and see if it
differs or not.
Basically, i have 4 tests and 1 control.
Tester
Test1 Test2 Test3 Test4 Control
20 25 1510 17
The inference is at alpha=0.05. they are all independent. I am trying to
find if
Hi all,
I want to develop a choice based questionnaire and I need to generate a set
of choice tasks
The design is comprised of four attributes (5*3*3*3) with the following
5price
3brand
3service
3installation
I need to determine the minimum no of choice tasks which can be used
CrazyJoe keizer_61 at hotmail.com writes:
I am very new to R. I am trying to perform an Anova Test and see if it
differs or not.
Basically, i have 4 tests and 1 control.
Tester
Test1 Test2 Test3 Test4 Control
20 25 1510 17
You can't make any
Hi
[EMAIL PROTECTED] napsal dne 15.05.2007 16:39:20:
Hi,
I am very new to R. I am trying to perform an Anova Test and see if it
differs or not.
Basically, i have 4 tests and 1 control.
Tester
Test1 Test2 Test3 Test4 Control
20 25 1510 17
The
Not sure which one you want, but the following should cover it:
R f - function(x) c(x=missing(x), y=exists(y))
R f(1)
x y
FALSE FALSE
R f()
x y
TRUE FALSE
R y - 1
R f()
xy
TRUE TRUE
R f(1)
x y
FALSE TRUE
Andy
From: Talbot Katz
Hi.
I'm having
Hi, Andy.
Thank you for the quick response! Unfortunately, none of these are exactly
what I'm looking for. I'm looking for the following: Suppose object y
exists and object z does not exist. If I pass y as the value of the
argument to my function, I want to be able to verify, inside my
Thank you Guys.
Let say that from Test1 to control i have multiple data
Tester
Test1 Test2 Test3 Test4 Control
20 25 1510 17
. . . . .
. . . . .
40 20 1535 45
Is this
Hello,
I am trying to print the lm
coefficients using grid graphs (code below), and am getting all the
coefficients
on top of each other.
My aim is to put them as legend,
showing both coefficients and mean values separated by colors as shown in
xyplot.
testData - data.frame(f1 =
Just need a bit more work:
R f - function(x) exists(deparse(substitute(x)))
R f(y)
[1] FALSE
R y - 1
R f(y)
[1] TRUE
R f(z)
[1] FALSE
Andy
From: Talbot Katz
Hi, Andy.
Thank you for the quick response! Unfortunately, none of
these are exactly
what I'm looking for. I'm looking for
[EMAIL PROTECTED] napsal dne 15.05.2007 17:32:35:
Thank you Guys.
Let say that from Test1 to control i have multiple data
Tester
Test1 Test2 Test3 Test4 Control
20 25 1510 17
. . . . .
. . .
Hello I'm trying to create a barplot with a couple of stacked positive
values and with one negative value for each group.
example:
trees-c(20,30,10)
shrubs-c(12,23,9)
veg-c(2,3,4)
soil-c(-100,-123,-89)
example1-t(cbind(trees,shrubs,veg))
barplot(example1)
#this works so far
#but now:
If your data is inside a data.frame names 'a' with data from each group in a
different columns,..., you can use stack and perform the anova with:
l - aov(values~ind,data=stack(a))
anova(l)
Christophe Pallier
On 5/15/07, CrazyJoe [EMAIL PROTECTED] wrote:
Thank you Guys.
Let say that from
justin bem wrote:
Hi,
If it was me I would have done this
- First reshape the data frame to get some thing like
header measure1 measure3 measure3
12800012.471.482.23 ...
Since you have same number of measure for all subject. The you define
By the end of the day I will have a vignette completed for MiscPsycho.
This vignette lays out the mathematical details for the primary
functions in the package and provides substantive examples on how to use
these functions in a sample session.
This vignette will ultimately end up being
Basically, i have 4 tests and 1 control.
For example.
Tester
Test1 Test2 Test3 Test4
20 25 1510
30 45 10 15
.. .. .. ..
.. .. .. ..
15 23 1345
The inference is at
Hi.
Thanks once more for the swift response. This solution works pretty well.
The only small glitch is if I pass the function an argument with the same
name as the function argument. That is, suppose x is the argument name in
my user-defined function, and that object x does not exist. If I
Try this modification:
chk - function(x) exists(deparse(substitute(x)), parent.env(environment()))
ab - 1
chk(ab)
[1] TRUE
exists(x)
[1] FALSE
chk(x)
[1] FALSE
On 5/15/07, Talbot Katz [EMAIL PROTECTED] wrote:
Hi.
Thanks once more for the swift response. This solution works pretty
Dear All,
Hope I am not bumping into a FAQ, but so far my online search has been fruitless
I need to read some data file using R. I am using the (I think)
standard command:
data_150-read.table(y_complete06000, header=FALSE)
where y_complete06000 is a 6000 by 40 table of numbers.
I am puzzled at
Another thing to watch out for is that an argument to a function can be
an expression (or even literal constants), instead of just the name of
an object. exists() wouldn't really do the right thing. I'm not sure
how to properly do the exhaustive check.
Andy
From: Gabor Grothendieck
Try this
If it's a matrix, use scan(). If the columns are not all the same type,
use the colClasses argument to read.table() to specify their types,
instead of leaving it to R to guess. That will speed things up quite a
lot.
Andy
From: Lorenzo Isella
Dear All,
Hope I am not bumping into a FAQ, but
Estimado, te agradeceria me envies el tutorial en espaniol para R. Estoy
dando mis primeros pasos con esta aplicacion.
Desde ya muchas gracias
Danilo Ceschin Ph.D
IGBMC
1 rue Laurent Fries
67404 ILLKIRCH CEDEX - FRANCE
tel 33 3 88 65 3457
email [EMAIL PROTECTED]
Maybe this:
chk2 - function(x) {
chr - deparse(substitute(x))
e - parse(text = chr)
structure(exists(chr, parent.env(environment())),
is.name = length(e) == 1 is.name(e[[1]]))
}
chk2(1) # structure(FALSE, is.name = FALSE)
ab - 1
chk2(ab+1) # structure(FALSE,
Hello,
There are a number of resources (in Spanish) at the bottom of the page
http://es.wikipedia.org/wiki/Lenguaje_de_programación_R
Regards,
Carlos J. Gil Bellosta
http://www.datanalytics.com
On Tue, 2007-05-15 at 18:49 +0200, Danilo Ceschin wrote:
Estimado, te agradeceria me envies el
Hi,
I'm working out d' and therefore I'd like to aggregate two variables
at once (hits and false alarms). In the raw data frame is subject,
stimulus_presence, and response information. I aggregate this into a
table now currently by separating out the target_presence=true data
I think parent.frame() is what is wanted, not parent.env(environment()) in
your suggested solution:
Consider this: (which does **not** however handle the arbitrary expressions
as argument issue):
foo1 - function(z){
cat(exists(deparse(substitute(z)),parent.frame()),
If you execute the same script several times and the data file does not
change, it may be a good idea to save it as an R object:
if (file.access('mydata.obj',0)==0) {
load('mydata.obj')
} else {
a-read.table(mydata.csv,...)
save(a,file='mydata.obj')
}
It
Right. Thanks. It should have been parent.frame(). Here is the correction:
chk3 - function(x) {
chr - deparse(substitute(x))
e - parse(text = chr)
structure(exists(chr, parent.frame()),
is.name = length(e) == 1 is.name(e[[1]]))
}
chk3(1) # structure(FALSE, is.name
I'm using R2HTML package to generate reports, but don't know how to extract
R codes from .rnw files using Stangle. It seems Stangle only works on
..tex file that has R codes embedded.
Thanks,
Tao
_
Hotmail.
On Tue, 15 May 2007, Lorenzo Isella wrote:
Dear All,
Hope I am not bumping into a FAQ, but so far my online search has been
fruitless
I need to read some data file using R. I am using the (I think)
standard command:
data_150-read.table(y_complete06000, header=FALSE)
where
Prof Brian Ripley wrote:
On Tue, 15 May 2007, Lorenzo Isella wrote:
Dear All,
Hope I am not bumping into a FAQ, but so far my online search has been
fruitless
I need to read some data file using R. I am using the (I think)
standard command:
data_150-read.table(y_complete06000,
On Tue, 2007-05-15 at 17:54 +0200, Paul Magdon wrote:
Hello I'm trying to create a barplot with a couple of stacked positive
values and with one negative value for each group.
example:
trees-c(20,30,10)
shrubs-c(12,23,9)
veg-c(2,3,4)
soil-c(-100,-123,-89)
On 5/15/07, Marc Schwartz [EMAIL PROTECTED] wrote:
On Tue, 2007-05-15 at 17:54 +0200, Paul Magdon wrote:
Hello I'm trying to create a barplot with a couple of stacked positive
values and with one negative value for each group.
example:
trees-c(20,30,10)
shrubs-c(12,23,9)
Talbot Katz wrote:
I'm having trouble testing for existence of an object inside a function.
No, you are having trouble testing for existence of an object
_before_ the function is called :-)
Suppose I have a function:
f-function(x){
...
}
and I call it with argument y:
f(y)
How can I get the value of the aspect ratio that is used in a lattice
plot? In a levelplot for instance, the native units per cm of my x and
y axes are different, and I need to know the aspect ratio so that I can
correctly plot vectors. I know how to set the aspect in a high-level
lattice
Danilo,
http://cran.r-project.org/other-docs.html#nenglish
Manuel Castejón Limas, Joaquín Ordieres Meré, Fco. Javier de Cos Juez, and
Fco. Javier Martínez de Pisón Ascacibar. Control de Calidad. Metodologia para
el
analisis previo a la modelización de datos en procesos industriales.
have a look at the examples in ?aov.
Also not that npk is a dataframe in this example
--- CrazyJoe [EMAIL PROTECTED] wrote:
Thank you Guys.
Let say that from Test1 to control i have multiple
data
Tester
Test1 Test2 Test3 Test4 Control
20 25 1510 17
.
On 5/15/2007 3:06 PM, Alberto Monteiro wrote:
Talbot Katz wrote:
I'm having trouble testing for existence of an object inside a function.
No, you are having trouble testing for existence of an object
_before_ the function is called :-)
Suppose I have a function:
f-function(x){
...
Enrico,
prop.test is for testing proportions two at a time. If you want to test
for differences between 4 proportions simultaneously (rather than two at a
time), try a logistic regression model (from which you can get confidence
intervals for each of your groups).
Cody Hamilton, PhD
Staff
Duncan Murdoch wrote:
Try this:
f - function(x) x + 1
f(y.does.not.exist)
y.does.not.exist
The error message is (almost) the same, and it happens when
parsing the line. There's no way to change f to change this.
That description is true in some languages, but not in R. R doesn't
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: Tuesday, May 15, 2007 12:52 PM
To: Salvatore Enrico Indiogine
Cc: R-help@stat.math.ethz.ch; [EMAIL PROTECTED]
Subject: Re: [R] confidence intervals on multiple comparisons
Tao Shi wrote:
I'm using R2HTML package to generate reports, but don't know how to extract
R codes from .rnw files using Stangle. It seems Stangle only works on
..tex file that has R codes embedded.
Stangle is meant for Rnw files, not for html ...
Uwe Ligges
Thanks,
Tao
Hello,
I have been trying to build a package for R to use on windows. I have
been able to build it with out problems except for one thing. I am
creating a zip file to be installed by the R gui.
I have four directories under the main dir. I have data, man, R, and
src. The problem that I have
An apology: it takes roughly a couple of minutes on my laptop, running
Debian. I had been running some other simulations for quite some time
and, though it looks odd for Linux, the subsequent work I did with R was
slowed down.
Many thanks
Lorenzo
Peter Dalgaard wrote:
Prof Brian Ripley
Boxwood Means, a financial services consultancy specializing in the
small balance mortgage sector and REIT analysis, is seeking a Data
Analyst.
Boxwood maintains several large databases of real estate transactions
around the nation as well as financial information on publicly traded
equities.
On 5/15/07, Waichler, Scott R [EMAIL PROTECTED] wrote:
How can I get the value of the aspect ratio that is used in a lattice
plot? In a levelplot for instance, the native units per cm of my x and
y axes are different, and I need to know the aspect ratio so that I can
correctly plot vectors.
It should be
lmer(success ~ yearF + (1 | bandnumb), data=quality, family = binomial,
method = PQL)
Reza
On 5/15/07, Stefan Grosse [EMAIL PROTECTED] wrote:
the lme4 function you want is probably lmer()
type
?lmer
btw. your R is very old we are at 2.5.0 ...
Stefan
Amelie LESCROEL
Dr. Ripley,
Thanks for you reply and hints. I was able to install R on the Unix system
(SunOs 5.8) by issuing the following commands:
./configure --with-readline=no --with-iconv=no
make
It may lose some functionality by specifying these options, I guess. But I
use Unix R only when I need to
On 5/15/07, Seyed Reza Jafarzadeh [EMAIL PROTECTED] wrote:
It should be
lmer(success ~ yearF + (1 | bandnumb), data=quality, family = binomial,
method = PQL)
It has to be method = PQL (the quotes are important) except that the
PQL method is no longer an option in lmer. Beginning with
Hello R users
I have a dataset that has different types of records with different
dates and times pertaining to each. I would like to have a bar graph of
a count of the types(ie. The number of types) of recods by hour grouped
by year. So the count of the types would be the y axis, the hour on
Thanks for your solution, it worked perfectly, it was exactly what I
wanted. I do have two more questions and hope you can help. I have
another analysis exactly like the last one except it is done by month
instead of year. When I graph it using barchart it makes the months go
in alphabetical
Here is my data. I tried table but it doesn't do what I want it to do
when it graphs. I want a count of the types (R for one graph and A for
another) by hour grouped by year. Hope that helps.
ID,,MM,DD,HH,MM,Type
YEG,2002,01,01,01,24,A
YEG,2002,01,01,02,40,R
YEG,2002,01,01,05,39,R
On 5/15/07, Spilak,Jacqueline [Edm] [EMAIL PROTECTED] wrote:
Hello R users
I have a dataset that has different types of records with different
dates and times pertaining to each. I would like to have a bar graph of
a count of the types(ie. The number of types) of recods by hour grouped
by
On 5/15/07, Spilak,Jacqueline [Edm] [EMAIL PROTECTED] wrote:
Here is my data. I tried table but it doesn't do what I want it to do
when it graphs. I want a count of the types (R for one graph and A for
another) by hour grouped by year. Hope that helps.
ID,,MM,DD,HH,MM,Type
I need to read two different random lines at a time from a large
ASCII file (120 x 296976) containing space delimited 0-1 entries.
The following code does the job and it's reasonable fast for my needs:
lineNumber = sample(120, 2)
line1 = scan(filename, what = integer,
On Tue, 2007-05-15 at 16:02 -0700, Juan Pablo Lewinger wrote:
I need to read two different random lines at a time from a large
ASCII file (120 x 296976) containing space delimited 0-1 entries.
The following code does the job and it's reasonable fast for my needs:
lineNumber =
I've searched for the answer to this in the help list archive, but wasn't
able to get the answer to work.
I'm interested in converting a row of a data.frame into a vector.
However, when I use as.vector(x,[1,]) I get another data.frame, instead of a
vector. (On the other hand, when I use
How can I get the value of the aspect ratio that is used in
a lattice
plot? In a levelplot for instance, the native units per cm of my x
and y axes are different, and I need to know the aspect
ratio so that
I can correctly plot vectors. I know how to set the aspect in a
probably something like,
unlist(x[1,])
HTH.
H.
Andrew Yee [EMAIL PROTECTED] 5/15/2007 4:37 PM
I've searched for the answer to this in the help list archive, but wasn't
able to get the answer to work.
I'm interested in converting a row of a data.frame into a vector.
However, when I use
On 5/15/07, Waichler, Scott R [EMAIL PROTECTED] wrote:
How can I get the value of the aspect ratio that is used in
a lattice
plot? In a levelplot for instance, the native units per cm of my x
and y axes are different, and I need to know the aspect
ratio so that
I can correctly
If the data frame has factors and numeric vectors, there is a question
on what form you want the row vector to be in. Only a data frame (list)
can have a mixture of the two.
Consider:
dat - data.frame(x=1:3, y=4:6, z=letters[1:3])
(r1 - dat[1,])
x y z
1 1 4 a
class(r1)
[1] data.frame
(r1
I am using R 2.5 on a Linux Redhat platform. I can successfully run some
example *.Rnw files through Sweave and generate pdf files. When I try my
own example file, test.Rnw:
\documentclass[a4paper]{article}
\title{Test Sweave Example}
\author{Thomas Adams}
\begin{document}
\maketitle
In this
On 15/05/2007 9:22 PM, Thomas Adams wrote:
I am using R 2.5 on a Linux Redhat platform. I can successfully run some
example *.Rnw files through Sweave and generate pdf files. When I try my
own example file, test.Rnw:
\documentclass[a4paper]{article}
\title{Test Sweave Example}
Duncan,
Thank you for your help — that certainly did the trick. With Sweave
being new to me, I somehow missed the obvious…
Regards,
Tom
Duncan Murdoch wrote:
On 15/05/2007 9:22 PM, Thomas Adams wrote:
I am using R 2.5 on a Linux Redhat platform. I can successfully run
some example *.Rnw
How can I get the Log - Rank p value to be output?
The chi square value can be output, so I was thinking if I can also have the
degrees of freedom output I could generate the p value, but can't see how to
find df either.
(survtest - survdiff(Surv(time, cens) ~ group, data = surv,rho=0))
Call:
On Tue, 15 May 2007, Eglin, Jason wrote:
Hello,
I have been trying to build a package for R to use on windows. I have
been able to build it with out problems except for one thing. I am
creating a zip file to be installed by the R gui.
I have four directories under the main dir. I have
On Wed, 2007-05-16 at 11:54 +1000, Murray Pung wrote:
How can I get the Log - Rank p value to be output?
The chi square value can be output, so I was thinking if I can also have the
degrees of freedom output I could generate the p value, but can't see how to
find df either.
(survtest -
Hi,
I'd like to reshape a sparse matrix generated from the Matrix package. I can't
seem to do it with the command
dim(A) - c(6,9)
which works perfectly with the base package matrices, but with the sparse
matrices it errors with
Error in dim(A) = c(6, 9) : dim- : invalid first argument
Hi friends,
I need some help regarding generalized linear mixed model of unbalanced
data.
1. Is their any package for applying Monte-Carlo Newton-Raphson (MCNR) or
Monte-Carlo EM (MCEM) to estimate fixed and random effects?
2. My data is unbalanced (groups having unequal number of
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