On Mon, 2010-08-23 at 17:51 -0400, ivo welch wrote:
> quizz---what does this produce?
Henrique has provided an answer to the question, but...
>d=data.frame( a=1:1000, b=2001:3000, z= 5001:6000 )
>attach(d); c <- (a+b)>25; detach(d)
...this is ugly and will potentially catch you out one d
On Tue, 24 Aug 2010, jim holtman wrote:
I sounds as if your data is encode as UTF-8. You may need to specify
the fileEncoding parameter on the read.table function.
Not UTF-8 ... that's a BOM mark in UCS-2. We don't have the
sessionInfo() output that we asked for and this is an area where
a <- 1
b <- 2
c <- 3
ls()[-a] # set minus to all the objects you want to retain
rm(list = ls()[-a] # will remove all the objects - except a
ls() # presto
--
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On 24 Aug 2010, at 02:20, Ben Bolker wrote:
>> Please, is there an R function /package that allows for 3D stairway plots
> like the attached one ?
>> In addition, how can I overlay a parametric grid plot??
>
> Not exactly, that I know of, but maybe you can adapt
>
> library(rgl)
> demo(hist3d)
2010/8/24 500600 :
>
> a <- 1
> b <- 2
> c <- 3
>
> ls()[-a] # set minus to all the objects you want to retain
>
> rm(list = ls()[-a] # will remove all the objects - except a
>
> ls() # presto
Only because a=1 and a is the first item in the list! Not because you
are doing '-a'! If a is 0 then
Hello,
I upgraded my Mac R version to the newest 2.11.1, then I ran the option to
update all packages but there was an error related to fetching one of those and
the process stopped. I retried updating all packages but nothing happens.
Although all my course project scripts work perfectly is th
Hi,
I have 1000 monthly time series (just a year) and I want to cluster them.
for instance (x):
jan 2010 feb 2010 mar 2010 apr 2010 ...
ts 1: 12300 12354550 1233 12312 ...
ts 2:23423232 2323 232323 ...
...
My approach is applying clara
Dear list members,
I need to create a table from a huge list object,
this list consists of matrices of the same size (but with different
content).
The resulting n tables should contain the same rows from all matrices.
For example:
n <- 23
x <- array(1:20, dim=c(n,6))
huge.list <- list()
for (i
try something like this:
n <- 23
x <- array(1:20, dim = c(n, 6))
huge.list <- rep(list(x), 1000)
out <- lapply(1:4, function (i) {
t(sapply(huge.list, "[", i = i, j = 1:6))
})
out[[1]]
out[[4]]
I hope it helps.
Best,
Dimitris
On 8/24/2010 11:29 AM, Vladimir Mikryukov wrote:
Dear list
On Aug 24, 2010, at 11:29 AM, Vladimir Mikryukov wrote:
> Dear list members,
>
> I need to create a table from a huge list object,
> this list consists of matrices of the same size (but with different
> content).
>
> The resulting n tables should contain the same rows from all matrices.
>
> Fo
Hi!
I personally use the function xlsReadWrite::write.xls(), which doesn't
need Perl to be installed. It might have less functionalities than
WriteXLS, but it is easier to use.
HTH,
Ivan
Le 8/24/2010 00:50, Erich Neuwirth a écrit :
> Search for an older message with the subject line
>
> [R]
Thanks a lot!! It sure helped.
and many thanks to all other repliers!
On Tue, Aug 24, 2010 at 3:40 PM, Dimitris Rizopoulos <
d.rizopou...@erasmusmc.nl> wrote:
> try something like this:
>
> n <- 23
> x <- array(1:20, dim = c(n, 6))
> huge.list <- rep(list(x), 1000)
>
> out <- lapply(1:4, functio
thanks, everyone. I did not even know about transform() and with(),
but they look quite useful.
actually, what I had intended to state was that Boolean variables in
select parts of subset statements, especially when mixed with other
variables that are just part of the data frame, leads to unexpec
On 08/24/2010 11:44 AM, Tim Elwell-Sutton wrote:
Hi
I'd appreciate some help with plotting odds ratios. I want to rotate the
labels on the x-axis by 45 degrees.
The usual way of doing this, using text - e.g. text(1, par('usr')[3]-2.25..)
- gives no result when the y-axis is a log scale.
I gue
Hi ALL
I want to run a program on my office computer, but for some reason I
have to run it later when I am not in the office anymore.
Is there a way to do it? Can I run a program now but in the beginning I
say that the rest should be run 3 hours later for example?
Thanks a lot
Leila
or use
#not checked
rm(setdiff(ls(),c("a", "b"))
On Aug 24, 2010, at 4:55 AM, Barry Rowlingson wrote:
2010/8/24 500600 :
a <- 1
b <- 2
c <- 3
ls()[-a] # set minus to all the objects you want to retain
rm(list = ls()[-a] # will remove all the objects - except a
ls() # presto
Only beca
have a look at Sys.sleep(), i.e., you probably need
Sys.sleep(3*3600)
I hope it helps.
Best,
Dimitris
On 8/24/2010 3:01 PM, l.mohammadikhankahdani wrote:
Hi ALL
I want to run a program on my office computer, but for some reason I
have to run it later when I am not in the office anymore.
Is
veepsirtt gmail.com> writes:
> Warning: dependency ‘caTools’ is not available
I'm not quite sure why this is complaining when you supposedly
just loaded it ... do you really have it installed?
> trying URL 'http://www.math.mcmaster.ca/~bolker/R/src/contrib/Rmail_1.0.zip'
> Error in download.f
ArtemisIrvine hotmail.com> writes:
Try options(error=recover) ?
__
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, mini
hello,
all my attempts to get a plot with tick marks on both sides of axis failed -
can someone please help me with this?
thanks,
kay
-
Kay Cichini
Postgraduate student
Institute of Botany
Univ. of Innsbruck
--
View this message in context
Readers,
According to the documentation for the function 'plotmath' there is no
apparent possibility to add the percent sign (%) to a plot function,
e.g.
plot(a[,1]~b[,2],ylab=expression(x~%),xlab=expression(z))
How to achieve this please?
yours,
rh...@conference.jabber.org
r251
mandriva2009
veepsirtt gmail.com> writes:
>
>
> I could not install Rmail Package .
> I got the following errors.
> Then how to do.please
>
> library("caTools")
> >
install.packages("Rmail",contriburl="http://www.math.mcmaster.ca/~bolker/R/src/contrib";)
try
install.packages("caTools")
install.packages
Thanks for constrctive comments. I was very careful when I wrote the code. I
wrote many functions and then wrapped up to get a single function.
Originally, I used optim() to get MLE, it was at least 10 times slower than
the code based on Newton method. I also vectorized all objects whenever
possib
You've tried:
plot(a[,1]~b[,2],ylab=expression(x~'%'),xlab=expression(z)) ?
On Tue, Aug 24, 2010 at 10:37 AM, e-letter wrote:
> Readers,
>
> According to the documentation for the function 'plotmath' there is no
> apparent possibility to add the percent sign (%) to a plot function,
> e.g.
>
> p
On Aug 24, 2010, at 9:31 AM, Kay Cichini wrote:
hello,
all my attempts to get a plot with tick marks on both sides of axis
failed -
can someone please help me with this?
The path I took to getting a glimmmer of the answer was to look at ?
axTicks and then notice in the "..." arguments d
On linux or unix you could use the command 'at' to start R in case that the
entire job can be started from the command line (e.g. 'R --vanilla <
my-R-script').
Tim
On Tue, Aug 24, 2010 at 03:01:29PM +0200, l.mohammadikhankahdani wrote:
> Hi ALL
>
> I want to run a program on my office computer, b
On 24/08/2010, Henrique Dallazuanna wrote:
> You've tried:
>
> plot(a[,1]~b[,2],ylab=expression(x~'%'),xlab=expression(z)) ?
>
This is successful for me, thank you. This instruction should be added
to the documentation for 'plotmath'.
__
R-help@r-projec
On Aug 24, 2010, at 9:37 AM, e-letter wrote:
Readers,
According to the documentation for the function 'plotmath' there is no
apparent possibility to add the percent sign (%) to a plot function,
Where did you see an assertion made???
e.g.
plot(a[,1]~b[,2],ylab=expression(x~%),xlab=expressi
Hello,
I am trying to see whether there has been a significant difference in whether
people experienced damages from wildlife in two different years. I therefore
have two columns:
year 1:
yes
no
no
no
yes
yes
no
year 2:
no
yes
no
yes
I wanted to do a chisq.test, but if I enter it this way:
c
Hi all,
all.equal is generally very useful when you want to find the
differences between two objects. It breaks down however, when you
have two long strings to compare:
> all.equal(a, b)
[1] "1 string mismatch"
Does any one know of any good text diffing tools implemented in R?
Thanks,
Hadley
...i'm aware of par("tcl") and par("tck") but i needed tick marks on both
sides of the axis - meaning to cross the axis perpendicularly - admittedly i
was clumsy not just calling axis twice, like:
plot(1)
axis(2,labels=T,tcl=0.25)
axis(2,labels=F,tcl=-0.25)
yours,
kay
-
--
If you are going to be doing this a lot, you may want to consider
making a little function. Here is an example 'do not remove'
function. It defaults to the global environment, but you can always
change it. Also, if you try to keep a variable name that does not
exist, it will throw an error and n
On Aug 24, 2010, at 10:12 AM, Marino Taussig De Bodonia, Agnese wrote:
Hello,
I am trying to see whether there has been a significant difference
in whether people experienced damages from wildlife in two different
years. I therefore have two columns:
year 1:
yes
no
no
no
yes
yes
no
year
On 24-Aug-10 14:16:55, Hadley Wickham wrote:
> Hi all,
> all.equal is generally very useful when you want to find the
> differences between two objects. It breaks down however,
> when you have two long strings to compare:
>
>> all.equal(a, b)
> [1] "1 string mismatch"
>
> Does any one know of an
On Aug 24, 2010, at 4:12 PM, Marino Taussig De Bodonia, Agnese wrote:
> Hello,
>
> I am trying to see whether there has been a significant difference in whether
> people experienced damages from wildlife in two different years. I therefore
> have two columns:
>
> year 1:
> yes
> no
> no
> no
Restoring the broken email chain. I can certainly understand why
Nabble users are often guilty of doing so at the beginning of their
participation, but after they have been around the list for a while, I
would think they should "get the message". If not then:
#
PLEASE do read the pos
Hi list, I have a data set - something like this
dfr <- data.frame(A=factor(letters[1:25]),B=runif(25),
C=sample(LETTERS[1:4],25,replace=TRUE))
and I want to create a dotplot:
library(lattice)dotplot(A ~ B | C, data=dfr,
scales=list(y=list(relation="free")))
but this puts uneven
Hi All,
I want to know is there any package in R which can plot multivariate control
charts ??
Thanks in advance,
Vikrant.
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.c
Simple one, have read and googled, still no luck!
I want to create several empty vectors all of the same length.
I would like multiple empty vectors (vec1, vec2, vec3) and want to create them
all in one line.
I've tried
vec1,vec2,vec3 <- vector(length=5)
and
c(vec1,vec2,vec3) <- vector(length=
Good morning,
I am trying to find a S-Plus code which shows how to find a
confidence interval using a bootstrapping on a corrected correlation value
of a two bivariate variables.
If you happen to know one, please shows me.
I am greatly appreciated your help.
Have a wonderful d
There is the stringMatch function in the MiscPsycho package.
> stringMatch('Hadley', 'Hadley Wickham', normalize = 'no')
[1] 8
> stringMatch('Hadley', 'Hadley Wickham', normalize = 'yes')
[1] 0.4285714
It uses Levenshtein distance to tell you how much they differ by, either
normalized or not. S
Hi,
This is one way (I chose integer but it can be any mode):
vec1 <- vec2 <- vec3 <- vector("integer", 5)
Best regards,
Josh
On Tue, Aug 24, 2010 at 4:58 AM, Maas James Dr (MED) wrote:
> Simple one, have read and googled, still no luck!
>
> I want to create several empty vectors all of the s
On Aug 24, 2010, at 7:36 AM, ivo welch wrote:
thanks, everyone. I did not even know about transform() and with(),
but they look quite useful.
actually, what I had intended to state was that Boolean variables in
select parts of subset statements, especially when mixed with other
variables that
thanks a lot.
sorry - i don't know what you mean by "restoring broken email chain",
could not get an answer by re-reading the posting guide, either.
yours,
kay
-
Kay Cichini
Postgraduate student
Institute of Botany
Univ. of Innsbruck
--
V
Thanks for all suggestions from Roman (?= 500600[via R]), Barry and Jim.
It seems that ls()[-objectname] didn't work even on numeric matrices and
user made functions. I will work with Jim's advice on using grep() and gc()
to see whether it works.
--
View this message in context:
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On 2010-08-23 11:03, Mark Breman wrote:
Hello everyone,
I am reading a HTML table from a website with readHTMLTable() from the XML
package:
library(XML)
moose = readHTMLTable("http://www.decisionmoose.com/Moosistory.html";,
header=FALSE, skip.rows=c(1,2), trim=TRUE)[[1]]
moose
Do you mean something like this?
>
n <- 5
>
(vec1 <- matrix(rep(1, n)))
[,1]
[1,]1
[2,]1
[3,]1
[4,]1
[5,]1
>
(vec2 <- matrix(rep(2, n)))
[,1]
[1,]2
[2,]2
[3,]2
[4,]2
[5,]2
>
(vec3 <- matrix(rep(3, n)))
[,1]
[1,]3
[2,]3
[3,]3
You can also do meta.summaries() - from rmeta package - followed by a
plot() on the resulting object.
Or for a much more flexible plot try forestplot() function, also from
rmeta package, but this requires a bit of work to set it up.
Regards, Adai
On 24/08/2010 05:50, C.H. wrote:
The correc
Hi,
I downloaded a package with .gz extension, which presumably works for Unix,
when I installed it directly on windows, it has the error msg:
Error in gzfile(file, "r") : cannot open the connection
In addition: Warning messages:
1: In unzip(zipname, exdir = dest) : error 1 in extracting from zip
Hi Cheng,
Check out the keep() function in package:gdata.
And to be sure the "removed" objects are really removed from system
memory i think you need to run gc().
hth,
Karl
On 8/23/2010 9:00 PM, Cheng Peng wrote:
How to remove all R objects in the RAM except for a few specified ones?
rm(l
None of this would work if the list is long. Isn't this an obvious
task for a loop, explicit or implicit?
e.g.
for(i in 1:100)assign(paste("vec",i,sep=""), vector("integer",5))
or probably better because it creates a list structure:
## warning, untested. You may have to fool with the syntax a b
Hello there,
I am using the Tps function to fit a spline to my data which is measurements
of a sedimentary layer on an xy plane to see how the thickness of the layer
changes. I was wondering if it was possible to change the Tps code so that
no minus values are calculated by the spline as these a
On 08/24/2010 07:27 AM, Doran, Harold wrote:
> There is the stringMatch function in the MiscPsycho package.
>
>> stringMatch('Hadley', 'Hadley Wickham', normalize = 'no')
> [1] 8
>> stringMatch('Hadley', 'Hadley Wickham', normalize = 'yes')
> [1] 0.4285714
>
> It uses Levenshtein distance to tel
Dear r-help,
I'm having trouble formatting xy labels for plot in r. I want to make the
following y-label
"benzene (mug-3)"
where mu is greek and -3 is superscript and benzene is held in an array.
I tried this,
myname = c("benzene", "etc")
plot (c(0:10), ylab = bquote(.(myname[1])~~(mu~g ~m^-3)))
Dear all, I was doing an experiment to disprove some theory therefore
performing lot of random simulation. Goal is to show the audience that
although something has very rare chance to occur but it doesn't mean that
event would be impossible.
In this case after getting that rare event I need to
Great. It is more clearer for me. Thanks all.
2010/8/24 Michael Dewey
> At 02:40 24/08/2010, rusers.sh wrote:
>
>> Hi all,
>> rmvnorm()can be used to generate the random numbers from a multivariate
>> normal distribution with specified means and covariance matrix, but i want
>> to specify the c
Hello All,
Using the standard "summary" function in 'R', I ran across some odd
behavior that I cannot understand. Easy to reproduce:
Typing:
summary(c(6,207936))
Yields::
Min. *1st Qu. MedianMean 3rd Qu.Max.*
6 *51990 104000 104000 156000 207900*
None of t
Hi R-help,
I am trying to use 'samr' for 10 pre and post paired samples to test
whether
post is different from pre (i.e., the location shift for the delta of
(post-pre)).
However, I got an error message saying
> samr.obj<-samr(d, resp.type="Two class paired", nperms=100,
random.seed=100)
pe
In addition to the other suggestions you may also want to look at the subplot
function in the TeachingDemos package (the version in Hmisc is a copy of a
prior version of this one) as well as the my.symbols or panel.my.symbols
functions in the same package.
--
Gregory (Greg) L. Snow Ph.D.
Stati
On Aug 24, 2010, at 12:39 PM, Jennifer Hains wrote:
Dear r-help,
I'm having trouble formatting xy labels for plot in r. I want to
make the following y-label
"benzene (mug-3)"
where mu is greek and -3 is superscript and benzene is held in an
array.
I tried this,
myname = c("benzene", "etc
BTW, can you recommend a book on statistical simulations? I want to know
more on how to generate random numbers from distributions, how to generate
the theoretical models,...
Thanks a lot.
2010/8/24 Michael Dewey
> At 02:40 24/08/2010, rusers.sh wrote:
>
>> Hi all,
>> rmvnorm()can be used to ge
On Aug 24, 2010, at 1:06 PM, Mike Williamson wrote:
Hello All,
Using the standard "summary" function in 'R', I ran across some odd
behavior that I cannot understand. Easy to reproduce:
Typing:
summary(c(6,207936))
Yields::
Min. *1st Qu. MedianMean 3rd Qu.Max.*
6 *51
On Tue, Aug 24, 2010 at 8:45 AM, Maas James Dr (MED) wrote:
> Thanks Bert, will have a look. I'm originally a Fortran programmer so tend
> to think in loops ... so yes expect it may be job for loops, just tried to
> avoid it
because several references say not to use loops in R.
-- Yes, an un
Hello-
A basic question which has nonetheless floored me entirely. I have a
dataset which looks like this:
Type ID DateValue
A 116/09/2020 8
A 1 23/09/2010 9
B 3 18/8/20107
B 1 13/5/20106
There are two Types, whi
I'm currently running into a little trouble with the kfilter method,
and would love some clarification if you are able to offer it. When
trying to run kfilter, I've been running into errors that seem to
result from having mismatched dimensions. Specifically, the dimension
of my observations is 2,
Hello J.R.M. Hosking,
charToRaw() works perfectly, thank you:
> charToRaw(as.character(moose[1, "V3"]))
[1] 24 38 38 30 2c 33 37 30 c2 a0
gsub("[[:space:]]", "", ...) did not remove them, but now I know what they
are (hex: c2 a0) I can remove them with gsub() by:
> gsub("[$,\xc2\xa0]", "", as.
summary.default uses the signif function to round for display purposes.
In ?summary, we can see the digits argument is used to control
the value passed to signif.
> lapply(1:6, function(x) summary(c(6, 207936), digits = x))
[[1]]
Min. 1st Qu. MedianMean 3rd Qu.Max.
6e+00 5e+04
I'm relatively new to R, but I'm attempting to do a non-linear maximum
likelihood estimation (mle) in R, with the added problem that I have a
non-linear constraint.
The basic problem is linear in the parameters (a_i) and has only one
non-linear component, b, with the problem being linear when
On 2010-08-24 11:06, Mike Williamson wrote:
Hello All,
Using the standard "summary" function in 'R', I ran across some odd
behavior that I cannot understand. Easy to reproduce:
Typing:
summary(c(6,207936))
Yields::
Min. *1st Qu. MedianMean 3rd Qu.Max.*
6 *5199
On Aug 24, 2010, at 1:19 PM, Chris Beeley wrote:
Hello-
A basic question which has nonetheless floored me entirely. I have a
dataset which looks like this:
Type ID DateValue
A 116/09/2020 8
A 1 23/09/2010 9
B 3 18/8/20107
B
An answer to 1)
> x = data.frame(Type=c('A','A','B','B'), ID=c(1,1,3,1), Date =
c('16/09/2010','23/09/2010','18/8/2010','13/5/2010'), Value=c(8,9,7,6))
> x
Type ID Date Value
1A 1 16/09/2010 8
2A 1 23/09/2010 9
3B 3 18/8/2010 7
4B 1 13/5/2010 6
> x$
The only problem with this is that Chris's unique individuals are a
combination of Type and ID, as I understand it. So Type=A, ID=1 is a
different individual from Type=B,ID=1. So we need to create a unique
identifier per person, simplistically by uniqueID=paste(Type, ID,
sep=''). Then, using th
On Aug 24, 2010, at 1:59 PM, Abhijit Dasgupta, PhD wrote:
The only problem with this is that Chris's unique individuals are a
combination of Type and ID, as I understand it. So Type=A, ID=1 is a
different individual from Type=B,ID=1. So we need to create a unique
identifier per person, sim
Dear expeRts,
I would like to have four plots appearing in one figure. The minimal example
shows this. However, the four figures are not properly aligned. Why? If I
comment out the scales=... arguments, then it works, but I would like to use
this... :-)
Cheers,
Marius
library(lattice)
set.
The paste-y argument is my usual trick in these situations. I forget
that tapply can take multiple ordering arguments :)
Abhijit
On 08/24/2010 02:17 PM, David Winsemius wrote:
On Aug 24, 2010, at 1:59 PM, Abhijit Dasgupta, PhD wrote:
The only problem with this is that Chris's unique individu
I have wondered this in the past too so thanks for the question.
On Aug 24, 2010, at 10:11 AM, Bogaso Christofer wrote:
Dear all, I was doing an experiment to disprove some theory therefore
performing lot of random simulation. Goal is to show the audience that
although something has very rare cha
Dear R Helpers,
I am a newbie and recently got introduced to R. I have a large database
containing the names of bank branch offices along-with other details. I am into
Operational Risk as envisaged by BASEL II Accord.
I am trying to express my problem and I am using only an indicative da
Hello guys,
I would to plot a bar line between to curves like in excel as shown in the
following image. but i do not know how I can do that in R.
http://r.789695.n4.nabble.com/file/n2337089/excel_Plot.png
any help would appreciate.
--
View this message in context:
http://r.789695.n4.nabble
> On Aug 24, 2010, at 10:11 AM, Bogaso Christofer wrote:
>
> Dear all, I was doing an experiment to disprove some theory therefore
> performing lot of random simulation. Goal is to show the audience that
> although something has very rare chance to occur but it doesn't mean that
> event would be
Dear all,
I'm trying to compile FORTRAN code to be used in R, failing miserably. I got
a simple code function to try to figure out what is going wrong. Here is the
code (available on the web) :
subroutine bar(n, x)
integer n
double precision x(n)
integer i
do
The following will attach (as an attribute) the current global value
of .Random.seed to the value of the evaluated 'expr' argument. If you
supply
the initial.Random.seed argument then it will use that when evaluating
the expression (and also attach it to the result) so you can repeat the
'unusual'
Hi,
the actual thread on "multiple assignments ?" made up an interesting point for
me.
If I have a matrix with g rows and x + y columns where columns 1 - x contain
values of group1 and columns x+1 to y contain the values of group2.
Now I want to compute a vector of length g that holds the p-valu
Hi, has there been a solution to this issue? I am encountering the same
problem on a Mac with OSX 10.6.4. The problem persists when I try to install
lme4 from the source (see below), and my R version is up to date according
to R's update check.
Thanks for any help,
Daniel
--
session
On Aug 24, 2010, at 1:40 PM, abotaha wrote:
Hello guys,
I would to plot a bar line between to curves like in excel as shown
in the
following image. but i do not know how I can do that in R.
http://r.789695.n4.nabble.com/file/n2337089/excel_Plot.png
any help would appreciate.
?segments
On Aug 24, 2010, at 3:13 PM, Daniel Malter wrote:
Hi, has there been a solution to this issue? I am encountering the
same
problem on a Mac with OSX 10.6.4. The problem persists when I try to
install
lme4 from the source (see below), and my R version is up to date
according
to R's update
Thanks for your solution, however, which values would I then provide someone
else so that they can produce the same results?
On Aug 24, 2010, at 12:06 PM, William Dunlap wrote:
The following will attach (as an attribute) the current global value
of .Random.seed to the value of the evaluated 'exp
You can certainly use apply()
#make up some data
x <- 10; y <- 10; g <- 5
set.seed(1969)
dat <- matrix(rnorm((x + y) * g), ncol = x + y)
# apply() the t.test function to each row fo the matrix and extract
just the p value
results <- apply(dat, 1, function(dat) {
t.test(x = dat[1:x], y = d
Hello,
I am using R to train a logistic regression model and save the resulting
model to disk. I am then subsequently reloading these saved objects, and
using predict.glm on them in order to make predictions about single-row data
frames that are generated in real-time from requests arriving at an
> -Original Message-
> From: r.ookie [mailto:r.oo...@live.com]
> Sent: Tuesday, August 24, 2010 12:25 PM
> To: William Dunlap
> Cc: r-help@r-project.org
> Subject: Re: [R] How to obtain seed after generating random number?
>
> Thanks for your solution, however, which values would I then
On 08/24/2010 09:01 PM, Marie-Hélène Ouellette wrote:
> Dear all,
>
> I'm trying to compile FORTRAN code to be used in R, failing miserably. I got
> a simple code function to try to figure out what is going wrong. Here is the
> code (available on the web) :
>
>
>
> subroutine bar(n, x)
>
Does anyone know where I can download the latest version of Sweave.sty? I have
looked all over the site http://www.stat.umn.edu/~charlie/Sweave/ with no luck.
__
R-help@r-project.org mailing list
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PLEASE do rea
Here is one:
http://svn.r-project.org/R/trunk/share/texmf/tex/latex/Sweave.sty
-Matt
On Tue, 2010-08-24 at 15:40 -0400, r.ookie wrote:
> Does anyone know where I can download the latest version of Sweave.sty? I
> have looked all over the site http://www.stat.umn.edu/~charlie/Sweave/ with
> no
On 08/24/2010 09:40 PM, r.ookie wrote:
> Does anyone know where I can download the latest version of
> Sweave.sty? I have looked all over the site
> http://www.stat.umn.edu/~charlie/Sweave/ with no luck.
I don't think there is anything newer than what is inside the latest
version of R. Look under
Hi:
This looks to be a step in the right direction, but the resulting
panels are all the same size. Perhaps you can build on it...
# Solution based on an R-help post by Deepayan Sarkar:
# http://tolstoy.newcastle.edu.au/R/e2/help/06/09/1579.html
dotplot(A ~ B | C, data=dfr, scales=list(y=list(re
On Tue, 2010-08-24 at 12:40 -0700, r.ookie wrote:
> Does anyone know where I can download the latest version of
> Sweave.sty? I have looked all over the site
> http://www.stat.umn.edu/~charlie/Sweave/ with no luck.
It is provided in the R sources as Sweave is an integrated part of R.
You can get i
Dear
all,
I have
written some R source program with many thousands of lines. I didnât insert
line breaks automatically or manually for the long lines. But now I would like
to edit the source code in Emacs/ESS to make it more formal as a package. One of
the major problems here is how to break
Hello,
I am using R to train a logistic regression model and save the resulting
model to disk. I am then subsequently reloading these saved objects, and
using predict.glm on them in order to make predictions about single-row data
frames that are generated in real-time from requests arriving at an
Hello,
As recommended by N. Gunther, I'm writing you because I have some
difficulties to create a script to track computer performance, as done in
Holtman, 2005 (Visualization techniques for analysing patterns in system
performance).
I would like to do a leveplot, with the kind of file here enclo
Dear R users,
I'm building linear models that use poly and has a lot of product terms. My
model summaries look messy. I know that the summary could use aliases for
the variable names, but I don't know how to get it to do that. Please help
Thanks,
--
View this message in context:
http://r.789695.
I am trying to get R to use the ACML BLAS, specifically the 'mp' version for
multithreading.
I have installed acml-4-4-0-gfortran-64bit
I added /opt/acml4.4.0/gfortran64_mp/lib to $LD_LIBRARY_PATH.
I ran: sudo ./configure --with-blas="-L/opt/acml4.4.0/gfortran64_mp/lib
-lacml_mp"
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