I think you would also get a speedup by instantiating your list at
size. For example:
a <- vector("list", 10)
so that memory is already set aside and there is not incremental
copying and increasing (although R may be clever and manage to avoid
this for you). It is really rather difficult to
Hi,
Compare this:
a<-list()
system.time(for(i in 1:10){
a[[i]]<-i*1000
})
system.time(a2 <- as.vector(1000 * (1:10), "list"))
all.equal(a, a2)
## > system.time(for(i in 1:10){
## + a[[i]]<-i*1000
## + })
##user system elapsed
## 28.660.02 28.84
## > system.time(a2 <- as
hi
I'm simulating some experiment by "for" function.
The problem is that it takes too much time.
for example,
a<-list()
for(i in 1:1){
a[[i]]<-i*1000
}
is there anyway to speed up the process?
I heard there is solution but i don't have idea.
Can anyone help me?
--
View this message i
Hi Sarah Goslee
Thanking you for replying to my doubt.
I downloaded the multtest package from CRAN and also went through the
package specific PDF file, they have given a list of functions which
includes the mt.transformV, mt.checkothers etc, that I am trying to access
which they call as the inter
Hi,
On Fri, Mar 16, 2012 at 12:17 AM, AGGARWAL, NEERAJ (NEERAJ)
wrote:
> Hi All,
>
> We have a product that performs ETL on files and finally load the database.
> We want to give the web based interface displaying graphs (that are generated
> using R-project forcasting) from this database.
> But
Hi All,
We have a product that performs ETL on files and finally load the database.
We want to give the web based interface displaying graphs (that are generated
using R-project forcasting) from this database.
But R-Project is stand alone desktop based software and it takes commands at
command l
On Thu, Mar 15, 2012 at 10:25 PM, Richard M. Heiberger wrote:
> ## I would like help in using variable values in plotmath expressions
> ## in lattice
>
> x <- 1:10
> y <- 1:10
> pval <- .95
>
> plot(y ~ x, ## works as I want in base graphics
> main=substitute(list(alpha * " = " * group("",list
Hi Rich,
I'm not sure if you have a reason outside this toy example to want to
be using list() and group(), but if not I would go with this:
xyplot(y ~ x, main = substitute(expression(alpha == p), list(p = pval)))
Hope this helps,
Josh
On Thu, Mar 15, 2012 at 7:25 PM, Richard M. Heiberger wro
Then save to a variable, and maybe then to a file?
Are you wanting it in long form? You can use reshape if so.
Note that a reproducible example and sample result always helps prompt more
specific answers.
---
Jeff Newmiller
## I would like help in using variable values in plotmath expressions
## in lattice
x <- 1:10
y <- 1:10
pval <- .95
plot(y ~ x, ## works as I want in base graphics
main=substitute(list(alpha * " = " * group("",list(p),"")),
list(p=pval)))
plot(y ~ x, ## doesn't work as intended: "pval" is d
Hi Camille,
Probably by adding these lines:
fpf <- unlist(performance(pred,"fpr")@y.values)
fnf <- unlist(performance(pred,"fnr")@y.values)
pLhood <- tpf/fpf # Positive Likelihood Ratio
nLhood <- fnf/tnf # Negative Likelihood Ratio
par(mfrow=c(1,2))
plot(x, pLhood, t='l', xlab='Value', ylab
> Date: Thu, 15 Mar 2012 14:00:46 -0700
> Subject: Re: [R] Get Details About Clusters
> From: peter.langfel...@gmail.com
> To: anxu...@hotmail.com
> CC: r-help@r-project.org
>
> On Thu, Mar 15, 2012 at 1:48 PM, A J wrote:
> >
> > Hi everybody!
> > Anybody knows how can I get detalied informati
Here you go:
library(mgcv)
y.list <- list(y1=runif(10), y2 <- runif(10), y3 <- runif(10))
y.list <- list(y1=runif(20))
x1 <- 1:20
x2 <- c(11, 15, 17, 2, 18, 6, 7, NA, 12, 10,21, 25, 27, 12, 28, 16,
17, NA, 12, 10)
y.gam <-lapply(y.list, function(y)gam(y~s(x1,x2, k=10), na.action = na.exclude))
y.f
Bill et. al:
1. This is new to me. Thanks.
2. As I read the man page, this is not guaranteed to work if the model
fitting function does not contain sufficient interrupts. Is that
correct?
-- Bert
On Thu, Mar 15, 2012 at 4:14 PM, William Dunlap wrote:
> There is a setTimeLimit function in base.
On Thu, Mar 15, 2012 at 7:51 PM, Adam Zeilinger wrote:
> Hello,
>
> I am trying to construct two matrices, F and V, composed of partial
> derivatives and then find the eigenvalues of F*Inverse(V). I have the
> following equations in ryacas notation:
>
>> library(Ryacas)
>> FIh <- Expr("betah*Sh*I
Hello,
I am trying to construct two matrices, F and V, composed of partial
derivatives and then find the eigenvalues of F*Inverse(V). I have the
following equations in ryacas notation:
> library(Ryacas)
> FIh <- Expr("betah*Sh*Iv")
> FIv <- Expr("betav*Sv*Ih")
> VIh <- Expr("(muh + gamma)*Ih
Dear R users,
I need to draw a barplot with 2 Y axis. I have 3 days each of wich having 2
groups (and error bar for each of them). The height of the 3rd day is too tall
compared to others. That's why I have to use a second Y axis for that. I am
using "barplot2" function of "gplots" library (t
Hi
When I run my script using ggplot and geom_smooth I get messages that I
would like to suppress:
p <- ggplot(dataSubset)
p <- p + aes(x = as.Date(factor(key),format="%Y%m%d")) + geom_line()
p <- p + geom_smooth(span=0.2,se=FALSE,size=0.7)
The messages look like this:
geom_smooth: method="auto"
Dear Michael,
Thanks. I tried it and I was not able to run it on the gam line (sorry my
programming is not that great). Since I am ussing the fitted vallues for
another program I actually want a blank value in place of the missing data.
Many thanks in advance
Regards
Minti
_
To add to the above, as far as I know Cran's Ubuntu repository is only
available for LTS ( Ubuntu 10.04 and soon 12.04, whenever they put the
repository together) and if you are using non LTS's versions of Ubuntu
Michael Rutter's ppa is definitely the way to go, it is very up to date.
--
View thi
On 15-Mar-2012 Filoche wrote:
> Hi everyone.
>
> Based on a dependent variable (y), I'm trying to generate some
> independent variables with a specified correlation. For this
> there's no problems.
> However, I would like that have all my "regressors" to be
> orthogonal (i.e. no correlation among
There is a setTimeLimit function in base. It could be encapsulated into
the following to limit the time spent on an expression:
timeOut <- function (expr, ...) {
on.exit(setTimeLimit())
setTimeLimit(...)
expr
}
E.g., with the following slow way to compute Euler's phi
f <- functio
Ramiro Barrantes precisionbioassay.com> writes:
> I have a program that consists of a loop fitting a function
> over many models. Sometimes the fitting on a
> particular model takes minutes to converge. Is there a way that
> I can limit the amount of time that R spends
> on a given model:
On Thu, Mar 15, 2012 at 2:24 PM, Ramiro Barrantes
wrote:
> Hello,
>
> I have a program that consists of a loop fitting a function over many models.
> Sometimes the fitting on a particular model takes minutes to converge. Is
> there a way that I can limit the amount of time that R spends on
On Mar 15, 2012, at 4:38 PM, G See wrote:
my.df <- data.frame(IDX=1:42, x=rnorm(1:42))
my.df[my.df$IDX %in% c(17, 42), ]
IDX x
17 17 -0.5110095
42 42 -1.0686427
subset(my.df, IDX %in% c(17,42) )
(I find subset to give clearer code than: df[which(df$colname cond>) , ], plus it c
I have a dataframe from an On-Board Survey with weights ("expwgt") and
variables for up to 8 used lines: VEH1 through VEH8. The lines are labeled
"MT-..1" through "MT-902". I want to know how many transfers there are between
MT-802 and MT-901. That is, when one of them is VEHx and the other is
Hello,
I have a program that consists of a loop fitting a function over many models.
Sometimes the fitting on a particular model takes minutes to converge. Is
there a way that I can limit the amount of time that R spends on a given model:
say if my line is:
fittingFunction( func, model.1
Rui:
Try reading it again. Orthogonal polynomials are generated (subject to
the caveats regarding machine precision stated therein). Note,
especially, the "raw" argument.
Cheers,
Bert
On Thu, Mar 15, 2012 at 1:06 PM, Rui Barradas wrote:
> Hello, again.
>
>>
>> This is not really working. Here w
I have figured it out.
It turns out that it is not a R problem. libfontconfig.so.1 links to
libfreetype, it happens that there are two sets of libfreetype files in my
system, one from Ubuntu and another from xampp. When the php script is run
to invoke R it uses the libfreetype files in /opt/lampp
Hello, again.
>
> This is not really working. Here what I have for the moment.
>
Right, I've read only half of the description line of function 'poly'. The
other half states that
"These are all orthogonal to the constant polynomial of degree 0."
But not pairwise orthogonal.
You can look for
Thank you everyone for your precious advice.
I'll take time to look at it and try to make it work.
Regards,
Phil
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Is there a way to make V-Optimal Histogram in R?
Thanks,
Aaron
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__
R-help@r-project.org m
Str() found the problem; there were 2 blanks at the end of the name.
Why does this occur when I read in with trim.factor.names = TRUE ?
>
> OBDataSumm <- read.spss("P:/Data/OBSurveys/OBSurvey-2010-2011/Final Delivery,
> Metro On-Board O-D
> Survey/LAMTA_OD_WEIGHTED_DATA_SETS_012512/LA
On Thu, Mar 15, 2012 at 03:47:24PM -0500, G See wrote:
> On second thought, I must not understand your question because your
> code works fine for me.
>
> > my.df[ which(my.df$IDX == c(17, 42)), ]
>IDX x
> 17 17 -0.5110095
> 42 42 -1.0686427
Hi.
This seems to work due to the fact
On Thu, Mar 15, 2012 at 1:48 PM, A J wrote:
>
> Hi everybody!
> Anybody knows how can I get detalied information about clusters after using
> hclust?
> The issue is that if I have some items in different clusters, I would like to
> get the cluster where each item is placed.
> Taking into account
Hi everybody!
Anybody knows how can I get detalied information about clusters after using
hclust?
The issue is that if I have some items in different clusters, I would like to
get the cluster where each item is placed.
Taking into account that my data set is too large, it is not useful to have t
On second thought, I must not understand your question because your
code works fine for me.
> my.df[ which(my.df$IDX == c(17, 42)), ]
IDX x
17 17 -0.5110095
42 42 -1.0686427
> my.df[my.df$IDX == c(17, 42), ]
IDX x
17 17 -0.5110095
42 42 -1.0686427
Can you provide samp
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Rich Shepard
> Sent: Thursday, March 15, 2012 1:19 PM
> To: r-help
> Subject: Re: [R] 64 bit version for linux and build huge matrix
>
> On Fri, 16 Mar 2012, Shengyun Peng wrote:
> my.df <- data.frame(IDX=1:42, x=rnorm(1:42))
> my.df[my.df$IDX %in% c(17, 42), ]
IDX x
17 17 -0.5110095
42 42 -1.0686427
Garrett
On Thu, Mar 15, 2012 at 3:25 PM, Ed Siefker wrote:
> I would like to subset by dataframe by matching all rows that have any value
> from a list of valu
On Mar 15, 2012, at 3:19 PM, Rich Shepard wrote:
> On Fri, 16 Mar 2012, Shengyun Peng wrote:
>
>> I downloaded R from http://cran.cnr.berkeley.edu/, with the linux version.
>> But after compile, I don't know how to start 64 bit R, as there is only
>> one R executable under bin folder, not alike
I would like to subset by dataframe by matching all rows that have any value
from a list of values. I can get it to work if I have exactly one value,
I'm not
sure how to do it with a list of values though.
This works and gives me exactly one line:
my.df[ which( mydf$IDX==17)), ]
I would like to
Hi,
On Thu, Mar 15, 2012 at 4:02 PM, Shengyun Peng wrote:
> Hello:
>
>
> I downloaded R from http://cran.cnr.berkeley.edu/, with the linux version.
> But after compile, I don't know how to start 64 bit R, as there is only one R
> executable under bin folder, not alike windows version which have
On Fri, 16 Mar 2012, Shengyun Peng wrote:
I downloaded R from http://cran.cnr.berkeley.edu/, with the linux version.
But after compile, I don't know how to start 64 bit R, as there is only
one R executable under bin folder, not alike windows version which have
two executables.
If your system
Hello:
I downloaded R from http://cran.cnr.berkeley.edu/, with the linux version. But
after compile, I don't know how to start 64 bit R, as there is only one R
executable under bin folder, not alike windows version which have two
executables.
Other than that, I want to build a matrix 655
We received a generous gift to support the Journal of Statistical Software
(www.jstatsoft.org) from the DC Area R Users Group. If you think the Journal is
a worthy cause, then support it through the Statistics Computing Support Fund at
https://giving.ucla.edu/Standard/NetDonate.aspx?SiteNum=107
Hi Pete,
I have got the same problem on Windows Vista, but I found a solution from
the docu:
setInternet2(use = TRUE)
seems to be necessary before download.file to make R be able to use some
special dll from internet explorer.
Ulrich
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On Thu, Mar 15, 2012 at 10:48:48AM -0700, Filoche wrote:
> Hi everyone.
>
> Based on a dependent variable (y), I'm trying to generate some independent
> variables with a specified correlation. For this there's no problems.
> However, I would like that have all my "regressors" to be orthogonal (i.e
The line I gave you will read them in, store them in an object called
a list (which is just a generic holding structure, like a struct in C
or a list in Python) and, once it's got them all in one list, rbind
the whole list together to make one "super"-data.frame. If you want to
keep them separately
Hi there.
This is not really working. Here what I have for the moment.
library(ecodist)
x <- 1:100
y1 <- corgen(x=x, r=.85, epsilon=.01)$y
y2 <- corgen(x=x, r=.5, epsilon=.01)$y
y3 <- corgen(x=x, r=.25, epsilon=.01)$y
a = poly(cbind(y1, y2, y3), degree=1)
cor(a[,1], a[,2])
In that case, the c
Try adding na.action = na.exclude to your call to gam.
Michael
On Thu, Mar 15, 2012 at 1:00 PM, Mintewab Bezabih
wrote:
> Dear R users,
>
> I have problems handling missing values. THe problem is that after running my
> codes the result I get just skips the missing values. What I want is for th
Beltrand was also on the mark, suggesting you add Michael Rutter's ppa to
your repository sources.
In both cases (adding the CRAN Ubuntu repositories or Michael Rutter's
ppa), an additional package repository is added to your system's packages.
apt then checks that repository along with the other
I don't use xlsReadWrite, but I've found XLConnect rather handy for
things like this: once you're going, you can just loop over all sheets
like so:
do.call("rbind", lapply(1:50, function(n) readWorksheet(wb, sheet = n,
OtherArgumentsGoHere)))
which will gather them all in a list (from lapply) and
Dear R experts,
I am trying to import some data from some Excle files into R. My Excle file
contains about 50 sheets.
One solution I can think about is to convert my Excle file into csv file
first and then load data into R using 'read.csv'.
But it seems to me that 'read.csv' only supports readin
Hello,
>
> However, I would like that have all my "regressors" to be orthogonal (i.e.
> no correlation among them.
>
?poly
poly(cbind(x1, x2, x3), degree=1)
Hope this helps,
Rui Barradas
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Thanks for the fast reply.
I'm using ggplot2 (it was a typo in my code example). However, using exactly
the same example as I mentioned before, it only shows the bars for cyl=6&8.
How can the behaviour be different on differing machines? I just checked for
updates and I have the latest ggplot2.
Hello,
I am performing cross-validation on a number of survival models fitted with the
coxph function and would like to use the likelihood ratio test on the left-out
cases as the criteria for comparison of PCA, SPCA, ridge, lasso, etc. Is there
an easy way to do this? I think predict.coxph mi
Hi everyone.
Based on a dependent variable (y), I'm trying to generate some independent
variables with a specified correlation. For this there's no problems.
However, I would like that have all my "regressors" to be orthogonal (i.e.
no correlation among them.
For example,
y = x1 + x2 + x3 where
Hello,
>
> What I want to do with each iteration of the loop I want to create another
> uniquely named
> groupedData object - I just don't know how to assign a unique variable
> name.
>
The best way should be to create a list of groupedData objects. Along the
lines of
library(nlme)
n <- 100
On 03/15/2012 09:51 AM, Alexander wrote:
Hi Martin,
thanks for your quick answer. I didn't know that '.' could be missleading.
Is there any standard way to name function for S4 objects? "get","set"
etc..?
Hi Alexander -- it's usually better to include the original email in the
reply, to provid
On Thu, Mar 15, 2012 at 12:58 PM, Rui Barradas wrote:
> Hello,
>
>>
>> Hi Emily,
>>
>> Yes (see below), but you might be better off by writing a simple
>> function. Here are examples both ways (usually eval parse is highly
>> discouraged).
>>
>> Cheers,
>>
>> Josh
>>
>
> Yes, eval/parse is discou
abline(h = 0) will do what you want. abline(h = c(0,1,2)) would draw
additional lines at y = 1 and y = 2. abline(v = 0) would draw a vertical
line at x = 0. All of these lines will be plotted on top of whatever you
plotted previously, so you may have to replot your points or lines so that
they're n
abline(h=0)
On Thu, Mar 15, 2012 at 1:05 PM, Michael wrote:
> How to add grid lines at specific position to a plot?
>
> Hi all,
>
> [Disclaimer: I have done extensive and intensive searching on Google and
> asked around but couldn't find a solution to this problem.
> Please help me instead of jus
Hi Martin,
thanks for your quick answer. I didn't know that '.' could be missleading.
Is there any standard way to name function for S4 objects? "get","set"
etc..?
I saw your example, and I was wondering, why get.par(ext) put out "Son1",
and not the same as get.par(new("Son1", name="Son1", par=3))
Dear R users,
I have problems handling missing values. THe problem is that after running my
codes the result I get just skips the missing values. What I want is for the
missing values to appear in my results as well. I have put a workable example
below and as you could see the missing value in
Hello,
>
> Hi Emily,
>
> Yes (see below), but you might be better off by writing a simple
> function. Here are examples both ways (usually eval parse is highly
> discouraged).
>
> Cheers,
>
> Josh
>
Yes, eval/parse is discouraged but there's a way of using it,
that is less troublesome, to c
How to add grid lines at specific position to a plot?
Hi all,
[Disclaimer: I have done extensive and intensive searching on Google and
asked around but couldn't find a solution to this problem.
Please help me instead of just pointing out how simple and stupid my
question is. God bless good folks!
Try this:
Go to your favorite CRAN mirror.
Click on "R Binaries" then "linux" then "ubuntu" and send the link
with directions to install current ubuntu binaries to your IS people.
Which, you'll note, is exactly what Jeff Newmiller suggested.
Sarah
On Thu, Mar 15, 2012 at 12:50 PM, Scott Raynaud
On Wed, Mar 14, 2012 at 9:47 PM, Alex Miller wrote:
> Dear R Users,
>
> I am trying to plot a matrix (a Digital Elevation Model) using wireframe
> [lattice] and color that matrix based on a separate/independent matrix of
> the same resolution
This makes no sense. the values in DEM are the z-coord
Hi Dorothea,
I've copied the ape maintainer on this, because I believe you may have
found a bug.
Your analysis works without warning in ape 2.8, but gives warnings in
the current ape_3.0.1.
But what makes me think that this is a problem with the new version is
that the example given in ?mantel.t
My IS people insist that the latest version of R avaialble via apt-get is
2.13.1. Anything later they claim will have to be compiled. True?
Will I have to compile every time I update R? Seems like a lot of work.
Surely there's a way around it.
- Original Message -
From: Jeff Newmiller
Hi Alexander --
On 03/15/2012 07:57 AM, Alexander wrote:
Hi,
I would like to create some S4 classes as follows
setClass("Father",representation(name="character"))
setClass("Son1",contains="Father",representation(par="numeric"))
setClass("Son2",contains="Father",representation(par="logical"))
S
Hi, okay thanks. I've attempted the commands you suggested (I have provided all
of the dput results - I am not sure how to provide a "smaller" example). I hope
this is ok.
> sessionInfo()
R version 2.14.2 (2012-02-29)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United
One possibility: update to ggplot2. The original ggplot isn't even on
CRAN any longer.
When I tried your example with ggplot2, the empty bar for 4 was
plotted as you'd expect.
Thanks for the small reproducible example.
Sarah
On Thu, Mar 15, 2012 at 11:47 AM, Bart6114 wrote:
> Hello,
>
> When p
Thank you.
Works perfectly
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https://stat.et
Great. Thanks very much for that... and for pointing out the obvious error!
Much appreciated.
Tom
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Sorry - I suspect this is a very basic query.
I have a data frame structured "flat" (i.e. separate observations for a given
subject ID on separate lines). Using the nlme library I create a groupedData
object suitable for further analysis thus:
dataset <- groupedData(VARIABLE ~ TIME|ID, data=sfa
Hello,
When plotting a barchart with ggplot it drops the levels of the factor for
which no counts are available.
For example:
library(ggplot)
mtcars$cyl<-factor(mtcars$cyl)
ggplot(mtcars[!mtcars$cyl==4,], aes(cyl))+geom_bar()
levels(mtcars[!mtcars$cyl==4,])
This shows my problem. Because no cou
Many thanks,
"parse" is exactly what I was looking for!!
Hannes
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On Mar 15, 2012, at 10:55 AM, capy_bara wrote:
Dear all,
I am confused about how to create an expression. I use a package
(rsbml)
which uses expressions and seems to make a difference if there is a
quote
around the expression or not.
For example, package works with expressions such as
e
You want to assign your call to boxplot as an object that contains the
plot information
set.seed(1)
b <- matrix(rgamma(100,(1:4)/2,.5),nc=4)
(bxp <- boxplot(b))
Now you can use the info in bxp for placement, e.g.:
text(1:length(bxp$names),bxp$stats[3,],round(bxp$stats[3,],2),pos=3)
By the way,
Hi,
You don't provide a reproducible example, but I suspect what you need is
parse():
> parse(text = "A + B")
expression(A + B)
Sarah
On Thu, Mar 15, 2012 at 10:55 AM, capy_bara wrote:
> Dear all,
>
> I am confused about how to create an expression. I use a package (rsbml)
> which uses express
Hi,
You didn't copy the main r-help list on this email, so I've added them back it.
Beyond that, I don't see anything grossly wrong. You don't provide
enough information to reproduce the problem: mantel.test() works with
the 7x7 parts you provided.
So you'll need to provide a small reproducible
My colleague asked me to thank those of you who responded.
Kind regards,
Sean
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Hi,
I would like to create some S4 classes as follows
setClass("Father",representation(name="character"))
setClass("Son1",contains="Father",representation(par="numeric"))
setClass("Son2",contains="Father",representation(par="logical"))
Son1<-new("Son1")
Son1@name<-"Son1"
Son1@par<-3
Son2<-new("S
Dear all,
I am confused about how to create an expression. I use a package (rsbml)
which uses expressions and seems to make a difference if there is a quote
around the expression or not.
For example, package works with expressions such as
> expression(A + B)
but not with
> expression("A + B")
I
On 15/03/2012 8:53 AM, mrzung wrote:
hi
I'm making some program and it need to be hidden.
it's not commercial purpose but it is educational,
so i do want to hide the code of function.
for example,
if i made following function:
a<-function(x){
y<-x^2
print(y)
}
i do not want someone to type
Sorry. I am still using the 9-11 March 2011 version of course2.pdf.
On Wed, Mar 14, 2012 at 5:52 PM, David Winsemius wrote:
>
> On Mar 14, 2012, at 4:09 PM, John Smith wrote:
>
> With most current version of R and RMS, the 4 curves are drew in
>> 4 separate panels. Can anyone show me how can I
On Mar 15, 2012, at 10:23 AM, MSousa wrote:
Sorry, I ll explain better.
For example in the next dataset,
x<-data.frame(v_source=3,v_destine=3,dist=4)
x<-rbind(x,data.frame(v_source=7,v_destine=7,dist=0))
x<-rbind(x,data.frame(v_source=6,v_destine=6,dist=0))
x<-rbind(x,data.frame(v_source=4
Hi,
I use R to plot a graph with 2 frames. Because the limit of pixel
or others, the screen fail to show the graph. So I have to draw a
point every 10 frames. So the total of x axis becomes 2000.
So when imaging the picture, I still hope that the total of x-axis is
2. I need to find a way
Sorry, I ll explain better.
For example in the next dataset,
x<-data.frame(v_source=3,v_destine=3,dist=4)
x<-rbind(x,data.frame(v_source=7,v_destine=7,dist=0))
x<-rbind(x,data.frame(v_source=6,v_destine=6,dist=0))
x<-rbind(x,data.frame(v_source=4,v_destine=4,dist=56))
x<-rbind(x,data.frame(v
hi
I'm making some program and it need to be hidden.
it's not commercial purpose but it is educational,
so i do want to hide the code of function.
for example,
if i made following function:
a<-function(x){
y<-x^2
print(y)
}
i do not want someone to type "a" and take the code of the function.
I obtain this message:
Error: C stack usage is too close to the limit
Marc Marí-Dell'Olmo
CIBER Epidemiología y Salud Pública
Servei de Sistemes d'Informació Sanitària (SeSIS)
Agència de Salut Pública de Barcelona
Pl. Lesseps 1. 08023 Barcelona
Tel. 93 2027775 | Fax. 93 3686943
www.aspb.cat
Hi All,
I am using the glmer function to adjust a simple model (mod1) with Poisson
response variable and random intercept. When I use logLik(mod1) I obtain a
value different from if I calculate the log-likelihood using the pdf.
Someone could explain if I have an error?
This is the code for the
Hello there,
I was wondering if anyone might be able to help me as I'm pretty new to R.
I'm trying to create a boxplot from a data table returned from an sproc. I
have the following code, which generates the plot as I'd like it:
library("RODBC");
conn <- odbcConnect("datawarehouse"); # connect to
Sorry, I ll explain better.
For example in the next dataset,
x<-data.frame(v_source=3,v_destine=3,dist=4)
x<-rbind(x,data.frame(v_source=7,v_destine=7,dist=0))
x<-rbind(x,data.frame(v_source=6,v_destine=6,dist=0))
x<-rbind(x,data.frame(v_source=4,v_destine=4,dist=56))
x<-rbind(x,data.frame(v
Hi,
I made a mistake, effectively Pascal your method given to the first message
is correct. I didn't use the good data, so it's sure that it could not work
!
But, I have an another question : Can I obtain a same graph except that the
y-axis represents the likelihood ratio ?
All the best,
Camill
hi ,I am trying to using wavelet package.
require(wavelets)
wt <- dwt(aa, n.levels=4, boundary="reflection", fast=TRUE)
plot(wt)
and we can get the figure as the attachment.
aa is a data array that came from the one day obvervation in 1HZ and has
the length of 86400 .
Can anyone tell me what i
Dear all,
I have data in the following format :
X-axisY-axis
<010%
0-20 20%
20-4030%
40-6040%
. and so on.
I want to plot a bar graph of the above. Also I would want to add a
trendline passing either through the
Thanks for the point in the right direction. I now have a great solution.
> library(GenomicRanges)
> system.time({
+ snplist<-with(snp, GRanges(CHR, IRanges(POS, POS)))
+ locations<-with(targets, GRanges(CHR, IRanges(START, STOP)))
+ olaps<-findOverlaps(snplist, locations)
+ })
user system el
On Mar 15, 2012, at 6:08 AM, Mauro Rossi wrote:
Dear R users,
first I take this opportunity to greet all the R community for
your continuous efforts.
I wrote a function to calculate the pdf and cdf of a custom
distribution (mixed gamma model).
The function is the following:
pmixedgamma
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