Hi Carlos:
I think you got a encoding problem.
Maybe is esier to convert it.
I don't know how to convert in Mac OS, but
in linux you can use "iconv" that converts many codes
to other.
Is the original file form a windos$ OS system?
Maybe the encoding is in windows-1256 and you need
to convert to
Hello,
In R under Mac OS X 10.5.4 I've had problems when I've tried to read a
data.frame with characters including tildes and accents.
For instance Floreña is changed to Flore\x96a and Ranchería is changed to
Rancher\x92a
In the code:
section<-read.table('Sectiondic.txt',sep='\t',header=T,str
dears R-users,
I'm interested in model selection problem, and i have faced some problems
that i would like to ask for help.
well,
this is a very small example with 4 variable (just one var. is the response
- z) with 100 individuals
i would like to do a stepwise search, for the "best" model, and a
Suppose I want to have a regexp match against a string, and return all
the matching substrings in a vector of strings.
regexp <- "[ab]+"
strlist <- c( "abc", "dbabddadd", "aaa" )
matches <- gregexpr(regexp,strlist)
With this input, I'd want to return list( list("ab"), list("ab", "a"),
li
Hi all,
I want to specify "arial" font in bitmap output but I found that the default
is "helvetica" and "arial" is not supported.
How can I add "arial" in my bitmap output file?
Thanks,
Hyunchul
[[alternative HTML version deleted]]
__
R-help
I'm trying to control the number of labels in the x-axis of my plot. My code
is the following:
graph1<-barplot(total_skew)
axis(1,at = graph1,labels=raw_date[1:length(total_skew)], las=2)
however, the length of my "total_skew" parameter is a few thousand elements,
and all the labels on the x-ax
Hi, all
I need to specify a font (for example, type=helvetica and size=10) in bitmap
output (for example, dev.print(bitmap, 'test.png').
How can I do this?
Thanks.
Hyunchul
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
Hi all,
I want to specify "arial" font in bitmap output but I found that the default
is "helvetica" and "arial" is not supported.
How can I add "arial" in my bitmap output file?
Thanks,
Hyunchul
[[alternative HTML version deleted]]
__
R-help
Hi, all
I need to specify a font (for example, type=helvetica and size=10) in bitmap
output (for example, dev.print(bitmap, 'test.png').
How can I do this?
Thanks.
Hyunchul
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
Try this:
regexp <- "[ab]+"
strlist <- c( "abc", "dbabddadd", "aaa" )
library(gsubfn)
s <- strapply(strlist, regexp)
s
# compactly show 1st few in ea component
str(s)
See gsubfn home page at
http://gsubfn.googlecode.com
On Sun, Aug 10, 2008 at 5:00 PM, Stavros Macrakis <[EMAIL
by the way if you are interested in beta testing R-PLus, email [EMAIL
PROTECTED]
--- On Sun, 8/10/08, S Ellison <[EMAIL PROTECTED]> wrote:
> From: S Ellison <[EMAIL PROTECTED]>
> Subject: Re: [R] RPro
> To: [EMAIL PROTECTED]
> Cc: r-help@r-project.org
> Date: Sunday, August 10, 2008, 12:53 P
by the way if you are interested in testing R-PLus, email [EMAIL PROTECTED]
--- On Sun, 8/10/08, S Ellison <[EMAIL PROTECTED]> wrote:
> From: S Ellison <[EMAIL PROTECTED]>
> Subject: Re: [R] RPro
> To: [EMAIL PROTECTED]
> Cc: r-help@r-project.org
> Date: Sunday, August 10, 2008, 12:53 PM
> >>
I am a beta tester for R-PLus from XLSolutions Corp and it's pretty cool. I
understand R-PLus will be free but let's see -
--- On Sun, 8/10/08, S Ellison <[EMAIL PROTECTED]> wrote:
> From: S Ellison <[EMAIL PROTECTED]>
> Subject: Re: [R] RPro
> To: [EMAIL PROTECTED]
> Cc: r-help@r-project.
Because of machine memory restrictions I think I need to go with a vector by
vector approach. When I concatenate I get:
> m <- melt(t, id.var=c("DayOfYear","Category","SubCategory","Sku"),
> measure.var=c("Quantity"))
Error: cannot allocate vector of size 7.8 Mb
Kevin
hadley wickham <[EMA
Hi,
I'm doing anova on a matrix of multivariate data where I want to assess the
effect of each column (element).
My matrix is 86 rows x 31 columns. I've created a grouping factor of length
86 containing group assignments of 6 types.
Then I run:
x<- aov(matrix~grouping.factor)
summary(aov.fit.r
I stand corrected. I thought I checked this a long time ago, but apparently
not. tsdiag.Arima DOES NOT use the fact that the series it is testing (or
diagnosing, if you will) are residuals from an ARIMA fit.
I keep a list of R time series bloopers here:
http://www.stat.pitt.edu/stoffer/tsa2/R
Dear List,
recently tried to reproduce the results of some custom model selection
function after updating R, which unfortunately failed. However, I
ultimately found the issue to be that testing with pchisq() in drop1()
seems to have changed. In the below example, earlier versions (e.g. R
2.4.1) pro
Hi
Peter Cowan wrote:
> Paul,
>
> The real case is a function where I'd want to allow end users to pass
> an arbitrary plotting function to be a sub plot within a larger plot
> (e.g. the code that does the density plots would actually be passed as
> a parameter. I adapted my example to reflect
Hi
Peter Cowan wrote:
> Paul,
>
> On Wed, Aug 6, 2008 at 1:40 PM, Paul Murrell <[EMAIL PROTECTED]> wrote:
>>> I'm trying to write a function that produces a main plotting region
>>> with several square plots along the right side. Ideally the size of
>>> right side plots will scale only with the
Hi
Peter Cowan wrote:
> Hello all,
>
> I'm trying to write a function using the gridBase package. I'd like
> to push several base subplots to a larger plot constructed with grid.
> However, I'm having trouble getting consistent results when running
> the function when the plotting window (quart
On Mon, Aug 11, 2008 at 08:23:19AM +1200, Gareth Campbell wrote:
> Hey team,
>
> If I have a matrix:
>
> 1, 2,
> 3, 4,
> 4, 0,
> 1, 3,
> 0, 3
>
> 2 columns.
>
> I want to write an if command that looks at (in this case) row 3 and looks
> to see if either [3,1] or [3,2] has a zero in it. IF it
I'm new to R and very excited about its possibilities. But I'm
struggling with some very simple things, probably because I haven't
found the correct documentation. Here's a simple example which
illustrates several of my problems.
Suppose I want to have a regexp match against a string, and return
Paul,
The real case is a function where I'd want to allow end users to pass
an arbitrary plotting function to be a sub plot within a larger plot
(e.g. the code that does the density plots would actually be passed as
a parameter. I adapted my example to reflect this. I expect most of
my end user
Hello all:
Is there a way to detect in R if a dataset is normally distributed or skewed
without graphically seeing it? The reason I want to be able to do this is
because I have developed and application with Visual Basic where Word,Access
and Excel "talk" to each other and I want to integrate R
Paul,
That is exactly what I was looking for. Thank you!
Peter
On Sun, Aug 10, 2008 at 1:41 PM, Paul Murrell <[EMAIL PROTECTED]> wrote:
> Hi
>
>
> Peter Cowan wrote:
>> Paul,
>>
>> On Wed, Aug 6, 2008 at 1:40 PM, Paul Murrell <[EMAIL PROTECTED]> wrote:
I'm trying to write a function that p
Hey team,
If I have a matrix:
1, 2,
3, 4,
4, 0,
1, 3,
0, 3
2 columns.
I want to write an if command that looks at (in this case) row 3 and looks
to see if either [3,1] or [3,2] has a zero in it. IF it does have a zero I
want the zero to be placed in another matrix in the same position. I know
Hi Kevin,
I think easiest way would be to create a single dataset with both
years in, and then work from that:
t2008$year <- 2008
t2007$year <- 2007
tall <- rbind(t2007, t2008)
mall <- melt(tall,
id.var=c("DayOfYear","Category","SubCategory","Sku", "year"),
measure.var=c("Quantity")
cast(mall,
This is to announce that we plan to release R version 2.7.2 on Monday,
August 25, 2008.
Release procedures start Friday August 15 (when we get back from useR).
The source tarballs will be made available daily (barring build
troubles) and the tarballs can be picked up at
http://cran.r-project.or
Dan,
Thank you very much for your comments and function definition code!! Responses
like yours make the r-help community amazing!!
Kurt
> Date: Sun, 10 Aug 2008 12:08:42 +0100
> From: [EMAIL PROTECTED]
> To: [EMAIL PROTECTED]
> CC: r-help@r-project
>>> Stephan Kolassa <[EMAIL PROTECTED]> 08/10/08 8:41 PM >>>
>I like RSeek:
>http://www.rseek.org/
The burning question is why an audio compay would be running a web
search engine for an open source statistics package...??!
***
This
Hi,
I like RSeek:
http://www.rseek.org/
And of course searching the R-help list archives, e.g., via
http://www.nabble.com/R-f13819.html
Good hunting!
Stephan
Carl Witthoft schrieb:
One thing I'll say: it's going to be much easier to Google for
references to "Rstat" than to "R" .
I've been t
e1 <- function(x,b,t){
d<-(x)*(b^t)
plot(d)
}
e1(2, 2,seq(from=0, to=6, by=1))
Is there a way to do this with a change in time. I would like to use
differential equations. I am trying to model
a population with an initial value, fecundity per time step, and a
death rate. The ab
The first three values are:
> filter <- c(0.1, 0.5, 1, 0.5)
> init <- 1:4
> filter %*% init + 1
[,1]
[1,] 7.1
> filter %*% c(7.1, init[1:3]) + 2
[,1]
[1,] 6.71
> filter %*% c(6.71, 7.1, init[1:2]) + 3
[,1]
[1,] 9.221
On Sun, Aug 10, 2008 at 12:47 PM, Sergey Goriatchev <[EMAIL PR
Thanks Dan. You did much more than just answer my question.
Sincerely,
Dan Davison wrote:
>
> On Sun, Aug 10, 2008 at 06:00:21PM +0100, Dan Davison wrote:
>> On Sun, Aug 10, 2008 at 09:02:59AM -0700, warthog29 wrote:
>> >
>> > Hi,
>> > I would like to use the R's outer function on y below s
Am 10.08.2008 um 18:53 schrieb stephen sefick:
the xaxt="n" needs to be inside of the plot command otherwise you can
never plot the x axis- I think.
Ah, ok, thanks!
Now it works fine.
Here a complete example of what I wanted to plot:
par(bty="n", mar=c(10, 10, 3, 3))
plot(c(1,2), c(50,60),
On Sun, Aug 10, 2008 at 06:00:21PM +0100, Dan Davison wrote:
> On Sun, Aug 10, 2008 at 09:02:59AM -0700, warthog29 wrote:
> >
> > Hi,
> > I would like to use the R's outer function on y below so that I can subtract
> > elements from each other. The resulting dataframe is symmetric, save for the
>
On 08/10/08 12:57, Carl Witthoft wrote:
> One thing I'll say: it's going to be much easier to Google for
> references to "Rstat" than to "R" .
> I've been tempted to start a movement to rename "R" something like
> "Ratistics" (or "RatStatPack" :-) ) just so it's locatable via search
> engines.
On Sun, Aug 10, 2008 at 09:02:59AM -0700, warthog29 wrote:
>
> Hi,
> I would like to use the R's outer function on y below so that I can subtract
> elements from each other. The resulting dataframe is symmetric, save for the
^^
outer() returns a matri
the xaxt="n" needs to be inside of the plot command otherwise you can
never plot the x axis- I think.
On Sun, Aug 10, 2008 at 12:51 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> j = c(5,6)
> f = c(1,2)
> plot(f,j, xaxt="n")
> axis(1, at=c(1,2), labels=c("group1", "group2"))
>
> On Sun, Aug 10, 2
One thing I'll say: it's going to be much easier to Google for
references to "Rstat" than to "R" .
I've been tempted to start a movement to rename "R" something like
"Ratistics" (or "RatStatPack" :-) ) just so it's locatable via search
engines.
That said, does anyone have any techniques for G
j = c(5,6)
f = c(1,2)
plot(f,j, xaxt="n")
axis(1, at=c(1,2), labels=c("group1", "group2"))
On Sun, Aug 10, 2008 at 12:09 PM, Jörg Groß <[EMAIL PROTECTED]> wrote:
>
> Am 10.08.2008 um 17:46 schrieb stephen sefick:
>
>> Would you provide reproducible code because form your example it looks
>> like y
Hello,
I thought I understood filter() with the help from Prof. Grothendieck,
but I guess I did not.
For example, how does this work:
filter(1:10, c(0.1, 0.5, 1, 0.5), "recursive", init=c(1,2,3,4))
Time Series:
Start = 1
End = 10
Frequency = 1
[1] 7.1 6.71000 9.22100 15.87710 21.458
Hi,
I would like to use the R's outer function on y below so that I can subtract
elements from each other. The resulting dataframe is symmetric, save for the
negative signs on the other half of the numbers. I would like to get only
half of the dataframe. Here is the code I wrote (it is returning o
Am 10.08.2008 um 17:46 schrieb stephen sefick:
Would you provide reproducible code because form your example it looks
like you are trying to label a scatterplot with just two lables - this
is your boxplot from yesterday? If so look into the ?boxplot and it
will tell you some answers
No, what
On 8/10/2008 11:35 AM, Angelo Scozzarella wrote:
Hi,
How can I make an ANOVA if I haven't got all data set but I know the
numbers of subjects for each group, the mean and di standard deviation
for each group?
See anova.mean() in the HH package:
http://finzi.psych.upenn.edu/R/library/HH/
This is what Rscript and R CMD BATCH are for.
Your problem is that you forgot to print the objects in your function:
auto-printing only occurs at the top level.
On Sun, 10 Aug 2008, Laura Poggio wrote:
Dear all,
I wrote a simple script in order to put together some functions and method
to be
For each time point is sums the value prior to it, the
value at the time point itself and the value at the next
time point. For the first timepoint there is no prior
value so its NA. For the second timepoint we have
1+2+3=6. For the third timepoint we have 2+3+4=9 and
so on.
> filter(1:10, c(
Would you provide reproducible code because form your example it looks
like you are trying to label a scatterplot with just two lables - this
is your boxplot from yesterday? If so look into the ?boxplot and it
will tell you some answers
On Sun, Aug 10, 2008 at 11:29 AM, Jörg Groß <[EMAIL PROTECTE
I don't know of a way to do it without the data
On Sun, Aug 10, 2008 at 11:35 AM, Angelo Scozzarella
<[EMAIL PROTECTED]> wrote:
> Hi,
>
> How can I make an ANOVA if I haven't got all data set but I know the
> numbers of subjects for each group, the mean and di standard deviation for
> each group
Hi,
How can I make an ANOVA if I haven't got all data set but I know the
numbers of subjects for each group, the mean and di standard deviation
for each group?
Thanks
Angelo Scozzarella
__
R-help@r-project.org mailing list
https://stat.ethz.c
Hello,
I cannot understand what filter() function in package stat is doing.
For example, what does filter(1:100, c(1,1,1)) mean?
Could someone please explain? Help file is not enough for me.
Thanks in advance.
Sergey
__
R-help@r-project.org mailing lis
Hi,
I want to customize the x-axis.
I tried that:
par(bty="n", xaxt="n")
so that the x-axis is supressed.
then I tried to plot two variables and add a customized x-axis:
plot(x, y)
axis(1, at=c(1,2), labels=c("group1", "group2"))
but the axis is not added to the plot.
I don't understand why..
I don't think the %in% works at all. It's non-standard and I think an
incorrect model specification. Here is an example where we can see that
the transpose of the model matrix for the random effects is different
when we compare what would be the same model if it "worked".
I don't think there is a
and I think the WPS to R bridge is doing quite well, and the SAS to R bridge
is likely to debut very very soon. Phil Rack from www.minequest.com is
leading this.
WPS is an ideal mix for R because it is great for data mining ,cleaning and
manipulation thus leaving R for the cleaned, data for the st
Dear all,
I wrote a simple script in order to put together some functions and method
to be executed on various files
I am trying to have to possibility to call the script changing few
parameters in order to use the different files.
I succeeded partly using the function method.
However in my script
Dear R users,
is there a way to plot "predicted" curves from a Cox model with a time
dependent covariates using "predict", in which other significant covariates
are included with specific values ?
I use cluster(id) in this time-dep. Cox model for rows belonging to the same
patient, and this
Try auto-printing, e.g.
three
[[1]]
a b
1 x
2 y
3 z
[[2]]
a b
4 q
5 r
6 s
or change print.data.frame *in the base namespace* by fixInNamespace.
When you said you 'reloaded that function' I suspect that in fact you
source()-d it into a different place, your workspace.
On Sun, 10 A
This should be an easy one, but I could not find the answer in the
obvious places.
one <- data.frame(a=c(1,2,3),b=c("x","y","z"))
two <- data.frame(a=c(4,5,6),b=c("q","r","s"))
> print(one)
a b
1 1 x
2 2 y
3 3 z
> print(one,row.names=F)
a b
1 x
2 y
3 z
So far, so good, but how do I do this
On Sat, Aug 09, 2008 at 08:53:00PM -0400, Kurt Newman wrote:
>
> Resending. Previous message was truncated. Sorry for possible confusion.
>
>
> > From: [EMAIL PROTECTED]
> > To: r-help@r-project.org
> > Date: Sat, 9 Aug 2008 18:25:47 -0400
> > Subject: [
On Sun, Aug 10, 2008 at 02:44:00PM +1200, Gareth Campbell wrote:
> I have a vector:
> alleles.present<-c("D3", "D16", ... )
>
> The alleles present changes given the case I'm dealing with - i.e. either
> all of the alleles I use for my calculations are present, or some of them.
>
> Depending on w
Patrick, you have misundertood me, I mean that Dan's solution (which is
also your solution) are both:
a) more clear and elegant
b) more time efficient. I've checked it with my working 1*3000 element
matrixes. The improvement in speed with your solution is evident.
I do not advise at all to use
Try with plotmeans in gplots
Regards,
Gianandrea
Jörg Groß wrote:
>
> Hi,
>
> I want to plot the mean of a variable and add the standard deviation
> as a line going above and below the mean (like the whiskers in a
> boxplot) but I don't know how to add these residual-lines.
>
> Is there a
That may be a better solution, but I don't think
it is clearly a better solution.
I presume you mean that your computation is the
most time efficient. That seems believable to me.
It is not the most human efficient -- it will take
some one reading the code non-trivial effort to
understand it.
W
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