Hi Scott,
I am not familiar with S-Plus (though many aspects are quite similar
to R). I will say that your function looks approximately correct. I
am not familiar with the ss.rand function. I searched, and found some
things that I suspect are similar in the packages MBESS, but without
knowing
On Tue, 4 Oct 2011, Sarah Goslee wrote:
You asked for pointers, and didn't provide a reproducible example, so
I
offered a pointer.
Sarah,
I did not realize that your pointer was to the factor component of
the
subset() command.
I think the most parsimonious thing for me
Hi bstudent,
I've had the same problem and I wish there was a definitive answer as this
seems to be the #1 problem with the package and pgmm would be awesome for
economists if we could figure out how to work it! I'm no expert on GMM, but
from what I've gathered from other posts, the problem may
?expand.grid
Am 05.10.2011 00:21, schrieb darkgaze:
I don't quite know how to word what I want, but if I have
(1, 2, 3); (a, b, c); (x, y)
I want:
1 a x
1 b x
1 c x
1 a y
1 b y
1 c y
2 a
...
and so forth
What is the appropriate command?
Best,
Don
--
View this message in context:
Dear all,
I think the coxph and survfit.coxph can give the cumulative hazard of cox
model.
But is there any method to calculate the hazard
Lambda(t)=lambda_0(t)*exp{beta*X(t)}?
Any suggestion will be great help.
Thank you very much!
Koshihaku
--
View this message in context:
Hi. I am trying to install using the following. Can someone suggest what is
wrong? I am using Windows 7 64bit, and R 2.10.1
('C:\Users\Bill\Desktop\DMwR_0.2.1.zip', repos=NULL )
Warning in install.packages(C:UsersBillDesktopDMwR_0.2.1.zip, repos = NULL) :
argument 'lib' is missing: using
my=function(x){
len=1
for(i in 1:len){
y[i]=x[i]
}
g=1
w=NULL
t=NULL
for(i in 1:len)w[i]=x[i+len]
for(i in 1:len)t[i]=x[i+2*len]
for(i in 1:len)g=g*dnorm(y[i])*dnorm(w[i])*dnorm(z[i])
return(g)
}
cuhre(6,1,my,rep(-100,6),rep(100,6))
Error in crff(match.call(), integrand, cuhre, libargs, ...) :
Dear all,
Your advices was a great help to my study.Thank you very much!
--
View this message in context:
http://r.789695.n4.nabble.com/Is-the-output-of-survfit-coxph-survival-or-baseline-survival-tp3861919p3873512.html
Sent from the R help mailing list archive at Nabble.com.
When R was invented, nearly all of the core R functions were
written to produce exactly the same answers as those returned by S-Plus.
Some very minor exceptions were made for time series functions, for
example, where better algorithms in R produced slightly better fits.
There may be
Hello,
i think you are speaking of a general workflow environment which can execute
R methods and arrange
different statistical methods in a flow (graph).
Here is a list of links of such (OpenSource) programms.
*Knime:*
http://www.knime.org/ http://www.knime.org/
*RapidMiner*
Dear William,
Please use a more informative subject line (as the posting guide asks you to
do).
Backslash have a special function. In file paths you need to use either double
backslashes or forward slashes.
'C:\\Users\\Bill\\Desktop\\DMwR_0.2.1.zip'
'C:/Users/Bill/Desktop/DMwR_0.2.1.zip'
Hi William
Try double backslashes \\ or / (see FAQ 2.16:
http://cran.r-project.org/bin/windows/base/rw-FAQ.html)
install.packages(C:\\Users\\rusa\\DMwR_0.2.1.zip, repos = NULL)
Regards,
Sina
--
View this message in context:
http://r.789695.n4.nabble.com/no-subject-tp3873600p3873738.html
Sent
On 09/22/2011 01:54 PM, Eugene Kanshin wrote:
Hello,
I need to convert dataframe from:
ID T0 T1 T2
A1 2 3
B4 5 6
C7 8 9
to:
ID Variable Value
A T0 1
A T1 2
A T2 3
B T0 4
B T1 5
B
I had thought that the problem might
be:
return(ne=ne1)
since R doesn't support that any more.
But when I tried it, I got results
(just without the name on the output).
Better would be to change that line to:
list(ne=ne1)
('return' is seldom necessary in either
R or S+.)
I'd suggest putting
On 10/05/2011 03:02 AM, William Claster wrote:
Hi. I am not that good at R but I was wondering if there is either a tool or
a strategy for testing many different models in R in a batch. I have used
something in Weka called the Experimenter interface which helps with doing
this kind of
Hi,
What happens when you update R under win7 to 2.12.1? And take a look at
the posting guide [1] for tips on what kind of information you need to
provide. Especially a sessionInfo() under both Mac and Windows would be
useful
cheers,
Paul
[1] http://www.R-project.org/posting-guide.html
On
On Wed, Oct 5, 2011 at 2:53 AM, Scott Raynaud scott.rayn...@yahoo.com wrote:
I'm trying to convert an S-Plus program to R. Since I'm a SAS programmer I'm
not facile is either S-Plus or R, so I need some help. All I did was convert
the underscores in S-Plus to the assignment operator -.
On 10/05/2011 04:27 AM, andrewH wrote:
Dear folks,
I’m trying to build a function to create and make available some variables I
frequently use for testing purposes. Suppose I have a function that takes
some inputs and creates (internally) several named objects. Say,
fun1 - function(x, y,
Hi Ista,
thanks for you reply.
If I understod correctly you run your R within Eclipse but as the Lunch Type
you use Rterm rather than RJ.
I changed my configuration so that R is now lunched as Rterm and NOT as RJ
and I also removed the
quite=FALSE from my configuration. Unfortunately, I still
Dear R Users,
at the moment I am trying to optimize an R script.
testvec - c(0,1,0,1,1,1,1,0,0,1,0,1,0)
sum.testvec - vector()
tempsum - 1
for (e in 1:length(testvec)){
sum.testvec[e] - tempsum+testvec[e]
tempsum - sum.testvec[e]
}
final.sum - c(1,sum.testvec)
Is there an option to do
You can vectorize it using cumsum.
cumsum(c(1, testvec))
all.equal(final.sum, cumsum(c(1, testvec)))
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Namens Chris82
Verzonden: woensdag 5 oktober 2011 11:50
Aan: r-help@r-project.org
Hi,
I am having troubles sourcing a file from our local network from R.
It looks like this file are not properly accessed by 'source', even they
can be downloaded with download.file. (See below my settings and some
tests I did). I ended up with a work around, but I would like to
understand
On 10/05/2011 09:52 AM, Heverkuhn Heverkuhn wrote:
Hello R users,
I have a plot type=b with x axis at=(1:36),
I would like to increase the distance between x tick-marks 8 and 9, and not
connect the points x=8 and x=9.
I can do the second thing, setting type=p and then drawing the lines, but
On Tue, 2011-10-04 at 23:41 -0400, Robert A'gata wrote:
AsOf(A,B) should return
A B
2011-09-0110 1.1
2011-09-0915 1.1 # (because latest value B prior to
2011-09-09 is 1.1)
2011-09-1020 1.5
2011-09-1525 1.7
How do I write
On Wed, 5 Oct 2011, Renaud Gaujoux wrote:
Hi,
I am having troubles sourcing a file from our local network from R.
It looks like this file are not properly accessed by 'source', even they can
be downloaded with download.file. (See below my settings and some tests I
did). I ended up with a
On Tue, 4 Oct 2011, Eduardo M. A. M. Mendes wrote:
Dear R-Users
I have come across the error that apparently has nothing to do with
command itself. Here is the error
(w - matrix (or vector) e testXaxis - dates).
plot(data.frame(testXaxis,w),col=blue,ylab=Q, [m3/s],xlab=Data,
+
From the help page ?file I -- had -- read the following:
For ‘url’ the description is a complete URL, including scheme
(such as ‘http://’, ‘ftp://’ or ‘file://’). Proxies can be
specified for HTTP and FTP ‘url’ connections: see ‘download.file’.
From the internet.info messages it seems that the
On Wed, Oct 05, 2011 at 12:44:12PM +0200, Renaud Gaujoux wrote:
Is source supposed to work through a proxy?
This worked for me:
Sys.setenv(http_proxy=http://192.168.0.252:8118;)
source(http://pc5.socio.gu.se:84/enkel-kurva.r;, echo = T)
my.vectory = c(1,30,2,3,3,4)
my.vectorx =
Dear All,
I have trouble generizising some code.
index - 0
sapply(list(c(1,2,3),c(1,2),c(1)),function(x){x[max(length(x)-index,0)]})
Will yield a wished for vector like so:
[1] 3 2 1
But in this case (trying to select te second to last element in each vector
of the list)
index - 1
dear all,
I want to make a dotplot with ratings from Items in 6 ItemsGroups.
I reordered the items by rating within each group.
I plotted the items by rating conditional on ItemGroup.
The ordering works as I wanted but my y-aches labels (items) within
each ItemGroup are now unequally spaced,
On Wed, 5 Oct 2011, Renaud Gaujoux wrote:
From the help page ?file I -- had -- read the following:
For ‘url’ the description is a complete URL, including scheme
(such as ‘http://’, ‘ftp://’ or ‘file://’). Proxies can be
specified for HTTP and FTP ‘url’ connections: see ‘download.file’.
So
Dear list memebers,
I am stuck with using regular expressions.
Imagine I have a vector of character strings like:
test - c('filename_1_def.pdf', 'filename_2_abc.pdf')
How could I use regexpressions to extract only the 'def'/'abc' parts of these
strings?
Some try from my side yielded no
Hello!
library(gsubfn)
test - c('filename_1_def.pdf', 'filename_2_abc.pdf')
gsubfn((.+_)([a-z]+)(\\.pdf), \\2, test)
Cheers!!
Albert-Jan
~~
All right, but apart from the sanitation, the medicine, education, wine, public
On 05/10/2011 13:45, Prof Brian Ripley wrote:
On Wed, 5 Oct 2011, Renaud Gaujoux wrote:
From the help page ?file I -- had -- read the following:
For ‘url’ the description is a complete URL, including scheme
(such as ‘http://’, ‘ftp://’ or ‘file://’). Proxies can be
specified for HTTP and FTP
On Wed, 5 Oct 2011, Renaud Gaujoux wrote:
On 05/10/2011 13:45, Prof Brian Ripley wrote:
On Wed, 5 Oct 2011, Renaud Gaujoux wrote:
From the help page ?file I -- had -- read the following:
For ‘url’ the description is a complete URL, including scheme
(such as ‘http://’, ‘ftp://’ or
I have this kind of matrix, with thousands of cases.
A 2 apple
A 2 peach
A 3 peach
B 1 pear
B 4 peach
B 4 beef
B 7 beef
C 1 peach
D 2 apple
D 5 peach
I have to distinguish, from the other rows, the rows with peach and this
is not a problem.
I also have to discriminate the rows with
I have a matrix like this
0.05 0.13 1.2 0 0 0 0 0 red
0 0 0 0 0 0 0 0 white
0 0.06 0 0 0 0 0 0 blue
If only 1 number in the first 8 columns is more than 0, in a new variable I
write 1, if they're all 0 or less, I write 0, so
0.05 0.13 1.2 0 0 0 0 0 red 1
0 0 0
I am trying to install or load pasilla package on R. i am getting the
following error. Please let me know how to install pasilla on R.
biocLite(pasilla)
Using R version 2.13.2, biocinstall version 2.8.4.
Installing Bioconductor version 2.8 packages:
[1] pasilla
Please wait...
Installing
Hello,
I wanted to parse some information from a text, where fields are tab
separated.
When I copy the text into an R session (under emacs) like:
mystring - field1 field2 field3
the tab character is replaced by a single space!
For ex, if I type mystring, I get:
field1 field2 field3
The tabs
Hi! Im new to R and I need to interpolate a shapefile using kriging. I've been
able to plot/read the shapefile using the package maptools or rgdal. I've
searched the internet for sample codes but most of the kriging codes that I've
found done in R is done using txtfiles or CSVs. An example
Hi there,
I am running simple univariate analyses via CAPER and when trying to obtain a
summary of my model using:
anova.pgls(model1)
It works for my analyses containing continuous explanatory variables but not
for those containing categorical explanatory variables (both have a
Hello,
I'd do:
ave(testvec, FUN=cumsum)+1
But in R everything can be done in a trillion different ways. ;-)
Cheers!!
Albert-Jan
~~
All right, but apart from the sanitation, the medicine, education, wine, public
order,
Hope I did this right. I repeated what I'd done before:
1) Opened script
2) Selected run all (this produced my inital post
Then as suggested I:
3) Typed ls()
4) Saw that the function was present and issued sshc(100,10)
Here's what I got:
ls()
[1] c.searchd convex Epower nef
I would appreciate help in knowing how to repeat categorical variable code
given in column=A, by the number in a matching column=B.
For example, I have a categorical variable code attributed to a household=A
and want to replicate the code for all member of the household, as given in
column=B. I
On Wed, 5 Oct 2011, Petr PIKAL wrote:
Hm. I seldom use such approach. In your original request you said you want
split your data to smaller data frames based on sites
Petr,
I need the additional information in the database, too.
From what we know it is difficult to say if there is some
Hello everyone,
I would like assistance with updating a snippet I have written to do a
recursive out-of-sample portfolio optimization.
The trouble I am having is with the fact that the return on the riskless asset
is time varying and so is different in each period.
This is what I have
Hi,
I have multiple three dimensional arrays.
Like this:
x1 - array(rnorm(1000, 1, 2), dim=c(10, 10, 10))
x2 - array(rnorm(1000, 1, 2), dim=c(10, 10, 10))
x3 - array(rnorm(1000, 1, 2), dim=c(10, 10, 10))
Now I would like to compute the mean for each corresponding cell.
As a result I want to
On 10/05/2011 07:44 AM, Scott Raynaud wrote:
Hope I did this right. I repeated what I'd done before:
1) Opened script 2) Selected run all (this produced my inital post
Then as suggested I:
3) Typed ls() 4) Saw that the function was present and issued
sshc(100,10)
Here's what I got:
ls()
Hi
I am testing a package, and after I make changes, I have to close R and open
R again to load the new version (same version number) of the package I am
working on. So my question:
is there a function which removes a package, i.e
library(myPackage)
Package is loaded
unlibrary(myPackage)
On 05.10.2011 13:44, Scott Raynaud wrote:
Hope I did this right. I repeated what I'd done before:
1) Opened script
2) Selected run all (this produced my inital post
Then as suggested I:
3) Typed ls()
4) Saw that the function was present and issued sshc(100,10)
Here's what I got:
ls()
I think we need more details about your setup. It works as I described
(prints errors then returns to R prompt) in all configurations I've
tried, including th Linux terminal, Emacs ESS, Eclispse statET console
setup, I couldn't get RJ workign).
What operating system are you using? R version?
On 10/05/2011 01:31 AM, sridhar wrote:
I am trying to install or load pasilla package on R. i am getting the
following error. Please let me know how to install pasilla on R.
biocLite(pasilla)
Using R version 2.13.2, biocinstall version 2.8.4.
Installing Bioconductor version 2.8 packages:
[1]
So source() always reads a URL using the internal method, because it
reads them chunk by chunk, and I suppose the other methods of
download.file (wget, etc...) do not support (?).
I guess the only way of finding out where the reading process gets stuck
is to get into the C code and add more
Hi Jannis,
just use the backreferences in gsub, see ?gsub, - replacement
test - c('filename_1_def.pdf', 'filename_2_abc.pdf')
gsub(.*_([A-z]+)\\.pdf, \\1, test)
hth.
Am 05.10.2011 13:56, schrieb Jannis:
Dear list memebers,
I am stuck with using regular expressions.
Imagine I have a
(x1+x2+x3)/3
I'm not aware of a pmean function but it wouldn't be hard to homebrew one if
you are comfortable with the ... argument
I'll draft one up and send it along
Michael Weylandt
On Oct 5, 2011, at 9:00 AM, Martin Batholdy batho...@googlemail.com wrote:
Hi,
I have multiple three
Hi Rainer,
for better or worse unlibrary actually is done by detach in R,
?detach
#first example
cheers
Am 05.10.2011 15:04, schrieb Rainer M Krug:
Hi
I am testing a package, and after I make changes, I have to close R and open
R again to load the new version (same version number) of the
Hi Rainer,
On Wed, Oct 5, 2011 at 9:04 AM, Rainer M Krug r.m.k...@gmail.com wrote:
Hi
I am testing a package, and after I make changes, I have to close R and open
R again to load the new version (same version number) of the package I am
working on. So my question:
is there a function which
On Wed, Oct 5, 2011 at 3:15 PM, Eik Vettorazzi e.vettora...@uke.de wrote:
Hi Rainer,
for better or worse unlibrary actually is done by detach in R,
?detach
#first example
cheers
ARG
I looked at detach, tried it, didn't work, because I overlooked the package:
part.
Thanks a lot,
On Wed, 5 Oct 2011, Rich Shepard wrote:
First thing this morning I'm upgrading to 2.13.2 and hoping that this
fixes an issue that just showed up yesterday afternoon: not being able to
access function help pages. For example, I tried ?subset and ?split because
I thought the latter is really
The problem with that function is that it does not really separate the
2parts of the graph but it inserts , when style is gap, a blank strip that
cover axis and points. So for example if a insert it at 8 and I set the gap
of length 5 , it would cancel al the point from 8 to 10.
Thanks
On Oct 5,
..all the point from 8 to 13.
On Wed, Oct 5, 2011 at 8:28 AM, Heverkuhn Heverkuhn heverk...@gmail.comwrote:
The problem with that function is that it does not really separate the
2parts of the graph but it inserts , when style is gap, a blank strip that
cover axis and points. So for example
As promised
### Untested
pmean - function(...){
dotArgs - list(...)
l - length(dotArgs)
if( l == 0L ) stop(no arguments)
temp - dotArgs[[1]]
if ( l 1L ) {for(i in 2L:l) {temp - temp + dotArgs[[i]]}}
temp/l
}
Clunky but gets the job done. Its still too early for me to
Hi
I would appreciate help in knowing how to repeat categorical variable
code
given in column=A, by the number in a matching column=B.
For example, I have a categorical variable code attributed to a
household=A
and want to replicate the code for all member of the household, as given
in
Hi
On Wed, 5 Oct 2011, Petr PIKAL wrote:
Hm. I seldom use such approach. In your original request you said you
want
split your data to smaller data frames based on sites
Petr,
I need the additional information in the database, too.
But you do not loose them, your data frame is
Hi Ista,
it's weird I don't know why this happens. I tried so many different ways but
now finally
I found a solution. I wrote a little shell script:
pdflatex -halt-on-error body.tex
bibtex body.aux
pdflatex -halt-on-error body.tex
pdflatex -halt-on-error body.tex
which does the job. So my
Hi Petr,
Thank you for the reply. Unfortunately my repetition is not uniform, but
dependent on values given in column B, which varies by each row.
Does this make it any clearer?
Not much.
let say
A - letters[1:10]
B - sample(1:3, 10, replace =TRUE)
A
[1] a b c d e f g h i j
B
On Wed, Oct 5, 2011 at 12:44 PM, Scott Raynaud scott.rayn...@yahoo.com wrote:
Hope I did this right. I repeated what I'd done before:
1) Opened script
2) Selected run all (this produced my inital post
Then as suggested I:
3) Typed ls()
4) Saw that the function was present and issued
On Wed, 5 Oct 2011, Petr PIKAL wrote:
But you do not loose them, your data frame is cut according to sites
variable and put into a list
I know this, Petr. But adding them to the database table ensures that the
information is there, too.
This brings up another question, but I should put
Dennis,
thanks for your help. I've read your email and the references you gave
and things are more clear to me.
Best,
Marcus
On Tue, Oct 4, 2011 at 19:28, Dennis Murphy djmu...@gmail.com wrote:
Hi:
INB4: if I have a nested design with treatment A and treatment B
within A, F-values are
On Wed, Oct 5, 2011 at 7:56 AM, Jannis bt_jan...@yahoo.de wrote:
Dear list memebers,
I am stuck with using regular expressions.
Imagine I have a vector of character strings like:
test - c('filename_1_def.pdf', 'filename_2_abc.pdf')
How could I use regexpressions to extract only the
It looks like this code was written for S+ 4.5 (aka '2000')
or before, which was based on S version 3. Try changing
return(name1=value1, name2=value2)
to
return(list(name1=value1, name2=value2))
In S+ from 5.0 onwards return(name=value) or return(name1=value1,
name2=value2) throws away the
Hi guys
I have vectors x - c(1,2,3,4) and y - c(4,3,9) and would like to generate a
matrix which has 3 rows (length(y)) and 4 columns (length(x)), and each row is
the corresponding y element repeated length(x) times.
4,4,4,4
3,3,3,3
9,9,9,9
Thanks.
Fernando Álvarez
mat - matrix(ncol = length(x), nrow = length(y))
for(i in 1:length(x)) { mat[,i] = y}
HTH,
Samuel
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of fernando.cabr...@nordea.com
Sent: 05 October 2011 17:11
To: r-help@r-project.org
matrix(rep(y, each=length(x)), nrow=length(y), byrow=TRUE)
or less explicitly
matrix(y, nrow=length(y),ncol=length(x))
Michael
On Wed, Oct 5, 2011 at 12:11 PM, fernando.cabr...@nordea.com wrote:
Hi guys
I have vectors x - c(1,2,3,4) and y - c(4,3,9) and would like to generate a
matrix
Hi,
I am still struggling with three dimensional arrays.
Now I would like to convert a three dimensional array into a data-frame with
the coordinate-columns: x, y, z and a value-column.
And I definitely don't want to loop over every element, since this would be
very resource intensive for
One more version: somewhere in the middle of the explicitness scale,
matrix(rep(y, times = length(x)), nrow=length(y))
On Wed, Oct 5, 2011 at 12:17 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
matrix(rep(y, each=length(x)), nrow=length(y), byrow=TRUE)
or less explicitly
reshape::melt does this I think
Michael
On Wed, Oct 5, 2011 at 12:20 PM, Martin Batholdy
batho...@googlemail.com wrote:
Hi,
I am still struggling with three dimensional arrays.
Now I would like to convert a three dimensional array into a data-frame with
the coordinate-columns: x, y, z
Hi:
There are a few ways to do this. If you only have a few arrays, you
can simply add them and divide by the number of arrays. If you have a
large number of such arrays, this is inconvenient, so an alternative
is to ship the arrays into a list and use the Reduce() function. For
your example,
L
I think you only have to change the multi-argument
returns to call list. You can remove the name from
the single argument return, as it will be ignore
return(name=value) - return(value)
return(n1=v1, n2=v2) - return(list(n1=v1, n2=v2))
(I say I think because I don't have easy access to
S+
m - matrix( rep( y, length( x ) ), length( y ), length( x ) )
On Wednesday 05 October 2011 18:11:18 fernando.cabr...@nordea.com wrote:
Hi guys
I have vectors x - c(1,2,3,4) and y - c(4,3,9) and would like to generate a
matrix which has 3 rows (length(y)) and 4
columns (length(x)), and
You can use [1] on the output of FUN to ensure that
exactly one value (perhaps NA from numeric(0)[1]) is
returned. E.g.
index - 1
sapply(list(c(1,2,3),c(1,2),c(1)),function(x){x[max(length(x)-index,0)][1]})
[1] 2 1 NA
I'll also put in a plug for vapply, which throws an
error if FUN
Hi,
I am not sure I understand your question.
Are you asking how to find the rows that satisfies the condition of the
second row?
Something like:
(let's say the data.frame is called X)
X[,1] == A X[,2] == 2 X[,3] == peach
?
Contact
On Sun, Jun 12, 2011 at 2:43 PM, bstudent marc.ruet...@gmx.de wrote:
Hello,
although I searched for a solution related to my problem I didn´t find one,
yet. My skills in R aren´t very large, however.
For my Diploma thesis I need to run a GMM estimation on a dynamic panel
model using the pgmm
On Tue, Oct 4, 2011 at 1:25 PM, Charles McClure cmccl...@atrcorp.com wrote:
I am new to R and have recently tried Tinn-R with very mixed and unexpected
results. Can you point me to a Tinn-R tutorial on the web or a decent
reference book?
In my experience, TINN-R does not work so well, and
Hi,
I have this sample-code (see above) and I was wondering wether it is possible
to speed things up.
What this code does is the following:
x is 4D array (you can imagine it as x, y, z-coordinates and a time-coordinate).
So x contains 50x50x50 data-arrays for 91 time-points.
Now I want to
Thanks for the response, Paul! But I thought these dumped the variables into
the global environment. Is that not correct? I want to make them available
in the calling environment, without making them available in the global
environment, unless that is where the function is called. This is my bow
*Paul,
I use Tinn-r and i didn't see anything hard to configure it. You have to
install the R, then the Tinn-r. After you have to open Tinn-r go to R
Configure Permanent . This procedure will open the Rprofile.site file, if
you want you can change some parameters and save. After that you have
Hello,
I want to plot one of the trees from a cforest object:
data(iris)
cf - cforest(Species, data=iris)
From the docs, cf@ensemble contains a list of BinaryTrees:
ensemble: Object of class list, each element being an object of class
BinaryTree ../../party/help/BinaryTree%2dclass. So I want to
Dear list,
I am unsure how to structure my model, i have tried something and it makes
sense but i am unsure if i am interpreting it correctly?
i have a continuous response variable - the observed quantity of evolutionary
history - EH
Then i have a number of species which have a hierarchical
On Wed, Oct 5, 2011 at 4:54 PM, Scott Raynaud scott.rayn...@yahoo.com wrote:
It seems I have things set up correctly. I suspect that the arguments
sshc(100,10) are the isuue. It seems that the 100,10 is not necessary since
the code itself specifies the arguments. It runs and produces a power
I took the original code, changed all return()
calls of the form return(n1=v1,n2=v2) to
return(list(n1=v1,n2=v2)) and then sshc(10,100)
chugged away and produced some plots and returned
something with no errors. It took a couple of minutes.
I also changed T-TRUE and F-FALSE, as that makes
the
I corrected your code a bit and put it into a function, f0, to
make testing easier. I also made a small dataset to make
testing easier. Then I made a new function f1 which does
what f0 does in a vectorized manner:
x - array(rnorm(50 * 50 * 50 * 91, 0, 2), dim=c(50, 50, 50, 91))
xsmall -
Hi everybody,
I used the krige.conv command (geoR package) to create a new data set. The
input was a matrix with three spatial coordinates (x, y, z) in the first
three columns and the value of a variable in the last column. The output
is... a weird sequence of numbers. How can I make this output
On Oct 5, 2011, at 8:14 AM, R. Michael Weylandt michael.weyla...@gmail.com
wrote:
(x1+x2+x3)/3
I'm not aware of a pmean function but it wouldn't be hard to
homebrew one if you are comfortable with the ... argument
I'll draft one up and send it along
pmean - function(lis)
Hi all,
I realise that the convention is to provide a working example of my problem
but the data are of a sensitive nature so I'm not able to do that in this
case.
I need to query a database for multiple search terms:
db - structure(list(ind = c(ind1, ind2, ind3, ind4), test1 = c(1,
2, 1.3,
Hi,
I want to ask which way is more effective to further analyse (multiple
comparisons) a mixed model repeated measures anova with 2 fixed factor and 1
random?
anova(lme(expr~treatment*age,random=~1|trial, data)
Is searching for an effect of one factor in each of the subsamples defined
by the
Hello,
I am looking for a code in R for the variance ratio test statistic (the
Lo and Mackinlay version or any other versions).
Does anybody have such a code they can share or know a library in which
I can find this function?
Basically I have a number of time series which I need to check for
dear R-Community
is there a function which sums data stepwise
exp:
2
1
4
5
Desired result
2 = 2
2+1 = 3
2+1+4 = 7
2+1+4+5 = 12
Is there a built in function for this?
Thx
Dom
--
View this message in context:
http://r.789695.n4.nabble.com/stepwise-sum-tp3874606p3874606.html
Sent from the
Dear all,
I’m analyzing this dataset containing biodiversity indices, measured over
time (Week), and at various contaminant concentrations (Treatment). We have
two replicates (Replicate) per treatment.
I’m looking for the effects of time (Week) and contaminant concentration
(Treatment) on
Ok, I chased down all the problems. This is my last output:
sshc(100,10)
[1] 0.8000 0.7908 0.7844 0.7773 0.7785 0.7989
[1] 5.37 10.29 13.27 13.04 9.66 3.54
[1] old.abs.dev= 0.0701944484789673
[1] abs.dev= 0.034407699378335
[1] 0.8000 0.8030 0.8057 0.8041 0.8035 0.8180
[1] 5.37 10.87
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