Hi all,
I have a problem with CPU usage while running the Rgui.exe
problem is
while I am running scripts on Rgui its taking 100% of CPU is there any
posibility to reduce the cpu consumption
are any package I can use to reduce CPU consumption
can any one help me out from this problem because
whi
Hi Doug,
you can add the fitted curve using the following general paradigm:
## Plotting the data
plot(p~z)
## Defining grid of z values
## (100 values ensures a smooth curve in your case)
zValues <- seq(min(z), max(z), length.out = 100)
## Adding predicted values corresponding to the grid val
How to determine restricted variable in SVAR and SVEC? There are some values
which set to be zero and others set to be NA.. How to determine values that set
to be 0? Thanks
Regards,
Arif
_
How to determine restricted variable in SVAR and SVEC? There are some values
which set to be zero and others set to be NA.. How to determine values that set
to be 0? Thanks
Regards,
Arif
_
k
On 10/09/2009 12:24 AM, Michael Friendly wrote:
est a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 a13 a14 a15
13 23 NA NA NA NA NA NA NA NA NA NA NA NA
24 NA 65 NA NA NA NA NA NA NA NA NA NA NA
35 NA NA 102 7 NA NA NA NA NA NA NA NA NA
46 NA NA 4 120 18 NA NA
I used *lapply* and *for* to call the function *GetC* respectively. The
output is different!!! I can't figured out how this happen! It's so wierd.
> x=lapply(ipi.go,GetC, ont = "BP", org=cell.cycle)
> unlist(x)
*IPI00011654 IPI00013683
"cell cycle" "cell cycle"*
> categorate=list()
> for (i in 1
HI there. nice discussion. I'm the developer for NppToR. First NppToR
should not make any changes to your workflow in Notepad++. It should be as
unobtrusive as possible. In fact I use it with several languages daily with
no problems. If is does get in the way I need to know about it, and please
Dear R list,
The code below puts qq-plots for two of three groups on the one
plot. However the legend includes all three groups, ie the auto.key
ignores the subset instruction. Is there an easy way to get around
this, so that only those groups plotted are included in the legend?
y1<-rnorm(
I have done some more tinkering this morning. It seems it's the first polar
plot that has the unusual size. If I add a third plot the last two grids
match ok. So I added a further kludge to solve my problem, I added a dummy
plot before the two 'real' plots and this works ok:
polar.plot(0,0,labels
Unfortunately, I need the ability to switch between languages within one editor
and I like the flexibility and simplicity offered via the NotePad++ to support
other key languages within my current workflow. Thus, I am attempting to limit
editor switching and build up skills within one editor.
Jason,
if you install NppToR, it generates the XML file for you, I think.
That is one way to get it.
Or, you could use Npptor as is, it not only gives R syntax
highlighting, but can also send commands to the Rgui (simply by
pressing F8). I have been using it for quite a while and I really like
it
Thank you very much for the reply, but this doesn't appear to be the XML file
to load into NotePad++ to use as a User Defined Language.
It appears to be an stand-alone executable that interacts with NotePad++ and
the R GUI.
I guess I will keep looking for a bit and maybe breakdown and build t
This may be a bug in objects of class "instantiatedProtoMethod". I
tried it with the devel version of proto and got no error so you could
try that. I will send it to you offline for your try.
On Thu, Oct 8, 2009 at 1:02 PM, baptiste auguie
wrote:
> Dear all,
>
> In mucking around with ggplot2,
Thanks for the reply Jim,
I didn't think of that but polar.plot seems to have a couple of quirks. The
first is if I try your idea using c(rep(maxlength,40)) as the lengths it
does not work, the grid is not even displayed, no error info is given. I
solved that by adding an additional point to the v
Douglas M. Hultstrand wrote:
Hello,
I have a non-linear function (exponential) that I am trying to display
the line with the data in a plot, is there a command similar to abline()
for the function I created, if not what is the best way to display the
fitted line. Example code below.
###
Try setting lty with par.settings:
xyplot(v3~v2,data=xx,
xlab='',
# group=v1,type='l', lty=1:3 ,
groups=v1, type='l',
par.settings = list(superpose.line = list(lty = 1:3)),
auto.key = list(text=levels(as.factor(xx$v1)) , title='',
cex.title=.75,points=F,lines = T,cex=1,
Danielle Dandreaux wrote:
I have been using R the past couple of years to run models on data we
are collecting. I recently got a new computer and updated to a new
version of R (2.60 -> 2.90). Since the update, I cannot get my syntax to
run. I have tried copying the file it is looking for into man
Try this:
RSiteSearch('dbf")
On Thu, Oct 8, 2009 at 4:17 PM, Michael Yutzi wrote:
>
> I have a heavy DATA saved in dbf format.
>
> What I want is to bring that data to R with SQL statements. Like: I want
> columns 1, 4, 5 and only when column 4 > 30.
>
> Sorry asking it here instead of keep sear
I am trying to recreate the analysis of the recent Kirwan et al paper in
Ecology [9(80), 2032-2032]. The SAS code is available but am having
troubles getting the R equivalents.
The SAS code is:
*Model 5 - Species-specific interaction coefficients;
PROC GLM; where N=150;
MODEL YIELD=G1 G2
Hi,
I know what the following correlation methods ("pearson", "kendall",
"spearman") are through wiki. But I'm wondering if there is a good
book that discuss why these metrics are defined and when to use one
coefficient versus the others.
cor.test(x, y,
alternative = c("two.sided", "less
The problem is your loop condition.
For example, when i reaches (n-3), a[i + 5, ] is out of bounds.
-k
On Thu, Oct 8, 2009 at 4:14 PM, Ashta wrote:
> Hi all,
> I have a matrix named x with N by C
> I want to select every 5 th rrow from matrix x
> I used the following code
> n<- nrow(x)
> >
I see !
Thank you everyone for the responses.
-k
On Thu, Oct 8, 2009 at 4:55 PM, David Scott wrote:
> Henrique Dallazuanna wrote:
>
>> Change the breaks argument:
>>
>> t1 <- hist(1:5, 0:5)
>> t1$counts
>>
>> On Thu, Oct 8, 2009 at 4:47 PM, Khanh Nguyen wrote:
>>
>>> Hi all,
>>>
>>> I have
Henrique Dallazuanna wrote:
Change the breaks argument:
t1 <- hist(1:5, 0:5)
t1$counts
On Thu, Oct 8, 2009 at 4:47 PM, Khanh Nguyen wrote:
Hi all,
I have a question about hist()
1)
t1 <- hist(c(1,2,3,4,5))
t1
$breaks
[1] 1 2 3 4 5
$counts
[1] 2 1 1 1
why is there 2 counts for 1? And sho
On Oct 8, 2009, at 4:40 PM, Marc Carpentier wrote:
Hello.
I'd like to plot only one component smooth function of a gam model
(for example the second) (library mgcv). So, I did :
plot(my.gam, select=2, ask=F)
But plot.gam doesn't seem to understand the "ask" option, so I can't
deactivate th
Hello.
I'd like to plot only one component smooth function of a gam model (for example
the second) (library mgcv). So, I did :
plot(my.gam, select=2, ask=F)
But plot.gam doesn't seem to understand the "ask" option, so I can't deactivate
the interactive plotting. I tried and failed by forcing it t
Bill:
Defensive programming seems to me to be a wise policy, so thanks for the
helpful tip.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: Thursday, October 08, 2009 12:01 PM
To: Bert Gunter; Erik Ive
On Oct 8, 2009, at 4:18 PM, David Winsemius wrote:
On Oct 8, 2009, at 4:14 PM, Ashta wrote:
Hi all,
I have a matrix named x with N by C
I want to select every 5 th rrow from matrix x
I used the following code
n<- nrow(x)
for(i in 1: n){
+ b <- a[i+5,]
b
}
Error: subscript out of bounds
I've been happily building a file with odfWeave, and just as the hour
draws nigh for it to be sent off, odfWeave or XML throws the following
catastrophic error:
...this is the tail of entirely uneventful processing of input file.
31 : term xml(label=LR_Fall_Model_Results)
32 : term verb
> x[seq(1, nrow(x), by=5), ]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of David Winsemius
Sent: Thursday, October 08, 2009 4:19 PM
To: Ashta
Cc: R help
Subject: Re: [R] row selection
On Oct 8, 2009, at 4:14 PM, Ashta wrote:
>
Hello,
Im trying to get a Oracle database to R, but I'm having problems with most
of SQL sintax. Suppose that Im trying to bring a table "TABLE" with columns
"COL1","COL2",.. from schema "SCHEMA". What am I doing wrong?
library(RODBC)
channel<-odbcConnect("xxx",uid="xxx",pwd="xxx")
sqlQuery(
I have a heavy DATA saved in dbf format.
What I want is to bring that data to R with SQL statements. Like: I want
columns 1, 4, 5 and only when column 4 > 30.
Sorry asking it here instead of keep searching in manuals, but it seems that
there are too many ways of doing it. So what's the appropri
Doran, Harold wrote:
>
> I am trying to use the adapt function in the package adapt. To make sure
> I am using it correctly, I am trying a toy example that should yield a
> result of 2/3.
>
> Suppose the function is f(x,y) = x*y^2 and I want to integrate over f as
>
> Int_0^1 Int_0^2 x*y^2 dx
On Oct 8, 2009, at 4:14 PM, Ashta wrote:
Hi all,
I have a matrix named x with N by C
I want to select every 5 th rrow from matrix x
I used the following code
n<- nrow(x)
for(i in 1: n){
+ b <- a[i+5,]
b
}
Error: subscript out of bounds
What did you expect when "i" in your loop counter b
On Oct 8, 2009, at 3:28 PM, Danielle Dandreaux wrote:
I have been using R the past couple of years to run models on data we
are collecting. I recently got a new computer and updated to a new
version of R (2.60 -> 2.90). Since the update, I cannot get my
syntax to
run. I have tried copying the
Hi all,
I have a matrix named x with N by C
I want to select every 5 th rrow from matrix x
I used the following code
n<- nrow(x)
> for(i in 1: n){
+ b <- a[i+5,]
>b
}
Error: subscript out of bounds
Can any body point out the problem?
[[alternative HTML version deleted]]
__
Or the right argument:
t1 <- hist(1:5, right=false)
t1[1:2]
$breaks
[1] 1 2 3 4 5
$counts
[1] 1 1 1 2
The issue has to do with what hist() does with the points that fall
right on the break points.
-Ista
On Thu, Oct 8, 2009 at 3:58 PM, Henrique Dallazuanna wrote:
> Change the breaks argument:
On 9/10/2009, at 8:47 AM, Khanh Nguyen wrote:
Hi all,
I have a question about hist()
1)
t1 <- hist(c(1,2,3,4,5))
t1
$breaks
[1] 1 2 3 4 5
$counts
[1] 2 1 1 1
why is there 2 counts for 1?
RTFM --- include.lowest = TRUE and right = TRUE by default.
And should the counts be '1 1
On Oct 8, 2009, at 3:47 PM, Khanh Nguyen wrote:
Hi all,
I have a question about hist()
1)
t1 <- hist(c(1,2,3,4,5))
t1
$breaks
[1] 1 2 3 4 5
$counts
[1] 2 1 1 1
why is there 2 counts for 1? And should the counts be '1 1 1 1 1' ?
It depends on the way the bins are defined.
Had you forced
On Oct 8, 2009, at 3:39 PM, Douglas M. Hultstrand wrote:
Hello,
I have a non-linear function (exponential) that I am trying to
display the line with the data in a plot, is there a command similar
to abline() for the function I created, if not what is the best way
to display the fitted li
Change the breaks argument:
t1 <- hist(1:5, 0:5)
t1$counts
On Thu, Oct 8, 2009 at 4:47 PM, Khanh Nguyen wrote:
> Hi all,
>
> I have a question about hist()
>
> 1)
>> t1 <- hist(c(1,2,3,4,5))
>> t1
> $breaks
> [1] 1 2 3 4 5
>
> $counts
> [1] 2 1 1 1
>
> why is there 2 counts for 1? And should the
On 9/10/2009, at 6:52 AM, Folkes, Michael wrote:
Hello all,
In order to get the desired order of panels within lattice I
realized I needed to define my factor levels.
I now see how to properly do it (attached code), but:
1. I don't understand why I have to use factor() on a variable that
i
For simply counting frequency of discrete data, table , xtabs, and ftable will
work.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Khanh Nguyen
> Sent: Thursday, October 08, 2009 2:47 PM
> To: r-help@r-project.org
> Subjec
Hi all,
I have a question about hist()
1)
> t1 <- hist(c(1,2,3,4,5))
> t1
$breaks
[1] 1 2 3 4 5
$counts
[1] 2 1 1 1
why is there 2 counts for 1? And should the counts be '1 1 1 1 1' ?
Is there any other function to count frequency of discrete data?
Thanks.
-k
[[alternative HTML vers
Hello,
I have a non-linear function (exponential) that I am trying to display
the line with the data in a plot, is there a command similar to abline()
for the function I created, if not what is the best way to display the
fitted line. Example code below.
### Example
z <- c(1,10,15,20,25,30,
I have been using R the past couple of years to run models on data we
are collecting. I recently got a new computer and updated to a new
version of R (2.60 -> 2.90). Since the update, I cannot get my syntax to
run. I have tried copying the file it is looking for into many different
directories to t
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of William Dunlap
> Sent: Thursday, October 08, 2009 12:01 PM
> To: Bert Gunter; Erik Iverson; Kajan Saied; r-help@r-project.org
> Subject: Re: [R] sort (all columns of) a matrix
..
Dear R help lists,
Hello!
I ran WinBUGS under R using the function bugs, but I kept getting the following
error message.
Error in FUN(X[[3L]], ...) :
.C(..): 'type' must be "real" for this format
I can not find the reason.
I am freaking out.
Could any one have a look at the model codes
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Bert Gunter
> Sent: Thursday, October 08, 2009 11:41 AM
> To: 'Erik Iverson'; 'Kajan Saied'; r-help@r-project.org
> Subject: Re: [R] sort (all columns of) a matrix
>
> Right. My
Right. My guess is that Kajan wants:
a[do.call(order,data.frame(a)),]
## this generalizes to an arbitrary number of columns
## do.call() is a very powerful and useful R feature worth learning about
Yet another reason why the posting guide asks for a simple, proper,
reproducible example.
-- Be
I assume you want to sort the whole dataset (matrix) by ordering one column
in ascending order (and order all other columns appropriately).
a<-matrix(a<-c(1,3,4,6,6,4,6,56,4,64,86,39,4,2),length(a),2)
a
#sort by first column
a[order(a[,1]),]
#sort by second column
a[order(a[,2]),]
#both give t
Try this:
a[order(a[,1], a[,2]),]
On Thu, Oct 8, 2009 at 1:50 PM, Kajan Saied wrote:
> Dear R-Help Team,
>
> I have been trying to sort (all columns of) a matrix:
>
>> a<-matrix(a<-c(1,3,4,6,6,4,6,56,4,64,86,39,4,2),length(a),2)
>> a
> [,1] [,2]
> [1,] 1 1
> [2,] 3 3
> [3,]
You need to define "sorted" in the context of a matrix. By each row
individually, by each column individually, or by column using the whole matrix,
or by row using the whole matrix?
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Beh
Untested, but
a.sorted <- apply(a,sort,2)
?
Kajan Saied wrote:
Dear R-Help Team,
I have been trying to sort (all columns of) a matrix:
a<-matrix(a<-c(1,3,4,6,6,4,6,56,4,64,86,39,4,2),length(a),2)
a
[,1] [,2]
[1,]11
[2,]33
[3,]44
[4,]66
[5,]6
Harold,
Try this:
ff <- function(x) x[1]*x[2]^2
adapt(2, lo = c(0,0), up = c(1,2), fun = ff)
The answer should be 4/3, since the first variable is integrated from 0 to 1
and the second variable (second-degree) is integrated from 0 to 2.
The answer is 1/3, if you do,
adapt(2, lo = c(0,0), u
Dear R-Help Team,
I have been trying to sort (all columns of) a matrix:
> a<-matrix(a<-c(1,3,4,6,6,4,6,56,4,64,86,39,4,2),length(a),2)
> a
[,1] [,2]
[1,]11
[2,]33
[3,]44
[4,]66
[5,]66
[6,]44
[7,]66
[8,] 56 56
[9,]4
devol wrote:
>
> Dear all,
>
> could you please advice whether it is possible somehow to modify an
> external (from the point of some function view) variable by some
> function-internal operators. For example
>
>> var=1
>> foo<-function(var){var=var+1}
>> foo(var)
>> var
> [1] 1
>
> but th
Bert Gunter wrote:
If I understand your intent, I believe you can get what you want much faster
(no interpreted loops and linear times) by looking at this slightly
differently.
First of all, the choice of columns is unimportant, as indexing can be used
to create a data frame containing only the
Hi Ben,
It looks like you're getting into trouble because you've
converted the body of a for-loop into a function.
You don't need to do that with foreach, and some of
your problems will be solved by reverting to something
more like your original for-loop, I suspect.
However, it also looks like yo
Hello all,
In order to get the desired order of panels within lattice I realized I needed
to define my factor levels.
I now see how to properly do it (attached code), but:
1. I don't understand why I have to use factor() on a variable that is already
a factor, and why levels() alone doesn't work
hi all,
It's not clear to me how (or if) I can pass multiple values for lty to a key in
xyplot?
I've tried: lines=list(lty=1:3), to no avail.
Do I need to use something other than auto.key?
(Deepayan, if you're out there, I have your book and must admit the answer
isn't jumping out at me.)
thank
Thanks all of you for your help!!!
2009/10/8 Bert Gunter :
> I am uncertain what you mean. Possibly...
>
> x <- 1
>> f <- function()x<<-2
>> f()
>> x
> [1] 2
>
> HOWEVER, this is very dangerous and most unwise as it depends on scoping
> rules and where f is called to determine exactly which "x" is
Bert, Jim, Dimitris and Joris,
Thank you all very much for your prompt help and suggestions.
After trying the ideas out, I have decided to go with Bert's approach
since it is by far the fastest of the lot.
Thanks again!
Rama Ramakrishnan
On Oct 8, 2009, at 12:49 PM, Bert Gunter wrote:
I am uncertain what you mean. Possibly...
x <- 1
> f <- function()x<<-2
> f()
> x
[1] 2
HOWEVER, this is very dangerous and most unwise as it depends on scoping
rules and where f is called to determine exactly which "x" is being assigned
the value of 2. So I think you would do well to reconsider
On Oct 8, 2009, at 1:14 PM, devol wrote:
Dear all,
could you please advice whether it is possible somehow to modify an
external (from the point of some function view) variable by some
function-internal operators. For example
var=1
foo<-function(var){var=var+1}
foo(var)
var
[1] 1
Use fun
On Oct 8, 2009, at 1:14 PM, Peng Yu wrote:
On Thu, Oct 8, 2009 at 10:57 AM, Steve Lianoglou
wrote:
Hi Peng,
On Oct 8, 2009, at 11:21 AM, Peng Yu wrote:
Hi,
I installed R on mac (see below). When I start it from a terminal by
the following command, the current working directory is always '
On Oct 8, 2009, at 1:13 PM, Albyn Jones wrote:
Quoting David Winsemius :
In insurance situation there is typically a cap on the covered
losses and there is also typically an amount below which it would
not make sense to offer a policy. So a minimum and a maximum are
sensible assumptions
Hi,
with assign,
foo <- function(var){
assign("var", var+1, envir = .GlobalEnv)
}
var =1
foo(2)
var
# [1] 3
HTH,
baptiste
2009/10/8 devol :
>
> Dear all,
>
> could you please advice whether it is possible somehow to modify an
> external (from the point of some function view) variable by
Is it possible to do what you mentioned somehow outside of the
function. I mean that there's some function and the only thing I can
do is to point on the variable to be modified inside the function
without any possibilities to modify the very function.
2009/10/8 Henrique Dallazuanna :
> See assign
See assign, you can use '<<-' assignment:
foo <- function(var) var <<- var + 1
On Thu, Oct 8, 2009 at 2:14 PM, devol wrote:
>
> Dear all,
>
> could you please advice whether it is possible somehow to modify an
> external (from the point of some function view) variable by some
> function-interna
... without answering my previous question, I have just found a
fortune-hate workaround,
getIcon <- function(geom){
st <- paste("Geom", firstUpper(geom),"$icon", sep="")
eval(parse(text=st))
}
getIcon("abline")
I'm still curious about the get() behaviour though.
Best,
baptiste
__
On Thu, Oct 8, 2009 at 10:57 AM, Steve Lianoglou
wrote:
> Hi Peng,
>
> On Oct 8, 2009, at 11:21 AM, Peng Yu wrote:
>
>> Hi,
>>
>> I installed R on mac (see below). When I start it from a terminal by
>> the following command, the current working directory is always '~' no
>> matter where my current
Dear all,
could you please advice whether it is possible somehow to modify an
external (from the point of some function view) variable by some
function-internal operators. For example
> var=1
> foo<-function(var){var=var+1}
> foo(var)
> var
[1] 1
but the goal is to get the var equal to 2 in t
I am trying to use the adapt function in the package adapt. To make sure
I am using it correctly, I am trying a toy example that should yield a
result of 2/3.
Suppose the function is f(x,y) = x*y^2 and I want to integrate over f as
Int_0^1 Int_0^2 x*y^2 dxdy
Where the limits of integration for y
Quoting David Winsemius :
In insurance situation there is typically a cap on the covered
losses and there is also typically an amount below which it would
not make sense to offer a policy. So a minimum and a maximum are
sensible assumptions about loss distributions in may real modeling
s
Dear all,
In mucking around with ggplot2, I've hit the following snag,
library(ggplot2)
# this returns a grob, OK
GeomAbline$icon()
# lines[GRID.lines.9]
# this returns the function icon, OK
GeomAbline$icon
# proto method (instantiated with ): function (.)
# linesGrob(c(0, 1), c(0.2, 0.8))
#
On Oct 8, 2009, at 12:53 PM, Albyn Jones wrote:
Maithili
I find it really hard to believe that a beta distribution would be a
reasonable probability model for loss data. There would have to be
an upper bound on the size of losses. What is the process that
generates the data. Is there
On Oct 8, 2009, at 11:59 AM, Peng Yu wrote:
Hi,
My Mac OS is Leopard. I downloaded
R-GUI-5496-2.9-leopard-Leopard64.dmg from http://r.research.att.com/.
I run R from the image disk, but it always "quit unexpectedly".
It is hard to say. If you were really trying to run R from the image,
the
Maithili
I find it really hard to believe that a beta distribution would be a
reasonable probability model for loss data. There would have to be an
upper bound on the size of losses. What is the process that
generates the data. Is there any natural upper bound? Why is there a
lower b
If I understand your intent, I believe you can get what you want much faster
(no interpreted loops and linear times) by looking at this slightly
differently.
First of all, the choice of columns is unimportant, as indexing can be used
to create a data frame containing only the columns of interest.
I'd like to generate on a single device multiple plots, each of which
contains two plots. Essentially, I've got sub-plots which consist of two
tracks, the upper one displaying gene expression data, and the lower one
showing gene position along the chromosome. I'd like to display four of
these sub-
Apologies for my overhasty and thus inaccurate reply. I of course assumed that
model2 is a submodel of model1. But thanks to Giovanni: a closer look at your
equations shows that this is only the case if at == bt (and of course same
innovations/ disturbances et, etat) and xt e.g. something like x
Ok, nevermind.
all I had to do was unlist(myList).
delnatan wrote:
>
> Hi ,
>
> I want to convert a my list:
>
>> myList
> [[1]]
> [1] 1 2 3 4 5
> [[2]]
> [1] 6 7 8
> [[3]]
> [1] 9 10 11
>
> into something like c(1,2,3,4,5,6,7,8,9,10,11)
>
> I realized that this is possible by
>
>> c(d
If you want do.call, you can use c indeed of cbind:
do.call(c, list(1:5, 6:8, 9:11))
On Thu, Oct 8, 2009 at 12:13 PM, delnatan wrote:
>
> Hi ,
>
> I want to convert a my list:
>
>> myList
> [[1]]
> [1] 1 2 3 4 5
> [[2]]
> [1] 6 7 8
> [[3]]
> [1] 9 10 11
>
> into something like c(1,2,3,4,5,6,7,8,
On Oct 8, 2009, at 12:23 PM, Steve Lianoglou wrote:
You probably need to install a 64bit version of R. The one
distributed "normally" through CRAN isn't 32bit.
I mean to say that the one distributed through cran isn't *64 bit*,
sorry.
-steve
--
Steve Lianoglou
Graduate Student: Computat
drlucyasher wrote:
>
> I am attempted to examine the temporal independence of my data set and
> think I need an unordered multinomial logistic regression (or logit model)
> with repeated measures to do so.
>
> What I have thought of so far is
>
> glm(Location~ Pen| Bird*AMPM*Time, family=mu
Hi,
On Oct 8, 2009, at 11:59 AM, Peng Yu wrote:
Hi,
My Mac OS is Leopard. I downloaded
R-GUI-5496-2.9-leopard-Leopard64.dmg from http://r.research.att.com/.
I run R from the image disk, but it always "quit unexpectedly".
What could cause the problem? Is it because that I have a 32 bit R
insta
Hi,
My Mac OS is Leopard. I downloaded
R-GUI-5496-2.9-leopard-Leopard64.dmg from http://r.research.att.com/.
I run R from the image disk, but it always "quit unexpectedly".
What could cause the problem? Is it because that I have a 32 bit R
installed already installed in the same machine.
Regards
Hi Peng,
On Oct 8, 2009, at 11:21 AM, Peng Yu wrote:
Hi,
I installed R on mac (see below). When I start it from a terminal by
the following command, the current working directory is always '~' no
matter where my current directory is in the terminal. I'm wondering if
there is a way to make the
Dear useRs and developeRs,
In the last few months we have had an increasing number of requests
about deSolve and dynamic modelling. We have also received several
requests to establish a SIG of our own about such things. We think
this is a good idea -- here it is!
---
On Oct 8, 2009, at 11:13 AM, delnatan wrote:
Hi ,
I want to convert a my list:
myList
[[1]]
[1] 1 2 3 4 5
[[2]]
[1] 6 7 8
[[3]]
[1] 9 10 11
into something like c(1,2,3,4,5,6,7,8,9,10,11)
?unlist
I realized that this is possible by
c(do.call("cbind",myList))
But only when list[[1]
Hi ,
I want to convert a my list:
> myList
[[1]]
[1] 1 2 3 4 5
[[2]]
[1] 6 7 8
[[3]]
[1] 9 10 11
into something like c(1,2,3,4,5,6,7,8,9,10,11)
I realized that this is possible by
> c(do.call("cbind",myList))
But only when list[[1]] through list[[3] have equal length of vectors within
them
I am attempted to examine the temporal independence of my data set and think
I need an unordered multinomial logistic regression (or logit model) with
repeated measures to do so. The data in question is location of chickens.
Chickens could be in any one of 5 locations when a snapshot sample was
ta
Gabor Grothendieck wrote:
Try this (possibly after scaling the rows or columns to 1):
library(gplots)
with(as.data.frame.table(as.matrix(jevons[-1])), balloonplot(Var1, Var2, Freq))
Thanks, Gabor --
That's not bad as a first cut; I can work with that.
--
Michael Friendly Email: frien
Dear all,
I didn't get any suggestion for my querry concerning the NLTM package so I am
re-posting it hoping that someone can give me any clue. I apologize for doing
so. Please find attached a copy of my previous email:
Dear R users,
I have a question concerning the nltm package. Before postin
Dear All,
I have temperature series for 10 different sea levels, where there are 3000
observations for each level. I am planning to use multiple time
series modelling on these series, with/ without correlation structure for the
residuals.
Is there any package that can be used to do this operat
Hi,
I installed R on mac (see below). When I start it from a terminal by
the following command, the current working directory is always '~' no
matter where my current directory is in the terminal. I'm wondering if
there is a way to make the gui R on mac inherent the current directory
from the term
<57d3c6e9-402b-4991-a53b-71e25af78...@comcast.net>
Content-Type: text/plain; charset="iso-8859-1"
Content-Transfer-Encoding: quoted-printable
MIME-Version: 1.0
No worries there =3B-)) But thanks for the reminder all the same!
=20
Monica
> CC: pisican...
Hi,
On Oct 8, 2009, at 11:06 AM, wesley mathew wrote:
Dear All,
I am a researcher, working at the Department of Software Engineering
at
University of Minho, Portugal. I am working on in the area of
Distributed
data mining and attempting to develop distributed data mining system
in „R‰
En
It should also be remembered that there will often be multiple
intersections of such density estimates.
I hope this is not being done in support of a data-driven
discretization of two group comparisons. Such practices are to be
eschewed (as they are even worse than obfuscation).
--
David
Dear All,
I am a researcher, working at the Department of Software Engineering at
University of Minho, Portugal. I am working on in the area of Distributed
data mining and attempting to develop distributed data mining system in R
Environment.
I have an enquiry.
Suppose I have 3 different nodes (m
1 - 100 of 138 matches
Mail list logo
