Dear friends, I'm examining the characteristics of two models that both
fit the sodium concentration in 16 pigs quite well under treatment or
control conditions. The more complicated model is by anova better than
the less complicated model. To take it further I have generated
replicate data us
Hi Everyone,
i have a data frame like this
labels starts ends priorities
1 firsttask 37987 38049 1
2 secondtask 38019 38112 2
3 thirdtask 38049 38144 3
4 fourthtask 38081 38207 4
5 fifthtask 38112 38239 5
now i want to apply a date for
Hi Polemon --
polemon wrote:
> On Mon, Aug 24, 2009 at 2:20 PM, polemon wrote:
>
>> Hello, I plan to use R with my cluster with OpenMPI.
>> I need the packaged 'snow' and 'Rmpi' for that, however, I get an error
>> while downloading and installing them:
>> When I do a:
>> install.packages("R
My prior solution was not correct.
If the idea is to combine each row of x with each row
of y then convert the matrices to data frames and
perform and outer join with SQL like this:
library(sqldf)
X <- as.data.frame(x)
Y <- as.data.frame(y)
as.matrix(sqldf("select * from X, Y", method = "raw"))
Hello
Your attachement didn't seem to get through.
You can simulate data using rnorm() or any of the r*() functions [1].
You can also use it to add noise to a custom function that you use to
generate your specific data.
Liviu
[1] http://www.statmethods.net/management/functions.html
On 8/25/09,
Does anyone have an idea about this, it's driving me insane!
Hello everyone. I have overlayed 1611 points on top of a tif file.
tifclassoverlay<-overlay(CRS11, classifiedmodrobin)
overlaydf<-as.data.frame(tifclassoverlay)
tifol<-remove.na.rows(overlaydf)
tifol is 854 rows long. tifol has p
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Michael Kogan
> Sent: Saturday, August 22, 2009 11:45 AM
> To: r-help@r-project.org
> Subject: [R] Help on comparing two matrices
>
> Hi,
>
> I need to compare two matrices with
It depends.
If there are patterns in the names, you can make use of pattern
argument of ls(). For example,
> x1="a";x2="b";x3="c"
> ls(pattern="x[1-2]")
[1] "x1" "x2"
> ls(pattern="x[^1-2]")
[1] "x3"
# to remove x1-x3
>rm(list=ls(pattern="x[1-2]"))
More generally, if you want to remove all butx,x
Hi all,
I am currently woking with hundreds of objects in workspace and whenever I
invoke ls() to observe the names of the objects, there are too much of
unnecessary variables.
For example, if I only require say 3 or 4 objects from hundreds of objects
in workspace, are there any methods that may
Hi. Am brand new to R and to mailing lists - have never posted anywhere
before, so hope I do this right.
Am using R 2.9.1 with lattice graphics (just installed, fully up to date).
Am doing trellis xyplot with y (emp=employment), x (yearmo=a time measure)
and conditioning variable (indf - factor d
My two cents: this is a hard problem to do, period (not just in R).
I would second the recommendation of the Dormann et al paper listed
below; also see Zuur, Alain F., Elena N. Ieno, Neil J. Walker, Anatoly A.
Saveliev, and Graham M. Smith. Mixed Effects Models and Extensions in
Ecology with R. 1s
Dear All
I know that you do not have to help me but please do, i am new to R as a CPI
compiler, i just need to do a sample to see which sampling method best works in
different situations, therefore since this is for practice purposes nobody will
finance a real project thats why i need you to h
Dear R users,
I am trying to fill in arrays (5 different according to distinct "id")
from objects produced from arbitrary data set below.
a <-
data.frame(id=rep(c("idA1","idA2","idA3","idA4","idA5"),2),pro=c("bb","uu","ee","tt","uu","gg","tt","bb","gg","ee"),sal=rpois(10,2))
idpro sa
On Aug 24, 2009, at 6:50 PM, Edward Chen wrote:
Thank you so much for your reply. I apologize for not making my
question clearer.
The problem I have right now is not just a matrix. I have a plot in
which the X and Y are both calculated by other functions before the
plot. After reviewing th
On Aug 24, 2009, at 8:19 PM, AllenL wrote:
Dear R gurus,
Trying to loop a graphing function and output a jpeg file each loop.
It
works fine for a single run (ie. not looping) but when I loop the
output is
nothing but white space.
Here is my code, somewhat abridged:
{
jpeg(filename=<"na
Hi Tim,
I don't believe there is a satisfactory solution in R - at least yet -
for non-normal models. Ultimately, this should be possible using lmer()
but not in the near-term. One possibility is to use glmPQL as described
in:
Dormann, F. C., McPherson, J. M., Araújo, M. B., Bivand, R., Bolliger,
On Aug 24, 2009, at 7:41 PM, Felix Andrews wrote:
2009/8/25 David Winsemius :
On Aug 24, 2009, at 7:01 PM, Felix Andrews wrote:
2009/8/25 David Winsemius
On Aug 24, 2009, at 11:00 AM, Rick wrote:
I would like to plot two variables against the same abscissa
values.
They
have different
Dear R gurus,
Trying to loop a graphing function and output a jpeg file each loop. It
works fine for a single run (ie. not looping) but when I loop the output is
nothing but white space.
Here is my code, somewhat abridged:
{
jpeg(filename=<"name for this loop.jpeg">)
xyplot(AbvBioAnnProd~Ye
2009/8/25 David Winsemius :
> On Aug 24, 2009, at 7:01 PM, Felix Andrews wrote:
>
>> 2009/8/25 David Winsemius
>>>
>>> On Aug 24, 2009, at 11:00 AM, Rick wrote:
>>>
I would like to plot two variables against the same abscissa values.
They
have different scales. I've found how to mak
Hrm. I have to admit, I don't entirely understand how to use the scaling,
and that seems like a lot of unneeded extra code. It is what it is, though.
The documentation about scaling is somewhat obtuse. Do you have a clear
explanation of what it is and how to use it in this instance? Perhaps e
On Aug 24, 2009, at 7:01 PM, Felix Andrews wrote:
2009/8/25 David Winsemius
On Aug 24, 2009, at 11:00 AM, Rick wrote:
I would like to plot two variables against the same abscissa
values. They
have different scales. I've found how to make a second axis on the
right
for labeling, but not
2009/8/25 David Winsemius
>
> On Aug 24, 2009, at 11:00 AM, Rick wrote:
>
>> I would like to plot two variables against the same abscissa values. They
>> have different scales. I've found how to make a second axis on the right
>> for labeling, but not how to plot two lines at different scales.
>
>
On 24-Aug-09 21:56:06, zrl wrote:
> Dear List:
> I am trying to find a command in R which is like the unix command
> "less" or "more" to show the data in a object of R.
> did anyone can help me on this?
>
> Is there a collection of such unix-like commands in R?
>
> Thanks.
>
> -ZRL
There is a p
On 25/08/2009, at 10:17 AM, Peng Yu wrote:
Hi,
I did a search but I was able to find how to generate a random matrix.
Can somebody let me know how to do it?
Uhhh, generate some random numbers and then arrange them in a matrix?
?matrix
?runif
?rnorm
?rgamma
.
.
.
cheers,
Hi,
I did a search but I was able to find how to generate a random matrix.
Can somebody let me know how to do it?
Regards,
Peng
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://ww
Dear List:
I am trying to find a command in R which is like the unix command "less" or
"more" to show the data in a object of R.
did anyone can help me on this?
Is there a collection of such unix-like commands in R?
Thanks.
-ZRL
[[alternative HTML version deleted]]
___
On 25/08/2009, at 3:00 AM, Rick wrote:
First of all, thanks to everyone who answers these questions - it's
most helpful.
I'm new to R and despite searching have not found an example of what I
want to do (there are some good beginner's guides and a lot of complex
plots, but I haven't found th
On Aug 24, 2009, at 4:10 PM, John Kane wrote:
I am trying to come up with a way of shading-in a grid for a simple
pattern
So far I can draw a square where I want but I cannot seem to draw a
complete grid. I am just drawing them along the diagonal!!
Clearly I am missing something simple b
On Aug 24, 2009, at 4:10 PM, John Kane wrote:
I am trying to come up with a way of shading-in a grid for a simple
pattern
So far I can draw a square where I want but I cannot seem to draw a
complete grid. I am just drawing them along the diagonal!!
Clearly I am missing something simple b
On Aug 24, 2009, at 4:01 PM, Michael Kogan wrote:
David: Well, e.g. the first row has 2 ones in your output while
there were no rows with 2 ones in the original matrix. Since the row
and column sums can't be changed by sorting them, the output matrix
can't be equivalent to the original one
On Aug 24, 2009, at 11:00 AM, Rick wrote:
First of all, thanks to everyone who answers these questions - it's
most helpful.
I'm new to R and despite searching have not found an example of what I
want to do (there are some good beginner's guides and a lot of complex
plots, but I haven't found
I am trying to come up with a way of shading-in a grid for a simple pattern
So far I can draw a square where I want but I cannot seem to draw a complete
grid. I am just drawing them along the diagonal!!
Clearly I am missing something simple but what?
Any suggestions gratefully accepted.
Exam
David: Well, e.g. the first row has 2 ones in your output while there
were no rows with 2 ones in the original matrix. Since the row and
column sums can't be changed by sorting them, the output matrix can't be
equivalent to the original one. But that means nothing, maybe it's
intended and just
On Mon, 24-Aug-2009 at 08:00AM -0700, Rick wrote:
>
> First of all, thanks to everyone who answers these questions - it's
> most helpful.
>
> I'm new to R and despite searching have not found an example of what I
> want to do (there are some good beginner's guides and a lot of complex
> plots, but
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of rami jiossy
> Sent: Monday, August 24, 2009 12:01 PM
> To: R-Help
> Subject: [R] create list entry from variable
>
>
> Hi;
>
> assume i<-10
>
> how can i create a list having
Yep;
great thanks :)
Date: Mon, 24 Aug 2009 16:11:20 -0300
Subject: Re: [R] create list entry from variable
From: www...@gmail.com
To: sra...@hotmail.com
CC: r-help@r-project.org
Try this:
l <- list(i + 1)
names(l) <- i
On Mon, Aug 24, 2009 at 4:01 PM, rami jiossy wrote:
Hi;
assume i
Try this:
l <- list(i + 1)
names(l) <- i
On Mon, Aug 24, 2009 at 4:01 PM, rami jiossy wrote:
>
> Hi;
>
> assume i<-10
>
> how can i create a list having key=10 and value=11
>
> list(i=11) generates a list with
>
> 'i'
> [1] 11
>
> and not
>
> 10
> [1] 11
>
> any help?
>
> Thanks
>
> _
Bert -
I took a look at that page just now, and I'd classify my problem as
spatial regression. Unfortunately, I don't think the spdep library fits my
needs. Or at least, I can't figure out how to use it for this problem. The
examples I have seen all use spdep with networks. They build a graph,
co
?round
Bert Gunter
Genentech Nonclinical Biostatisics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Mehdi Khan
Sent: Monday, August 24, 2009 11:52 AM
To: Erik Iverson
Cc: r-help@r-project.org
Subject: Re: [R] Unique command not de
Hi;
assume i<-10
how can i create a list having key=10 and value=11
list(i=11) generates a list with
'i'
[1] 11
and not
10
[1] 11
any help?
Thanks
_
Facebook.
:ON:WL:en-US:SI_SB_facebook:082009
[[alternative HTM
Hi Emma,
>>
R gives you the tools to work this out.
## Example
set.seed(7)
TDat <- data.frame(response = c(rnorm(100, 5, 2), rnorm(100, 20, 2)))
TDat$group <- gl(2, 100, labels=c("A","B"))
with(TDat, boxplot(split(response, group)))
summary(aov(response ~ group, data=TDat))
Regards, Mark.
e
Duplicated did not work, I agree with Erik. Is there any way I can specify a
tolerance limit and then delete?
On Mon, Aug 24, 2009 at 11:41 AM, Erik Iverson wrote:
> I really don't think this is the issue. I think the issue is that some
> columns of the data.frame, specifically V1, V2, and V4 s
Have you looked at the "Spatial" task view on CRAN? That would seem to me
the logical first place to go.
Bert Gunter
Genentech Nonclinical Biostatisics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of timothy_hand...@nps.gov
Sent: M
I really don't think this is the issue. I think the issue is that some columns
of the data.frame, specifically V1, V2, and V4 should be checked versus R FAQ
7.31.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Don McKenzie
Sen
duplicated()
> test.df
V1 V2 V3 V4 V5 V6 V7
1 -115.380 32.894 195 162.940 D 8419 D
2 -115.432 32.864 115 208.910 D 8419 D
3 -115.447 32.773 1170 264.570 D 8419 D
4 -115.447 32.773 1170 264.570 D 8419 D
5 -115.447 32.773 1170 264.570 D 8419 D
6 -115.447 32.773 1170
Hello everyone, when I run the "unique" command on my data frame, it deletes
the majority of duplicate rows, but not all of them. Here is a sample of my
data. How do I get it to delete all the rows?
6 -115.38 32.894 195 162.94 D 8419 D
7 -115.432 32.864 115 208.91 D 8419 D
8 -115.447 32.773
Hello folks,
I have some data where spatial autocorrelation seems to be a serious
problem, and I'm unclear on how to deal with it in R. I've tried to do my
homework - read through 'The R Book,' use the online help in R, search the
internet, etc. - and I still have some unanswered questions. I'd g
On Aug 24, 2009, at 12:55 PM, Edward Chen wrote:
Hi all,
Is there a quick way to display or recall data points from a
specific region
on the plot? For example I want the points from x>5 and y>5?
Thank you very much!
Your question is pretty light on specifics but assuming that you have
co
Dear Mike:
I don't know.
1. What specific error message do you get?
2. Your example is too long for me to parse, especially with
the color being stripped before I saw it.
3. Have you tried using "debug(cro.etest.grab)", then
walking through t
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Budi Mulyono
> Sent: Monday, August 24, 2009 3:38 AM
> To: r-help@r-project.org
> Subject: [R] hdf5 package segfault when processing large data
>
> Hi there,
>
> I am currently
whizvast wrote:
>
> Hi, Adrian-
>
> If you use "overwrite=T" parameter, you will overwrite the entire table,
> not each record. this is the essence of my problem and i still haven't
> found out right solution. i am thinking of writing my own MySQLwriteTable
> function...
>
> Thank you for your
I am assuming that you want to desplay all the data and highlight the subset.
Set up a vector to indicate the breakdown of the data and you can do it fairly
easily in ggplot2 if you treat the vector as a factor.
library(ggplot)
mydata <- data.frame(x=1:21, y= -10:10)
z <- ifelse(mydata[,1]>5 & m
> -Original Message-
> From: Marc Schwartz [mailto:marc_schwa...@me.com]
> Sent: Monday, August 24, 2009 9:57 AM
> To: Daniel Nordlund
> Cc: r help
> Subject: Re: [R] Combining matrices
>
>
> On Aug 24, 2009, at 11:46 AM, Marc Schwartz wrote:
>
> >
> > On Aug 24, 2009, at 11:16 AM, Dani
Try this:
kronecker(cbind(x, y), rep(1, 3))
On Mon, Aug 24, 2009 at 12:16 PM, Daniel Nordlund wrote:
> If I have two matrices like
>
> x <- matrix(rep(c(1,2,3),3),3)
> y <- matrix(rep(c(4,5,6),3),3)
>
> How can I combine them to get ?
>
> 1 1 1 4 4 4
> 1 1 1 5 5 5
> 1 1 1 6 6 6
> 2 2 2 4 4 4
>
I may be misunderstanding the question but would
cor(d1, use='complete.obs') or some other variant of "use" help?
--- On Mon, 8/24/09, Christian Meesters wrote:
> From: Christian Meesters
> Subject: [R] robust method to obtain a correlation coeff?
> To: "r-help@r-project.org Help"
> Received:
On Aug 24, 2009, at 11:46 AM, Marc Schwartz wrote:
On Aug 24, 2009, at 11:16 AM, Daniel Nordlund wrote:
If I have two matrices like
x <- matrix(rep(c(1,2,3),3),3)
y <- matrix(rep(c(4,5,6),3),3)
How can I combine them to get ?
1 1 1 4 4 4
1 1 1 5 5 5
1 1 1 6 6 6
2 2 2 4 4 4
2 2 2 5 5 5
2
Hi all,
Is there a quick way to display or recall data points from a specific region
on the plot? For example I want the points from x>5 and y>5?
Thank you very much!
--
Edward Chen
[[alternative HTML version deleted]]
__
R-help@r-project.org
jlwoodard wrote:
>
>
> Each of the above lines successfully excludes the BLUE subjects, but the
> "BLUE" category is still present in my data set; that is, if I try
> table(Color) I get
>
> RED WHITE BLUE
> 82 151 0
>
> How can I eliminate the BLUE category completely so I can do
On Aug 24, 2009, at 11:16 AM, Daniel Nordlund wrote:
If I have two matrices like
x <- matrix(rep(c(1,2,3),3),3)
y <- matrix(rep(c(4,5,6),3),3)
How can I combine them to get ?
1 1 1 4 4 4
1 1 1 5 5 5
1 1 1 6 6 6
2 2 2 4 4 4
2 2 2 5 5 5
2 2 2 6 6 6
3 3 3 4 4 4
3 3 3 5 5 5
3 3 3 6 6 6
The num
have you tried:
fits <- lm(a~b)
fstat <- sapply(summary(fits), function(x) x[["fstatistic"]][["value"]])
it takes 3secs for 100K columns on my machine (running on batt)
b
On Aug 23, 2009, at 9:55 PM, big permie wrote:
Dear R users,
I have a matrix a and a classification vector b such that
ogbos okike schreef:
Hi,
I am trying to use the image function to do a color plot. My matrix columns
are labeled y and x. I tried >image(y, x) but I had error message ("Error in
image.default(y, x) : increasing 'x' and 'y' values expected").
Could anybody please tell me how to add these increasin
Your question is a little vague. Do you just want to know how often z falls
in one the three classes? If so, you could either code an indicator variable
(e.g. z.cat) that expresses the three categories and then do table(z.cat).
Alternatively, you could just do
sum(z>0&z<1000)
sum(z>1&z<3000)
s
Try this;
do.call(rbind, lapply(split(x, seq(nrow(x))), cbind, y))
On Mon, Aug 24, 2009 at 1:16 PM, Daniel Nordlund wrote:
> If I have two matrices like
>
> x <- matrix(rep(c(1,2,3),3),3)
> y <- matrix(rep(c(4,5,6),3),3)
>
> How can I combine them to get ?
>
> 1 1 1 4 4 4
> 1 1 1 5 5 5
> 1 1 1
[Note R-Devel is the wrong list for such questions. R-Help is where this
should have been directed - redirected there now]
On Mon, 2009-08-24 at 17:02 +0100, Corrado wrote:
> Dear R-experts,
>
> I have a question on the formulas used in the gam function of the mgcv
> package.
>
> I am trying to
Inchallah Yarab wrote:
>
> i want to do a table summerizing the number of variable where z is in
> [0-1000],],[1000-3000], [> 3000]
>
You can use "cut" to create a new vector of labels and tabulate the result.
Options control closed/open endpoints (see ?cut):
> z <- c(100,1500,1200,500,3500,
On Aug 24, 2009, at 10:59 AM, Inchallah Yarab wrote:
hi,
i want to use the function table to build a table not of frequence
(number of time the vareable is repeated in a list or a data
frame!!) but in function of classes
I don t find a clear explnation in examples of ?table !!!
example
You need to create a factor that indicates which group the values in 'z' belong
to. The easiest way to do that based on your situation is to use the 'cut'
function to construct the factor, and then call 'table' using the result
created by 'cut'. See ?cut and ?factor
-Original Message
Hello, I am sorry, I have this problem before and Uwe send me the answer but I
misplaced it and can not find it.
writing a model for BRugs
> library(BRugs)
Loading required package: coda
Loading required package: lattice
Welcome to BRugs running on OpenBUGS version 3.0.3
> setwd("c:/tmp")
Error i
If I have two matrices like
x <- matrix(rep(c(1,2,3),3),3)
y <- matrix(rep(c(4,5,6),3),3)
How can I combine them to get ?
1 1 1 4 4 4
1 1 1 5 5 5
1 1 1 6 6 6
2 2 2 4 4 4
2 2 2 5 5 5
2 2 2 6 6 6
3 3 3 4 4 4
3 3 3 5 5 5
3 3 3 6 6 6
The number of rows and the actual numbers above are unimportant,
Try
lapply(abc, function(x) x*3)
Peter Ehlers
Brigid Mooney wrote:
I apologize for what seems like it should be a straighforward query.
I am trying to multiply a list by a numeric and thought there would be a
straightforward way to do this, but the best solution I found so far has a
for loop.
On Aug 24, 2009, at 10:58 AM, Brigid Mooney wrote:
I apologize for what seems like it should be a straighforward query.
I am trying to multiply a list by a numeric and thought there would
be a
straightforward way to do this, but the best solution I found so far
has a
for loop.
Everything e
Inline below.
Bert Gunter
Genentech Nonclinical Biostatisics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of David Winsemius
Sent: Monday, August 24, 2009 8:53 AM
To: David Winsemius
Cc: r-help@r-project.org Help; ted.hard...@manch
hi,
i want to use the function table to build a table not of frequence (number of
time the vareable is repeated in a list or a data frame!!) but in function of
classes
I don t find a clear explnation in examples of ?table !!!
example
x y z
1 0 100
5 1 1500
6
Mcdonald, Grant wrote:
>
> Dear sir,
>
> I am fitting a glm with default identity link:
>
>
>
> model<-glm(timetoacceptsecs~maleage*maletub*relweight*malemobtrue*femmobtrue)
>
> the model is overdisperesed and plot model shows a low level of linearity
> of the residuals.
>
> >> I don't
I apologize for what seems like it should be a straighforward query.
I am trying to multiply a list by a numeric and thought there would be a
straightforward way to do this, but the best solution I found so far has a
for loop.
Everything else I try seems to throw an error "non-numeric argument to
On Aug 24, 2009, at 11:38 AM, David Winsemius wrote:
On Aug 24, 2009, at 11:26 AM, (Ted Harding) wrote:
On 24-Aug-09 14:47:02, Christian Meesters wrote:
Hi,
Being a R-newbie I am wondering how to calculate a correlation
coefficient (preferably with an associated p-value) for data like:
d[
On Aug 24, 2009, at 11:26 AM, (Ted Harding) wrote:
On 24-Aug-09 14:47:02, Christian Meesters wrote:
Hi,
Being a R-newbie I am wondering how to calculate a correlation
coefficient (preferably with an associated p-value) for data like:
d[,1]
[1] 25.5 25.3 25.1 NA 23.3 21.5 23.8 23.2 24.2 22
On 24-Aug-09 14:47:02, Christian Meesters wrote:
> Hi,
> Being a R-newbie I am wondering how to calculate a correlation
> coefficient (preferably with an associated p-value) for data like:
>
>> d[,1]
> [1] 25.5 25.3 25.1 NA 23.3 21.5 23.8 23.2 24.2 22.7 27.6 24.2 ...
>> d[,2]
> [1] 0.0 11.1 0
First of all, thanks to everyone who answers these questions - it's
most helpful.
I'm new to R and despite searching have not found an example of what I
want to do (there are some good beginner's guides and a lot of complex
plots, but I haven't found this).
I would like to plot two variables a
Great idea - thx!
On Mon, Aug 24, 2009 at 9:30 AM, hadley wickham wrote:
> Sorry, we had some problems with the initial sending of our weekly digest
> which resulted in a rather empty email. Here is the correct version:
>
> CRAN (and crantastic) updates this week
>
> New packages
>
Hi,
Being a R-newbie I am wondering how to calculate a correlation
coefficient (preferably with an associated p-value) for data like:
> d[,1]
[1] 25.5 25.3 25.1 NA 23.3 21.5 23.8 23.2 24.2 22.7 27.6 24.2 ...
> d[,2]
[1] 0.0 11.1 0.0 NA 0.0 10.1 10.6 9.5 0.0 57.9 0.0 0.0 ...
Apparent
Hello,
I'm trying to tackle a problem that would require the implementation of a
recurrent NN. However, even though the CRAN is very big, I can’t seem to
find a package for this. Does anybody here know if one exits?
BR,
John
--
View this message in context:
http://www.nabble.com/Recurrent-neu
I like it. Thanks.
2009/8/24 hadley wickham :
> Sorry, we had some problems with the initial sending of our weekly digest
> which resulted in a rather empty email. Here is the correct version:
>
> CRAN (and crantastic) updates this week
>
> New packages
>
>
> * atm (0.1.0)
> Charlotte
Sorry, we had some problems with the initial sending of our weekly digest
which resulted in a rather empty email. Here is the correct version:
CRAN (and crantastic) updates this week
New packages
* atm (0.1.0)
Charlotte Maia
http://crantastic.org/packages/atm
An R package for creat
CRAN (and crantastic) updates this week
New packages
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On Mon, 24 Aug 2009 05:22:17 -0700 (PDT) rajclinasia
wrote:
R> my.gantt.info<-read.csv("C:/Documents and
R> Settings/balakrishna/Desktop/one.csv").
R>
R> and for create gantt chart i used below code.
R>
R> gantt.chart("my.gantt.info").
This again is why others have pointed you first to have a
To drop empty factor levels from a subset, I use the following:
a.subset <- subset(dataset, Color!='BLUE')
ifac <- sapply(a.subset,is.factor)
a.subset[ifac] <- lapply(a.subset[ifac],factor)
Mike
> dataset
Color Score
1 RED10
2 RED13
3 RED12
4 WHITE22
5 WHITE27
6 WHIT
On 8/24/2009 2:06 AM, Petr PIKAL wrote:
> Hi
>
> r-help-boun...@r-project.org napsal dne 23.08.2009 17:29:48:
>
>> On 8/23/2009 9:58 AM, David Winsemius wrote:
>>> I still have problems with this statement. As I understand R, this
>> should be impossible. I have looked at both you postings and ne
Hi,
I am trying to use the image function to do a color plot. My matrix columns
are labeled y and x. I tried >image(y, x) but I had error message ("Error in
image.default(y, x) : increasing 'x' and 'y' values expected").
Could anybody please tell me how to add these increasing 'x' and 'y' values.
T
On Mon, Aug 24, 2009 at 2:20 PM, polemon wrote:
> Hello, I plan to use R with my cluster with OpenMPI.
> I need the packaged 'snow' and 'Rmpi' for that, however, I get an error
> while downloading and installing them:
> When I do a:
> install.packages("Rmpi", dependencies=T)
>
> I get this er
Dear all,
I have encountered a weird behaviour in R
survival package which seems to me to be a bug.
The weird behaviour happens when I am using
100 variables in the ridge function when calling
coxph with following formula Surv(time = futime,
event = fustat, type = "right") ~ ridge(X1, X2,
X
On Aug 24, 2009, at 8:22 AM, rajclinasia wrote:
hi every one,
i have a excel sheet like this
labels starts ends
1 first task 1-Jan-04 3-Mar-04
2 second task 2-Feb-04 5-May-04
3 third task 3-Mar-04 6-Jun-04
4 fourth task 4-Apr-04 8-Aug-04
5 fifth task 5-May-04 9-Sep-04
now i co
Clarification:
Lm is much better than the base forecast from lme level=0,
Level=1 produces a much tighter fit than lm.
I was expecting that level=0 would produce something very close to lm, but it
does not.
[[alternative HTML version deleted]]
_
2009/8/24 Lucas Sevilla García :
>
> Hi! I'm a beginner with this webpage so, I don't know if I'm sending my
> question to the correct site. Anyway, I'm working with R and I need to import
> and export ENVI files, (*.HDR files). A colleague told me that there is a
> package to import/export envi
Your code is not reproducible, we do not have rfc, y, zVals nor
NoCols.
It's much easier to reproduce: just type in the first example from
the "image" help page
x <- y <- seq(-4*pi, 4*pi, len=27)
r <- sqrt(outer(x^2, y^2, "+"))
image(z = z <- cos(r^2)*exp(-r/6), col=gray((0:32)/32))
t
Hello r-help,
I am using lme with two specs for the variance func
varComb(varFixed(~1/n)),varPower(~Age))
this produces worse forecasts than the lm model with simple
weights=n
I think due to the fact that the lme spec works on variance inside the group. I
need to show it that 1/n scales the v
On Aug 24, 2009, at 8:21 AM, Anne Skoeries wrote:
Hi there,
a text Window is supposed to map the shortcuts for copying and
pasting (, ) automatically.
I'm working under Mac OS X and my text window doesn't really map
these functions automatically - it works fine under Windows.
That is a b
Moumita Das ha scritto:
How to NATURAL sort a vector or data frame* by row* , in ascending order ?
V1 V2V3 V4
i1 5.00e-01 1.036197e-17 4.825338e+16 0.
i104.001692e-18 1.365740e-17 2.930053e-01 0.76973827
i
On Aug 24, 2009, at 8:11 AM, maram salem wrote:
Hi all,
I've a trivial question. If (q) is a continous variable,actually a
vector of 1000 values. how to calculate the probability that q is
greater than a specific value, i.e. P(q>45)??
sum(q>45)/1000 # if no NA's in vector
sum(q>45, na.rm
maram salem wrote:
Hi all,
I've a trivial question. If (q) is a continous variable,actually a vector of 1000
values. how to calculate the probability that q is greater than a specific value,
i.e. P(q>45)??
Do you want to estimate any distribution or do you just want the
empirical informati
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