Dear R users,
We have encountered a slight problem when using the lmer()
function:
1. Data description: 11 locations; Nt: monthly mosquito population
density from 1994-2005 in each location.
2. Question: to examine the degree of spatial heterogeneity in the
system by testing model
Hi
On 17 Dec 2006 at 10:04, Duncan Murdoch wrote:
Date sent: Sun, 17 Dec 2006 10:04:37 -0500
From: Duncan Murdoch [EMAIL PROTECTED]
To: Webmaster [EMAIL PROTECTED]
Copies to: r-help@stat.math.ethz.ch
Subject:Re: [R]
Hi
I used write() the other day to save some results.
It seems that write() does not record the full precision of
the objects being written:
write(pi,file=~/f,ncolumns=1)
pi.saved - scan(~/f)
Read 1 item
dput(pi)
3.14159265358979
dput(pi.saved)
3.141593
pi-pi.saved
[1] -3.464102e-07
Hello sir:
a data with 2 columns:
id x
a 1
b 2
c 3
I wanna get such kind of plot:
x: a b c
y:1 2 3
But the plot command doesn't permit string character as x.
How can I get it ?
Thanks a lot !
My best
__
R-help@stat.math.ethz.ch mailing list
On Mon, 18 Dec 2006, Robin Hankin wrote:
Hi
I used write() the other day to save some results.
Why not save()? It is the only way to preserve the results exactly.
It seems that write() does not record the full precision of
the objects being written:
write(pi,file=~/f,ncolumns=1)
On 18 Dec 2006, at 08:50, Prof Brian Ripley wrote:
On Mon, 18 Dec 2006, Robin Hankin wrote:
Hi
I used write() the other day to save some results.
Why not save()? It is the only way to preserve the results exactly.
It seems that write() does not record the full precision of
the objects
On Mon, 18 Dec 2006, Robin Hankin wrote:
On 18 Dec 2006, at 08:50, Prof Brian Ripley wrote:
On Mon, 18 Dec 2006, Robin Hankin wrote:
Hi
I used write() the other day to save some results.
Why not save()? It is the only way to preserve the results exactly.
It seems that write()
Dear list members,
I am facing some problems using the aggregate() function.
I want to calculate a sum and a mean of one variable over the
combination of 12 factors with the aggregate() function to avoid loops
but it doesn't work (or the job is far too long, it exceeds 2 hours). It
works with
XinMeng wrote:
Hello sir:
a data with 2 columns:
id x
a 1
b 2
c 3
I wanna get such kind of plot:
x: a b c
y:1 2 3
But the plot command doesn't permit string character as x.
How can I get it ?
What sort of plot do you want? For a barplot() of x with bars labeled
by id you
Hello,
I have a large data set 320.000 rows and 1000 columns. All the data has the
values 0,1,2.
I wrote a script to remove all the rows with more than 46 missing values. This
works perfect on a smaller dataset. But the problem arises when I try to run it
on the larger data set I get an error
I've rolled up R-2.4.1.tar.gz a short while ago. This is a maintenance
release and fixes a number of mostly minor bugs. See the full list
of changes below.
You can get it (in a short while) from
http://cran.r-project.org/src/base/R-2/R-2.4.1.tar.gz
or wait for it to be mirrored at a CRAN site
Joachim Claudet wrote:
Dear list members,
I am facing some problems using the aggregate() function.
I want to calculate a sum and a mean of one variable over the
combination of 12 factors with the aggregate() function to avoid loops
but it doesn't work (or the job is far too long, it
Dear R Help,
Why is it that if you try to return more than one objects using return(),
it says it is 'deprecated'? So how do I return more than 1 objects back to
the parent function?
Thanks,
Tim
__
R-help@stat.math.ethz.ch mailing list
Peter Dalgaard wrote:
Alternatively, rewrite aggregate() and send us a patch ;-)
It is not necessarily all that hard. Here's a rough idea
IX - as.data.frame(by)
OO - do.call(order,IX)
Y - x[OO,]
g - cumsum(!duplicated(IX))
FF - unique(IX)
cbind(FF, sapply(split(x,g),FUN))
(completely
odfWeave: avoid zip.exe version 2.1 on Windows
Problem:
After weaving with odfWeave the .odt file can not be opened with
OpenOfice.org 2.1 Writer. Writer instead launches the Filter Selection
but none of the formats offered works.
My system:
R-2.4.0
odfWeave 0.4.4 package from
Well, put it in a list and return that list.
It's all written in ?return by the way.
[EMAIL PROTECTED] wrote:
Dear R Help,
Why is it that if you try to return more than one objects using return(),
it says it is 'deprecated'? So how do I return more than 1 objects back to
the parent
Hi Mat,
You can more-or-less do what you want with the xlabs argument to
biplot. It only takes characters, rather than the normal plot symbol
codes, though. If you need symbols, you can cheat with things like
+ o x. I usually use letters for the different groups within the data:
temp -
Now R 2.4.x defaults to CHM help, you may encounter a problem with a
fairly recent Windows security patch described in
http://www.helpscribble.com/chmnetwork.html
http://support.microsoft.com/kb/896358
This may apply if you have R itself or some library on a networked drive
or share. Apart
On 12/18/2006 6:06 AM, [EMAIL PROTECTED] wrote:
Dear R Help,
Why is it that if you try to return more than one objects using return(),
it says it is 'deprecated'? So how do I return more than 1 objects back to
the parent function?
Put them in a list, e.g.
return(a,b)
should be coded
--- XinMeng [EMAIL PROTECTED] wrote:
Hello sir:
a data with 2 columns:
id x
a 1
b 2
c 3
I wanna get such kind of plot:
x: a b c
y:1 2 3
But the plot command doesn't permit string character
as x.
How can I get it ?
Thanks a lot !
My best
It is not clear exactly what
Hi All,
I'm working on a class example that demonstrates one way to deal with
factors and their labels. I create a function called myLabeler and apply
it with lapply. It works on the whole data frame when I subscript it as
in lapply( myQFvars[ ,myQFnames ], myLabeler ) but does not work if I
Dear Prof. Ripley:
Thanks very much. That works. I got stuck on the help page for
dumpMethods and failed to check See Also.
Best Wishes,
Spencer Graves
Prof Brian Ripley wrote:
Do you want E (type 'E') or its methods (getMethods(E) works for me)?
On Sun, 17 Dec 2006,
You get a list, not a data.frame. Try,
as.data.frame(lapply( myQFvars, myLabeler ))
Gabriel
At 05:10 PM 12/18/2006, Muenchen, Robert A (Bob) wrote:
Hi All,
I'm working on a class example that demonstrates one way to deal with
factors and their labels. I create a function called myLabeler and
Dear All,
I looked at help(par), but could not figure out which setting controls the
distance between the x-axis values and the x-axis title. Any pointer would be
appreciated!
Thanks in advance,
--
Wolfgang Viechtbauer
Department of Methodology and Statistics
University of Maastricht,
Bob,
This is I think exactly what one wants to have happen. The first four
observations are still women. Both the labels and the underlying integers
should change. (If you want to give all the people sex changes, try Relevel in
the Epi package.
mydata$afterthechange -
On Mon, 18 Dec 2006, Robin Hankin wrote:
Hi
I used write() the other day to save some results.
It seems that write() does not record the full precision of
the objects being written:
write(pi,file=~/f,ncolumns=1)
pi.saved - scan(~/f)
Read 1 item
dput(pi)
3.14159265358979
On Mon, 18 Dec 2006, Viechtbauer Wolfgang (STAT) wrote:
Dear All,
I looked at help(par), but could not figure out which setting controls
the distance between the x-axis values and the x-axis title. Any pointer
would be appreciated!
mgp: looking at An Introduction to R may help you find
--- Viechtbauer Wolfgang (STAT)
[EMAIL PROTECTED] wrote:
Dear All,
I looked at help(par), but could not figure out
which setting controls the distance between the
x-axis values and the x-axis title. Any pointer
would be appreciated!
Thanks in advance,
?mpg probably
Is this what you
Thanks to all who responded so quickly! Yes, I totally overlooked par(mpg).
Exactly what I was looking for.
--
Wolfgang Viechtbauer
Department of Methodology and Statistics
University of Maastricht, The Netherlands
http://www.wvbauer.com/
-Original Message-
From: John Kane
Hello,
I have some problems trying to write up the formula for lmer.
I have 43 subjects ( random factor) which were seen twice ( Visit : repeated
measure - fixed). on each visit the patient performed a graded effort exercise
( effort : repeated measures, ordered, fixed 4 levels).
So Subject is
Folks,
Please help with a graphics problem, I am running R2.4.0 on Windows XP.
In much earlier version of R (1.3? about December 2001)
I could
par(mfrow=c(1,1))
plot(1,1,xlab=X,ylab=Y)
plot1-recordPlot()
plot(2,2,xlab=X2,ylab=Y2)
plot2-recordPlot()
par(mfrow=c(2,1))
plot1
plot2
Iris --
I hope the following helps; I think you have too much data for a
32-bit machine.
Martin
Iris Kolder [EMAIL PROTECTED] writes:
Hello,
I have a large data set 320.000 rows and 1000 columns. All the data
has the values 0,1,2.
It seems like a single copy of this data set will be at
In addition to my off-list reply to Iris (pointing her to an old post of
mine that detailed the memory requirement of RF in R), she might
consider the following:
- Use larger nodesize
- Use sampsize to control the size of bootstrap samples
Both of these have the effect of reducing sizes of trees
Hi all,
I have to recode some values in a dataset. for example changing all zeros to
. or 999 would be also ok. does anybody know how to do this? thanks in
advance. lars
--
View this message in context:
http://www.nabble.com/-R--Replacing-values-tf2841687.html#a7934402
Sent from the R help
Folks,
Please help with a graphics problem, I am running R2.4.0 on Windows XP.
In much earlier version of R (1.3? about December 2001)
I could
par(mfrow=c(1,1))
plot(1,1,xlab=X,ylab=Y)
plot1-recordPlot()
plot(2,2,xlab=X2,ylab=Y2)
plot2-recordPlot()
par(mfrow=c(2,1))
plot1
On Mon, 2006-12-18 at 10:58 -0800, downunder wrote:
Hi all,
I have to recode some values in a dataset. for example changing all zeros to
. or 999 would be also ok. does anybody know how to do this? thanks in
advance. lars
R has its own missing value designator, which is NA. A . or 999
You can do something like this for a scatter plot:
x - c(a,b,c)
y - c(1,2,3)
xnum - rep(1:length(x))
plot(x=xnum, y=y, xlab=x, xaxt=n)
axis(side=1, at=xnum, labels=x)
This fakes a numerical axis and suppresses the y-axis labels that you then
draw with the axis function the way that you want
Using the survival library, it is possible to get a plot of all the subjects
in a sample and it is possible to get a plot of all subgroups in the same
plot. How does one get a separate plot for each subgroup?
plot(survfit(Surv(time,death==1)~group),col=1:10)
The above results in a hideous
On Mon, 18 Dec 2006, Farrel Buchinsky wrote:
Using the survival library, it is possible to get a plot of all the subjects
in a sample and it is possible to get a plot of all subgroups in the same
plot. How does one get a separate plot for each subgroup?
There is an example in ?survfit of how
Yes. Thank you. Worked Brilliantly.
Here it is for any future reference:
for(i in
1:10)plot(survfit(Surv(time,death==1)~group)[i],xlab=days,main=levels(group)[i])
On 12/18/06, Thomas Lumley [EMAIL PROTECTED] wrote:
On Mon, 18 Dec 2006, Farrel Buchinsky wrote:
Using the survival library, it
Thanks Duncan,
yes it is surface3d{rgl} I am trying. Unfortunately your solution
doesn't work with the data (I get a subscript out of bounds error, after
converting to integer and assigning the matrix to z via z[cbind(data$x,
data$y)] - data$z).
As outlined previously, I have a dataframe with
Dear all,
I have a matrix of repeating columns in R, for example a matrix X is
[,1] [,2] [,3] [,4]
[1,] 1 1 1 1
[2,] 1 1 2 2
I want to store unique columns of the matrix X in a new matrix Y.
Therefore, Y will be
[,1] [,2]
Dear Roman,
You can use unique(X, MARGIN=2). See ?unique for details.
I hope this helps,
John
John Fox
Department of Sociology
McMaster University
Hamilton, Ontario
Canada L8S 4M4
905-525-9140x23604
http://socserv.mcmaster.ca/jfox
Hello,
I am new to R, and I am trying to figure out how to use it for a
random effects model. I am using version 2.4.0, and I also have the
book Applied Linear Regression by Sanford Weisberg.
I have four variables: Swimmer, Sex, Swim, and Difference. Swimmer
identifies the number assigned
If these are the only variables, you are missing Type. Swim and Type are
both conditioning variables that you need to provide. Maybe your intention
is to use Sex instead of Type.
Try:
?xyplot
And look under x as the 1st argument for xyplot.
Rene
-Original Message-
From: [EMAIL
Hi Experts,
I have a problem in Dates.
I have a zoo object called 'intra'. And the class of index(intra) is
(Chron Dates Time). I need to put the index of this zoo object into a
data frame. So I used,
idat-data.frame(Datetime=as.POSIXlt(index(intra),GMT))
But I get the values of
Hi Experts,
I have a problem in Dates.
I have a zoo object called 'intra'. And the class of index(intra) is
(Chron Dates Time). I need to put the index of this zoo object
into a data frame. So I used,
idat-data.frame(Datetime=as.POSIXlt(index(intra),GMT))
But I get the
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