On 16 Sep 2005, at 17:21, Sundar Dorai-Raj wrote:
My guess is he wants this:
c1 - factor(c)
d1 - factor(d)
m - lmer(a ~ b + (1|c1:d1)+(1|c1))
which assumes d1 is nested within c1.
Take a look at Section 3 in the MlmSoftRev vignette:
library(mlmRev)
vignette(MlmSoftRev)
Ah, that
I have a vector of over 7,000 chron dates in the format mm/dd/yy. I need
to increment each date in the vector by a standard number of months.
Lapply with seq.dates is working OK; this increments the vector x by 3
months:
dates(unlist(lapply(x, function(g) seq.dates(g, by=months, length=4)[4])))
I don't know if this is any faster but it does not use any loops or apply
so maybe it is. x is assumed to be a vector chron dates. Note that
the question is not well specified for days near the end of the month
and I have not addressed that.
with(month.day.year(x), {
year - ifelse(month
On 9/18/05, Yan Wong [EMAIL PROTECTED] wrote:
On 16 Sep 2005, at 17:21, Sundar Dorai-Raj wrote:
My guess is he wants this:
c1 - factor(c)
d1 - factor(d)
m - lmer(a ~ b + (1|c1:d1)+(1|c1))
which assumes d1 is nested within c1.
Take a look at Section 3 in the MlmSoftRev
I have a couple of other comments. You can write the nested grouping
factors as Sundar did without having to explicitly evaluate the
interaction term and drop unused levels. The lmer function, like most
modeling functions in R, uses the optional argument drop.unused.levels
= TRUE when creating
Announcement of a major revision of package tuneR
The package tuneR for the analysis of sound has been updated, improved,
and extended.
Most notably the functionality of the orphaned package
sound has been merged into tuneR. Because tuneR's representation and
API is very different, it was not
To elaborate a bit on Harold's reply, the use of a spaghetti plot
for more than 4 or 5 groups is not recommended. Consider the
longitudinal data sleepstudy in the Matrix package. A spaghetti
plot can be produced by
data(sleepstudy, package = Matrix)
library(lattice)
xyplot(Reaction ~ Days,
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Dear Group Members,
Forgive me if I am a little bit out of subject. I am looking for a good
way to test the homogeneity of two variance-covariance matrices using R,
prior to a Hotelling T² test. You’ll probably tell me that it is better
to use a robust version of T², but I have no precise
Hi,
In order to make a movie-like animation of different graphs
with replayPlot inside a fast loop:
is there a way to avoid the appearance of some
white stripes/streaks between the different calls to replayPlot?
thanks in advance
Johan VH
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subject: trimmed mean in R seems to round the trimming fraction
to [EMAIL PROTECTED]
Consider the following example of 10 numbers. 10% trimmed mean is correct
but you can see that the result is the same for many trimming fractions
till 0.20!
For example 13% trimmed mean should use
[EMAIL PROTECTED] wrote:
subject: trimmed mean in R seems to round the trimming fraction
to [EMAIL PROTECTED]
Consider the following example of 10 numbers. 10% trimmed mean is correct
but you can see that the result is the same for many trimming fractions
till 0.20!
For example
Or the longer version help(Boston)
Cheers
Francisco
From: Gabor Grothendieck [EMAIL PROTECTED]
Reply-To: [EMAIL PROTECTED]
To: Jun Ding [EMAIL PROTECTED]
CC: r-help@stat.math.ethz.ch
Subject: Re: [R] Variable descriptions of a built-in dataset
Date: Sun, 18 Sep 2005 01:33:15 -0400
?Boston
On
I have not seen a reply to this, so I will offer a comment.
1. You are doing something very wise in reading Pinheiro and Bates
(2000) Mixed-Effects Models in S and S-Plus.
2. Unfortunately, beyond this, I don't really understand the
question; if others had
Dear R Users,
There exists a non-central hypergeometric distribution function in the
(MCMCpack) package, and a hypergeometric distribution function in the (stats)
package.
Is there a function for sampling from an extended hypergeometric distribution?
Thanks,
Narcyz
This message is
?order
Martin Lam wrote:
Hi,
I was wondering if someone know how to sort a data set
by column.
I've tried sort() but without luck. I would think
there should be a function for it somewhere. An
example with the iris data set would be appreciated.
Thanks,
Martin
If you want to sort a data frame according a variable, the use it like
datfram[order(datfram$var),]
to sort all the data.frame, not just a variable
Spencer Graves wrote:
?order
Martin Lam wrote:
Hi,
I was wondering if someone know how to sort a data set
by column.
I've tried sort()
Hi R-users,
I'm running R under Unix and producing postscript output of graphs.
Soemtimes, some tick axis labels disappear from my output.
eg if I have a vector
months - c(J,F,M,A,M,J,J,A,S,O,N,D)
Then the M and one or 2 other letters are dropped from the axis.
This seems to be a size
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