before or immediately
after the first plot statement (the type='n' one) in the above code
fragment, the resulting graph still only shows one column of data.
Have I misinterpreted the instructions or the functionality of new=TRUE?
Thank you,
Paul Lemmens
similar result with using points
Yes I new that, but I wanted to try and go without an if() for
deciding between the first and consecutive columns.
Thnx for helping out!
Paul Lemmens
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I always use -loc for the y-axis of a
similar plot for a completely different data set.
Kind regards,
Paul Lemmens
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PLEASE do read the posting guide http://www.R
On 7/30/06, Kartik Pappu [EMAIL PROTECTED] wrote:
Hello All,
I have a device that spews out experimental data as a series of text
files each of which contains one column with several rows of numeric
data. My problem is that for each trial it gives me one text file
(and I run between 30 to
Dear all,
I have carried out a pairwise comparison study that I want to analyze
using the BradleyTerry package to establish a rank order of my
stimuli. However, BT does not handle ties between stimuli, so I need
to find those in my data before I can use that model.
The code below goes from the
, sep=), skip=1)))
rm(datafiles)
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Hoi Dan,
--On donderdag 16 september 2004 13:55 +0100 Dan Bolser
[EMAIL PROTECTED] wrote:
Is there an R cookbook?
Yes there is (sort of): StatsRus http://www.ku.edu/~pauljohn/R/Rtips.html
kind regards,
Paul
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for Anova
Phrased differently, ?Anova says Calculates type-II or type-III
analysis-of-variance tables for model objects produced by 'lm' and 'glm',
so it's not suitable for the aovlist that aov() with Error()-term returns.
How can I compute Type III SS for such objects?
kind regards,
Paul
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Paul
, that several online communities have
dedicated discussion groups for R. One of them is Orkut.com. The name of
the other one slips my mind, but if you search the archives for my name and
orkut, then you'll probably find those emails quickly.
regards,
Paul
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-0.018
I mean how to select the first four columns.
subset(df, select=-c(V3,V4,V5,V6))
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,
Paul
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Phonenumber
toolbox.r now contains:
my.subset - function(x, drop.unused.levels=FALSE, ...) {
subsetted - subset(x, ...)
if (drop.unused.levels) {
subsetted[] - lapply(subsetted,
function(x) if (is.factor(x)) factor(x) else x)
}
subsetted
}
Thnx for all the help!
--
Paul Lemmens
NICI
numbers) for
which positions in dat Diagnosis==0.
cdat - dat[dat$Diagnosis==0,]
This OTOH, uses the above vector to index the rows of dat, indeed selecting
those rows from dat that have Diagnosis==0. This is assigned to cdat. You
could have done
cdat - dat[csubset,]
as well
HTH
--
Paul
- subset(dd, c==1, drop=TRUE)
table(dd$c)
1 2
5 0
So to lose the second level of dd$c, in method B I still need to 'dd$c -
dd$c[, drop=TRUE]', while the manual seems to imply that with the drop
argument to subset() this would not be necessary.
Could you comment?
kind regards,
Paul
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Paul Lemmens
Dear Peter,
--On woensdag 16 juni 2004 17:06 +0200 Peter Dalgaard
[EMAIL PROTECTED] wrote:
Paul Lemmens [EMAIL PROTECTED] writes:
Hello!
If I read ?subset, the workings of the argument drop (to me) seem to
imply equivalence of A and B (R 1.9.0):
# A
dd - data.frame(rt=rnorm(10), c=factor(gl(2,5
/will
function as the low-entry level help that was discussed around a year ago.
kind regards, Paul
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[,!Besch]
attach(x)
x[-Besch]
...
Try
x2 - subset(x, select=-Besch)
kind regards,
Paul
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on (if it were the type I/III
difference then the exp interactions would have been different as well,
right?). I cannot deduce the reason of the difference. Can anybody help
with this?
thank you for your time,
regards,
Paul Lemmens
--
Paul Lemmens
NICI, University of Nijmegen ASCII
$facts[2])
# This won't work here. How do I continu?
# Or perhaps also
# data.tmp - data$resp[data$facts[1] == 'i']
}
thank you,
Paul Lemmens
P.S:
str(wery)
`data.frame': 150 obs. of 11 variables:
$ V1 : int 1 1 1 1 1 1 1 1 1 1 ...
$ V2 : int 1 2 3 4 5 6 7 8 9 10 ...
$ V3 : int 960 1060
extension
regards,
Paul
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) that the
only way to have if() know that an else will be present is to put it on the
same line as the closing curly brace of the if() (compound) statement. But
if I look at some code from, e.g., aov and lm, I see plenty violations of
that rule.
regards,
Paul
--
Paul Lemmens
NICI, University
)
is a multiple of 5.
kind regards,
Paul
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summary. I've already memorized the is.na()
construct, so I should be safe for the time being :
kind regards,
Paul
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the Introduction to R (p21) helpfull?
regards,
Paul
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(*) Such a remark will probably lead to some kind of reprimand because it's
probably somewhere within the 10e6 manual pages but I'm trying my luck here.
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the plot region?
When I am trying to put a legend on a plot where I am using lines, R
just
ignores it. I can do it with boxplot or plot, but just not with lines.
Am I
doing something wrong? Maybe I am just making a mistake?
--
Paul Lemmens
NICI, University of Nijmegen ASCII Ribbon
as values?
xbar1 - tapply(as.vector(x), gl(2,4), mean);
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it off and
print() calls display immediately. Lengthy output becomes slower,
though.
If you don't want to depend on you (or other people) turning of the
buffering, use something like
cat(this or that); flush.console.
regards,
Paul
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Paul Lemmens
NICI, University of Nijmegen ASCII
Hoi Thomas,
--On woensdag 28 mei 2003 7:16 -0700 Thomas Lumley
[EMAIL PROTECTED] wrote:
On Wed, 28 May 2003, Paul Lemmens wrote:
Hi!
Apologies for sending the mail without any code. Apparently somewhere
along the way the .R attachments got filtered out. I have included the
code below as clean
number of means calculated.);
}
else
{
subject.bin.means
}
}
-- Forwarded Message --
Date: dinsdag 27 mei 2003 14:53 +0200
From: Paul Lemmens [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: [R] Numbers that look equal, should be equal
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