This won't be as quick as Bill's elegant solution, but it's a one-liner:
apply(d, 1, function(x), match(1, x))
See ?match.
-Peter Ehlers
On 2010-10-22 10:36, David Herzberg wrote:
Bill, thanks so much for this. I'll get a chance to test it later today, and
will post the outcome.
David
Whoops, got an extra comma in there somehow; should be:
apply(d, 1, function(x) match(1, x))
-Peter Ehlers
On 2010-10-24 08:17, Peter Ehlers wrote:
This won't be as quick as Bill's elegant solution, but it's a one-liner:
apply(d, 1, function(x), match(1, x))
See ?match.
-Peter
with it?
-Peter Ehlers
On 2010-10-26 07:50, David Herzberg wrote:
Thank you - I will try this solution as well.
Sent via DROID X
-Original message-
From: Petr PIKALpetr.pi...@precheza.cz
To: David Herzbergdav...@wpspublish.com
Cc: Adrienne Woottenamwoo...@ncsu.edu,
r-help@r
' will no longer
be paired. Another reason to prefer dataframes.
#check out the evidence.something's wrong with loess()???
There's nothing wrong with loess; it just needs more than a
single intercept and slope to plot its predictions.
-Peter Ehlers
par(mfrow=c(1,2))
plot(a,b)
lines(lowess(a,b
- stack(DF)
DF
## use xtabs() to tabulate the values
xtDF - xtabs( ~ ., DF)
## transpose so that columns are the heights to
## feed to barplot()
heights - t(xtDF)
barplot(heights, col=2:4)
-Peter Ehlers
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))
With lattice:
xyplot(Trait ~ Age | Group, data = x)
(I suspect that both Group and Age should really be factors,
which would improve the plot.)
-Peter Ehlers
Le 10/27/2010 11:21, Rosario Garcia Gil a écrit :
Hello
I have a data set summarized like this:
File name= Height
Group Ind Age Trait
1
)
-Peter Ehlers
Thanking in advance.
Regards
Vincy
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only the target result.
Use unname(x) or as.vector(x) on the result.
-Peter Ehlers
Regards,
David S. Herzberg, Ph.D.
Vice President, Research and Development
Western Psychological Services
12031 Wilshire Blvd.
Los Angeles, CA 90025-1251
Phone: (310)478-2061 x144
FAX: (310)478-7838
email
?
-Peter Ehlers
Thanks, Jason
GLD.Close
2010-10-01128.91
2010-10-04128.46
2010-10-05130.99
2010-10-06131.81
2010-10-07130.37
2010-10-08131.66
2010-10-11132.29
2010-10-12131.96
2010-10-13134.07
2010-10-14134.75
2010-10-15133.68
2010-10-18
problem.
That would be the best way for this (common) situation since
the ordering is not likely to be wanted in any other way.
For playing with different orderings, lattice provides
the index.cond argument described in the '...' part of
help(xyplot).
-Peter Ehlers
Hope it helps,
Rainer
to lead to some 'extension' of the xlims. If you
want to avoid that, you'll have to specify different height/width
values, something like:
x11(height = 7, width = 7 * (100 - 0)/(70 + 80))
and then place your plot call. You might also want to set the
xaxs and/or yaxs parameters.
- Peter Ehlers
AND
This indicates that you haven't looked at (or absorbed) the
help page for logic operators. It's pretty well explained
there that both and perform AND, but the first is for
elementwise comparison which is what your application
clearly requires.
-Peter Ehlers
Thanks a lot.
Martin
On 11/18
are directed by ?plot.
Peter Ehlers
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.
If you want your tick marks to point inward, you might
also have a look at par(tcl = ).
Peter Ehlers
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to lengthen the segment.
Maybe this post will help you:
https://stat.ethz.ch/pipermail/r-help/2010-July/246599.html
Peter Ehlers
Thank you
Felipe Parra
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to the maintainer of sm.
Peter Ehlers
Thank you in advance,
Lucía
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(-.7,1.2))
points(eventtime, impact, pch=95, font=5, cex=2, col=4)
text(eventtime, impact, label, pos = 1 + 2*(impact 0))
})
abline(h=0, lwd=2)
axis(1, pos=0, lwd=2, lwd.ticks=1)
Peter Ehlers
--
mrgomel
---
Below is my example code in R:
the_data-
structure(list
,...)
})
or, putting the grid call inside the panel.xyplot:
xyplot(value ~ date, data=X,
panel=function(x,y,...) {
panel.xyplot(x, y,
grid=list(v=-1, h=-1,
col=3, x=x, y=y), ...))
})
Peter Ehlers
Thanks.
- Elliot
) is in the posting guide,
but it seems that nobody wants to read that quite
brief document.
Peter Ehlers
From: Jorge Ivan Velezjorgeivanve...@gmail.com
Cc: r-help@r-project.org
Sent: Thu, November 25, 2010 4:46:37 PM
Subject: Re: [R] overlap cdf plots and add
even the median.
Is there a way to apply a generic function in the manner I described?
You could use the abind function in the abind package to
convert your list to a 3d array and then use apply on that:
require(abind)
xa - abind(x, along=3)
apply(xa, 1:2, mean, trim=0.3)
Peter Ehlers
= list(
xlab = 3
# axis.xlab.padding = 3
# axis.bottom = 3
)))
Comment/uncomment to suit.
Peter Ehlers
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how you are using lm()
and that will generate the error?
Peter Ehlers
--
dr. katharina manderscheid
soziologisches seminar
universität luzern
kasernenplatz 3
6000 luzern 7
tel. ++41 41 228 4657
web: http://www.unilu.ch/deu/dr.-katharina
predictors;
etc, until the problem appears.
It's almost certain that the problem is not with R.
Peter Ehlers
thanks,
katharina
Von: Peter Ehlers [ehl...@ucalgary.ca]
Gesendet: Donnerstag, 25. November 2010 21:54
An: Manderscheid Katharina
Cc: 'r-help@r
]
or
summary(rt.aov)[[2]][[1]][1,Pr(F)]
Peter Ehlers
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and the origin is
detached, but throgut pulling the window, I can get the one like fig_1.
Now, I want to know how to use the code to obtain directly the formation in
fig.1(orgin together)?
Just use xaxs='i', yaxs='i' and xlim=c(3,8), ylim=c(3,8).
Peter Ehlers
thanks
kevin
)
This can't be right - your parentheses don't match.
And if you have your graphical parameters *inside* the
envelope function, you certainly won't get what you want.
I just tried the example on the help page for plot.envelope
with main/xlab/legend settings and had no problem.
Peter Ehlers
But somehow
numerical format. How can I get
that?
You'll need a greater penalty; see ?options.
Peter Ehlers
Thanks in advance.
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PLEASE
:-(
xyplot. - function(u) {
xyplot(u[,2] ~ u[,1],
par.settings = standard.theme(color=FALSE))
}
see ?xyplot.ts
Peter Ehlers
Cheers,
Marius
On 2010-11-27, at 12:00 , ottorino wrote:
Il giorno sab, 27/11/2010 alle 00.35 +0100, Marius Hofert ha scritto:
The reason why I would
that it is probably not R but W XP.
It's not likely to be R.
How are you reading your pdf docs? Maybe you
have some multiple clipboards software running that's
confusing things.
Peter Ehlers
Cheers
Troy
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= transparent),
clip = list(panel = off
(You have to modify the standard.theme parameters.)
See ?modifyList and also:
http://www.mail-archive.com/r-help@r-project.org/msg64699.html
Peter Ehlers
Cheers,
Marius
On 2010-11-27, at 13:16 , Peter Ehlers wrote:
On 2010-11-27 03:31
across something that is known, or it is meant to
behave this way, I couldn't find anything.
Hmm, how hard have you looked?
I doubt that I'll be the first to remind you to check the FAQ.
Peter Ehlers
Cory Rieth
R.version() output:
platform x86_64-apple-darwin9.8.0
arch x86_64
os
to predict and the second column contains the 'curveid',
aka the grouping variable (if you have one). In your
case, the grouping reduces to a single group, i.e. a
vector of all '1's.
This could all be made considerably clearer in the documentation.
Peter Ehlers
Brant Inman
, here's the next step:
v - sample(50, 100, replace = TRUE)
result - diff(v) / v[-length(v)]
Peter Ehlers
On Sun, Nov 28, 2010 at 7:29 PM, ericericst...@aol.com wrote:
Just starting to learn R so excuse me if this is a simple question. I'm
wondering how I get the percent difference
-major order, transpose first:
ttmp - t(tmp)
ttmp[, p] - vec
(ans - t(ttmp))
Peter Ehlers
TIA. Bryan
*
Bryan Hanson
Professor of Chemistry Biochemistry
DePauw University, Greencastle IN USA
sessionInfo()
R version 2.12.0 (2010-10-15)
Platform: x86_64-apple-darwin9.8.0/x86_64
(experiment1, experiment2),
fill = c(red, green))
Does anyone know how I can surpress these labels for the second boxplot?
Perhaps all you need is a bit more care in typing: naxt???
Peter Ehlers
Thanks in advance,
Karin
__
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. Here's
how you can get the same result from both methods:
1. use na.action = na.pass in aggregate.formula;
this will duplicate your x1 result.
2. use d - d[complete.cases(d), ] in your x1 calculation;
this will duplicate your x2 result.
Peter Ehlers
thanks very much
david freedman, atlanta
/default/langref_sect266.htm
and here for python
http://adorio-research.org/wordpress/?p=262
Check swp() in pkg:ggm.
Peter Ehlers
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to be something
to do with the way rasterImage and pdf play together. But this
works for me: insert the following line
do.call(clip, as.list(par('usr')))
before your final map call.
Peter Ehlers
Thanks,
Ben
On Nov 29, 2010, at 2:03 PM, Ben Tupper wrote:
Hello,
Below is a function
this
xyplot(bbb ~ aaa | ddd * fff)
Peter Ehlers
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and provide commented, minimal, self-contained
be really grateful if anyone could tell me if there is a simple
method of achieving this.
You probably want to look at ?mtext and especially its
arguments 'outer', 'side', 'adj' and 'line'.
But make sure that there's enough space in the margin with
par(oma = c()).
Peter Ehlers
Many thanks
Sam
anything wrong, really. But do check
range(x)
Should be quite a range; that's what 'heavy tails' means.
To see the fit, try this:
hist(x, freq=FALSE, breaks=5000, xlim=c(-6,6))
curve(dcauchy(x), col = 2, add = TRUE)
Peter Ehlers
x- rchisq(1, df = 4)
hist(x, freq = FALSE, breaks=100
,
panel=function(...){
panel.rect(90,0,98,1000,col='bisque',border=NA)
panel.histogram(...,col='transparent',lwd=2)
}
)
Again, you can define bar colours with a colour vector.
Lattice is more customizable albeit a little harder
to learn.
Peter Ehlers
Thanks,
Jason
Try fitdistr() in pkg MASS.
-Peter Ehlers
On 2010-07-12 11:17, Oscar Rodriguez wrote:
Dear R community:
Sorry if this question has a simple answer, but I am a new user of R.
Do you know a command, or package that can estimate the weibull distribution's
mean, standard deviation
611
2 84 473
You could consider using the Fligner-Killeen test in base R.
See ?fligner.test.
-Peter Ehlers
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object; it's a list as
class(compo) or reading the help page for decompose.graph
would tell you.
-Peter Ehlers
Thanks
chakri
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PLEASE do read the posting guide http
.
As to your example: there is no sense at all in doing a
test on such data (other than to satisfy some hypothetical
fanatical journal editor).
-Peter Ehlers
My version for R is 2.11.1 (2010-05-31) running on x86_64 GNU/Linux
(RHEL).
Thanks in advance for any help with this.
Govind
your groups on. SPSS uses the
mean, car uses the median.
-Peter Ehlers
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and provide commented
to the boxes scaled with 0.1.
Am I missing something?
Maybe you're looking for the 'outcex' parameter?
See ?bxp.
-Peter Ehlers
Thanks in advance!
Robert
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PLEASE do
(), but be warned: any non-numeric columns will
cause trouble.
-Peter Ehlers
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and provide commented
I would change that first dataOnly in
by(...) or lapply(...) to dataOnly[,-3].
In fact, if the dataframe mydata is suitably
subset, then, because of the as.matrix() in
function(x), both the by() and lapply() methods
will work fine with mydata.
-Peter Ehlers
On 2010-07-15 15:42, Phil Spector
)
KLdiv(mydata)
KL seems to be not defined. Can somebody explain what is going on?
Thanks,
Ralf
Ralf,
You can adjust the 'eps=' argument. But I don't know
what this will do to the reliability of the results.
KLdiv(mydata, eps = 1e-7)
-Peter Ehlers
invalid data in some of your files.
Perhaps missing values coded as '*' or as 'N/A'. If so, set
the na.strings= argument.
-Peter Ehlers
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and it would read out ben
Check out HWidentify or HTKidentify in pkg:TeachingDemos.
-Peter Ehlers
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
. And pcaMethods has slplot().
Please, folks, do tell us what packages you are using!
[Now, I'm having trouble with the pemqzrsTawYY function;
can anybody help? :)]
-Peter Ehlers
On 2010-07-17 16:32, Dennis Murphy wrote:
In which package would one find slplot? It's not in any of the 1800+
packages on my
You can always use the identify() function which will
leave the id of the point on the plot; see ?identify.
But you do eventually have to sort out your Mac problem.
Sorry, I can't help with that.
-Peter Ehlers
On 2010-07-18 7:43, James Platt wrote:
This is exactly what I want as I will have
=' argument. But I don't know
what this will do to the reliability of the results.
KLdiv(mydata, eps = 1e-7)
-Peter Ehlers
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'???')
-Peter Ehlers
Regards,
Sean.
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Don't thank me, thank the authors of sos (Spencer Graves,
Sundar Dorai-Raj, and Romain Francois) and Jonathan Baron
for hosting the search site.
-Peter Ehlers
On 2010-07-18 23:43, Sean Carmody wrote:
Thanks Peter...I hadn't some across the sos package before, but I'm
sure I'll be putting
, ...){
panel.xyplot(x, y, ...)
panel.text(x, y, labels = iris[,5], cex = 0.5)
panel.abline(h = 0, v = 0)
}
)
-Peter Ehlers
Thank you kindly,
--
Shawn Way, PE
MECO, Inc.
(p) 281-276-7612
(f) 281-313
dotplot(test)
useful, as well.
-Peter Ehlers
Thank you kindly,
--
Shawn Way, PE
MECO, Inc.
(p) 281-276-7612
(f) 281-313-0248
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.
Otherwise, I think you can make your inherited code work
with something like
Usr - par('usr')
Pin - par('pin')
dux - Usr[2] - Usr[1]
duy - Usr[4] - Usr[3]
uin - c(dux,duy)/Pin
to replace
uin - par(uin)
But you'll also have to fiddle with atan()
-Peter Ehlers
On 2010-07-19 9:20, Michael Friendly
)
)
ddply(ma, .(variable), function(x) f(x, v = x[[value]]))
#
Another option is to use data.frame in place of summarise:
#
f - function(x,v) data.frame(
mean = mean(v),
sd = sd(v),
skewness = skewness(v),
mean.gt.med = mean.gt.med(v)
)
#
-Peter Ehlers
On 2010-07-21
undesirable side
effects, but I haven't given it much thought. I haven't
had any problems with this version of legend().
-Peter Ehlers
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On 2010-07-22 10:43, Wu Gong wrote:
Hi William,
I'm curious about that you used d[]- lapply(d, factor...
Could you please tell me if there are any differences between d[] and d?
Thank you.
Why not try it both ways and inspect the result.
d[] -
d -
Obvious?
-Peter Ehlers
= data1, data2 = data2)
lapply(L, t.test)
-Peter Ehlers
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and provide commented, minimal, self
(bquote(paste(.(a), : , bar(zeta),
Boxplot from 2001 to 2009, sep = )))
It seems that setting main=... where ... contains
bquote() works with plot(), but not with boxplot().
-Peter Ehlers
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On 2010-07-22 14:40, David Winsemius wrote:
On Jul 22, 2010, at 4:24 PM, Peter Ehlers wrote:
On 2010-07-22 11:44, Marcus Liu wrote:
Hi everyone, I am plotting a boxplot with main title as main =
bquote(paste(.(ts.ind[s]), : , bar(zeta), Boxplot from 2001 to
2009, sep = )) but it doesn't
On 2010-07-22 15:21, David Winsemius wrote:
On Jul 22, 2010, at 5:01 PM, Peter Ehlers wrote:
On 2010-07-22 14:40, David Winsemius wrote:
On Jul 22, 2010, at 4:24 PM, Peter Ehlers wrote:
On 2010-07-22 11:44, Marcus Liu wrote:
Hi everyone, I am plotting a boxplot with main title as main
calling bwplot().
(BTW: is your : meant to be |?)
-Peter Ehlers
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Another suggestion: compare
-1:4
with
-(1:4)
-Peter Ehlers
On 2010-08-12 12:28, Amit Patel wrote:
Hi
I am trying to calculate the row sums of a matrix i have created
The matrix ( FeaturePresenceMatrix) has been created by
1) Read csv
2) Removing unnecesarry data using [-1:4,] command
'DATA$labs' when
using the data= argument?
JJ
I don't think that's expected behaviour, nor do I think that it occurs.
There must be something else going on. Can you produce this with a
small reproducible example?
-Peter Ehlers
On Wed, Aug 18, 2010 at 11:29 AM, David Winsemiusdwinsem
what gives and it
may well be a design decision that I'm not aware of. I can't see
anything in the help page that refers to this effect.
-Peter Ehlers
On Wed, Aug 18, 2010 at 5:52 PM, David Winsemiusdwinsem...@comcast.netwrote:
On Aug 18, 2010, at 6:45 PM, Peter Ehlers wrote:
On 2010-08-18
or matrices. These can be given as named arguments.
Surely, you checked?
-Peter Ehlers
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and provide
to understand aspect ratio even
better.
Try this:
x11(width = 10, height = 5)
plot(X,Y,pch=+,col=blue,xlim=c(-2.5,2.5),ylim=c(-2.5,2.5))
(using Ted's X,Y) and compare with the 'asp' version.
-Peter Ehlers
On Aug 19, 2010, at 3:05 PM, (Ted Harding) wrote:
Spencer, you came up with your example just
(quantile) shows that there are quite a few ways to define
quantiles and that R defaults to 'type=7'.
-Peter Ehlers
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SSweibull(Time + 0.01, ...). But a better way would
be to use
init - getInitial(Level~ SSweibull(Time,Asym,Drop,lrc,pwr))
(which will often give the converged values). You can
follow up with
fm - nls(Level ~ Asym-Drop*exp(-exp(lrc)*Time^pwr), start = init)
-Peter Ehlers
kc
On Thu, Aug 26
Just a small fix to my solution; inserted below.
On 2010-08-27 3:51, Peter Ehlers wrote:
On 2010-08-26 15:52, Marlin Keith Cox wrote:
I agree. I typically do not use non-linear functions, so am seeing the
art in describing functions of non-linear plots. One last thing. I
tried
to use a self
it to look at potentially interesting parts.
If I remember correctly, there's also the xysplom() function
in pkg:HH.
-Peter Ehlers
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device. But even then, there's bound to be someone
who'll want a pairs plot for 1000 variables.
As usual with R, improvements are always welcome.
You could submit appropriate code and, if it is deemed useful,
I'm fairly sure that a better pairs() function will become
part of Rx.xx.x.
-Peter Ehlers
it useful to add the curve you get with your start values
to the scatterplot to see how reasonable the values are:
plot(Joules~Days)
curve(8 + 9 * exp(-.229 * x), add=TRUE)
After you get a convergent solution, you can add a curve with
the model values.
-Peter Ehlers
function?
Best, Niels
Looks like that's fixed in R2.12.0.
(It's always a good idea to check the development version.)
-Peter Ehlers
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One comment on the function: I see that it uses T/F instead of
TRUE/FALSE in a number of places. You'll save yourself some
headaches if you replace those 'T/F's.
-Peter Ehlers
On 2010-09-08 1:25, DrCJones wrote:
Hi,
How does R automatically load functions so that they are available from
your version of R produces different results from mine. I
don't know if I should open a bug report. It would be good if someone
You're doing the right thing by asking here first before reporting.
It would definitely not be a good idea to report a (non-)bug
in an outdated version of R.
-Peter
a matter of taste, but I would use \341 and \361.
However, these are still not scalable, AFAICS.
-Peter Ehlers
On Sat, Sep 11, 2010 at 10:01 AM, baptiste auguie
baptiste.aug...@googlemail.com wrote:
What do people use to show angle bracketsin R graphics? Have I
missed something obvious
message or to coerce to
POSIXct. Note that
all.equal(as.POSIXct(x), as.POSIXct(x))
yields TRUE, as does using as.Date(x).
-Peter Ehlers
My system:
R version 2.11.1 (2010-05-31)
x86_64-pc-mingw32
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3
. Is there a way that the above line could be
modified so that it would use the group 3 itself, rather than the
corresponding charecter?
Wrap your pch values in as.character().
-Peter Ehlers
thanks again,
k
On Mon, Sep 13, 2010 at 5:00 PM, David Winsemiusdwinsem...@comcast.netwrote:
On Sep
=' argument to key():
key(divide = 1, etc)
See the 'key' section in ?xyplot.
-Peter Ehlers
Thanks very much!
John
- Original Message
From: David Winsemiusdwinsem...@comcast.net
To: array chiparrayprof...@yahoo.com
Cc: r-help@r-project.org
Sent: Mon, September 13, 2010 4:05:04
(which.max(score
-Peter Ehlers
On Wed, 15 Sep 2010, Kevin Burnham wrote:
How can I get the outlier in this boxplot of Score to be represented by
the corresponding value in SubNo?
score=c(6,6,7,14,5,7,6,8)
SubNo=1:8
mydata=data.frame(SubNo, score)
boxplot(mydata$score)
Thanks!
Kevin
for me.
As to trafo and contrMat, try this:
?trafo ## in pkg:coin
?contrMat ## in pkg:multcomp
-Peter Ehlers
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PLEASE do read the posting guide http://www.R
(~ . + .(x)))
print(fo)
}
(using update.formula() as illustration)
-Peter Ehlers
On Sat, Sep 18, 2010 at 9:30 AM, Benjamin Godlovebgodl...@oberlin.edu wrote:
Hi,
First let me say I am a big fan of R and appreciate all your time and
effort.
The update() function does not seem to work
('testfile.nc',write=TRUE)
t - ncobj$dim[['t']]
c - var.def.ncdf('c','hrs',t,missval=-999)
ncobj - var.add.ncdf(ncobj,c)
close(ncobj)
tmp - open.ncdf('testfile.nc',write=TRUE)
put.var.ncdf(tmp,c,c(.1,.2,.3,.4,.5))
close(tmp)
-Peter Ehlers
__
R
getting error:
chol(dat['A1'])
Error in chol.default(dat[A1]) : non-numeric argument to 'chol'
Can somebody point me where I am doing wrong?
You need another set of brackets:
chol(dat[['A1']])
-Peter Ehlers
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On 2010-09-21 5:51, Nikhil Kaza wrote:
example(factor)
iris1$Species- factor(iris1$Species, drop=T)
will get you what you need.
Hmm, doesn't work for me. ?factor does not list a 'drop='
argument.
-Peter Ehlers
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North
=c(0,0),type='p',
points=list(col=1:2,pch=0:1,cex=2,lwd=2),
text=list(lab=c('A','B'),cex=1.5,font=2)))
Any suggestions?
You can add this line to your xyplot call:
par.settings = list(grid.pars = list(lwd = 2)),
-Peter Ehlers
Thanks
John
- Original Message
From: array
with
setting lex=2, but that would thicken all lines including the
axes and tick marks.
I'm not sure that my advice was the best; it's just what
occured to me at the moment. Perhaps Deepayan will weigh in
with the definitive solution.
-Peter Ehlers
- Original Message
From: Peter Ehlersehl
=TRUE, axes=FALSE)
box(); axis(1)
axis(2, at=1:4, lab=NA)
mtext(levels(df$SPECSHOR), side=2, at=1:4, line=1, cex=0.8)
-Peter Ehlers
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.
-Peter Ehlers
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and provide commented, minimal, self-contained, reproducible code.
,
vertical = TRUE, at = 1:3 + 0.1,
axes = FALSE,
pch = 21, bg = orange, cex = 1.5)
axis(1, at = 1:3, labels = c(0.5, 1, 2))
axis(2)
box()
legend(2, 9, c(Ascorbic acid, Orange juice),
pch = 21, pt.bg = c(yellow, orange), pt.cex = 1.5)
-Peter Ehlers
On 2010-09-26 4:48
augment the data frame appropriately instead.
-Peter Ehlers
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and provide commented, minimal, self
(key.right = 0.8))
You might then have to adjust the space on the right with
par.settings = list(layout.widths = list(key.right = 0.8,
right.padding = 2))
-Peter Ehlers
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