s, but here's what you posted as your 'last line of code':
lines(modb, predict(nls.2009, lines(as.numeric(x)=modb)))
Perhaps just a typo: lines -> list???
In any case, the newdata in predict should have a variable 'Year'.
Peter Ehlers
I've been using this code
nk that lines(as.numeric(x)=modb)
would qualify as newdata.
It's usually a bad idea to shove too much stuff into a single command
and a good idea to use str() often.
This 'exact' code worked in the past?
Peter Ehlers
The model is fine, but it's the plotting of the mo
On 2011-07-12 07:03, Sam Steingold wrote:
[snip]
the totally unnecessary semi-colons)
then why are they accepted?
optional syntax elements suck...
They're accepted because they *can* be useful (multiple
statements on one line).
Is there *any* language that can *not* be abused?
, then you should provide a
reproducible example that illustrates the problem.
Peter Ehlers
Regards,
Dieter
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uld be very much
appreciated.
I don't know anything about a Brice-Model, but I doubt that you have
it right. 4*pi*390*Na is about 3*10^27 and that doesn't seem right
to me.
In addition, do put your data into a data.frame (containing
variables r and D) and learn to use dput for post
NULL.
Katia probably just needs to do do two things:
1. extract array X from list X (e.g. with X <- X$X).
2. learn to read the documentation of functions she's trying to use.
Peter Ehlers
Y<- X
dim(Y)<- c(64,64,21,300)
Y[ 1, 1, 1, ]
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l.xYplot
mn <- with(bin.data, tapply(y, x, mean))
M <- Cbind(mn, lower = mn - .1, upper = mn + .1) ## NB: capital 'C'
stripplot(jitter(y, factor =.6) ~ x,
data = bin.data,
ylab = "",
panel = function(...) {
panel.stripplot(..., jitter=TRUE, factor
0)
title(bquote( x %<=% .(x) ))
?plotmath
Peter Ehlers
Thanks
John
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide co
st"
m.out.base<- matchit(formula=f, data=A, method=m)
B<- match.data(m.out.base)
An<- nrow(A)
Bn<- nrow(B)
Cn<- An - Bn
C<- ??
Can't you just use
idx <- setdiff(rownames(A), rownames(B))
C <- A[idx, ]
Peter Ehlers
From
ta frame a matrix? Just like the test
data ¡§gasoline¡¨ in your PLS package:
library(pls)
data(yarn)
is.vector(density) returns TRUE
is.matrix(NIR) returns TURE (NIR is a matrix)
This is all perfectly well described in ?data.frame where
you will also find reference to the use of I() to
accompli
logCFU ~ t | microorg + tratam,
strip = function(..., which.given, par.strip.text)
strip.default(...,
which.given = which.given,
par.strip.text = list(
font = myfonts[which.given])))
Peter Ehlers
Thank you for your help.
Kenn
egrand <- function (x, mu){
to
{integrand <- function (mu, x){
Peter Ehlers
Thank you!
Hannah
f1<- function(x)
{integrand <- function (x, mu){
dnorm(x, mean=mu, sd=1)*dnorm(mu, mean=2, sd=1)
}
integrate(integrand, -Inf,
on(mu){
integrand <- function (x, mu){
dnorm(x, mean=mu, sd=1) * dnorm(mu, mean=2, sd=1)
}
integrate(integrand, -Inf, Inf, mu)[["value"]]
}
But then again, you could just evaluate dnorm(mu, 2, 1).
So I suspect that you want something different.
Ditto for f2.
Peter Ehl
i in seq_along(x) ) ok[i] <- any( apply(y - x[i], 1, prod) <= 0 )
x[ok]
Peter Ehlers
best,
salih
On Fri, Jun 24, 2011 at 4:12 PM, Dennis Murphy wrote:
Hi:
That leaves open several possibilities. Could you please supply a
small, reproducible example (i.e., one that someone can co
names() makes it the "data"
argument.
And it would be better to use the "data" argument than to attach/detach.
Peter Ehlers
This gets part of the way there using the spreadout function in
plotrix, but it apparently needs to be set up so that it only gets
applied within eac
://svn.r-project.org/R/trunk/src/library/stats/src/ks.c
for the relevant C code. Or download the source code from
your favourite CRAN mirror.
Peter Ehlers
Thanks,
-Steve Wolf
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h_Canada.1252
[3] LC_MONETARY=English_Canada.1252 LC_NUMERIC=C
[5] LC_TIME=English_Canada.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
Peter Ehlers
On 2011-06-22 22:14, Dennis Murphy wrote:
Hi:
As Uwe suggested...
pdf('testgraph.pdf')
osed code modification is an improvement on the existing
implementation of rle. Is it impertinent to suggest this R-modification to the
gurus at R?
Best wishes (in flame-war trepidation),
Well, it's not worth a flame, but ...
from the help page (see 'Details'):
"Missing valu
obtaining help.)
Try setting the 'pch' parameter to NA or to "".
Peter Ehlers
Cheers
Anna
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PLEASE do r
?par and check out the 'xpd' parameter.
Peter Ehlers
i don't mind if it looks "messy" and steps on the margin a bit.
- erik
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PLEASE do read
second example
in help('wireframe'):
wireframe(z ~ x * y, data = g, groups = gr,
scales = list(arrows = FALSE,
x = list(at = c(2, 5, 10)),
y = list(at = c(6, 10, 14),
lab = c('A', 'BBB', 'C'))
))
Peter
On 2011-06-04 11:11, Peter Ehlers wrote:
On 2011-06-03 13:34, Jason024 wrote:
I have a data frame like this:
col1 col2
r1 21
r2 43
r3 65
r4 87
r5109
r612 11
r714 13
r816 15
r918 17
r10 20 19
I want to modify this data
been struggling to make it to work. Any help is appreciated!
This seems made for within(); calling your data.frame 'd':
d.new <- within(d, {
col1 <- ifelse(col1 < 12, -1, col1)
col2 <- ifelse(col2 > 10, -1, col2)
;^2,'A'^3,'A'^4)),
par.settings = list(layout.heights = list(strip = 1.5)))
Peter Ehlers
Here is an example that comes up on a search with terms expression&
strip.default (which I thought was the correct argument to the strip
parameter but turns out I was not rememb
thanks
I suspect that you need to use the 'subscripts' argument.
Peter Ehlers
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___
p <- barplot(t(dd), beside=TRUE)
mp
lines(colMeans(mp), cc, col=2, lwd=3)
Peter Ehlers
Many thanks,
Galen
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PLEASE
; ~ .(N) ~
"bar"))), font=2, cex=1.2))
## => "font=2" is ignored (of course)
You could add
N <- as.character(N)
before your call to xyplot.
Peter Ehlers
Cheers,
Marius
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ied adding 0? Or you could use the seq() function.
I assume that you have a reason for wanting this.
Peter Ehlers
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PLEASE do read
uot;^(.*?)\\..*$","\\1",y, perl=TRUE)
}
newx <- unlist(L)
Peter Ehlers
Matt
On Sun, May 29, 2011 at 6:44 PM, Ian Gow wrote:
Not a new approach, but some benchmark data (the perl=TRUE speeds up Jim's
suggestion):
x<- c('18x.6','12x.9','302x.3&
Someone have a straightforward silver bullet?
No silver bullet, but this seems to work:
plot(y ~ x, data=dat, type="n")
points(y ~ x, data=dat, panel.first=bgfun())
(I think that plot.formula may need a fix but
offhand I can't see whether that's easy or hard.)
Peter Ehlers
T
called lines() without
opening a graphics device.
Peter Ehlers
On Tue, May 24, 2011 at 9:22 AM, Steve Lianoglou<
mailinglist.honey...@gmail.com> wrote:
Hi,
On Mon, May 23, 2011 at 11:41 PM, Rekha wrote:
Hello All,*
*I want to draw a histogram with density curve. *
*For that simply i c
different strings it works correctly> upex("bb")
[... snip ...]
Try the plyr package:
require(plyr)
ddply(testframe, .(Name), function(x) {
boxplot.stats(x[["Value"]])[["stats"]][5]})
Peter Ehlers
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o be of the same type). This is documented in ?apply.
I am not sure if this is indended behaviour or not, but is there an elegant
way to apply 'as.list' to all of the dataframe rows without coercing
everything to strings?
This may not be elegant, but why not just use a loop:
L
this case I don't think that combineLimits answers Harold's
request.
Peter Ehlers
Regards
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email: home mac...@northnet.com.au
At 01:29 18/05/2011, you wrote:
On 2011-05-17 06
;:
barplot(t(d[-1]), names.arg=d[,1], beside=TRUE)
Give a careful reading to the definition of the 'height' argument on
the help page.
Peter Ehlers
James
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PLEASE
he dollar habit.
Replace your
example$x
with
example[[x]]
or with
example[, x]
Peter Ehlers
Within the real function, I will continue do calculations on the column of
data. My problem is that I am either getting a character expression or NULL
from my function.
Thanks for your he
(-3,3), c(-3,3), c(-60,90))
)
There's a comment on ?xyplot in the scales section:
"When relation is "free", xlim or ylim can be a list, ..."
Peter Ehlers
Thanks
Harold
sessionInfo()
R version 2.12.0 (2010-10-15)
Platform: i386-pc-mingw32/i386 (32-bit
hasn't yet been introduced to the with() function,
which is linked to on the ?attach page.
Note also this sentence from the ?attach page:
" attach can lead to confusion."
I can't remember the last time I needed attach().
Peter Ehlers
[... no
>= which.max(.df$depth)),]
--> .df[seq_len(nrow(.df))>= which.max(.df$depth),]
(Thanks for providing a simple reproducible example.)
Peter Ehlers
Thanks in advance.
Sam
[[alternative HTML version deleted]]
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I had meant to copy the list on this; must have hit 'Reply'
instead of 'Reply All'.
P Ehlers
Original Message
Subject: Re: [R] rbind with partially overlapping column names
Date: Mon, 16 May 2011 11:14:11 -0600
From: Peter Ehlers
To: Jonathan Flowers
for anyone to do that manually, though not for me. Can anyone
please help me ?
Example:
set.seed(2718)
m <- matrix(sample(1:9, 100, TRUE), 10, 10)
is.na(m) <- sample(100, 20)
d <- as.data.frame(m)
d[rowSums(is.na(d)) / nrow(d) <= 0.2,]
d[colSums(is.na(d)) / ncol(d) <= 0.
On 2011-05-12 07:16, George Locke wrote:
thanks for reading the manual for me :X
For a bit more reading, you could check out ?title.
You could replace your mtext() calls with
title(ylab='Y axis', cex.lab=1.5, line=4, font.lab=2)
Peter Ehlers
2011/5/12 Prof Brian Ripley:
rley, pch=c(1,2), col="black")
No need to specify the panel function. (To see what's wrong with
the panel function, add '...' to the arguments.)
Peter Ehlers
Many thanks in advance.
Esther
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have expected the lines to plot and the axis labels
not to be visible. Are you telling the whole story?
_Minimal reproducible_ code is always a good thing.
Peter Ehlers
Thanks.
--Chris Ryan
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ed.
I see that you've had the solution you seek in terms of round(),
but what exactly is wrong with the 'digits' argument? It works
for me in the cases you cite. (Note that format = "f" is handled
differently from other formats, according to ?formatC.)
Peter Ehlers
Is
ith R 2.13.
plot(1:4, xaxt='n')
axis(1, at=2:3, lab=c('a', 'b'),
col.ticks=3, col.axis=2, lwd=0, lwd.ticks=1)
par(xpd = TRUE)
abline(v = 4)
Peter Ehlers
Nick Sabbe
--
ping: nick.sa...@ugent.be
link:<http://biomath.ugent.be/> http://bio
eters', that
doesn't make sense to me. Can you provide a sensible example?
In any case you would no doubt have to look at the covariance
matrix of the parameter estimates (with vcov()).
Peter Ehlers
str(summary(yourmodel))
Cheers,
Josh
On Tue, Apr 26, 2011 at 11:21 AM, Schatzi wrote:
Put all your p.values into a vector and write that out.
While you're at it, check ?str to see how handy the
str() function is in identifying the pieces of an R object.
Peter Ehlers
Thank you very much for your time.
Bruce Kindseth
[[alternative HTML version de
mount, I would choose
Peter D's (f4) for its simplicity and transparency.
I kind of figured that you'd do your magic and pull
out a speed demon.
Peter Ehlers
f1<- function (x)
+apply(x, 1, function(row) sort(row, decreasing = TRUE)[2])
f2<- function (x)
+ -appl
u asked for more _efficient_ and
I just thought 'shorter code'. I should know
that whenver I think 'apply', I should think
'matrix'.
Peter Ehlers
I hope it helps.
Best,
Dimitris
On 4/26/2011 2:01 PM, Lars Bishop wrote:
Hi,
I need to extract the second largest element f
pply(a, 1, order, decreasing=T)[2,])
ans<- sapply(1:length(sec.large), function(i) a[i, sec.large[i]])
ans
Try
apply(a, 1, function(x) sort(x, decreasing=TRUE)[2])
Peter Ehlers
Thanks in advance for your help,
Lars.
[[alternative HTML versio
f, plot(Sq~FACETTE, type="n"))
I get boxplots in both cases. What should I do to get points instead of
boxes?
Thanks in advance for your help,
Ivan
Are you looking for stripchart()?
Peter Ehlers
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ht
o mimic the OP's code
df <- data.frame(x = "2008-01-01")
df$x <- as.POSIXlt(df$x, "%Y-%m-%d")
str(df)
#'data.frame': 1 obs. of 1 variable:
# $ x: POSIXlt, format: "2008-01-01"
Peter Ehlers
__
;] < 0.1, , drop=FALSE ])
should do what you want.
Peter Ehlers
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, m
as.Date (or even nothing) on your Date
variable, you'll find that ddply does what you want.
To see why it doesn't work with strptime, check
str(df) and then ?Posixlt. You've converted Date
values to lists.
My comment about cbind() is to warn you that your
Values variable, as you have
not R-specific, can such a quote be 'fortuned'?
Peter Ehlers
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented
t I think
that you meant
dat[ !ave(dat$index, list(dat$day), FUN=function(x) x==min(x)), ]
Here's another way, using the plyr package
require(plyr)
ddply(dat, .(day), .fun = function(x) subset(x, index != min(index)))
Peter Ehlers
--
David Winsemius, MD
West Hart
On 2011-04-23 08:03, David Neu wrote:
On Sat, Apr 23, 2011 at 10:37 AM, Peter Ehlers wrote:
On 2011-04-23 07:13, David Neu wrote:
On Sat, Apr 23, 2011 at 9:47 AM, David Winsemius
wrote:
On Apr 23, 2011, at 9:26 AM, David Neu wrote:
Hi,
I'd like to change the default orientati
(1, 3))
and I'd consider 'jitter'.
BTW, your method of generating 'y' seems overly complicated:
y <- sample(c("A","B","C"), 100,
replace=TRUE,
prob=c(1/2, 1/3, 1/6))
Peter Ehlers
___
means<- apply(permutations, 1, mean)
means
And you might note that
means <- rowMeansy(permutations)
is about 10-15 times faster (if speed matters).
Peter Ehlers
[...snipped...]
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don't need this very often or very general, then use
mat <- diag(0,4)
mat[1:2, 1:2] <- A
mat[3:4, 3:4] <- B
mat
If you do need more flexibility, you could use the 'assist'
package's function bdiag() which handles block-diagonal
matrices:
require(assist)
mat <
(what = GPFoblq, tracer = expression(maxit<- 2000), at = 10,
print = FALSE)
to override the maximum number of iterations, which works but is a
hassle. Anyone have other ideas/techniques?
Thanks,
Josh
Does
formals(GPFoblq)$maxit <- 2000
omega()
do what you want?
Peter
e.
You might also consider a more informative subject
line. "Help needed" is true for all questions (not
answers) on R-help. Just think, why is this list
called R-***help***?
Peter Ehlers
Best,
Shuangyan
>
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the rpanel package. Might give you some ideas.
Peter Ehlers
Andreas
Ivan Calandra schrieb:
Dear users,
I have looked on different sources and found different functions to
prompt the user to provide input. However, I couldn't find one that
does exactly what I'm looking for.
select.
Seconded.
Peter Ehlers
On 2011-04-19 14:11, Marc Schwartz wrote:
On Apr 19, 2011, at 4:08 PM, Rolf Turner wrote:
On 20/04/11 07:19, Steven Wolf wrote:
[...snip...]
It sounds to me like you don't understand lists. If you are going to use
R you really should understand them. They
ot in robustbase. Look for partial sorting here:
https://svn.r-project.org/R/trunk/src/main/sort.c
Peter Ehlers
thanks,
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ation.variance$model$variance.jackknife
or
z <- fit.simex[['extrapolation.variance']]
z[['model']][['variance.jackknife']]
Peter Ehlers
str(fit.simex)
List of 24
$ coefficients : Named num [1:2] -17.1 3
..- attr(*, "names")= chr [1:2]
ip]>
Now is there a quick way to alter this indexing of rows in case of
my "transformed_dataframe"? I mean, I would
rownames(newdata)<- 1:nrow(newdata)
or, perhaps a bit simpler
rownames(newdata) <- NULL
Peter Ehlers
__
R-help@
ou
may want to look at the histbackback() function in
pkg Hmisc.
Peter Ehlers
[...snip...]
On Sun, Apr 17, 2011 at 10:51 AM, Stratford, Jeffrey
wrote:
Hi everyone,
I'm looking to produce a side-by-side histogram of the number of trips
taken by jays with a particular number of aco
ot most, of
your predictors should be factors.
Peter Ehlers
I have attached the dataset, feel free to take a look at it.
So far, running it with alle the combinations did not take too long and
there seem to be some effects between the parameters. However, 2x2
combinations might suffice.
T
retty close to what I want, but here's where I'm
stuck: How do I move the x-axis labels to the _center_ of each bar? Right now,
they line up with the right-side of each bar.
Try to set the parameter: adj=.5 (see ?par). HTH, *S*
That won't work (try it).
Use
text(1:4-0.5,
On 2011-04-15 16:42, Bill Hyman wrote:
Hi friends,
Does anyone know how to make truncated y axis with cut mark (\\) in R plot? Many
thanks!
See axis.break() in the plotrix pkg.
Peter Ehlers
Bill
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https
insight...
Are you seriously contemplating up to 10-way interactions?
I hope that you have a great deal of data and much patience
as you attempt to interpret those interactions.
Peter Ehlers
Dorien
On 15 April 2011 18:07, Dieter Menne wrote:
dorien wrote:
I calculate an anova test
uld be to download the source
file.
There's nothing being 'protected by the programmer'.
This might also be a good time to check out ?methods.
Peter Ehlers
Soyeon
On Thu, Apr 14, 2011 at 1:18 PM, Jannis wrote:
Just type the name of your function without (). Or use fix(). Or
.08143996
You can get the results from wilcox.exact() with
pvalue(wilcoxsign_test(x ~ y, zero.method="Wilcoxon",
distribution="asympt"))
#[1] 0.05061243
and
pvalue(wilcoxsign_test(x ~ y, zero.method="Wilcoxon",
dist="exact"))
#[1] 0.05
out kludge to do what the
OP wants. ...
That might work. Or try this, replacing QQQ with the name of the object
you want to keep.
rm(list=ls()[!grepl('^QQQ$',ls())])
cur
package gdata has function keep() which does just what the OP asks for.
Peter Ehlers
On 2011-04-12 02:50, Peter Ehlers wrote:
On 2011-04-11 09:44, PhDGuy wrote:
Hello,
I am using the function simple.violinplot from the package UsingR.
I have some outliers in my dataset so that the distribution has very long
tails.
As a result, the y-axis of the output of simple.violinplot
zoom on the y-axis with a command such as
ylim=c(a,b), as in boxplot(x,ylim=c(a,b)). However, doing
simple.violinplot(x,ylim=c(a,b)) does not work. Is there any way out?
Check out the zoomplot() function in the TeachingDemos package.
Peter Ehlers
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ameterized.
How can I work around that?
Take out p3; it's redundant.
Peter Ehlers
Thanks,
Felix
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PLEASE do read the
e following works for me (note: I'm using Jim Lemon's
well-known penchant for eschewing the spacebar):
testlen<-rnorm(24)*2+5
testpos<-0:23+rnorm(24)/4
clock24.plot(testlen,testpos,show.grid=FALSE,line.col=3)
clock24.plot(testlen[7:19],testpos[7:19],point.col=4,rp.type="s",point.
Thibault,
Your questions indicate that you would benefit enormously
from reading 'An Introduction to R'.
A very useful function is str().
Understanding the concept of "factors" is crucial in R.
"Checking" anything with Excel is never much use.
Peter Ehlers
On 2011-04-10 12:19, Jeff Stevens wrote:
Many thanks, Peter. This works brilliantly, and I prefer to have the
labels assigned outside of panel function as well.
Cheers,
Jeff
Small add-on:
It's probably best to ensure that the labels are of
type character, not factor.
Peter
[...snip...]
___
1","f2")]), "", lab))
## now use lab2 in bwplot()
bwplot(dv ~ f1 | f2, data = df, ylim = c(0.5, 1),
panel = function(x, y, ..., subscripts) {
lab <- df$lab2[subscripts] # note the lab2
panel.bwplot(x, y, ...)
panel.text(x, 0.55, labels = lab)
}
)
Pet
ata = df, ylim = c(0.5, 1),
panel = function(x, y, ..., subscripts) {
at.y <- rep(0.55, nrow(df))
is.na(at.y) <- which(duplicated(df$lab))
panel.bwplot(x, y, ...)
panel.text(x, at.y[subscripts], labels = df$lab[subscripts])
}
)
I think that the alpha argument is too o
fill.col='yellow', cir.ind=NULL, bins=1)
symbols(0, 0, circles=1, inches=FALSE,
fg="red", add=TRUE)
You could also have a look at the draw.circle() function
in the plotrix package.
Peter Ehlers
__
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he percent intervals are evenly spaced?
Hopefully I am making sense here
How about giving us a reproducible example?
Code is better than mere description;
code + description is best.
Peter Ehlers
Thanks.
britt
[[alternative
On 2011-04-06 15:21, Marius Hofert wrote:
Haha, I found a hack (using the letter "l"):
plot(0,0,main=expression(italic(X)[1]^bolditalic("l")))
To my eye, this is a bit neater:
expression(italic(X)[1]^bold("/"))
Peter
Cheers,
Marius
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R-help
.
http://downloads.teradata.com/download/applications/teradata-r/1.0
I just tried this with Firefox 3.6.16 and had no problem.
Proxy server?
Peter Ehlers
Is there an alternative site or location for obtaining the necessary driver?
Thanks
WHP
[[alternative HTML version deleted
see inline;
On 2011-04-06 14:22, Peter Ehlers wrote:
On 2011-04-06 14:14, Marius Hofert wrote:
Dear expeRts,
I would like to create a plotmath-label of the form X_1^\prime. Here is how to
*not* do it [not nicely aligned symbols]:
plot(0,0,main=expression(italic(X*minute[1])))
plot(0,0,main
=expression(italic(X)[1]*minute))
Any suggestions?
Hmm ; your subject line is a clue:
expression(italic(X)[1]^minute)
Note the '^'.
Peter Ehlers
Cheers,
Marius
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xyplot(0 ~ 0, xlab = xlab)
Marius,
I always find paste a bit tricky with plotmath.
Maybe this will do what you want:
mylab <- expression( atop(lab==list(alpha==1, beta==2), bold(foo)) )
xyplot(0 ~ 0, xlab = mylab)
Peter Ehlers
Cheers,
Marius
On 2011-04-04, at 18:59 , David Winsemius wr
tation is somewhat
spotty. The help page for gap.barplot indicates that the
input 'y' should be 'data values'; not overly informative.
Peter Ehlers
Thanks,
Drew Steen
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able results, you can add the prettyfying
statements back in. If you still get errors, then either I'm
out-to-lunch (quite possible!) or you need to use debug() to
figure out what's going on.
Peter Ehlers
On 2011-04-01 14:59, David Winsemius wrote:
On Apr 1, 2011, at 5:38 PM, steve_fried..
Aren't you missing a set of parentheses?
I can't run your code since it's not reproducible, but to
my aging eyes it seems that you need a set of '{}' around
the contents of your loop:
for(j in 1:(varsize[4]-1)) { loop stuff }
Peter Ehlers
On 2011-04-01 11:48, ste
ransparent" in the boxplot() call.
But then again, the whole point of a boxplot is to
represent general distributional shape for which
the median is surely more effective.
Peter Ehlers
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use apply to speed up aclculation of the fisher.exact test?
apply(yourMatrix, 1, function(x) fisher.test(cbind(x, 100 - x)))
or, if you only want the P-value:
apply(yourMatrix, 1,
function(x) fisher.test(cbind(x, 100 - x))$p.value)
Peter Ehlers
e additive errors on the original scale.
But your model looks simple enough - why not run
it through both functions and see what the difference is.
Ultimately, everything depends on what assumptions
you're comfortable with.
Peter Ehlers
Stephen
On Thu, Mar 31, 2011 at 8:34 PM, Steven
that
includes intercepts. If the intercepts turn out to
be significantly nonzero, what will you do?
Peter Ehlers
I also tried
lm(body_length ~ sex*head_length-1)
lm(body_length ~ sex*head_length-sex-1)
But none of them worked.
Would anyone be able to help me? All I want to do is to compare th
Ruben,
One more thing you might try is to add a 'size' component to
the lines list. The default is size=5, but since you have
enought space in the plot, try size=8 or 9; the line types
will show up more clearly:
lines=list(pch=1:4, lty=1:4, type='b', size=8),
groups=z,layout=c(2,3),
pch=1:4,lty=1:4,col='black',type='b',
key=list(x = .65, y = .75, corner = c(0, 0),
title="title here", cex.title=.9, lines.title=3,
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