sort before testing:
vtest <- function(x, lookfor){
any(apply(x, 1, function(v)
{identical(sort(v), sort(lookfor))}))}
-Peter Ehlers
On 2010-03-27 2:46, Berend Hasselman wrote:
David Scott-6 wrote:
I am sure someone can come up with a clever way of do
7
[3,]13
vtest(ma,c(3,7))
[1] TRUE
vtest(ma,c(1,7))
[1] FALSE
Berend
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I find ddply() in package plyr handy for this sort of thing:
library(plyr)
f <- function(x) x[which.max( x[["score"]] ), ]
## x will be a subset of Dat according to ID
ddply(Dat, "ID", f)
-Peter Ehlers
On 2010-03-09 11:59, Ista Zahn wrote:
Hi Richard,
There a
ing something? Or do you just want:
qdata[order(qdata$flow, decreasing=TRUE), ]
-Peter Ehlers
But this sorts the data by flow value and then year I think giving:
day month year flow
[1,] 14 3 1947 222.40
[2,] 15 3 1947 NA
[3,] 18 3 1947 NA
[4,] 17 3 1
As you can see on ?effects help page, plot.eff() uses
lattice graphics. You can't mix those with traditional
graphics commands. This should work:
plot(allEffects(GSMOD), ask=FALSE)
trellis.focus("panel", 1, 1)
panel.points(y, x)
trellis.unfocus()
-Peter Ehlers
On 2010-03
I imagine it's in
https://svn.r-project.org/R/trunk/src/main/model.c
-Peter Ehlers
On 2010-03-05 12:43, Werner W. wrote:
Hi,
I would like to see how model.matrix expands factor column to a set of dummy
columns. I think that is done int .Internal(model.matrix(t, data)) which is
c
Your description of your data isn't clear to me.
What are the values in column 2, for example?
Are you trying to construct regression or classification trees?
A reproducible example would really help.
-Peter Ehlers
On 2010-03-08 20:40, valeriano.parravic...@unige.it wrote:
Hi,
I h
)
read_data$DEC<- with(read_data,
cut(Target, breaks=brks, labels=1:10))
But I still don't see why you want a list of separate data
frames. For most analyses, it's more convenient to just use the
factor variable to subset the data as needed.
-Peter Ehlers
Thank
Ditching T/F for TRUE/FALSE would get my vote, too.
-Peter Ehlers
On 2010-03-08 17:44, Rolf Turner wrote:
On 9/03/2010, at 11:17 AM, Mike Prager wrote:
Rolf Turner wrote:
I solved the problem by putting in a colClasses argument in my
call to read.csv(). But I really think that the read
DNAME <- paste(names(mf), collapse = " by ")
names(mf) <- NULL
y <- do.call("anova.welch", c(as.list(mf), list(nu)))
## include 'nu' in the parameters passed to
##anova.welch.default
y$data.name <- DNAME
y
}
(
ta)
next_decile = ...and so on...
bottom_decile = ...
I would just add a factor variable indicating to which decile
a particular observation belongs:
dat$DEC <- with(dat, cut(Target, breaks=10, labels=1:10))
If you really want to have separate data frames you can then
split on the decile:
edict(fit, data.frame(x=xx))
plot(y ~ x)
lines(yy ~ xx, col='red')
-Peter Ehlers
But I have the following message error (sorry, this is in German):
Fehler in qr(.swts * attr(rhs, "gradient")) :
Dimensionen [Produkt 3] passen nicht zur L�nge des Objektes [23]
Zus�tzlich:
Here's a variation on the theme:
boxplot(x[,i])
title(main =
bquote(.(mainlabel1)~~italic(.(predictor[i]))~~.(mainlabel2))
)
-Peter Ehlers
On 2010-03-08 8:46, Miguel Porto wrote:
Hello,
Try this way (not sure if it's the best way, but it works):
boxplot(x[,i],
main=
mldata, reflevel="1")
use
mlogit.model <- mlogit(brand~1|female+age,
data = mydata, ## note: 'mydata', not 'mldata'
reflevel = "1",
varying = NULL,
choice = "brand"
ly
where the error has crept in.
I tried also with R version 2.10.1 Patched (2010-01-05 r50896).
Same result. So I suspect it's the version of mlogit.
I should have cc'd Yves; doing so now.
-Peter Ehlers
On 2010-03-07 11:30, David Winsemius wrote:
Looks like a problem that the main
mlogit() which
appears to cause the problem. This works:
mdata<-mlogit.data(mydata, varying=NULL, choice="brand", shape="wide")
mlogit.model<- mlogit(brand~1|female+age, data = mdata, reflevel="1")
-Peter Ehlers
brand female age
1.1 TRUE 0 24
1
r.get()
-Peter Ehlers
Thanks for any tips.
Hadassa
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(x) sum(!is.na(x
# A B C D
# 4 2 3 3
For (b), you could either use reshape() to transform to wide format
or generate an appropriate matrix; in either case, follow with apply():
m <- with(data, tapply(result, list(instance, solver), function(x) x))
m
apply(m[,-1], 2, function(x) sum(x &
I don't know if this is too long for a fortune, but it sure
seems to be that it should be one.
Anyway, thanks for the chuckle, Greg.
-Peter Ehlers
On 2010-03-04 13:29, Greg Snow wrote:
Well, the HeadSlap package would of course require the esp package so that it could tell
the diffe
e found it that way.
But
libary(sos)
???"bind arrays"
would find it as the first item.
-Peter Ehlers
cheers,
Rolf Turner
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hard.
help.search('Bartlett') immediately leads you to bartlett.test.
But why on earth would you want to do a Bartlett test? You would
be well advised to read the reference quoted in ?fligner.test.
-Peter Ehlers
Thanks in advance
This was sent to me personally but was probably meant for
R-help.
Original Message
Subject: Re: [R] Kohonen Package
Date: Mon, 1 Mar 2010 14:45:13 +1000
From: Martin
To: ehl...@ucalgary.ca
Hi
Any idea if the kohonen package can produce umatrices with hexagons and
component p
o such a change.
I agree with Rolf. Indeed, I'm not fond of the use of T/F for
TRUE/FALSE at all.
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
Talk about asleep at the switch.
My sincere apologies to both Susanne and David for my
stupid message and to group for wasting everyone's time.
Ouch, that headslap hurt.
-Peter
On 2010-02-27 11:05, Peter Ehlers wrote:
David, Susanne,
There may be a misunderstanding here. As I understa
ple? Or do you just want to plot the Normal curve with mean
equal to b1 and SD equal to s1?
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other than inserting
delimiters before passing off to R.
But Susanne, why do you need to read your data in this
piece-meal fashion?
-Peter Ehlers
On 2010-02-27 10:46, David Winsemius wrote:
On Feb 27, 2010, at 11:47 AM, Balzer Susanne wrote:
Hei David,
Thanks for your quick response
or df=c(1,1,dfr).
For models with an intercept the first component of df should
always be 1. But this is discarded in the output matrix.
With two numerical predictors: y ~ x1 + x2,
you should find that asgn = c(0,1,2) leading to df = c(1,1,1,dfr).
-Peter Ehlers
Thank you.
Kevin
_
uot;, as.integer(n),
as.double(x),
rep(format,n),
out= rep(dummy, n), NAOK=TRUE,
PACKAGE="mvpart")$out
if (is.matrix(x)) matrix(temp, nrow=nrow(x))
#else temp
else matrix(te
o/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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University of Calgary
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How about taking the unusual step of reading 'An Introduction to R',
where, if you peruse the table of contents, you will quickly be led
to Chapter 8: Probability Distributions.
-Peter Ehlers
On 2010-02-26 7:23, Антон Морковин wrote:
Dear all,
how to calculate v
Jon Erik,
I don't know where you get 'varslist', but if it's from a text file,
then why not scan() it into a *vector* instead of using (I assume)
read.table() to create a data.frame. You'll save yourself much grief.
-Peter Ehlers
On 2010-02-24 20:19, David Winsemius
is not an
S4 object
Looks like the wrong summary method is being used.
Your sessionInfo() might help.
-Peter Ehlers
I'm running on Windows XP R 2.10.1
Does anyone have any ideas why this is failing ?
Thanks
Steve
Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortu
generated. Any tips or ideas beyond plotting a circle in the
margin?
Why not just use bold and a larger font size?
-Peter Ehlers
Benjamin
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 |
__
May Nov May Nov
Aug Feb Aug Feb
http://n4.nabble.com/file/n1566433/TimeSeries_Example1.jpg
The plotrix package has staxlab().
-Peter Ehlers
Respectfully,
Eric
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Peter Ehlers
University of Calgary
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And as long as the interval is symmetric about zero:
sum(abs(x) <= 2)
[1] 6
-Peter Ehlers
On 2010-02-22 9:18, Jorge Ivan Velez wrote:
Hi,
Here is a suggeston:
x<- c(-1.3, 1, -1.5, -1, 1.5, -2.5, 3, -0.5)
sum(x>=-2& x<=2)
[1] 6
HTH,
Jorge
On Mon, Feb 22, 2010 at
wo dataframes. Try instead:
a <- data.frame(a, b) ## no problem (assuming same number of rows)
or
a[, "b"] <- b ## R tells you that you're trying to do
something you shouldn't.
-Peter Ehlers
__
R-help@r-pr
x27;t work).
-Peter Ehlers
trying various values for the second vector element, but do not notice any
change. Consulting this nice tutorial
http://research.stowers-institute.org/efg/R/Graphics/Basics/mar-oma/index.htm
has not helped. Can anyone point me in the right direction?
Thank you.
--
Pe
t on the x-scale;
Now we need to know the relationship of 's' to 'd': 3*d/s = 2;
Hence 3*d + 3*d/2 = 1 => d = 2/9 (so your 0.22 was spot-on).
How do we know that 3*d/s = 2? That's the default box.ratio value.
See what happens if you add box.ratio=3 (or whatever) to y
lease give me some suggestions.
Try this:
rm(sample)
sample(c(0,1,2),1,prob=c(0.2,0.3,0.5))
-Peter Ehlers
Thank you
Best,
Jing
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PLEASE do read the posting
odont is in pkg nlme).
-Peter Ehlers
On 2010-02-20 19:40, 孟欣 wrote:
Hi all:
I have a question about the command "followup.plot" of epicale package.
As to the demo data "Orthodont", the command "followup.plot" works well.But if I delete some rows
of data(delete Male
).
and the cumalative plot as well?
And this by studying help(plot).
But it appears that you may have to work your way through some
introductory material on R first. There's plenty available.
Perhaps start with 'An Introduction to R'.
-Peter Ehlers
. As a check, I would also do the
confidence interval calculations 'by hand'.
Or you could provide a *reproducible* example, preferably
minimal (i.e. skip the xlab= , etc stuff) and someone might
try the code and tell you where the problems lie.
-Peter Ehlers
Thanks a lot,
fussel
__
On 2010-02-18 1:04, Jim Lemon wrote:
On 02/18/2010 05:31 PM, Peter Ehlers wrote:
I agree with Bill's advice, but if you want the easy way out,
try dotplot.mtb in package plotrix. Jim Lemon's done the job
for us.
In fact, Barry Rowlingson and Rolf Turner did the job, I just bask in
I agree with Bill's advice, but if you want the easy way out,
try dotplot.mtb in package plotrix. Jim Lemon's done the job
for us.
-Peter Ehlers
On 2010-02-17 23:23, bill.venab...@csiro.au wrote:
R can't do anything.
The question is whether you can do it with R. R puts yo
On 2010-02-16 10:18, e-letter wrote:
On 16/02/2010, Peter Ehlers wrote:
On 2010-02-16 9:21, e-letter wrote:
Readers,
I tried to the following commands:
plot(y~x,ylab=expression(A[1]~B[2],xlab=expression(C~D))
mtext(expression(A[1]~B[2]),"additional text",side=3,line=1)
Your plot
quot;additional text"),side=3,line=1)
-Peter Ehlers
I receive the text that I want, but the command terminal shows the
following response:
Warning message:
NAs introduced by coercion in: mtext(expression(A[1]~B[2]),"additional
text", side = 3,
What is my
models to be based on the same set of subjects.)
Finally: (Re-)read the help page and note the 'warning'.
-Peter Ehlers
### outputs from R console ###
pop<- step(
+ lm(pop.rate ~ as.numeric(year) + as.factor(policy) +
as.numeric
m' to 'model' in
function(x) sum(vcov(fm)*outer(x,x
?
-Peter Ehlers
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a
2001)
This link might be of some use:
http://sci.tech-archive.net/Archive/sci.stat.math/2005-07/msg00229.html
-Peter Ehlers
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PLEASE do read the posting guide http://www
se
rlaplace(3, location, dispersion)
With dispDiag, rlaplace is probably taking 's' to be
c(3,0,0,0,.20,0,0,0,.1) and is only using the first three elements.
(Untested)
In case you're not aware: there's also rlaplace() in the VGAM package.
-Peter Ehlers
=NA)
the output in the left margin is "c(0.5". Does anyone have an explanation? Is
this a bug?
Can this problem be solved?
But ylab is not suppressed. Add ylab="" to your (weird) plot call.
-Peter Ehlers
Regards,
Martin Ivanov
__
le saves typing.)
-Peter Ehlers
kMan wrote:
Dear Peter,
Ah, I see your point, Professor. The point at x=23.5 is carrying the model.
Allow me to clarify. I was making a similar point.
I was suggesting that the cube term could be left in the model, as you did,
but rather than dropping the data
yhat1 <- predict(fm1, list(x = xx))
yhat2 <- predict(fm2, list(x = xx))
plot(x,y)
lines(xx, yhat1, col="blue", lwd=2)
lines(xx, yhat2, col="red", lwd=2)
That's how much difference *one* point makes in a cubic fit!
I'm not much of a gambler, so I wouldn'
would consider to be the better test,
namely using the median as location measure. lawstat gives
you options: try it with location="mean".
Then switch to the Fligner-Killeen test (fligner.test() in
package 'stats'). The reference cited in ?fligner.test makes
for good reading.
kMan wrote:
I would use all of the data. If you want to "drop" one, control for it in
the model & sacrifice a degree of freedom.
You like to live dangerously.
-Peter Ehlers
Why the call to poly() by the way?
KeithC.
-Original Message-----
From: Peter Ehle
Try
text(.5, .5, bquote(bold(.(myText
-Peter Ehlers
Mark Heckmann wrote:
# I want to plot bold text. The text should depend on a variable
containing a character string.
plot.new()
text(.5, .5, expression(bold("Some text")))
# now I would like to do the same replacing &quo
(20),function(r) r>0.5))
I would like a single curve, the following example displays two curves.
qqmath(~x,groups=f,data=x,distribution=qunif,f.value=pp)
Here's one way:
qqmath(~x, data=x, type='b',
col=ifelse(x$f, 2, 4),
pch=ifelse(x$f, 17, 19),cex=2)
-Peter Ehlers
And this hasn't even addressed the relative abundance of x=0 data.)
-Peter Ehlers
David Winsemius wrote:
On Feb 13, 2010, at 1:35 PM, Rhonda Reidy wrote:
The following variables have the following significant relationships
(x is the explanatory variable): linear, cubic, exponenti
[combn() is now in utils]
-Peter Ehlers
Greg Snow wrote:
Here is one quick way using the combinat package:
library(combinat)
tmpfun <- function(x) {
+ tmp <- rep(1,5)
+ tmp[x] <- -1
+ tmp
+ }
combn(5,2, tmpfun)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] -1 -1
ow to do.
You can look inside the plot.Fstats function with
strucchange:::plot.Fstats
There you will find the equivalent of the following:
k <- fs$nreg
n <- fs$nobs
x <- fs$Fstats
pvals <- 1 - pf(x, k, (n - 2 * k))
which gives you the P-values.
-Peter Ehlers
Here a is an
Assuming that you are using the xts package, try this:
data(sample_matrix)
sample.xts <- as.xts(sample_matrix)
open <- as.vector(sample.xts[,1])
month <- as.Date(time(sample.xts))
plot(open, month, type="l")
-Peter Ehlers
JSmaga wrote:
Basically it works, but I use
Nice, Uwe.
Small correction: make that nrow=4:
x1 <- as.numeric(rbind(matrix(rep(x, each=2), nrow=4), NA))
-Peter Ehlers
Uwe Ligges wrote:
On 11.02.2010 22:38, Tim Clark wrote:
Dear List,
I am trying to plot several separate polygons on a graph. I have
figured out how to do it
Pat Burns makes a good point. -Peter
Original Message
Subject: Re: [R] Using seq_len() vs 1:n
Date: Fri, 12 Feb 2010 09:01:20 +
From: Patrick Burns
To: Peter Ehlers
References: <4b746aef.10...@ucalgary.ca>
If you want your code to be compatible with
S+, then &#
How about
t(outer(seq(.1,1,.1), 1:12, foo))
Convert to dataframe, etc.
-Peter Ehlers
Steven Worthington wrote:
Dear R users,
I have a function (simplified here) that accepts two arguments and performs
various calculations:
foo <- function(y, x) {
a <- y*
simulated/example data if you
can't show your real data).
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University of Calgary
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f n can be zero, there is
an advantage to using seq_len.
Is there ever a *dis*advantage?
Peter Ehlers
University of Calgary
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PLEASE do read the posting guide http://www.R-p
_character_ data that
you want to read into R where its storage mode should be 'raw'.
I don't know how to do that.
If the main purpose is to circumvent R's memory requirements,
then there have been plenty of posts on that issue.
-Peter Ehlers
Johan Jackson wrote:
"I suspec
clear
that you haven't read the colClasses description in
?read.table very carefully. The one thing R help pages are
pretty good at is careful definition of arguments.
I do hope that your day will improve.
-Peter Ehlers
[..]
--
Peter Ehlers
University of Calgary
___
s for lower= and upper= and see
if you can understand why -Inf, Inf won't work.
You can also plot your function with, e.g.
curve(f, 7, 9)
-Peter Ehlers
Charles Annis, P.E. wrote:
Here's a suggestion: Plot the function:
x <- seq(3, 13, length=101)
plot(x, y=dnorm(x, mean=8,sd=1)
In addition: Warning message:
In file(file, "r", encoding = encoding) :
cannot open file 'test.R': No such file or directory
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ther this is expected behavior, or whether it
ought to be reported as a bug.
I don't see the problem in R version 2.10.1 Patched (2010-01-05 r50896)
nor in R 2.11.0 Pre-release (Windows Vista).
-Peter Ehlers
Best,
Russell Pierce
--
Peter Ehlers
University of Calgary
_
teger vector?
use str();
3. must NB reside in a matrix?
-Peter Ehlers
Andrew Kosydar wrote:
Hi Dennis,
Thank you for your response. No, NB is not a matrix, and I have no
covariates. Here's a very small sample of the data:
effectNB
-0.0032001
-0.1208003
-0.0032002
-7.6
Could somebody let me
know what a 'call' class is and what 'language' is?
See Chapter 2 in the R Language Definition manual.
-Peter Ehlers
##
form1=skips ~ Panel * Opening
terms1=terms(form1)
str(attr(terms1, 'variables'))
class(attr(terms1, 'variab
egards
Our Thoughts have the Power to Change our Destiny.
Sunita
Sent from Pune, MH, India
On Tue, Feb 9, 2010 at 11:22 PM, Peter Ehlers wrote:
Here is a simple 3-step solution:
1. type ?barplot
2. find the section labelled 'Arguments'
3. carefully read what each argument means/does
es* "x1", etc;
as.data.frame(z) is a data.frame with *variables* named "x1" etc.
If you really want to use subset(), then
subset(z, z[, "x1"] < 0, select = <...>)
will work, but I wouldn't use it.
-Peter Ehlers
DonDiego wrote:
Hi,
I have a matrix
o be secretive about the error; just say what it was.)
-Peter Ehlers
Sunitap22 wrote:
Hello
(this might be a very simple question)
My data is as follows (table name is student)
YearStudentsPassed
1 2000300
2 2001360
3 2002450
4 2003450
5 2004270
6 2005280
7
id 'z' limits
So, are some your 'NaN's actually 'Inf's?
-Peter Ehlers
Uwe Ligges wrote:
On 07.02.2010 22:46, Andrew Wang wrote:
I have this data set that both x& y are ordered vectors of length
600& 700 respectively; z is a 600 by 700 matrix whose
e been unable to implement this test using library(coin) - is this possible?
Thanks. --Dale
Try this:
wilcoxsign_test(x ~ rep(M.0, length(x)), dist = 'exact')
--
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University of Calgary
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ht
d graph in the range 400 and Y in the
columns with values below 400 do not appear on the chart?
If I understand correctly, you should be able to just subtract 400
from atend and adjust the axis() command:
barplot(atend-400, las=1, xlab='Meses', ylab='Número de atendimentos&
11
Perhaps R should do a bit less coercing.
-Peter Ehlers
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)
-Peter Ehlers
IoanLoft wrote:
Hi all,
I have encountered a problem which appears to have defeated my (admittedly
nascent) R skills. I want to draw a spider plot with many cases (just over
300). I am primarily interested in the difference between 4 categories of
cases, and want to display them
t cases" approach to organization
of longitudinal data is regrettable."
http://n4.nabble.com/Hierarchical-data-sets-which-software-to-use-td1458477.html#a1470430
--
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University of Calgary
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I don't really know qcc, but it seems to me that you might
have to provide information about within-group variability.
But maybe I'm completely out to lunch on this.
-Peter Ehlers
vikrant wrote:
Thanks Bart and Peter for your help and the example is working for c chart as
well
retty well
an R-requirement.
-Peter Ehlers
wesley mathew wrote:
Dear All
Subject : Multiple Lines in Graph
Could you please help me to draw two lines in a graph.
plot(f1, t1, type ="b") and plot(f2,t2, type="b") where t1, t2, f1,and f2
are single dimensional matrix.
I have these
Stephen,
You probably should name your dataframe 'dat' and
replace the line
x <- subset(x, Creek=="fbms" & station==i)
with
x <- subset(dat, Creek=="fbms" & station==i)
-Peter Ehlers
stephen sefick wrote:
Both of the approx functions w
Tom,
t.test(MAE ~ type, data=data, subset=type %in% c('hpc','rfc'))
-Peter Ehlers
Thomas Adams wrote:
Dennis,
Thank you for the suggestion, but I get this error:
> t.test(MAE ~ type,data=data)
Error in t.test.formula(MAE ~ type, data = data) :
grouping factor mus
Somehow, in looking for those many examples, you missed the
'sleep' data example on the help page for t.test.
(BTW, I wouldn't consider your sample data to be "minimal"
or even close to minimal.)
-Peter Ehlers
Thomas Adams wrote:
I am trying to use t.test on the foll
some funny thing
in your startup.
Anyway, try deleting .Random.seed and re-creating it:
rm(.Random.seed)
runif(1)
-Peter Ehlers
Rosa Manrique wrote:
Dear friends,
I cannot load the pachkages labdsv...which i do not understand is linked to
'mgcv' package. Anyway neither the last is loa
Is this what you want:
singer1 <- subset(singer, voice.part == "Bass 1")
brks <- seq(65, 75, 2)
histogram( ~ height, data = singer1, breaks = brks)
or, slightly different:
histogram( ~ height, data = singer1, breaks = brks,
scales = list(x = list(at = brks)))
-Peter Eh
I suspect that you may have set qcc.options("cex") to
too large a value. Try lowering it with
qcc.options(cex=whatever)
-Peter Ehlers
vikrant wrote:
ok. I will give the example for which i m getting this error. The data for
plotting R chart and S chart is very huge. SO i will ta
ls package is what you want.
-Peter Ehlers
thank you
khazaei
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, min
han a bug.
I must admit, I've never run across this situation. Good of you
to spot it.
-Peter Ehlers
If you consider data.frame an unusual class I could accept your point but
if help page tells me that a function works for most classes I would not
expect that data.frame class shall b
ok by Bates and Watts.
As always, the most important questions are 1) why do you want
to compare models and 2) what will you do with the result of
the comparison.
-Peter Ehlers
Thanks in advance
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e:
median.data.frame <- function(x, ...) sapply(x, median, ...)
I think that it would be desirable to have similar behaviour
for both functions or at least a warning if median.default
is incorrectly applied to a data.frame object.
-Peter Ehlers
r-help-boun...@r-project.org napsal dne 04.02.20
s a dataframe? Use class(myData)
before the write.* call.
Use write.csv() (this won't fix the 'problem').
-Peter Ehlers
Many thanks in advance!
John Woodard
--
Peter Ehlers
University of Calgary
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7;):
[1] toString.default
Warning message:
In methods("toString") : function 'toString' appears not to be generic
-
Anna Lippel
--
Peter Ehlers
University of Calgary
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nyway, here are two suggestions:
1. to answer your question directly: use
base::toString(x)
to enforce use of that function.
2. try
methods('toString')
to see what other toString()'s you may have around.
-Peter Ehlers
anna wrote:
isn't there a way to specify from which
Philipp,
Check ?str which displays the structure of R objects.
And do use extractor functions when available:
coef(yourModel) instead of yourModel$coef
-Peter Ehlers
Philipp Rappold wrote:
Dear all,
I have a simple question: How can I retrieve a list with all properties
of an object
u get from
c("red","blue","purple","pink")["a4"]
4. assuming that 'Tanks' is a factor, see what you get from
c("red","blue","purple","pink")[Tanks]
-Peter Ehlers
Marlin Keith Cox wrote:
OK
Anna,
Try omitting pkg:RBloomberg.
If you really need to use that package, you will have to
install the non-CRAN package RDCOMClient from omegahat.
I still don't see why toString() wouldn't do its job, even
with RBloomberg loaded.
-Peter Ehlers
anna wrote:
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