2008/8/27 Nicky Chorley [EMAIL PROTECTED]:
2008/8/27 rr400 [EMAIL PROTECTED]:
Hi, i have never really used R before and i need to produce a density curve
for use in an assignment. Following the instructions in the manual provided
with my course i keep getting this error message:
Error in
You were very close to an answer :
as.vector(unlist(df[1,]))
Nael
On Wed, Aug 27, 2008 at 7:53 AM, Ronnen Levinson [EMAIL PROTECTED] wrote:
Hi.
How do I convert a one-dimensional array of characters to a character
vector? In the example below I am trying to get the result c(a,d).
Thank you for the hints,
finally I was looking for the substitution of
splitted$2
to
splitted[[as.character(counter)]]
I did not know that I can substitute f.e the $2 with [[2]] or/and
$y (second column) with [[2]]
Knut
__
R-help@r-project.org
Hi all,
Sorry to ask again but I'm still not sure how to get the full
variance-covariance matrix. Peter suggested a three-level treatment
factor. However, I thought that the censoring variable could only take
values 0 or 1 so how do you programme such a factor.
Alternatively, is there another
This was exactly the problem. It has worked a treat this time around.
Thank you so much!
--
View this message in context:
http://www.nabble.com/Error-producing-density-curve-tp19173403p19176280.html
Sent from the R help mailing list archive at Nabble.com.
As ``vi'' user I remarked a while ago,
that - after calling ``R'' -
three character commands like ``ctX'' will beep and do not work
whereas two character commands like ``fX'' work as expected.
Is there some help?
Best regards - Wolfram
__
Thank you so much, at least i got that part right. But what is the meaning of
the e-11 at the end??
Thanks again
You commands are correct and the interpretation is that the probability that
a normal random variable with mean 1454.190 and
standard deviation 162.6301 achieves a value of 417 or
N. Lapidus wrote:
You were very close to an answer :
as.vector(unlist(df[1,]))
I'd use as.character() there, for clarity. It's not easy to remember
what as.vector on a factor variable does.
Also notice that the original post has serious confusion about what the
data structures are:
df is
2008/8/27 rr400 [EMAIL PROTECTED]:
Thank you so much, at least i got that part right. But what is the meaning of
the e-11 at the end??
Exponential notation. It means 10^-11.
Regards,
Nicky Chorley
__
R-help@r-project.org mailing list
Laura Bonnett wrote:
Hi all,
Sorry to ask again but I'm still not sure how to get the full
variance-covariance matrix. Peter suggested a three-level treatment
factor. However, I thought that the censoring variable could only take
values 0 or 1 so how do you programme such a factor.
Hi,
anybody any hints how to get a barplot with both juxtaposed and stacked
bars?
/Stefan
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
Here is the exact code I have written which does the standard vs nt1 and
standard vs nt2 and also gives me the hazard ratio for nt1 vs nt2.
with - read.table(allwiths.txt,header=TRUE)
fix(arm)
function (data)
{
dummy - rep(0,2437)
for(i in 1:2437){
if(data$Arm[i]==CBZ)
Hi
[EMAIL PROTECTED] napsal dne 26.08.2008 17:06:27:
Hi,
I would like to have six dot plots on one page. I make the plots in a
'for loop' because it is six times the same graph but for different
subjects (species).
I tried it with par(mfrow=c(3,2), oma=c(3,3,2,0), mar=c(2,2,2,2));
Marc Schwartz wrote:
on 08/26/2008 07:31 PM (Ted Harding) wrote:
On 26-Aug-08 23:49:37, hadley wickham wrote:
On Tue, Aug 26, 2008 at 6:45 PM, Ted Harding
[EMAIL PROTECTED] wrote:
Hi Folks,
This tip is probably lurking somewhere already, but I've just
discovered it the
Laura Bonnett wrote:
Here is the exact code I have written which does the standard vs nt1
and standard vs nt2 and also gives me the hazard ratio for nt1 vs nt2.
with - read.table(allwiths.txt,
header=TRUE)
fix(arm)
function (data)
{
dummy - rep(0,2437)
for(i in 1:2437){
Hi,
what is the appropriate syntax to get the random error correct when
performing repeated measures anova with 'lme'.
let's say i have 3 independent variables, with 'aov', i would write
something like: aov(dep_var~(indep_var1*indep_var2*indep_var3) +
Hi JP,
I suggest that you read the literature cited in the help file, most
particularly
Pinheiro, J.C., and Bates, D.M. (2000) Mixed-Effects Models in S
and S-PLUS, Springer.
MASS (Modern Applied Statistics in S) also has some useful things in
it.
Cheers
Andrew
On Wed, Aug 27,
Hello,
I am trying to load the package SparseM. It seems that I have successfully
installed SparseM (version 0.78), but I did not succeed in loading the
SparseM package into R 2.7. Does anybody know a trick for loading
SparseM?
Thanks in advance,
Heike
library(SparseM,lib.loc=my.lib.loc)
I have been experimenting with interactive packages such iplots and
playwith. Consider the following sample dataset:
A B C D
1 5 5 9
3 2 8 4
1 7 3 0
7 2 2 6
Let's say I make a plot of variable A. I would like to be able to
click on a data point (e.g. 3) and have a pop-up window tell
--- On Wed, 27/8/08, saggak [EMAIL PROTECTED] wrote:
From: saggak [EMAIL PROTECTED]
Subject: How to learn R language?
To: r-help@r-project.org
Date: Wednesday, 27 August, 2008, 3:37 PM
Hi!
I am a post graduate in Statistics. I want to learn R language, but am very
confused as to how to
Duncan Murdoch wrote:
On 26/08/2008 7:54 AM, Jim Lemon wrote:
Hi again,
Oops, I meant the expected value of the swap is:
5*0.5 + 20*0.5 = 12.5
Too late, must get to bed.
But that is still wrong. You want a conditional expectation,
conditional on the observed value (10 in this case). The
I am sure that there are as many ways as people, but this is what I
did. MASS 4, Simple R (internet), and this list. I taught myself and
it may have been easier to have a guide as this would have cut the
learning curve down, but this is not essential. Find out what you
need to do, do the
Dear R-helpers,
I am desperately looking for a solution for how to print out the console
output to a standard printer. For example, I would like to print out the
summary.lm() output, the output of different ftable-functions etc. I use
R on a linux machine.
The only ways so far have been to
?capture.output
?sink
x - capture.output(runif(20))
x
[1] [1] 0.2730090246 0.4462614490 0.2382041477 0.9826505063
0.1556554718 0.3746872961
[2] [7] 0.6108254879 0.6617410595 0.6694177436 0.4650380281
0.0414420397 0.2307212995
[3] [13] 0.5338913775 0.9186298891 0.0006410333 0.8046684864
Hi!
Thanks, Jim, for Your quick answer. But my question was how to
(re)direct the output to a printer.
Kind regards,
Kimmo
jim holtman wrote:
?capture.output
?sink
x - capture.output(runif(20))
x
[1] [1] 0.2730090246 0.4462614490 0.2382041477 0.9826505063
0.1556554718 0.3746872961
Hi there,
I am a beginner with R. Anyone could help to explain me what the following
warning msg means?
Warning messages:
1: In any(predictorMatrix[j, ]) :
coercing argument of type 'double' to logical
Got 6 of it...although further calculations were done. Because of this
warning I am not
saggak wrote:
Hi!
I am a post graduate in Statistics. I want to learn R language, but am very
confused as to how to begin systematically. I need to learn R language from
Statistics point of view e.g. I need to fit distributions to data or run
regression analysis etc. No doubt there are so
Hi!
I am a post graduate in Statistics. I want to learn R language, but am very
confused as to how to begin systematically. I need to learn R language from
Statistics point of view e.g. I need to fit distributions to data or run
regression analysis etc. No doubt there are so many articles
I am looking at histogram breaks, and notice something odd:
foo - hist(runif(1),breaks=20)
length(foo$breaks)
[1] 21
This makes sense to me.
foo - hist(runif(1),breaks=200)
length(foo$breaks)
[1] 201
This also makes sense.
BUT
foo - hist(runif(1),breaks=250)
length(foo$breaks)
Dear all,
I am organizing a set of specific R code as package (to ease the
documentation and deployment of it to users).
Before doing so, I would like to know if there are written coding rules
for R (with functions, objects naming convention for example).
Thanks a lot,
Thomas
--
See
http://www1.maths.lth.se/help/R/RCC/
On Wed, Aug 27, 2008 at 9:36 AM, Thomas LOUBRIEU
[EMAIL PROTECTED] wrote:
Dear all,
I am organizing a set of specific R code as package (to ease the
documentation and deployment of it to users).
Before doing so, I would like to know if there are
Sorry for the double post - I accidentially hit my the send button...
We want to build a global optimizer for extremely expensive noisy objective
functions based
on physical measurements.
Our approach is based on optimizing kriging response surfaces surface models.
(as seen in
Dear Sagga,
How give a look at the introduction material on
Help / Manual (in pdf) / An introduction to R
HTH
miltinho astronauta
brazil
On 8/27/08, saggak [EMAIL PROTECTED] wrote:
Hi!
I am a post graduate in Statistics. I want to learn R language, but am very
confused as to how to begin
Hi Jean-Pierre,
A general comment is that I think you need to think more carefully about
what you are trying to get out of your analysis. The random effects
structure you are aiming for could be stretching your data a little thin.
It might be a good idea to read through the archives of the
Hi,
I've some problem to close a com Object that I've open.
example :
I create the Com Object Excel :
fxl = comCreateObject(Excel.Application)
To close I normally used
gc(verbose = FALSE) - silent
This is ok with my first Pc, the Com Object Excel is kill, but for my second
computer
Hello,
I have a basic question about the display() function, but I can't find the
answer easily.
I am running R 2.7.16.0 on Suse linux 11.0
with KDE 4.0
I have several desktops, that I use at the same time, called Desktop 1, Desktop
2, etc.
I want to open a X11() window in another
Hello,
I have this command:
x.axis - seq(from=0.5, to=4.5, length.out=13112)
How can I which of the x.axis components is the closest to a given
value, for example 3.2?
Best,
Dani
--
Daniel Valverde Saubí
Grup de Biologia Molecular de Llevats
Facultat de Veterinària de la Universitat Autònoma
I think the answer is in this line in hist.default():
breaks - pretty(range(x), n = breaks, min.n = 1)
length(pretty(c(1:1000), 100))
[1] 101
length(pretty(c(1:1000), 200))
[1] 201
length(pretty(c(1:1000), 250))
[1] 201
length(pretty(c(1:1000), 300))
[1] 201
length(pretty(c(1:1000),
Try this:
x.axis[which.min(abs(x.axis - 3.2))]
On Wed, Aug 27, 2008 at 10:12 AM, Dani Valverde [EMAIL PROTECTED] wrote:
Hello,
I have this command:
x.axis - seq(from=0.5, to=4.5, length.out=13112)
How can I which of the x.axis components is the closest to a given value,
for example 3.2?
Maybe this
which.min(abs(x.axis-3.2))
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dani Valverde
Sent: Wednesday, August 27, 2008 9:13 AM
To: R Help
Subject: [R] Closest value
Hello,
I have this command:
x.axis - seq(from=0.5, to=4.5,
Hi
[EMAIL PROTECTED] napsal dne 27.08.2008 15:12:58:
Hello,
I have this command:
x.axis - seq(from=0.5, to=4.5, length.out=13112)
How can I which of the x.axis components is the closest to a given
value, for example 3.2
E.g.
which.min((x.axis-3.2)^2)
Regards
Petr
Best,
Dani
--
Quick disclaimer: I'm new to the list and I'm not sure if I'm sending this
to the correct forum. If there is a better place for this questions, please
let me know!
Dear R List,
I'm interested in modifying a regression tree algorithm to use the
difference between a control and a dosed sample
What type of system are you running on? You can create a
textConnection and maybe 'pipe' the output to 'lpr' if you are running
on UNIX. On Windows there is probably a way of connecting to a
printer; you can save the file and then execute the PRINT command.
On Wed, Aug 27, 2008 at 7:51 AM, K.
Hello,
I was wondering whether anyone's using R for reliability
(RAMS/LCC) engineering?
Sincerely,
Wolfgang Keller
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Hi Stefan,
Could you be a bit more explicit? Do you have an example dataset that
you are trying to visualise?
Hadley
On Wed, Aug 27, 2008 at 4:38 AM, Stefan Uhmann
[EMAIL PROTECTED] wrote:
Hi,
anybody any hints how to get a barplot with both juxtaposed and stacked
bars?
/Stefan
I'm not a statistician so my approach may not make sense for you but I'd
suggests having a look at Bob Muenchen's R for SAS and SPSS users in pdf form
(very useful) or his new book with the same title (which I have not seen yet)
for a start. http://rforsasandspssusers.com/
If you want some
There are many books, but if I had to choose one that teaches R and teaches
statistics at the same time (Yes, you already know stats, so it will be that
much easier) I'd choose Peter Dalgaard's book, Introductory Statistics with
R. It's exceptionally well written, easy to follow, and will
Hey fellas:
I would like to integrate the following function:
integrand - function (x,t) {
exp(-2*t)*(2*t)^x/(10*factorial(x))
}
with respect to the t variable, from 0 to 10.
The variable x here works as a parameter: I would like to integrate the said
function for each value of x in
On Wed, 27 Aug 2008, David Cobey wrote:
Quick disclaimer: I'm new to the list and I'm not sure if I'm sending this
to the correct forum. If there is a better place for this questions, please
let me know!
Dear R List,
I'm interested in modifying a regression tree algorithm to use the
jose romero wrote:
Hey fellas:
I would like to integrate the following function:
integrand - function (x,t) {
exp(-2*t)*(2*t)^x/(10*factorial(x))
}
with respect to the t variable, from 0 to 10.
The variable x here works as a parameter: I would like to integrate the said
function
Here is one way:
integrand - function (t, x) {
exp(-2*t)*(2*t)^x/(10*factorial(x))
}
x - 0:44
ans - sapply(x, function(x) integrate(integrand, lower=0, upper=10, x=x))
cbind(x=x, integral=unlist(ans[1,]), abs.error=unlist(ans[2,]))
Ravi.
Peter Dalgaard wrote:
jose romero wrote:
Hey fellas:
I would like to integrate the following function:
integrand - function (x,t) {
exp(-2*t)*(2*t)^x/(10*factorial(x))
}
with respect to the t variable, from 0 to 10.
The variable x here works as a parameter: I would like to
I want to fit something like:
z = b0 + b1*x + b2*y
Since x, y,and z all have measurement errors attached, the proper way
to do the fit is with principal components analysis, and to use the
first component (called loadings in princomp output).
My dumb question is: how do I convert the princomp
Hi Heike
library(SparseM,lib.loc=my.lib.loc)
Error in packageDescription(pkg)$Version :
$ operator is invalid for atomic vectors
In addition: Warning message:
In packageDescription(pkg) : no package 'SparseM' was found
Error : .onLoad failed in 'loadNamespace' for 'SparseM'
Error:
Quick disclaimer: I'm new to the list and I'm not sure if I'm sending this
to the correct forum. If there is a better place for questions like these,
please let me know!
Dear R List,
I'm interested in modifying a regression tree algorithm to use the
difference between a control and a dosed
Dear all,
I have a dataframe with 132 columns and 100 rows. Every 2nd column is a
repeat measurement so that the columns could be titled, a a b b c c d d etc.
I would like to average the repeats such that I am left with a data frame of
66 columns (of means) and 100 rows.
I have been trying
hello,
gamma(a) works fine for a173, but gamma(173) returns
[1] Inf
Warning message:
value out of range in 'gammafn'
What I really need is the ratio
gamma(nu/2) / gamma((nu+1)/2)
is anyone aware of a way to re-express this in a form that pushes the
threshold past which a warning message is
- go with Bob and Peter's book. Too many sources can be confusing.
- Join the list, create filters for keywords of your specialty (like from
R -Help ) contains regression.
- Try use it for a live project .
Let the R begin
Ajay
On Wed, Aug 27, 2008 at 7:46 PM, John Kane [EMAIL
Hi Sam,
thank your for your reply. The lib location statement is correct. I have
installed SparseM locally because I do not have su rights.
Best, Heike
(BTW: An installation by our systems administrator into the R directory did
not work, too.)
On Wednesday 27 August 2008 19:00, Samuel
c(as.matrix(df[1,]))
will also work in your case.
Gasper Cankar
-Original Message-
From: N. Lapidus [mailto:[EMAIL PROTECTED]
Sent: Wednesday, August 27, 2008 9:06 AM
To: r-help@r-project.org
Subject: Re: [R] Converting 1-D array to vector
You were very close to an answer :
You can simply do
a - c(2,4,3,4,5)
b - c(4,3,4,5,2)
barplot(cbind(a,b))
or you can put a and b in matrix as help suggests.
Gasper Cankar
-Original Message-
From: Stefan Uhmann [mailto:[EMAIL PROTECTED]
Sent: Wednesday, August 27, 2008 11:39 AM
To: r-help@r-project.org
Subject: [R]
On 27-Aug-08 14:24:31, Charles Annis, P.E. wrote:
There are many books, but if I had to choose one that teaches R
and teaches statistics at the same time (Yes, you already know
stats, so it will be that much easier) I'd choose Peter Dalgaard's
book, Introductory Statistics with R. It's
On 8/27/2008 11:14 AM, [EMAIL PROTECTED] wrote:
hello,
gamma(a) works fine for a173, but gamma(173) returns
[1] Inf
Warning message:
value out of range in 'gammafn'
What I really need is the ratio
gamma(nu/2) / gamma((nu+1)/2)
is anyone aware of a way to re-express this in a form that
Try this:
# test data
DF - data.frame(a = 1:9, a = 21:29, b = 31:39, b = 41:49, check.names = FALSE)
t(apply(DF, 1, tapply, names(DF), mean))
It gives a matrix so use as.data.frame on that if you need a data frame
On Wed, Aug 27, 2008 at 9:47 AM, JonD [EMAIL PROTECTED] wrote:
Dear all,
I
All,
I'm witnessing some strange behavior when checking the values of one of my
variables. My guess is that it has something to do with the number of
significant digits being represented, but perhaps not as my variable was
created from other variables that only had one decimal place. See
This will work:
exp( lgamma(nu/2) - lgamma((nu+1)/2) )
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor, The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins
I have two questions for the group. One is very concrete, and is dangerously
close to a please do my homework posting. The second follows from the
first one but is more general. I would welcome the advice of experienced R
users.
As for the first one: I have a data frame with two variables
X Y
FAQ 7.31
On Wed, Aug 27, 2008 at 11:51 AM, David Afshartous
[EMAIL PROTECTED] wrote:
All,
I'm witnessing some strange behavior when checking the values of one of my
variables. My guess is that it has something to do with the number of
significant digits being represented, but perhaps not
On Wed, 27 Aug 2008, Giuseppe Paleologo wrote:
I have two questions for the group. One is very concrete, and is dangerously
close to a please do my homework posting. The second follows from the
first one but is more general. I would welcome the advice of experienced R
users.
As for the first
Hi,
I am using ggplot2 to generate graphs for a paper I am writing in two
column format. When I shrink the graphs to fit in a single column, the
graph is clear but the axis and tick labels are way too small.
I have increased the font sizes by manipulating the grid. However, when
I do this
Hi there,
I have two questions and believe that there is an extremely easy solution.
Being a beginner with R makes thinks a bit more complicated.
This is the code:
rpois(15,3)
n-15
DATA-cbind(D,rpois(15,3))
data-as.data.frame(DATA)
colnames(data)-c(D,X)
*# 1. question: is it possible to put
on 08/27/2008 09:26 AM David Cobey said the following:
I'm interested in modifying a regression tree algorithm to use the
difference between a control and a dosed sample as its dependent variable.
I was wondering if anyone knew where I could find code to implement a basic
chaid algorithm.
Thanks. Is there simple way around this for simple checking of single
values? The all.equal() mentioned in the FAQ doesn't seem appropriate.
X = c(1.2, 2)
X.new = X -1
X == 1.2
[1] TRUE FALSE
X.new == .2
[1] FALSE FALSE
On 8/27/08 11:47 AM, jim holtman [EMAIL PROTECTED] wrote:
FAQ
That's great, thanks!
Jon
Gabor Grothendieck wrote:
Try this:
# test data
DF - data.frame(a = 1:9, a = 21:29, b = 31:39, b = 41:49, check.names =
FALSE)
t(apply(DF, 1, tapply, names(DF), mean))
It gives a matrix so use as.data.frame on that if you need a data frame
--
Try this using 'sapply'
sapply(X.new, all.equal, .2) == TRUE
[1] TRUE FALSE
On Wed, Aug 27, 2008 at 12:13 PM, David Afshartous
[EMAIL PROTECTED] wrote:
Thanks. Is there simple way around this for simple checking of single
values? The all.equal() mentioned in the FAQ doesn't seem
Ajay ohri wrote:
- go with Bob and Peter's book. Too many sources can be confusing.
- Join the list, create filters for keywords of your specialty (like from
R -Help ) contains regression.
- Try use it for a live project .
I think this is really very good advice:
- 2 books (I would
Recently a course with this title, from Vose consulting, was announced
on the list. Does anyone know of any books/websites/downloadable
tutorials etc that cover the same ground.
ie not just quantitative risk analysis, but specifically on using R
for risk analysis and as an alternative to
Thanks Martin!
Valerie
Martin Morgan wrote:
Hi Valerie --
curlPerform is expecting the url to be an option that is named, i.e.,
myopts -curlOptions(netrc=1)
url - http://www.omegahat.org/RCurl/testPassword/index.html;
cat(curlPerform(url=url, .opts=myopts))
!DOCTYPE HTML PUBLIC
I am able to call R functions like min, mean, and mad from from a C/C++
embedded application, but when I try to get the standard deviation or variance
(sd, var) R segfaults. Can someone suggest how to resolve this?
Thanks and best regards,
EBo --
Hello
I wrote a simple program to modify a boxplot:
gdsbox - function(indvar){
boxplot(indvar~gds3lev,
main = paste('Boxplot of', substitute(indvar), for GDS groups),
names = c('1', '3', '4, 5, 6'))
}
If I attach the dataframe gdsgraph, this works fine. However, I've been warned
On Wednesday 27 August 2008, Josip Dasovic wrote:
Hello:
As someone making the move from STATA to R, I'm finding it difficult at
times to perform basic tasks in R, so forgive me if I've missed an obvious
and easily obtained solution to my problem. I've searched the help guides
and the
Try this:
merge(aggregate(x$DAY, x[, c(YEAR, MONTH)], length),
data.frame(YEAR = unique(x$YEAR), MONTH = 1:12), all = T)
On Wed, Aug 27, 2008 at 2:11 PM, Josip Dasovic [EMAIL PROTECTED] wrote:
Hello:
As someone making the move from STATA to R, I'm finding it difficult at
times to
The reason that taking the log fixes things is because log( 2 ) = -log( 1/2 ),
so log takes us to an additive difference.
I think the use of log(utility) is actually based on experiments where people
were asked what they were willing to risk for various potential gains and the
results were
To Ravi Varadhan and Peter Dalgaard:
Infinite thanks for your help and suggestions- they were very instructive, as i
have still to learn on the appropriate use of the apply family of functions.
jlrp
--- On Wed, 8/27/08, Ravi Varadhan [EMAIL PROTECTED] wrote:
From: Ravi Varadhan [EMAIL
Hello,
I would like to combine different label sizes in the same x-axis, for
example
1 2 3 4 5 6 7 8 9
How can I do it?
Best,
Dani
--
Daniel Valverde Saubí
Grup de Biologia Molecular de Llevats
Facultat de Veterinària de la Universitat Autònoma de Barcelona
Edifici V, Campus UAB
08193
On 8/27/2008 1:15 PM, Peter Flom wrote:
Hello
I wrote a simple program to modify a boxplot:
gdsbox - function(indvar){
boxplot(indvar~gds3lev,
main = paste('Boxplot of', substitute(indvar), for GDS groups),
names = c('1', '3', '4, 5, 6'))
}
If I attach the dataframe gdsgraph,
Hi,
I'm not sure I understand what you want, but perhaps:
x - 1:9
y - x
plot(x, y, ann=F, xaxt=n)
axis(1, lab=F) - positions
mtext(paste(positions), side = 1, line = 1, outer = F, at = test,
cex=1:length(positions))
HTH,
baptiste
On 27 Aug 2008, at 18:36, Dani Valverde wrote:
Hello,
Whoops, typo in my previous post:
x - 1:10
y - x
plot(x, y, ann=F, xaxt=n)
axis(1, lab=F) - positions
mtext(paste(positions), side = 1, line = 1, outer = F, at = positions,
cex=seq(1, 9, length=length(positions)))
Is this what you mean by label sizes?
On 27 Aug 2008, at 18:51, baptiste
-Original Message-
From: Mario [mailto:[EMAIL PROTECTED]
Sent: Monday, August 25, 2008 5:41 PM
To: Greg Snow
Cc: r-help@r-project.org
Subject: Re: [R] Two envelopes problem
Dear Greg,
The problem is that in your code you are creating a
distribution where there are only 5-10
Hello all and thanks in advance for any help or direction. I have
co-authorship data that looks like:
PaperAuthor Year
1 SmithKK JonesSD 2008
2 WallaceAR DarwinCA 1999
3 HawkingS2003
I would like:
Paper Author
Hi,
name of the data.frame is assumed to be dt.
cnt.tmp - strsplit(dt$Author, )
cnt - sapply(cnt.tmp, length)
paper.cnt - dt$Paper[cnt]
author - unlist(cnt.tmp)
year - dt$Year[cnt]
dt.new - data.frame(Paper = paper.cnt, Author = author, Year = year)
HTH,
Dong-hyun Oh
On Aug 27, 2008, at
On Tue, Aug 26, 2008 at 6:56 PM, Alex Karner [EMAIL PROTECTED] wrote:
Thanks Deepayan, works like a charm.
A followup question though--I'd like to produce the same data on four
panels with the final two zoomed in, i.e. plotted with shorter x and
y axes. Since I can't access panel.number in
Dear Grant,
Try this:
x=Paper,Author, Year
1,SmithKK JonesSD, 2008
2,WallaceAR DarwinCA, 1999
3,HawkingS, 2003
X=read.delim2(textConnection(x),sep=,,header=TRUE)
closeAllConnections()
author= strsplit(as.character(X$Author), )
l=lapply(author,length)
I'm sorry Grant, my bad. To reproduce what you'd like to do, my previous
code should be:
x=Paper,Author, Year
1,SmithKK JonesSD, 2008
2,WallaceAR DarwinCA, 1999
3,HawkingS, 2003
X=read.delim2(textConnection(x),sep=,,header=TRUE)
closeAllConnections()
author=
I'm trying to figure out how to write a r function that will calculate
the extreme spread of a group of points given their (x,y)
coordinates. Extreme Spread is the maximal Euclidean distance between
two points in a group
ex.spread = max{ sqrt [ (xi-xj)^2 - (yi-yj)^2 ] } for i not equal to
On Wed, 27-Aug-2008 at 02:42PM +0200, Groeneweg M (GEN) wrote:
| Hello,
|
| I have a basic question about the display() function, but I can't find the
answer easily.
|
| I am running R 2.7.16.0 on Suse linux 11.0
Are you sure of that? R-2.7.2 is the most recent.
| with KDE 4.0
Your
Dear R users,
I am trying to call a Perl subroutine from R . The subroutine returns an
arrray contaning three elements wihch are all strings. But the calling in R
return an integer which is 0. I have no idea how this could happen. Maybe
becasue I shouldn't use system() in R or I should load a
You need to do:
system(perl presentPerformance.pl,intern=TRUE)
It does pay to read the help, you know.
cheers,
Rolf Turner
On 28/08/2008, at 8:36 AM, kevinchang wrote:
Dear R users,
I am trying to call a Perl subroutine from R .
2008/8/27 Patrick Connolly [EMAIL PROTECTED]:
Your question is one relating to your window manager. The display
parameter for x11() does not relate to desktops. All the desktops are
part of the same display. Some very clever R-devel people might be
able to think of a way of making a
Steven Matthew Anderson adastra69 at mac.com writes:
I'm trying to figure out how to write a r function that will calculate
the extreme spread of a group of points given their (x,y)
coordinates. Extreme Spread is the maximal Euclidean distance between
two points in a group
ex.spread
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