Try:
unlist(mapply(seq, c(1,20,50), c(7,25,53)))
[1] 1 2 3 4 5 6 7 20 21 22 23 24 25 50 51 52 53
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Chris Oldmeadow
Sent: Tuesday, August 26, 2008 12:42 AM
To: r-help@r-project.org
Hi Rolf, Hi Mark, Hi List,
I have not digested Rolf's response yet. It may well answer my problems.
In the meantime I have some reproducible code which actually shows my
problem:
patches -
structure(list(URBAN_AREA = structure(c(2L, 19L, 23L, 2L, 19L,
23L, 2L, 19L, 23L, 2L, 19L, 23L),
On Tue, 26 Aug 2008, Rolf Turner wrote:
On 26/08/2008, at 4:50 AM, Jarrett Byrnes wrote:
I'm currently working on writing up some documentation for some of my code,
but am having the darndest time coding in equations. For example, the
equation in the following:
\details{ Calculated the R
On Mon, 25 Aug 2008, Wolfgang Huber wrote:
Hi Paul,
You could
label=codechunk1a,fig=TRUE,include=FALSE=
plot(1:10)
@
label=codechunk1b=
cat(figure1.eps caption goes here \n,file=readme.txt,append=T)
cat(figure1.pdf caption goes here \n,file=readme.txt,append=T)
@
You can suppress one of the
Dear R-help,
I have a character vector, some elements will be numeric, some not,
and some even empty. E.g.:
temp1 - c(abcd, 2 ,)
I'm only interested in the numeric elements, the rest I can just throw
away. It is easy enough to loop through the vector:
temp - try(eval(parse(text=temp1[1])),
Just use as.numeric. Non numeric will be NA. So the solution of your
problem is na.omit(as.numeric(temp1))
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and
That's great, thanks. I can live with the warnings!
Cheers,
Rob
On Tue, Aug 26, 2008 at 4:49 PM, ONKELINX, Thierry
[EMAIL PROTECTED] wrote:
Just use as.numeric. Non numeric will be NA. So the solution of your
problem is na.omit(as.numeric(temp1))
HTH,
Thierry
Hi there,
I try to understand the usage of environments but I'm not sure if I get it. I
wrote a test script like this:
testenv - new.env(environment())
myfun - function(x) {
print(testvar)
testenv$testvar_2 - 20
}
environment(myfun) - testenv
testenv$testvar - 10
R-help,
Whenever I try to save a plot with savePlot
the file is not stored in my hard disk with the selected
format. Several formats are set and none of them
works. I just get the file name with missing extension
and it can't be open with programs like Paint and Microsoft Photo
Editor
Th only one
Dear R help forum,
I am using the function 'coxph' to obtain hazard ratios for the comparison
of a standard treatment to new treatments. This is easily obtained by
fitting the relevant model and then calling exp(coef(fit1)) say.
I now want to obtain the hazard ratio for the comparison of two
Thank you Martin,
It turned out that my problem was that I used underscores _ in my variable
names. No problem for R, but Latex didn't like it.
Kind regards,
Eelke
Martin Elff wrote:
On Monday 25 August 2008 (15:36:50), Eelke wrote:
Hello,
I'm using memisc to output regression results to
Laura Bonnett wrote:
Dear R help forum,
I am using the function 'coxph' to obtain hazard ratios for the comparison
of a standard treatment to new treatments. This is easily obtained by
fitting the relevant model and then calling exp(coef(fit1)) say.
I now want to obtain the hazard ratio
Code:
options(contrasts)
$contrasts
factor ordered
contr.treatment contr.poly
I want to change the first entry ONLY, without retyping contr.poly. How do
I do it? I have tried various possibilities and cannot get anything to work.
I found out that the response to
The standard treatment is the same in both comparison.
How do you do a three-level treatment factor?
I thought you had to have a censoring indicator which took values 0 or 1 not
1, 2 or 3?
Thanks,
Laura
On Tue, Aug 26, 2008 at 11:05 AM, Peter Dalgaard
[EMAIL PROTECTED]wrote:
Laura Bonnett
Hi to all,
seems to be simple, but I do not find the solution:
What must I write for the splitted to get
splitted$3$x and splitted$3$x
y = c(rep(2,5),rep(3,5))
ma - data.frame(x = 1:10, y=y )
splitted - split(ma, ma$y)
for (counter in (min(ma$y):max(ma$y)))
{
splitted$x
}
Regards
Try this also:
paste(apply(matrix(x, 2), 2, paste, collapse = ' '), collapse = \n)
On Mon, Aug 25, 2008 at 8:36 PM, remko duursma [EMAIL PROTECTED] wrote:
Dear R-helpers,
I have a numeric vector, like:
x - c(1,2,3,4,5,6)
I make this into a string for output to a text file, separated by
I think you need assign, see ?assign for more details.
On Tue, Aug 26, 2008 at 6:02 AM, Antje [EMAIL PROTECTED] wrote:
Hi there,
I try to understand the usage of environments but I'm not sure if I get it.
I wrote a test script like this:
testenv - new.env(environment())
myfun -
You need type the extension of the file:
plot(rnorm(10))
savePlot(test.png, type=png)
savePlot(test.bmp, type=bmp)
On Tue, Aug 26, 2008 at 6:29 AM, Luis Ridao Cruz [EMAIL PROTECTED] wrote:
R-help,
Whenever I try to save a plot with savePlot
the file is not stored in my hard disk with the
On 26/08/2008 6:30 AM, David Epstein wrote:
Code:
options(contrasts)
$contrasts
factor ordered
contr.treatment contr.poly
I want to change the first entry ONLY, without retyping contr.poly. How do
I do it? I have tried various possibilities and cannot get anything
At 19:44 25/08/2008, Jorge Ivan Velez wrote:
Dear R-list,
I'd like to do a meta-analysis plot similar to
Since these plots are known as forest plots
?forestplot
might help you.
As it says you have to do a bit more work but you do get much more flexibility
install.packages('rmeta')
Hi Mario,
If I understand your problem statement, the random choice of one of the
two envelopes ensures that the probability of choosing one or the other
is 0.5. If you find 10 (units, I can't find the pound symbol on this
keyboard), there is an equal likelihood that the other envelope contains
5
Hi,
I have to match names where names can be recorded with errors or additions.
Now I am searching for a string search function which returns always the
closest match. E.g. searching for Washington it should return only
Washington but not Washington, D.C. But it also could be that the list
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.
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Another way to do it (based on summary data):
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/131342.html
--
Jonathan Baron, Professor of Psychology, University of Pennsylvania
Home page: http://www.sas.upenn.edu/~baron
__
R-help@r-project.org
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 answer depends
on the
Strictly, you need to type the extension _if your filename or path
include a period (.)_ but not otherwise.
The filename test alone will be saved as paste(test,type). So will,
for example, /plots/test. But filenames such as
test.it or ../plots/test will not include the extension
automatically.
jcarmichael jcarmichael314 at gmail.com writes:
Thank you for the GGobi reference, it is a very handy tool! My main goal,
however, is to be able to identify univariate outliers (with boxplots for
example), and I'm having a hard time getting the rggobi package to do that.
Hmmm... I
Sorry, chaps. I need one more:
dbDisconnect(con.in)
Error in sqliteCloseConnection(conn, ...) :
RS-DBI driver: (close the pending result sets before closing this connection)
I am pretty sure I have fetched everything there is to be fetched. I
am not sure what I need to do to say goodbye
On Tue, Aug 26, 2008 at 6:07 AM, Henrique Dallazuanna [EMAIL PROTECTED] wrote:
I think you need assign, see ?assign for more details.
On Tue, Aug 26, 2008 at 6:02 AM, Antje [EMAIL PROTECTED] wrote:
Hi there,
I try to understand the usage of environments but I'm not sure if I get it.
I wrote
I have to match names where names can be recorded with errors or
additions.
Now I am searching for a string search function which returns always
the closest match. E.g. searching for Washington it should
return only Washington but not Washington, D.C. But it also could be
that the list
Douglas Bates wrote:
As Henrique said, the canonical way of assigning a value within an
environment is the assign. A more obscure, but also more effective,
approach is evalq which quotes an expression then evaluates it in the
given environment. For example
env - new.env()
evalq({aa -
Okay, I see, there is no really easy way (I was wondering whether I can set
an environment as default for new created variables). Is there any difference
if I call
myenv$myvar - 10
or
assign(myvar,10, env=myenv)
?
Antje
Douglas Bates schrieb:
On Tue, Aug 26, 2008 at 6:07 AM, Henrique
I have the following code which produces the output below it
clus1 - svydesign(ids = ~schid, data = lower_dat)
items - as.formula(paste( ~ , paste(lset, collapse= +)))
rr1 - svymean(items, clus1, deff='replace', na.rm=TRUE)
rr1
mean SE DEff
W525209 0.719748 0.015606 2.4932
Antje wrote:
Okay, I see, there is no really easy way (I was wondering whether I
can set an environment as default for new created variables). Is there
any difference if I call
myenv$myvar - 10
or
assign(myvar,10, env=myenv)
?
No. And with(myenv, myvar - 10) is also the same. However, you
On 26-Aug-08 10:30:30, David Epstein wrote:
Code:
options(contrasts)
$contrasts
factor ordered
contr.treatment contr.poly
I want to change the first entry ONLY, without retyping contr.poly.
How do I do it? I have tried various possibilities and cannot get
d - structure(list(Site = structure(c(8L, 12L, 7L, 6L, 11L, 5L, 10L,
4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L,
1L, 9L, 8L, 12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L,
12L, 7L, 6L, 11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L, 12L, 7L, 6L,
11L, 5L, 10L, 4L, 3L, 2L, 1L, 9L, 8L,
Duncan: I think I see what you're saying but the strange thing is that if
you use the utility function log(x) rather than x, then the expected values
are equal. Somehow, if you are correct and I think you are, then taking the
log , fixes the distribution of x which is kind of odd to me. I'm sorry
That works perfectly, great.
Thanks a lot for that Richard!
Werner
- Ursprüngliche Mail
Von: [EMAIL PROTECTED] [EMAIL PROTECTED]
An: Werner Wernersen [EMAIL PROTECTED]
CC: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Gesendet: Dienstag, den 26. August 2008, 14:10:11 Uhr
Betreff: Re: [R]
Here's a solution with ggplot2
library(ggplot2)
ggplot(na.omit(d), aes(x = EPT.Taxa, colour = Site)) + geom_density()
ggplot(na.omit(d), aes(x = EPT.Taxa)) + geom_density() + facet_grid(Site
~ .)
HTH,
Thierry
ir.
...
To pick up on what Mark has said: it strikes me that this is related to the
simplex, where the bounded nature of the vector space means that normal
arithmetical operations (i.e. Euclidean) don't work---that is, they can be
used, but the results are wrong. Covariances and correlations for
Did you try:
savePlot(test.bmp, type = bmp)
Monica
Message: 118
Date: Tue, 26 Aug 2008 10:29:13 +0100
From: Luis Ridao Cruz
Subject: [R] savePlot() does not save plot with format set
To:
Message-ID:
Content-Type:
definitly need to look into ggplot- very cool.
thanks
On Tue, Aug 26, 2008 at 10:20 AM, ONKELINX, Thierry
[EMAIL PROTECTED] wrote:
Here's a solution with ggplot2
library(ggplot2)
ggplot(na.omit(d), aes(x = EPT.Taxa, colour = Site)) + geom_density()
ggplot(na.omit(d), aes(x = EPT.Taxa)) +
On Mon, 25 Aug 2008, Jorge Ivan Velez wrote:
My problem is that I don't have the raw data as rmeta _requires_ and, even
when I have my data set in the _same_ (?) format that summary(a), when I
tried plot(mydata) it doesn't work.
No, it doesn't _require_ that.
If you just want a forest plot
On Tue, 26 Aug 2008, ivo welch wrote:
Sorry, chaps. I need one more:
dbDisconnect(con.in)
Error in sqliteCloseConnection(conn, ...) :
RS-DBI driver: (close the pending result sets before closing this connection)
I am pretty sure I have fetched everything there is to be fetched.
Mark
My experience was similarly frustrating. Maybe formulating the
problem a bit differently could help to clarify it.
State it like this:
Someone chooses an amount of money x. He puts 2x/3 of it in one
envelope and x/3 in an other. There is no assumption about the
distribution of x.
If you
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)); but
this does not work for dot plots apparently.
Then I tried
On 8/26/2008 9:51 AM, Mark Leeds wrote:
Duncan: I think I see what you're saying but the strange thing is that if
you use the utility function log(x) rather than x, then the expected values
are equal.
I think that's more or less a coincidence. If I tell you that the two
envelopes contain X
I am working on embedding R into some visualization research programs. Can
any point me to a collection of embedded and standalone R/C/C++ examples? The
documentation is to terse for me to figure out how to develop this and I am
looking for some simple examples to study.
Thanks and best
On 8/26/2008 9:44 AM, Peter Dalgaard wrote:
Antje wrote:
Okay, I see, there is no really easy way (I was wondering whether I
can set an environment as default for new created variables). Is there
any difference if I call
myenv$myvar - 10
or
assign(myvar,10, env=myenv)
?
No. And with(myenv,
Hi Duncan: I think I get you. Once one takes expectations, there is an
underlying assumption about the distribution of X and , in this problem, we
don't have one so taking expectations has no meaning.
If the log utility fixing the problem is purely just a coincidence, then
it's surely an odd one
You might also want to look at existing visualisation applications
that connect with R:
* http://ggobi.org
* http://rosuda.org/mondrian
* http://rosuda.org/software/Gauguin/gauguin.html
to name a few.
Hadley
On Tue, Aug 26, 2008 at 10:31 AM, EBo [EMAIL PROTECTED] wrote:
I am working on
I'm replying in public to this,
since it relates to an R-help thread of April 15-16, 2008
(and am also adding back 'References:' to one of those postings
of mine).
Mark Segal mark at biostat ucsf
on Mon, 25 Aug 2008 11:25:05 -0700 writes:
Hi Martin,
You posted the following
Hi,
Simply re-write your integrand as follows:
integrand - function(X) {0.5 * cos(theta) * 2 / (exp(-pi*X/2 - X*theta) +
exp(pi*X/2 - X*theta)) }
Now the problem goes away.
theta - -1
integrate(integrand, -Inf, 1)
0.9842315 with absolute error 1.2e-05
This would work for all theta such
Hi Everyone
I just installed Yellow Dog Linux (v 5.02) on an old G4 Powerpc replacing
the Mac OSX. The system is running beautifully. I would like to install R on
this system. I am not sure how to go about doing it. I have installed R on
my Ubuntu system as well as on Mac OSX but there are
Hello,
the arm package is failing to install, it says I need gnu make but it's not
clear how I specify which make to use in the R environment. For example I
already have gmake installed. How to I instruct R to use it?
* Installing *source* package 'Matrix' ...
usage: make [-BeikNnqrstWX] [-D
Dear R users/developers,
simple comparison of code execution time of R 2.7.1 and R 2.7.2 shows a
dramatic slowdown of the newer version. Rprof() identifies .Call
function as a main cause (see the code below). What happened with R
2.7.2?
Kind regards
Marek Wielgosz
Bayes Consulting
Hi,
I have downloaded a R script from
http://www.wessa.net/rwasp_edauni.wasp#output.
This script produces a densityplot graphic, amongst others, when is
executed from the web page.
However, when I run it in my machine the *densityplot* function produces
any output, I mean a blank graphic.
But,
Thanks so much. It works well in my MCMC sampling. May I know why re-writing
the integrand can solve the problem? I have been thinking it was the skewness
of the distribution brought the error. Thanks!
Xia
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]; r-help@r-project.org
Subject:
Hi all,
I am trying to do a 3d plot where the x,y,z axes intersects with the origin
(0,0,0) using the plot3d() funtion in the rgl package without success. I
looked back at the past archives on this subject and someone suggested using
djmrgl package. I searched and found it, installed it but when
On Tue, 26 Aug 2008, Chris Oldmeadow wrote:
Hi,
I have a vector of start positions, and another vector of stop positions,
eg start-c(1,20,50)
stop-c(7,25,53)
Is there a quick way to create a sequence from these vectors?
new-c(1,2,3,4,5,6,7,20,21,22,23,24,25,50,51,52,53)
Vectorize the
Duncan: Just one more thing which Heinz alerted me to. Suppose that we
changed the game so that instead of being double or half of X,
we said that one envelope will contain X + 5 and the other contains X-5.
So someone opens it and sees 10 dollars. Now, their remaining choices
are 5 and 15 so
So if I put $10 and $20 into the envelopes, then told you that the
values were multiples of $10, it would be wrong for you to assess
probabilities on $100, $1,000,000 and so on? :-)
But what if you reasoned that there were far more multiples of 10 above
20 than below 20?
What if I was really
Duncan Murdoch [EMAIL PROTECTED] 26/08/2008 16:17:34
If this is indeed the case, switch; the expected gain is
positive because _you already have the information that you hold the
median value of the three possibilities_. The tendency when
presented
with the problem is to reason as if this is
Thank you for your suggestion, I will play around with it. I guess my concern
is that I need each test result to occupy its own cell rather than have
one or more in the same row.
Adaikalavan Ramasamy-2 wrote:
There might be a more elegant way of doing this but here is a way of
doing it
Dear R Gurus:
What would be the best way to evaluate a Latin square problem, please?
Does it work in Rcmdr, please?
thanks,
Edna Bell
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
On 8/26/2008 1:12 PM, [EMAIL PROTECTED] wrote:
Duncan: Just one more thing which Heinz alerted me to. Suppose that we
changed the game so that instead of being double or half of X,
we said that one envelope will contain X + 5 and the other contains X-5.
So someone opens it and sees 10
On 8/26/2008 1:05 PM, Chibisi Chima-Okereke wrote:
Hi all,
I am trying to do a 3d plot where the x,y,z axes intersects with the origin
(0,0,0) using the plot3d() funtion in the rgl package without success. I
looked back at the past archives on this subject and someone suggested using
djmrgl
Marek Wielgosz wrote:
Dear R users/developers,
simple comparison of code execution time of R 2.7.1 and R 2.7.2 shows
a dramatic slowdown of the newer version. Rprof() identifies .Call
function as a main cause (see the code below). What happened with R
2.7.2?
I don't see much of a
beautiful. now, i think i got it. the fact that the calculation works
in the additive case doesn't make the calculation correct. the expected
value calculation is kind of assuming that the person putting the
things in the envelopes chooses what's in the second envelope AFTER
knowing
I have a large table of data which i can read in using read.table.
I then want to conduct various tests on the data.
Is there a simple way to conduct these tests by specifying the column headers
which relate to different conditions of the experiment?
e.g. data1 - read.table(test.table,
Thanks. I revised the code but it doesn't make difference. The cause of the
error is just as stated in the ?integrate If the function is approximately
constant (in particular, zero) over nearly all
its range it is possible that the result and error estimate may be seriously
wrong. I have
And somewhat faster still (YMMV):
unlist(mapply(:, start, stop))
HTH,
Ray Brownrigg
Victoria University of Wellington
Christos Hatzis wrote:
Try:
unlist(mapply(seq, c(1,20,50), c(7,25,53)))
[1] 1 2 3 4 5 6 7 20 21 22 23 24 25 50 51 52 53
-Christos
-Original Message-
Try the following:
sech-function(X) 2/(exp(-X)+exp(X))
your.integrand - function(X) {0.5*cos(theta)*exp(X*theta)*sech(pi*X/2)}
my.integrand - function(X) {0.5 * cos(theta) * 2 / (exp(-pi*X/2 - X*theta)
+ exp(pi*X/2 - X*theta)) }
theta - -1
my.integrand(-800)
your.integrand(-800)
Do you
A very important point is missing here. If there is x in one envelope
and 2x in the other the expected gain is 3x/2. If the idea is to
switch after observing the second envelope the expected gain is again
3x/2. In the case being put x will be either 5 or 10. But x is a
parameter and in this
Hello,
I am trying to calculate pairwise Pearson correlation distances, and
using biodist package, function cor.dist.
I start with a table of 4 rows and about 10 columns. (each of 4 samples,
or rows have values in each of the 10 categories, no zeros or NAs).
I am getting an error message:
Try
t.test(data1[,1],data1[,2])
for the first two columns of data1.
--- On Tue, 8/26/08, Richard Emes [EMAIL PROTECTED] wrote:
From: Richard Emes [EMAIL PROTECTED]
Subject: [R] stats tests on large tables
To: r-help@r-project.org
Received: Tuesday, August 26, 2008, 11:52 AM
I have a
Got it. Thanks so much!
Xia
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
CC: r-help@r-project.org
Subject: RE: [R] Problem with Integrate for NEF-HS distribution
Date: Tue, 26 Aug 2008 14:50:18 -0400
Try the following:
sech-function(X)
2/(exp(-X)+exp(X))
your.integrand -
Hello,
Is there a way to output the number of parameters in my model (gls)? I
can count the number of estimates, but I'd like to use the number of
parameters as an R object.
Thanks, Mark
__
R-help@r-project.org mailing list
?extractAIC
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678Champaign, IL 61820
On Aug 26, 2008, at 2:27 PM, Mark Na wrote:
Can't thank you enough! This worked!
N'DOYE Souleymane [EMAIL PROTECTED] 8/26/2008 1:20 AM
Hi Roderick,
Let me suggest you to save your spss file in txt, and use the read.table or
read.csv fonction. That is how I proceed.
I hope it will help you,
Best regards
On Mon, Aug 25, 2008 at
I encountered an error that does not make sense to me given my reading
of the documentation and does not appear to be referenced in the list
archives or online. The error occurred when a function received a NULL
value rather than a numeric value as a parameter and then attempted to
use the
Is this of any help
http://cran.r-project.org/web/views/ExperimentalDesign.html
--- On Tue, 8/26/08, Edna Bell [EMAIL PROTECTED] wrote:
From: Edna Bell [EMAIL PROTECTED]
Subject: [R] Latin squares in R
To: [EMAIL PROTECTED]
Received: Tuesday, August 26, 2008, 1:55 PM
Dear R Gurus:
What
On 27/08/2008, at 3:52 AM, Richard Emes wrote:
I have a large table of data which i can read in using read.table.
I then want to conduct various tests on the data.
Is there a simple way to conduct these tests by specifying the
column headers which relate to different conditions of the
R Friends,
I'm running R2.7.1 on Windows XP.
I'm trying to get some lattice functionality which I have not seen
previously documented--I'd like to plot the exact same data in multiple
panels but changing the grouping variable each time so that each panel
highlights a different feature of the
On 26/8/08 9:40 AM, Doran, Harold wrote:
Computing the mean of the item by itself with svymeans agrees with the
sample mean
mean(lower_dat$W524787, na.rm=T)
[1] 0.8555471
Compare this to the value in the row 9 up from the bottom to see it is
different.
You might be omitting more cases due
[Rolf, this crosses with your reply. I will look at your email next.]
I pasted the wrong code last time. The following code is supposed to
illustrate my problem with lattice plotting character changes.
patches -
structure(list(URBAN_AREA = structure(c(2L, 19L, 23L, 2L, 19L,
23L, 2L, 19L,
Is this what you want:
y = c(rep(2,5),rep(3,5))
ma - data.frame(x = 1:10, y=y )
splitted - split(ma, ma$y)
for (counter in (min(ma$y):max(ma$y)))
+ {
+ cat(counter, :, splitted[[as.character(counter)]]$x, '\n')
+ }
2 : 1 2 3 4 5
3 : 6 7 8 9 10
On Tue, Aug 26, 2008 at 6:37 AM, Knut Krueger
On Tue, Aug 26, 2008 at 2:26 PM, Alex Karner [EMAIL PROTECTED] wrote:
R Friends,
I'm running R2.7.1 on Windows XP.
I'm trying to get some lattice functionality which I have not seen
previously documented--I'd like to plot the exact same data in multiple
panels but changing the grouping
Indeed, missing data are the problem. The function removes any row with missing
data and the means are based on the remaining rows. So, I wrote a function that
just loops over each variable individually and organizes the data as I need it.
-Original Message-
From: James Reilly on
jim holtman wrote:
Is this what you want:
y = c(rep(2,5),rep(3,5))
ma - data.frame(x = 1:10, y=y )
splitted - split(ma, ma$y)
for (counter in (min(ma$y):max(ma$y)))
+ {
+ cat(counter, :, splitted[[as.character(counter)]]$x, '\n')
+ }
2 : 1 2 3 4 5
3 : 6 7 8 9 10
But maybe this
Hi,
I have the following script:
t.R ---
grafica - function() {
v - read.csv('preprocessed/komolongma.ece.uprm.edu.active',sep=',')
x - as.ts(v$active)
bitmap(file=output.png)
densityplot(~x,col='blue',main='Density Plot')
dev.off()
}
grafica()
t.R ---
When I sourced it from
Lattice functions do not create plots. In this respect they are quite
different to functions like plot(...), hist(...) c, which do create plots.
If you want to see the plot from a lattice function, you need to print the
object it creates. It's the action of printing that creates the graphical
See FAQ 7.22
On 27/08/2008, at 11:07 AM, John Sanabria wrote:
Hi,
I have the following script:
t.R ---
grafica - function() {
v - read.csv('preprocessed/komolongma.ece.uprm.edu.active',sep=',')
x - as.ts(v$active)
bitmap(file=output.png)
densityplot(~x,col='blue',main='Density
Hi Folks,
This tip is probably lurking somewhere already, but I've just
discovered it the hard way, so it is probably worth passing
on for the benefit of those who might otherwise hack their
way along the same path.
Say (for example) you want to do a logistic regression of a
binary response Y on
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 hard way, so it is probably worth passing
on for the benefit of those who might otherwise hack their
way along the same path.
Say
If you look at your sech(pi*x/2) you can write it as
sech(pi*x/2) = 2*exp(pi*x/2)/(1 + exp(pi*x))
For x -15, exp(pi*x) 10^-20, so for this interval you can replace
sech(pi*x/2) by 2*exp(pi*x/2) and so the integral from -Inf to -15 (or even -10
- depends on your accuracy requirements) can be
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 hard way, so it is probably worth passing
on for the benefit of those who might
Thanks for the answers, this one worked:
library(Agricolae)
michal33 wrote:
Hi there,
I am trying to run LSD.test(model)
I used the following commands:
attach(model)
m1- glm(ttl.m ~ site+year, family=quasipoisson, data= model)
df-df.residual(m1)
MSerror-deviance(m1)/df
The
On 27/08/2008, at 12:32 PM, michal33 wrote:
Thanks for the answers, this one worked:
library(Agricolae)
No, it couldn't have. There is no such package as ``Agricolae''.
There ***is*** a package called ``agricolae''. :-)
cheers,
Rolf Turner
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 hard way, so it is probably worth passing
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