On 12/04/2011 08:21 PM, set wrote:
Hello R-users,
I've got a file with individuals as colums and the clusters where they occur
in as rows. And I wanted a table which tells me how many times each
individual occurs with another. I don't really know how such a table is
called...it is not a
I forgot to change the header, so I guess no one read my mail. That's why I'm
trying it again...
Hello everbody,
I am new to this mailing list and hope to find some help.
I'm trying to get into the spatstat package and encountered two problems.
First a graphical one:
There is an
Chapman Hall/CRC: The R Series
We are delighted to announce that our new series of books on R is up and
running, with two books already published and another nine forthcoming
(including three set to publish in 2012). We are keen to receive proposals for
books covering all aspects of the
neither the *DescribeDisplay* package, nor *ggplot* are available in 2.13.2.
ggplot's package name is ggplot2. It certainly _was_ available for 2.13.2.
S***
This email and any attachments
Hi,
I'm trying to assign iterative names to variable, but all my attempts have
failed. I have a loop, and for every iteration, I need to create a variable,
and I'd like to name them iteratively, such as:
for(i in 1:10)
{
x_i - c(values)
}
I need it to return ten variables: x_1, x_2, ...,
On 05.12.2011 11:00, bcdc wrote:
Hi,
I'm trying to assign iterative names to variable, but all my attempts have
failed. I have a loop, and for every iteration, I need to create a variable,
and I'd like to name them iteratively, such as:
for(i in 1:10)
{
x_i- c(values)
}
You actually
?assign
for (i in 1:10) assign(paste('x_', i, sep = ''), i)
but before you do this, consider using a 'list'; check the archives since this
is asked a lot.
Sent from my iPad
On Dec 5, 2011, at 5:00, bcdc bia@gmail.com wrote:
Hi,
I'm trying to assign iterative names to variable, but all
Dear Sarah and R users,
Sorry again for not being explicit enough. Here is my full problem
dat - read.table(E:/thin plate/thin plate.csv, header=T, sep=,)
dat
names(dat) - c(x1, x2, y)
library(fields)
plot (dat)
#load the fields package (containing Tps) and fit a thin plate spline
tpsfit -
Re: [R] Data alignment
Thanks for your suggestions. I will try them.
The - in my original post was actually only there to serve as a
separator
so that it is easier for you to see the data structure but apparently it
rather confused you... sorry :)
That is why dput is to be used. Try
Hi
I have been trying to merge datasets, one of which has a long format
(Adata) and one has a (different) long format (Bdata):
Adata Bdata
subject order bpm subject order trial agegroup gender
1 1 70.21 1
Hello!
Can anybody help to convert this table (in fact there are two adjacent
tables) into dataset?
Hello
The solution maybe there but i need to compute it into R if someone can
help me :
(Previously store score vector t in score matrix T; store loading vector p
in loading matrix P) that's done by nipals in chemometrics
After use of nipals function you have T and P matrix
First : Calculate the
Try
plot( dat$x1, dat$y )
or
plot( x1, y, data=dat )
Note that your example is still not reproducible because you have not supplied
the data or a shortened version of your data for us to work with. Also, your
problem is not with thin plates but with plot, and reading the help for that
Hi R users,
I really need help with subsetting data frames:
I have a large database of medical records and I want to be able to match
patterns from a list of search terms .
I've used this simplified data frame in a previous example:
db - structure(list(ind = c(ind1, ind2, ind3, ind4), test1
This is not a homework help list.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Dear Jeff and R users,
Thanks Jeff.
Your first plot suggestion worked. And when I stick dat$x1, dat$x2,dat$y,
tpsfit seems to work as well. But it gives me a new error message
tpsfit - Tps(cbind(dat$x1, dat$x2), dat$y, scale.type=unscaled)
Warning message:
In gcv.Krig(out, nstep.cv =
Sorry for stupid question but is there any way to just download file (excel
file) from the url to the disk with a certain name?
Now it seems that I should open it (via read.xls) and then write it by
write.xls. But I am pretty sure it should be something much easier?
[[alternative HTML
Dear all,
I need to extract the covariance matrix of my quantile regression estimation to
use in a test. My regression is:
qf2_1 - summary(rq(wb2 ~ apv2 + vol2, tau = phi2[1]), cov = TRUE)
I can extract the covaraince matrix by using: qf2_1 [3]. However, if I try to
use it in the test, it
Hi Antonio,
how about this:
txt2 - useRs may fly into JFK or laGuardia
chartr(a-zA-Z, A-Za-z, txt2)
Cheers.
Am 05.12.2011 08:11, schrieb Tonio:
Hello R-help list,
I am looking for way to toggle the case of the characters like a flip-flop;
that is from ''Hello'' to hELLO or vice versa.
On Mon, Dec 5, 2011 at 7:03 AM, Philipp Chapkovski chapkov...@gmail.com wrote:
Sorry for stupid question but is there any way to just download file (excel
file) from the url to the disk with a certain name?
Now it seems that I should open it (via read.xls) and then write it by
write.xls. But I
Thank you for your brevity you don't have to say more things to show who
you are.
Else if you don't understand what we are talking about, you shouldn't be
rude and insulting.
May be you are quite young please take time to read Nonaka and Takeuchi and
moreover Takanashi to uderstand the way to go
Dear all,
I am a PhD student in energy modelling. I am completely stuck with the
following problem and would be very grateful for any kind of help.
I have cells and each cell is assigned a number. I would like to form
clusters of cells where the sum of the cell numbers in each cluster must
not
I'm not at my desk so this is untested, but I'm pretty sure double brackets
will do the trick:
qf2_1[[3]]
M
On Dec 5, 2011, at 7:14 AM, Julia Lira julia.l...@hotmail.co.uk wrote:
Dear all,
I need to extract the covariance matrix of my quantile regression estimation
to use in a test.
Change which to which.min, but then you won't easily be able to get the
corresponding x,y indices. Look at the arrayInd() function to translate the
result to a particular x,y.
Michael
On Dec 4, 2011, at 7:04 PM, vamshi999 vamshi...@gmail.com wrote:
this is exactly what i wanted.
How to
Your data was scrubbed by the server. Use dput() to create a plain text
representation of dat you can put in the body of the email.
Michael
On Dec 5, 2011, at 6:54 AM, Mintewab Bezabih mintewab.beza...@economics.gu.se
wrote:
Dear Jeff and R users,
Thanks Jeff.
Your first plot
does this do what you want:
db - structure(list(ind = c(ind1, ind2, ind3, ind4), test1 = c(1,
+ 2, 1.3, 3), test2 = c(56L, 27L, 58L, 2L), test3 = c(1.1, 28,
+ 9, 1.2)), .Names = c(ind, test1, test2, test3), class =
+ data.frame, row.names = c(NA,
+ -4L))
terms_include - c(1,2,3)
terms_exclude
Thank you very much, assign was exactly what I needed! For my case, it worked
better than list! :)
Beatriz
--
View this message in context:
http://r.789695.n4.nabble.com/iterative-variable-names-tp4159888p4160262.html
Sent from the R help mailing list archive at Nabble.com.
Dear sir,
As far as my understanding goes, the book “dynamic linear models with
R” does everything based up on the non innovations form.
The problem is that I couldn’t find auxiliary residuals, which can be
used to check for the presence of outliers and structural breaks. R
packages “dlm” and
Yes, that seems like a sensible idea to me.
Terry Therneau
On Sat, 2011-12-03 at 12:00 +0100, r-help-requ...@r-project.org wrote:
The problem: There are no a priori groupings to run a classification
on
My solution:
This is a non-R code question, so I appreciate any thoughts. I have
Thanks Michael
here is my data
structure(list(x1 = c(409.5, 349, 385.5, 273, 543, 746, 198.75,
262.5, 320.5, 259, 399.5, 595.5, 374, 293, 330, 658, 299, 775,
559.75, 251, 402, 395.5, 345.5, 283.5, 452, 816.5, 266.5, 360.25,
386, 160.25, 360, 337, 326, 758, 342.5, 389, 347.5, 819.25, 355.25,
281,
On Sat, Dec 3, 2011 at 11:16 AM, Jim Lemon j...@bitwrit.com.au wrote:
On 12/03/2011 06:04 AM, Hadley Wickham wrote:
Hi all,
I was wondering if any one had scripts that they could share for
capturing the current version of R packages used for a project. I'm
interested in creating a project
On 11-12-04 8:57 PM, wchips wrote:
G'day everyone,
I've been trying to get an interactive OpenGL plot to work from within
Python 2.6 using Rpy2 and the persp3d function. The problem is that the plot
seems to freeze upon activation. All interactivity is lost even though it
works fine when run
No problem, almost all complicated objects that get returned are internally
lists so you have to use [[ to get at them as [ will just give you a list
again. Take a look at
? '[[' # those should be back ticks
for the official documentation.
Best,
M
On Dec 5, 2011, at 8:09 AM, Julia Lira
If you were to rtfm you'd see that your anova suggests that the slope
coefficients
are the same for your tau = .15 and tau = .30 models.
url:www.econ.uiuc.edu/~rogerRoger Koenker
emailrkoen...@uiuc.eduDepartment of Economics
vox: 217-333-4558
Given the following data, I am plotting log.det ~ norm.beta, where the
points depend on a parameter, lambda
(but there is no functional form).
I want to find the (x,y) positions along this curve corresponding to two
special values of lambda
lambda.HKB - 0.004275357
lambda.LW - 0.03229531
and
-Original Message-
Thank you very much, assign was exactly what I needed! For my
case, it worked better than list! :)
If assign(paste('x_', i)) works better for your case than using a list or array
, you must surely have a very strange case
S
Hi Michael,
I can get what appears to be a good interpolation with a regression spline
in a multivariate LM, playing around with the tuning parameter to leave 1
residual df. Try this:
library(splines)
mod - lm(cbind(log.det, norm.beta) ~ bs(lambda, df=4), data=pd)
summary(mod)
x -
Dear R People:
I'm working with R-2.14.0 created from source on a quad core Windows 7 machine.
Starting yesterday, nearly everything I run just sits. When I hit
escape or Ctrl C, nothing happens.
Has anyone run into this, please?
Any suggestions would be much appreciated.
Thanks,
Erin
--
Perhaps you can try the examples given in
http://www.jstatsoft.org/v18/i02 for getting a better idea about how
the NIPALS algorithm works. BTW, yes,it looks like a homework question
specially when your user name is zz dd void1...@gmail.com so you
can't be recognized.
On Mon, Dec 5, 2011 at 9:27
Hi,
On Mon, Dec 5, 2011 at 10:03 AM, Erin Hodgess erinm.hodg...@gmail.com wrote:
Dear R People:
I'm working with R-2.14.0 created from source on a quad core Windows 7
machine.
Starting yesterday, nearly everything I run just sits. When I hit
escape or Ctrl C, nothing happens.
Has
That really all depends on what you need; and I can't tell you what you need.
set wrote
I'm sorry, I made a mistake in my example. you're right. I don't really
know how a similarity alogrithm worksbut I'm willing to try that...are
there any good examples available?
Thank you
--
On Dec 5, 2011, at 8:02 AM, R. Michael Weylandt wrote:
Change which to which.min, but then you won't easily be able to get
the corresponding x,y indices. Look at the arrayInd() function to
translate the result to a particular x,y.
The arrayInd help page is shared by the which help and
Hi,
I'm trying to use StatEt IDE for Eclipse as my R editor, but I'm completely
lost. I've read all I could find online, made apparently all I had to do
(installing rj, configuraing StatEt, etc.) but still cannot make R running.
Below is the error log file.
Thank you so much for assistance.
Matteo
Hi All,
I'm having a problem with barplot:
mydata
[1,] 2 108 0 0 0 1 3 0 0 0 0 0 7 18 3 4 8 20 26 20 19 7 1 1
mycol = c(rep('yellow', 2), rep('white', 3), rep('orange',2), rep('white', 5),
rep('orange',3), rep('red',9))
barplot(mydata, col = mycol)
gives me an uniformly
Thank's Pedro Madrones,
you're right that i have removed my name of the email (because of Spam, and
datas that i can't remove on the net).
Ok, i explain a bit more :
X is a centred matrix.
PCA summarise all variation of X into a few new variables called scores
T.These new variables are linearly
On Dec 5, 2011, at 10:44 AM, Federico Calboli wrote:
Hi All,
I'm having a problem with barplot:
mydata
[1,] 2 108 0 0 0 1 3 0 0 0 0 0 7 18 3 4 8 20 26 20
19 7 1 1
mycol = c(rep('yellow', 2), rep('white', 3), rep('orange',2),
rep('white', 5), rep('orange',3),
x - c(2L, 108L, 0L, 0L, 0L, 1L, 3L, 0L, 0L, 0L, 0L, 0L, 7L, 18L,
3L, 4L, 8L, 20L, 26L, 20L, 19L, 7L, 1L, 1L)
mycol = c(rep('yellow', 2), rep('white', 3), rep('orange',2),
rep('white', 5), rep('orange',3), rep('red',9))
barplot(x, col = mycol)
Produces a multi-colored barplot on my machine so I
Thanks, John
That's very clever. I didn't realize that the splines package supports
multivariate regression
splines and that works well enough for my purposes; plus this solution
is very transparent.
best,
-Michael
On 12/5/2011 9:50 AM, John Fox wrote:
Hi Michael,
I can get what appears
Dear Michael,
Thank you very much for your suggestion.
It indeed worked!
All the best,
Julia
CC: r-help@r-project.org
From: michael.weyla...@gmail.com
Subject: Re: [R] extract cov matrix in summary.rq and use as a matrix.
Date: Mon, 5 Dec 2011 07:55:45 -0500
To:
I have a question that is more statistical than r-specific
I have to recode a variable into quartiles, and I have weighted individuals.
To calculate the quartiles that will serve me to recode the variable, should
I use the weight or not?
--
View this message in context:
What are all the methods and properties for the flexclust package's kcca object?
Neither http://cran.r-project.org/web/packages/flexclust/flexclust.pdf nor
http://www.stat.uni-muenchen.de/~leisch/papers/Leisch-2006.pdf appear to include
a comprehensive centralized list.
Mark Pundurs
Data Analyst
On 5 Dec 2011, at 15:58, R. Michael Weylandt wrote:
x - c(2L, 108L, 0L, 0L, 0L, 1L, 3L, 0L, 0L, 0L, 0L, 0L, 7L, 18L,
3L, 4L, 8L, 20L, 26L, 20L, 19L, 7L, 1L, 1L)
mycol = c(rep('yellow', 2), rep('white', 3), rep('orange',2),
rep('white', 5), rep('orange',3), rep('red',9))
barplot(x, col =
On Dec 5, 2011, at 10:19 AM, SilvaForever wrote:
I have a question that is more statistical than r-specific
I have to recode a variable into quartiles, and I have weighted
individuals.
To calculate the quartiles that will serve me to recode the
variable, should
I use the weight or not?
barplot() defaults to stacked bars for matrices and uses the colors
fresh for each bar: e.g.,
layout(1:2)
barplot(matrix(1:6, 2), col = c(red, blue,green))
barplot(matrix(1:6, 3), col = c(red, blue,green))
To get what you are looking for, try something like this:
x = matrix(1:6,1); colnames(x)
Howdy,
I have read that if you put a URL in the text of a plot being saved
into pdf, the result is a functional hyperlink. I am interested in
having text in a plot that is linked to a URL, but I would like the
text to be something other than the URL. Is this possible? Thank you.
- Fincher
Dear Researches,
sorry for the easy and common question. I am trying to justify the idea of
RandomForest don't require a transformations (e.g. logarithmic) of
variables, comparing this non parametrics method with e.g. the linear
regressions. In leteruature to study my phenomena i need to apply a
On 05-Dec-11 18:39:21, Justin Fincher wrote:
Howdy,
I have read that if you put a URL in the text of a plot
being saved into pdf, the result is a functional hyperlink.
I am interested in having text in a plot that is linked to
a URL, but I would like the text to be something other than
Tree based models (such as RF) are invriant to monotonic transformations in the
predictor (x) variables, because they only use the ranks of the variables, not
their actual values. More specifically, they look for splits that are at the
mid-points of unique values. Thus the resulting trees are
On Dec 5, 2011, at 1:39 PM, Justin Fincher wrote:
Howdy,
I have read that if you put a URL in the text of a plot being saved
into pdf, the result is a functional hyperlink.
Don't believe everything you read.
pdf(test.pdf)
plot(1,1,main=http://test.gov/some.htm;)
dev.off()
It seems I missed the context of this post -- who is you, and what
is something other than the URL?
I feel the tikzDevice package should be an option for the task.
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State
about the because they only use the ranks of the variables. Using a
leave-one-out, in each interaction the the predictor variable ranks change
slightly every time RF builds the model, especially for the variables with
low importance. Is It correct to justify this because there are random
Dear list members,
I have a really simple problem.
I connected to a DB and have the following query
adat - dbGetQuery(con, paste(select * from kmdata where SzeAZ=',
szeazok[i], ' order by datum, sep=))
now I have the data in the adat variable which is a list. In fact the
elements of the
You should see no differences beyond what you'd get by running RF a second time
with a different random number seed.
Best,
Andy
From: gianni lavaredo [mailto:gianni.lavar...@gmail.com]
Sent: Monday, December 05, 2011 2:19 PM
To: Liaw, Andy
Cc:
nevermind, typo (and some gray hair)
sorry
12/5/2011 8:21 PM keltezéssel, Kehl Dániel írta:
Dear list members,
I have a really simple problem.
I connected to a DB and have the following query
adat - dbGetQuery(con, paste(select * from kmdata where SzeAZ=',
szeazok[i], ' order by datum,
Dear R users,
besides the current R 2.14 I would like to install a second version of R
(2.12.2) on my Ubuntu system. The current version is easily installed as a
precompiled package from Cran but I am heavily fighting with the older version.
I tried to follow the instructions here:
First, some general suggestions:
To see the structure of an object I would recommend
the str() function or, for a more concise output,
the class() function.
I don't think most ordinary users should be using
is.list() and, especially, is.vector().
Now for the particulars.
dbGetQuery
For example, say I am plotting some data that is genomic and therefore
maps to a specific locus on the human genome, like a gene. I was
hoping to have the title of the plot display the gene name, but have
it be a link so that clicking on it would take you to those
coordinates on a public browser,
On 05/12/2011 3:10 PM, Justin Fincher wrote:
For example, say I am plotting some data that is genomic and therefore
maps to a specific locus on the human genome, like a gene. I was
hoping to have the title of the plot display the gene name, but have
it be a link so that clicking on it would
Sorry I experimented with tikzDevice using \href{}{} but failed.
You can probably try the old image hotspot technique in HTML. I
remember someone did this before in R, but I cannot find the link to
the work now. The key is your need to use the two functions grconvertX
and grconvertY.
Regards,
Hi
On 6/12/2011 9:10 a.m., Justin Fincher wrote:
For example, say I am plotting some data that is genomic and therefore
maps to a specific locus on the human genome, like a gene. I was
hoping to have the title of the plot display the gene name, but have
it be a link so that clicking on it
Hello there,
I have a matrix with some data and I want to split this matrix based on the
values in one column. Is there a quick way of doing this? I have looked at
cut but I am not sure how to exactly use it?
for example:
I would like to split the matrix a based on the spending such that the
I'd so something like
split(a, a$spending)
and you can include a round(a$spending, -2) or something similar if
you want to group by the 100's.
Michael
On Mon, Dec 5, 2011 at 3:37 PM, Diviya Smith diviya.sm...@gmail.com wrote:
Hello there,
I have a matrix with some data and I want to split
Just a clarification: I can't get round to work as I first expected so
if you want to do bins by 100's you'd probably want:
split(a, cut(a$spending, breaks = (0:5)*100))
Michael
On Mon, Dec 5, 2011 at 3:41 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
I'd so something like
Hi all,
I have the following code,
When I run the code, it never terminate this is because of the while loop i
am using. In general, if you need a loop for which you don't know in
advance how many iterations there will be, you can use the `while'
statement so here too i don't know the number how
I forgot to upload the R-code in last email, so heare is one
epiann - function(T0 = 1, N=1000, ainit=1, binit=1,rho = 0.99, amean = 3,
bmean=1.6, avar =.1, bvar=.1, f){
moving - 1
count - 0
Temp - T0
aout - ainit
bout - binit
while(moving 0){
Greetings:
I have check the 'Building R Extensions' manual and can find no advice
on this issue,
so I am asking --
I have created an R package consisting entirely of R source code, and
created an installer
using the R CMD build / R CMD check commands. The package installs
correctly, using
R
Hi there,
I am wondering how R calculates the acf too?
I have a data set of approx 1500 returns when I calculate the lag 1
autocorrelation in excel I get a value of -0.4 but in R its approximately
-0.18?
I have cross checked with PC give and PC give agrees with excel?
I'm sure its just some
Hello,
I have a data set which I am using to find a model with the most significant
parameters included and most importantly, the p-values. The full model is
of the form:
sad[,1]~b_1 sad[,2]+b_2 sad[,3]+b_3 sad[,4]+b_4 sad[,5]+b_5 sad[,6]+b_6
sad[,7]+b_7 sad[,8]+b_8 sad[,9]+b_9 sad[,10],
where
On Mon, Dec 5, 2011 at 11:47 AM, Jannis bt_jan...@yahoo.de wrote:
Dear R users,
besides the current R 2.14 I would like to install a second version of R
(2.12.2) on my Ubuntu system. The current version is easily installed as a
precompiled package from Cran but I am heavily fighting with
Thank you very much Michael. This is very helpful. However, if there is any
way to exclude zero length bins. Lets imagine that the matrix was as
follows -
a - data.frame(patient=1:7, charges=c(100,500,200,90,400,500,600),
age=c(0,3,5,7,10,16,19), spending=c(10, 60, 110, 200, 400, 450, 500))
Add the drop = TRUE command to split
?split
split(a, cut(a$spending, breaks = (0:5)*100), drop = TRUE)
Michael
On Mon, Dec 5, 2011 at 4:06 PM, Diviya Smith diviya.sm...@gmail.com wrote:
Thank you very much Michael. This is very helpful. However, if there is any
way to exclude zero length
Did you check the MASS reference given above in the thread? If you
want to see the source, it's here:
http://svn.r-project.org/R/trunk/src/library/stats/src/filter.c (best
I can tell)
Michael
On Mon, Dec 5, 2011 at 2:24 PM, Bazman76 h_a_patie...@hotmail.com wrote:
Hi there,
I am wondering how
That's solved.
Just a cor between X and ti values.
Thank you all.
2011/12/5 zz dd void1...@gmail.com
Thank's Pedro Madrones,
you're right that i have removed my name of the email (because of Spam,
and datas that i can't remove on the net).
Ok, i explain a bit more :
X is a centred matrix.
Thanks a lot Eik!
That is exactly what I was looking for.
Best
Antonio
Fra: Eik Vettorazzi e.vettora...@uke.de
Cc: r-help@R-project.org r-help@r-project.org
Sendt: 23:21 mandag den 5. december 2011
Emne: Re: [R] Toggle cASE
Hi Antonio,
how about this:
Dear Michael and R users,
I generated a more orderly looking data again and below are the data and the
codes I am trying to run.
many thanks
mintewab
dat - read.table(E:/thin plate/thin plate.csv, header=T, sep=,)
names(dat) - c(x1, x2, y)
library(fidelds)
plot (dat)
plot (dat$x1, dat$x2)
As David said, your minimal working example is neither minimal nor
working...consider this (admittedly quite clunky) example though:
time - seq(10, 20, length.out = 100)
S1 - 10*exp( cumsum(rnorm(100))/20)
S2 - 10*exp( cumsum(rnorm(100))/20)
plot(time, S1, ylim = range(c(S1,S2)), type = n)
Your code is not reproducible nor minimal, but why don't you put a
command print(acceptprob) in and see if you are getting reasonable
values. If these values are extremely low it shouldn't surprise you
that your loop takes a long time to run.
More generally, read up on the use of print() and
Yes, I checked the acceptprob, it is very high but in my view, the while
loop is not stopping, so there is some thing wrong in the use of while
loop. When I removed the while loop, it returned some thing but not the
result what I want. When i run the while loop separately, it never stops.
On
My data consists of a numeric (yy)
variable and a categorical (xx) variable, as shown below.
xx =
c(rep(C, 5), rep(D,5))
yy = rnorm(10, 0, 4)
xx1 =
as.integer(as.factor(xx))
plot(xx1, yy, ylim =
c(-13.5, 4), col=blue)
I
wish to generate a scatter plot of the data such
On 11-12-05 3:04 PM, Rick Reeves wrote:
Greetings:
I have check the 'Building R Extensions' manual and can find no advice
on this issue,
so I am asking --
I have created an R package consisting entirely of R source code, and
created an installer
using the R CMD build / R CMD check
On 11-12-05 4:03 PM, Peter Langfelder wrote:
On Mon, Dec 5, 2011 at 11:47 AM, Jannisbt_jan...@yahoo.de wrote:
Dear R users,
besides the current R 2.14 I would like to install a second version of R
(2.12.2) on my Ubuntu system. The current version is easily installed as a
precompiled
plot(xx1, yy, ylim = c(-13.5, 4), col=blue, xaxt=n)
axis(1, at=c(1, 2), labels=c(C, D))
Defining your own axes with axis() offers a great deal of flexibility.
And thank you for providing a small reproducible example.
Sarah
On Mon, Dec 5, 2011 at 5:43 PM, Juliet Ndukum jpnts...@yahoo.com wrote:
If you run
out- epiann(f = function(a,b) log(dnorm(a)*dnorm(b))), N = 10)
It takes less than 0.5 seconds so there's no problem I can see:
perhaps you want to look elsewhere to get better speed (like Rcpp or
general vectorization), or maybe your loglikihood is not what's
desired, but there's no
My data consists of numeric (yy) and categorical (xx) variables, as shown
below.
xx = c(rep(C, 5), rep(D,5)) xx [1] C C C C C D D D D
D yy = rnorm(10, 0, 4) yy [1] 2.7219346 -3.7142481 6.8716534 -0.9352463
0.4901249 3.8113247 [7] -2.6602041 1.7714471 4.7298233 0.8848188
xx1 =
Hi,
I have a dataset which includes monthly data for 17 years. I want to be able
to see the monthly data behavior for the period of all 17 years.
What my data looks like:
http://r.789695.n4.nabble.com/file/n4162534/data.jpg
I would like to represent the data in a graph such as the one below:
Thanks Duncan. The thing is, if you want to display a static R graph using
Rpy2 you need to bring your Python program to a halt immediately after
printing the graph to screen using something like:
raw_input(Press any key to continue ...)
, otherwise it flashes up on the screen for a fraction of
Hi, so I don't speak computer and I have no idea what this code is telling
the program to do, but I apparently need to be able to find and isolate
influencial observations. Problem, I have no idea what the error means and
where it may be from in the code.
error I get is below the code
{
## OLS
head, tail and fix commands don't really work well if I have large
matrix/array for which I would like to be able to scroll up and dow, left
and right ...
Could anybody please help me?
Thanks
[[alternative HTML version deleted]]
__
For example, fix does have scrolling, but could we have a grid based or
cell based viewer/editor just like Excel sheet or like Matlab's object
editor?
For some weird reason, fix can only show numbers a weird format...
On Mon, Dec 5, 2011 at 7:01 PM, Michael comtech@gmail.com wrote:
head,
And fix doesn't show the full content...
On Mon, Dec 5, 2011 at 7:07 PM, Michael comtech@gmail.com wrote:
For example, fix does have scrolling, but could we have a grid based or
cell based viewer/editor just like Excel sheet or like Matlab's object
editor?
For some weird reason, fix can
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