On Tue, Jul 31, 2012 at 2:43 AM, Elliot Joel Bernstein
elliot.bernst...@fdopartners.com wrote:
Is there an easy way to thin a lattice plot? I often create plots from
large data sets, and use the pdf command to save them to a file, but the
resulting files can be huge, because every point in the
On Mon, Jul 30, 2012 at 10:16 PM, Bert Gunter gunter.ber...@gene.com wrote:
David:
I think one needs to carefully parse the xyplot help, where it says:
To use more than one legend, or to have arbitrary legends not
constrained by the structure imposed by key, use the legend argument.
So I
Thanks Prof Ripley. I found the workaround quickly. I had read the
documentation while searching for connection string but did not look for
below specific issue. Lesson learnt.
Regards,
Alok
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Monday, July
Thanks very much for the reply Michael.I guess I have to combine some levels to
make categorical variable to a binary variable. Dushanthi Date: Sat, 28 Jul
2012 16:38:24 +0100
To: dushan...@bell.net; r-help@r-project.org
From: i...@aghmed.fsnet.co.uk
Subject: RE: [R] metafor package,
Rui,
You made that WAY too easy!
Thanks so much,
Matt Guzzo
On Mon, Jul 30, 2012 at 6:45 PM, Rui Barradas [via R]
ml-node+s789695n4638471...@n4.nabble.com wrote:
Hello,
Sorry for my earlier post, they show the problem of not providing
context. Fortunatelly, arun's post made me realize
Below is the string to parse and return the embedded number = 11
string = \t\t\t\t\tspan class=\compliment-count profile\11/span
On Jul 29, 2012, at 3:00 AM, r-help-requ...@r-project.org wrote:
char
[[alternative HTML version deleted]]
Hi,
Please check this article which details the method for comparison:
Statistica Sinica 17(2007), 1115-1137
TESTING FOR THE EQUALITY OF k REGRESSION
CURVES
Juan Carlos Pardo-Fern ́ndez, Ingrid Van Keilegom and Wenceslao Gonz
́lez-Manteiga
Also, check some examples
Hi,
When you cbind two zoo objects, I guess the index should match, otherwise it
gives warning messages:
x1zoo_f-zoo(rnorm(5,25),c(1,5,10,15,9))
xzoo-zoo(c(5,9,10,15),c(1,5,9,10,15))
cbind(xzoo,x1zoo_f)
xzoo x1zoo_f
1 5 24.85877
5 9 25.09264
9 10 25.79896
10 15 26.70625
15
Hello,
I tried your code in R 2.15 with Ubuntu 12.04. It looks okay to me.
datosdv-Ciencias Sociales y Jurídicas n:74 | 33.94%
print(datosdv)
#[1] Ciencias Sociales y Jurídicas n:74 | 33.94%
library(hwriter)
p=openPage('test.html')
hwrite(datosdv,p,br=TRUE)
#test.html output
Ciencias
Hello Rahul,
I can see comments for each line of the code which is not very hard to
understand.
From your earlier post, I think you were having some problems with the compute
method. Check this link
Hi Simon,
Thanks for your reply.
m - bam(Correct ~ cEnglishTotal + te(WSTResid, RavenResid) + s(Stimulus,
bs=re) + s(Subject, bs=re), data = dat, family = binomial)
# cEnglishTotal, WSTResid and RavenResid are continuous variables; Correct,
Stimulus and Subject are factors.
vis.gam(m,
I just followed the instructions on
CRANhttp://cran.r-project.org/bin/linux/ubuntu/README to install
R on an Ubuntu instance.
sudo apt-get install r-base
Why does it install an old version of R? Can I install version 15.1?
I changed my sources.list to be a current cran mirror. I believe that I
I got R to update to the latest version
The problem was the sources.list file. I had not fixed the sources.list
file correctly, which was apparent when I opened the file using
sudo nano sources.list
Opening the file from the command line turns on syntax highlighting (as
opposed to opening a file
Hello,
Inline
Em 31-07-2012 02:59, Wellington Silva escreveu:
Ok,
This really helped.
You've probably noticed, I'm a begginer using R.
And when you said that you tried with rnorm and the counts were not zero,
where did you use the rnorm?
Where the runif is, in its stead use
matrix(
Hello,
Try the following.
string - \t\t\t\t\tspan class=\compliment-count profile\11/span
gsub([^[:digit:]], , string)
Then use as.numeric or as.integer.
Hope this helps,
Rui Barradas
Em 31-07-2012 01:19, Shelby McIntyre escreveu:
Below is the string to parse and return the embedded number
Hi Michael,
1. I try this
xzoo - zoo()
and it does work. OK, I will read the reference you provide to find
out why not to do so
2. How can I create an array of zoo objects?
3. Do you mean by R's wisest virgil Pat Burns' R Inferno?
To Arun:
Even when indices
Hi,
Here i have a Matrix
MyMatrix -
NameAge
- ---
ANTONY27
IMRAN 30
RAJ 22
NAHAS 32
GEO 42
and here i have an array with Minimum and Maximum values.
MinMaxArray - data.frame(MIN = 25,MAX=35)
MIN
On Sun, 29 Jul 2012, Gene Leynes wrote:
I would really like some help with understanding the panel function, in
zoo. Thank you.
Have you looked at ?plot.zoo. Some of the features you ask about are
explained there. In particular, it is explained that arguments like col,
lty, etc. are
On Jul 31, 2012, at 02:50 , Ario Ario wrote:
Hi,
Thank you. I've tried to find the substitute of such packages with google,
but I couldn't find any. Could you please tell me how to find it? Sorry, I'm
a beginner user of this software.
First explain what you want to do and where you got
Hello,
Please learn how to use dput(), it's not your first post.
And try the following.
myMatrix - data.matrix(read.table(text=
NameAge
ANTONY27
IMRAN 30
RAJ 22
NAHAS 32
GEO 42
, header=TRUE))
MinMaxArray -
Hi,
I understand that to test the significance of correlation between two
PAIRED variables, the function, cor.test () can be used. However, in my
case, I have tested the correlation (i.e., Correlation Coefficient, r)
between two independent (i.e, different) variables, and now I wish to test
for
Hello,
Ok, final retouches.
1. Divide the par() call:
oldpar - par(mar=c(9, 4.1, 4.1, 2.1))
oldfont - par(font=3)
Then, just before legend()
par(oldfont) # back to normal font
2. Modify the text() call to
text(as.vector(bp), y = -6.5, cex=0.7,
pos = 2,
offset = -0.25,
If you have pearson correlation coefficients r calculated from n data points
elsewhere, you can use a t test on
t.r - r * sqrt( (n-2)/(1-r^2) )
You'll need to use pt() to get your p value from t.r, though.
S Ellison
I understand that to test the
On 31/07/12 21:13, Chintanu wrote:
Hi,
I understand that to test the significance of correlation between two
PAIRED variables, the function, cor.test () can be used. However, in my
case, I have tested the correlation (i.e., Correlation Coefficient, r)
between two independent (i.e, different)
On Mon, Jul 30, 2012 at 6:00 PM, jim holtman jholt...@gmail.com wrote:
try this:
(x - rep(letters,2))
[1] a b c d e f g h i j k l m n o p
q r s t u v w
[24] x y z a b c d e f g h i j k l m
n o p q r s t
[47] u v w x y z
values - c(aa, a, b, NA, d, zz)
repl - c(aa, A, B, NA, D, zz)
hard to judge what the problem is without an example and without the actual
error message that mlogit (which one? there are several functions with that
name in various packages) gives you.
best, Ingmar
On Mon, Jul 30, 2012 at 11:30 AM, Lee van Cleef l.van.cl...@gmx.net wrote:
Dear all,
does
On Fri, 2012-07-27 at 11:52 -0700, Gordon Holtgrieve wrote:
Hello,
I am using vegan to do an NMDS plot and I would like to suppress the labels
for the loading vectors. Is this possible? Alternatively, how can I avoid
overlap?
Hi Gordon,
You seem to be trying to fit the species scores as a
Dear all,
I am having a matrix that stores data information in the following format.
roofPart1$TimeStamps[1:5,]
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 2011 7 21 15 25 20.609
[2,] 2011 7 21 15 25 23.265
[3,] 2011 7 21 15 25 26.000
[4,] 2011 7 21 15 25 28.671
Thanks everyone for your replies. I didn't know about the ecdfplot
function, so I'll start using that instead of Ecdf. Why is Ecdf not a
lattice plot? The result certainly looks like other lattice plots, the
arguments are similar to other lattice plots. In fact, internally it seems
to just call
On Jul 31, 2012, at 02:58 , Jeff Newmiller wrote:
Amazingly, RTFM.
RTFWP, I think you mean. Sounds like you are referring to this one:
http://cran.r-project.org/bin/linux/ubuntu/
Linux typically uses repositories to download software from. If you follow
the instructions on CRAN specific
Hello,
Try the following.
mat - data.matrix(read.table(text=
20117 21 15 25 20.609
20117 21 15 25 23.265
20117 21 15 25 26.000
20117 21 15 25 28.671
20117 21 15 25 31.343))
secs - apply(mat, 1, function(x)
ISOdatetime(x[1], x[2], x[3],
Well, yes.
Terminology-wise, I guess one could say that it's a trellis plot in
the Hmisc package.
But I'd agree that this is nitpicking.
-- Bert
On Tue, Jul 31, 2012 at 6:13 AM, Elliot Joel Bernstein
elliot.bernst...@fdopartners.com wrote:
Thanks everyone for your replies. I didn't know about
a) There are various routes to modifying sources.list... Ubuntu Software Center
has a dialog box for it, and there are several other such tools.
b) I always copy-pasted from the instructions so I had no problems with fiddly
bits.
c) Not all mirrors are created equal... sometimes you need to
Hello,
Glad it helped.
E fico à espera do relatório.
Rui Barradas
Em 31-07-2012 14:40, Wellington Silva escreveu:
Rui,
This is a cientific initiation program.
The idea is to develop a code which can simulate data and calculate the ARL
later.
*So, a little bit later yesterday night, after
Hi all
I'm dealing with a supervised binary classification issue. I'd like to use
the GBM package to classify individuals as uninfected/infected. I have 15
times more uninfected than infected individuals.
I was wondering if GBM models suffer in the case of imbalanced class sizes?
I didn't find
I need to multiply to very large, nonsparse matrices, and so get the
error allocMatrix: too many elements specified.
Is there a way to set the limit for allocMatrix?
In my case, the two matrices, A and B, are nxm and mxp where m is
small, so I could subdivide each into blocks of submatrices
On Tue, Jul 31, 2012 at 2:40 AM, jpm miao miao...@gmail.com wrote:
Hi Michael,
1. I try this
xzoo - zoo()
and it does work. OK, I will read the reference you provide to find
out why not to do so
2. How can I create an array of zoo objects?
I'm not quite sure what
Hi!
The kernlab function kpca() mentions that new observations can be transformed
by using predict. Theres also an example in the documentation, but as you can
see i am getting an error there (As i do with my own data). I'm not sure whats
wrong at the moment. I haven't any predict functions
Hm.. seems like its a problem with loading it in the profile..
If i load it again in the console it works fine. Must have something to do with
the masking.
--
Changed load-order with the package which required stats that did the
masking, and it works now, so, nevermind.
On
I'm attempting to update to R 2.15.1, and I'm having trouble with a package
that depends on the Matrix package. I've created a dummy package
consisting only of a DESCRIPTION file that specifies the dependence on
Matrix, a NAMESPACE file, and an R directory, containing a single
function, square -
How about sending an email to the OP with a message like:
Hi,
Thanks for submitting a question to the R-help list.
We hope you did read the Posting Guide and submitted a reproducible example
of your code (by the use of dput, structure, ...).
Then there is no need to add the message to the end
Dear R-help members
I would be grateful if anyone could help me with the following problem:
1: I changed the font style of the plant names (along x-axis, under the bars)
to italic . How can I avoid that the font in the legend also changes to
italic?
2. I would like to place the plant names
Dear list members,
I have a large dataset with point data. Each data point has X, Y, number of
species present and values for one single explanatory variable. Each point
was generated from a polygon grid shapefile and they are at equal
distances.
I conducted GLS analyses introducing variance
Hello,
Try this:
string = \t\t\t\t\tspan class=\compliment-count profile\11/span
gsub(.*(11).*,\\1,string)
#[1] 11
A.K.
- Original Message -
From: Shelby McIntyre smcint...@scu.edu
To: r-help@r-project.org
Cc:
Sent: Monday, July 30, 2012 8:19 PM
Subject: [R] How can I parse this
Thanks Arun,
Yes, I have Windows 7. I have tried 2 versions of R, 2.14.1 and 2.15.x, but
it did not change anything.
Right now I can't try a different version of win.
Ramón
On Tuesday, July 31, 2012, arun kirshna [via R] wrote:
Hello,
I tried your code in R 2.15 with Ubuntu 12.04. It looks
Hi,
#Without indices
xzoo-zoo()
x1zoo_f-zoo(1:10,)
cbind(xzoo,x1zoo_f)
x1zoo_f
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
#With indices on one
x1zoo_f-zoo(1:5,1:10)
cbind(xzoo,x1zoo_f)
x1zoo_f
1 1
2 2
3
Hi,
Try this:
myMatrix-read.table(text=
Name Age
ANTONY 27
IMRAN 30
RAJ 22
NAHAS 32
GEO 42
,sep=,header=TRUE)
MinMaxArray - data.frame(MIN = 25,MAX=35)
myMatrix[myMatrix$Age=MinMaxArray$MAX myMatrix$Age=MinMaxArray$MIN,]
#
I have a dataset with the variables:
idstart stop event tempdiff straling
The data contains recurrent event, so for each id more than one interval is
created
and more than one event can occur.
I have fitted the next model:
mod1 = coxph(Surv(start,stop,event) ~ tempdiff3 +
Hello!
I want to realise an univariate time series analysis in R, can someone help
me for the first steps?
Thanks
--
View this message in context:
http://r.789695.n4.nabble.com/Univariate-Time-Series-Analysis-in-R-tp4638538.html
Sent from the R help mailing list archive at Nabble.com.
Hi
I trying to add a wind rose as a plot together with other plots.
Im unsing the roseavent function from the climatol package.
Ive tried par(mfrow=c(2,2))
but when plotting the windrose:
rosavent(windfreq_rose,4,3,ang=-3*pi/16,main=Windrose) it skips one
position, and when plotting the next one
Hello,
I am a new user of R. I am trying to create a vector of time series called
LTot from a data frame. The purpose is to call LTot element from a loop
to make multiple regressions over a large number of regressors. Here a piece
of R code
OGData - read.table(heartatk4R.txt, header=TRUE)
OGCol
Hello,
I have a data frame, one element in that data frame is a LIST, with each
element being a character string. I am trying to extract the first year listed
in each of those character strings. The character elements are typically csv,
but the position of the year can vary (think citations
A follow up on this issue...if I first build the package using R CMD build
temp, I can install the resulting .tar.gz file from within R using the
install.packages command, but I still can't install it from the command
line using R CMD INSTALL.
- Elliot
On Tue, Jul 31, 2012 at 11:33 AM, Elliot
Does anyone know of any X^2 tests to compare the fit of logistic models
which factor out the sample size? I'm dealing with a very large sample and
I fear the significant X^2 test I get when adding a variable to the model
is simply a result of the sample size (200,000 cases).
I'd rather use
On 31/07/2012 15:10, David Romano wrote:
I need to multiply to very large, nonsparse matrices, and so get the
error allocMatrix: too many elements specified.
Is there a way to set the limit for allocMatrix?
No: see ?'Memory-limits'.
There are also limits on individual objects. On all
On Jul 31, 2012, at 7:13 AM, override wrote:
Hello!
I want to realise an univariate time series analysis in R, can someone help
me for the first steps?
Thanks
http://cran.r-project.org/web/views/TimeSeries.html
**
The contents of this message do not reflect any
On Jul 30, 2012, at 8:48 PM, arun wrote:
Hi,
Please check this article which details the method for comparison:
Statistica Sinica 17(2007), 1115-1137
TESTING FOR THE EQUALITY OF k REGRESSION
CURVES
Juan Carlos Pardo-Fern ́ndez, Ingrid Van Keilegom and Wenceslao
Gonz ́lez-Manteiga
Also,
Hello,
Try the following.
x - c(text, text, 2001, text, text, 2000, text, 1999, text, text,
text)
extract.year - function(x, n = 4){
pattern - paste(.*([[:digit:]]{, n, }).*, sep=)
as.integer(sub(pattern, \\1, x))
}
extract.year(x)
The argument 'n' is the number of digits of year.
On Tue, Jul 31, 2012 at 9:13 AM, override hugos...@gmail.com wrote:
I want to realise an univariate time series analysis in R, can someone help
me for the first steps?
There's a lot of material on this subject if you just do a basic
search of Google or rseek.org. As already mentioned, the CRAN
DanielFV kalandru at hotmail.com writes:
[snip]
I conducted GLS analyses introducing variance structures and correlation
structures to improve the model.
I plotted variograms before and after adding the correlation structures but
I would like to be able to conduct a test that gives me
Dear R-Help,
I am using 'rms' package to draw nomogram. I wonder how is the Points
determined for each predictor in the model? Is it by the coefficient
estimate (beta) relative to the highest effect in the model or?
Thanks
Lin
--
View this message in context:
Hi,
I'm trying to use gmm package in order to calculate linear regression (I
need to use the gmm for other application and this is a prior test I'm
doing).
I've defined a function for linear regression with 2 variables (x[,1] holds
the y values, while x[,2:3] holds the x values):
function(tet,
On Jul 31, 2012, at 1:31 AM, Rui Barradas wrote:
Hello,
Please learn how to use dput(), it's not your first post.
And try the following.
myMatrix - data.matrix(read.table(text=
I think the data.matrix transformation is a bad idea. It forces the
numeric values to be character and the
At 01:18 31/07/2012, Dushanthi Pinnaduwage wrote:
Thanks very much for the reply Michael.
I guess I have to combine some levels to make categorical variable
to a binary variable.
Without knowing more about your outcome variable (you have not even
told us yet whether it is ordered or not)
On Tue, Jul 31, 2012 at 6:43 PM, Elliot Joel Bernstein
elliot.bernst...@fdopartners.com wrote:
Thanks everyone for your replies. I didn't know about the ecdfplot function,
so I'll start using that instead of Ecdf. Why is Ecdf not a lattice plot?
The result certainly looks like other lattice
On Jul 31, 2012, at 10:35 AM, M Pomati marco.pom...@bristol.ac.uk wrote:
Does anyone know of any X^2 tests to compare the fit of logistic models
which factor out the sample size? I'm dealing with a very large sample and
I fear the significant X^2 test I get when adding a variable to the
Hello,
This should be pretty simple but I cannot get it right. Please point to the
right code. Thanks.
last - read.csv(file.path(dataDir,plot1.csv), as.is=T,stringsAsFactors
= FALSE)
last
dater_wvht
18/6/2008 0.977
28/8/2008 0.773
3 8/11/2008 1.483
4
Hello,
You could use dput(), it's not your first post...
last - structure(list(date = c(8/6/2008, 8/8/2008, 8/11/2008,
8/13/2008, 8/14/2008, 8/18/2008, 8/20/2008, 8/27/2008,
8/28/2008, 8/31/2008, 9/2/2008, 9/3/2008, 9/4/2008,
9/5/2008, 9/8/2008, 9/11/2008, 10/12/2009, 10/14/2009,
10/19/2009,
On Tue, Jul 31, 2012 at 12:02 PM, Yolande Tra yolande@gmail.com wrote:
Hello,
This should be pretty simple but I cannot get it right. Please point to the
right code. Thanks.
last - read.csv(file.path(dataDir,plot1.csv), as.is=T,stringsAsFactors
= FALSE)
last
dater_wvht
1
Hello,
Sorry, I forgot the time series part of your question. You could use
instead one of
# 1. type = l gives a line plot
plot(r_wvht ~ date, data = last, type=l)
# 2. use time series object plot
library(zoo)
z - zoo(last$r_wvht, order.by=last$date)
plot(z)
Rui Barradas
Em 31-07-2012
Err... some fundamental problems here.
On Tue, Jul 31, 2012 at 9:57 AM, monthodon fousserea...@hotmail.com wrote:
Hello,
I am a new user of R. I am trying to create a vector of time series called
LTot from a data frame. The purpose is to call LTot element from a loop
to make multiple
Hello,
It only gives that error if you don't
last$date - as.Date(last$date, format=%m/%d/%Y)
You must have dates, not character values.
Try it, then make a zoo object, then plot it.
Rui Barradas
Em 31-07-2012 18:54, Yolande Tra escreveu:
Thank you everyone for the attempt to solve the
Marc, thank you very much for your help.
I've posted in on
http://math.stackexchange.com/questions/177252/x2-tests-to-compare-the-fit-of-large-samples-logistic-models
and added details.
Many thanks
Marco
--On 31 July 2012 11:50 -0500 Marc Schwartz marc_schwa...@me.com wrote:
On Jul 31,
Hello,
Try this:
list1-list(
text, text, 2001, text,
text, 2000, text,
1999, text, text, text)
l1-lapply(list1,function(x) gsub(\\D,,x))
[[1]]
[1] 2001
[[2]]
[1] 2000
[[3]]
[1] 1999
unlist(l1)
[1] 2001 2000 1999
A.K.
- Original Message -
From: jimi adams
Thank you everyone for the attempt to solve the problem
It is an irregular series and insert NAs when a date is missing
library(zoo)
z - zoo(last$r_wvht, order.by=last$date)
plot(z)
Error in plot.window(...) : need finite 'xlim' values
In addition: Warning messages:
1: In xy.coords(x, y,
Thanks.
This is an irregular time series. The line plot does not look good because
of the gap between 9/11/2008 and 10/12/2009. I think two plots would be
better.
How would you include the date on the x-axis. Right now it only gives one
tick mark, 2009.
Yolande
On Tue, Jul 31, 2012 at 2:01 PM,
Hello everyone,
the sources of package{plspm} give me some headaches.
rounding:
Firstly, rounding is used on different occasions where I would
not suspect it. For an example, please have a look at the sources of
plsreg1{plspm}. I wonder why to perform rounding within the
function, and not when
I see. I typically use a (one-sided) formula as the first argument to Ecdf,
but didn't even think about that distinction in putting together this
example.
Thanks again for your help.
- Elliot
On Tue, Jul 31, 2012 at 12:46 PM, Deepayan Sarkar deepayan.sar...@gmail.com
wrote:
On Tue, Jul 31,
Dear All,
using the example from the package scatterplot3d I created a 3d plot as follows:
x -rnorm(500,50,2)
y -rnorm(500,5,1)
z -rnorm(500,6,1)
scatterplot3d(x, y, z, highlight.3d=TRUE, col.axis=blue,col.grid=lightblue,
main=scatterplot3d - 1, pch=20)
I would like to ask if anyone could
Hello,
Ok, try this, then.
(I've renamed the function and included some numbers in the text strings.)
x - c(text, text, 2001, text, 1234, text, 2000, 1234, text, 1999,
text, 1234, text, text)
extract.first.year - function(x, n = 4){
pattern - paste(\\D*(\\d{, n, }).*$, sep=)
Hi all,
I want to find all the mappings of one graph in another graph, based on
their vertex labels
Is there any way to do this in igraph based on vertex labels.
(as far as i know Igraph allows the subgraph isomorphism based only on
vertex and edge colors)
Eg:
graph 1:
x(1) x(2)
x(2) y(3)
y(4)
On 12-07-31 2:54 PM, Andras Farkas wrote:
Dear All,
using the example from the package scatterplot3d I created a 3d plot as follows:
x -rnorm(500,50,2)
y -rnorm(500,5,1)
z -rnorm(500,6,1)
scatterplot3d(x, y, z, highlight.3d=TRUE, col.axis=blue,col.grid=lightblue,
main=scatterplot3d - 1,
Hello,
Adapted from the help page for plot.zoo
#library(zoo)
#z - zoo(last$r_wvht, order.by=last$date)
plot(z, xaxt = n)
tt - time(z)
ix - seq(1, length(tt), length.out=8)
axis(side = 1, at = tt[ix], labels = FALSE)
labs - format(tt, %Y-%b-%d)
axis(side = 1, at = tt[ix], labels = labs[ix], tcl
On Jul 31, 2012, at 10:25 AM, M Pomati wrote:
Marc, thank you very much for your help.
I've posted in on
http://math.stackexchange.com/questions/177252/x2-tests-to-compare-the-fit-of-large-samples-logistic-models
and added details.
I think you might have gotten a more statistically
See in-line comments below.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of monthodon
Sent: Tuesday, July 31, 2012 7:57 AM
To: r-help@r-project.org
Subject: [R]
On 2012-07-30 23:55, Deepayan Sarkar wrote:
(I wish that reading lattice help would be less like trying to follow
an Asimov robot story, but there are too many possible interactions to
document both comprehensively and accurately.)
I, Robot, having assessed said lattice help, determine it to
On Jul 31, 2012, at 7:21 AM, Jan Näs wrote:
Hi
I trying to add a wind rose as a plot together with other plots.
Im unsing the roseavent function from the climatol package.
Ive tried par(mfrow=c(2,2))
but when plotting the windrose:
rosavent(windfreq_rose,4,3,ang=-3*pi/16,main=Windrose) it
On Tue, Jul 31, 2012 at 8:23 AM, arun smartpink...@yahoo.com wrote:
Hi,
#Without indices
xzoo-zoo()
x1zoo_f-zoo(1:10,)
cbind(xzoo,x1zoo_f)
x1zoo_f
11
22
33
44
55
66
77
88
99
10 10
#With indices
Dear All,
I am running a r code on 32bit win, involving absolutely small numbers.
Although I tried sth like the ratio of numers like 10^(-100) and did not
have issue to get the correct answer,
but still a little concerned about it.
Could anyone give some suggestion or have any experience?
Best
On Tue, Jul 31, 2012 at 2:45 PM, Jie jimmycl...@gmail.com wrote:
Dear All,
I am running a r code on 32bit win, involving absolutely small numbers.
Although I tried sth like the ratio of numers like 10^(-100) and did not
have issue to get the correct answer,
but still a little concerned about
On Tue, Jul 31, 2012 at 03:45:14PM -0400, Jie wrote:
Dear All,
I am running a r code on 32bit win, involving absolutely small numbers.
Although I tried sth like the ratio of numers like 10^(-100) and did not
have issue to get the correct answer,
but still a little concerned about it.
Could
Thanks in advance.
Nowadays I just calculate the bandwidth h of cross validation in kernel
smoothing using R language.
And I just looked up the usage of function, which is lscv(x,..,
exact=FALSE)
My question is what does stand for and mean? do you mind
specifically explaining it
Hello all,
I apologize for the simplistic question, but I have been having some trouble
learning how to do mediation analysis in R. Ideally, I would like to use
Preacher's Bootstrapping test for mediation (Preacher Hayes, 2004). I have
attempted to use the mediate package to set this up, using
I am using VGAM R package for my research works.
In my study, I am trying to simulate 1000 datasets from Beta binomial
distribution for a given set of (n,prob,size,rho)
and then I need to estimate the parameters prob and rho of the simulated
1000 datasets and need to store the fitted
It does look better indeed. Thanks,
Yolande
On Tue, Jul 31, 2012 at 3:15 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Adapted from the help page for plot.zoo
#library(zoo)
#z - zoo(last$r_wvht, order.by=last$date)
plot(z, xaxt = n)
tt - time(z)
ix - seq(1, length(tt),
Hi,
Thanks for reading and any suggestions. I am slowly teaching myself R. I
am currently modeling critical thresholds or breakpoints/switchpoints in
ecological systems. I have using the strucchange package, but tweaking it
for my questions at hand.
What I am trying to figure out is how to
I should add that it's also explained in more technical detail in
4.1.2 and 4.3 of the R Language Manual.
-- Bert
On Tue, Jul 31, 2012 at 1:08 PM, chester123 chester...@live.cn wrote:
Thanks in advance.
Nowadays I just calculate the bandwidth h of cross validation in kernel
smoothing using R
Hi,
On Tue, Jul 31, 2012 at 4:08 PM, chester123 chester...@live.cn wrote:
Thanks in advance.
Nowadays I just calculate the bandwidth h of cross validation in kernel
smoothing using R language.
And I just looked up the usage of function, which is lscv(x,..,
exact=FALSE)
Do you mean
Hi All,
I have some data where I am doing fairly simple calculations, nothing more
than adding, subtracting, multiplying and dividing.
Im running into a problem when I divide one variable by another and when
theyre both 0 I get NaN. I realize that if you divide a non-zero by 0 then
you get
Forgot to cc the archives,
M
On Tue, Jul 31, 2012 at 3:27 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
Well, there aren't quite so many dots -- only 3 I'd imagine.
It's how R allows variadic functions -- that is, allows extra
arguments that don't correspond to named formal
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