Could someone or Richard explain to me what he meant by
"This also shows a singular Error(). We look at the data and see that
plot is identical to the three-way veget:fruit:block interaction."
It seems to me that I just needed to recoded the plots, in order to get rid
of the Error message. If t
On 26/08/2009, at 3:08 PM, Donald Braman wrote:
This is a simple problem that has stumped me: I'm trying to loop
through a
few dozen variable names in graphs. I've tried various approaches
like
this:
attach(mydata)
ivs <- c("oneiv", "anotheriv", "yetanotheriv")
dvs <- c("onedv", "anotherdv
This is a simple problem that has stumped me: I'm trying to loop through a
few dozen variable names in graphs. I've tried various approaches like
this:
attach(mydata)
ivs <- c("oneiv", "anotheriv", "yetanotheriv")
dvs <- c("onedv", "anotherdv", "yetanotherdv")
for (iv in ivs) {
for (dv in dvs) {
g
Is multinom the function you are looking for?
library(nnet)
library(MASS)
?multinom indicates that this
fits multinomial log-linear models. If you are looking for multiple
logistic regression
you may want to read up on glm or lrm from the Design package.
Could you elaborate on what you mean by a
> r<-rcorr(d[[1]]) #d is matrix containing observation
> r[[1]] #r values
age sex BMI
age 1.000 -0.30010322 -0.13702263
sex -0.3001032 1. 0.06300528
BMI -0.1370226 0.06300528 1.
> r[[2]] #Number of obervations
age sex BMI
age 100 100 100
sex
Hi,
On Tue, Aug 25, 2009 at 10:11 PM, bwgoudey wrote:
>
> I'm using the rcorr function from the Hmisc library to get pair-wise
> correlations from a set of observations for a set of attributes. This
> returns 3 matrices; one of correlations, one of the number of observations
> seen for each pair a
I'm using the rcorr function from the Hmisc library to get pair-wise
correlations from a set of observations for a set of attributes. This
returns 3 matrices; one of correlations, one of the number of observations
seen for each pair and the final of the P values for each correlation seen.
>From
Mark,
The paper published by Bentor is part of a published collection. I don't
know if it is available online. (I can, if you like, scan the relevant
2-3 pages and email them to you.)
Computer Based Horse Race Handicapping and Wagering Systems: A Report
William Bentor
Thanks!
-Noah
On 8/25/
Hi
Please preserve as get_num_cpus.c as follows.
>8>8>8>8>8>8>8>8>8>8>8>8
#include
#include
#ifdef WIN32
#include
#endif
#include
#include
SEXP get_num_cpus(void)
{
SEXP cpus;
PROTECT (cpus = allocVector (INTSXP, 1));
INTEGER(cpu
Hello,
I performed a boosting analisis with adabag package to obtain a classification
tree with the following set of commands:
Tesis.boost <- adaboost.M1(Captura~., data=Tesis2, mfinal=2)
> arb<-Tesis.boost$tree[[1]]
> post(arb, file ="")
> post(arb, file ="",title= "Arbol 1")
I would like
Dear All,
Please can anyone assist on installing 'ptproc', I downloaded it on the
contributors website and tried to install it manual but R wont unzip it.
Thank you
--
--
Prince Oyelola A. Adegboye
Interuniversity Institute for Biostatistics and statistical Bioinformatics
University of Hasselt
Hi Gustavo,
depends on how your thematic maps are stored.
Give a look on SPATIAL / CRAN at
http://cran.r-project.org/web/views/Spatial.html
If your data is raster-based, try to give a look on raster package.
bests
milton
On Tue, Aug 25, 2009 at 7:20 PM, dalposso wrote:
>
> how to create a cro
how to create a cross table to quantify the classes of two thematic maps?
--
View this message in context:
http://www.nabble.com/error-matrix%2C-cross-table%2C-tp25143926p25143926.html
Sent from the R help mailing list archive at Nabble.com.
__
R-hel
Thank you very much for your help.
To the R gurus: It will be better at the future to simplify this
options.
They are too cumbersome!!!
Thank you!!!
El mar, 25-08-2009 a las 18:16 -0400, David Winsemius escribió:
> On Aug 25, 2009, at 5:51 PM, David Winsemius wrote:
>
> >
> > On Aug 25, 20
Check out bquote as in this example:
https://stat.ethz.ch/pipermail/r-help/2009-August/209373.html
On Tue, Aug 25, 2009 at 4:30 PM, Kenneth Roy Cabrera
Torres wrote:
> Hi R users:
>
> I will like to have a legend with math symbols and also with
> the value of a variable.
>
> But I cannot obtain b
On Aug 25, 2009, at 5:51 PM, David Winsemius wrote:
On Aug 25, 2009, at 4:30 PM, Kenneth Roy Cabrera Torres wrote:
Hi R users:
I will like to have a legend with math symbols and also with
the value of a variable.
But I cannot obtain both at the same time (symbol + value of a
variable):
He
On Aug 25, 2009, at 3:52 AM, Uwe Ligges wrote:
I highly suggest to rethink your problem in a way that you store the
things labelled, e.g., TA1 to TA5 (as well as all the others) as the
elements of a list TA. This way you have just one object TA that can
easily be accessed by index operatio
On Aug 25, 2009, at 4:30 PM, Kenneth Roy Cabrera Torres wrote:
Hi R users:
I will like to have a legend with math symbols and also with
the value of a variable.
But I cannot obtain both at the same time (symbol + value of a
variable):
Here is a reproducible example:
m1<-5
m2<-12
I think I
David -
Here's the easiest way I've been able to come up with.
I'll provide some sample data to make things clearer (hint, hint):
dat = data.frame(matrix(rnorm(100),20,5))
dat[3,4] = NA
dat[12,3] = NA
scrs = factanal(na.omit(dat),factors=2,scores='regression')$scores
rownames(scrs) = rownam
On Aug 25, 2009, at 9:23 AM, David Winsemius wrote:
On Aug 25, 2009, at 6:25 AM, Inchallah Yarab wrote:
Thank you very much Peter it works!!
the second step is to calculate the sum of variable in each class!!!
how can i do this!!!
?table
?xtab
I read this incorrectly. Try instead;
> b
Ben Bolker wrote:
>
> My two cents: this is a hard problem to do, period (not just in R).
> I would second the recommendation of the Dormann et al paper listed
> below; also see Zuur, Alain F., Elena N. Ieno, Neil J. Walker, Anatoly A.
> Saveliev, and Graham M. Smith. Mixed Effects Models and E
Hi David,
>> I am doing a factor analysis on survey data with missing values. ... Is
>> there a way to subset
>> from my original data set that will work with factanal() and preserve the
>> original rows or that
>> will allow me to append the factor scores back onto the original dataset
>> with
On Tue, Aug 25, 2009 at 10:16 AM, w_poet wrote:
>
> Hi R community,
>
> I'm just starting out in R and have a basic question about xyplot and
> tables. Suppose I had a table of data with the following names: Height,
> Age_group, City. I'd like to plot mean Height vs Age_group for each City.
>
> Whe
Ha! Much better than mine and without PERL, to boot. Bravo Gabor!
-- Bert
Bert Gunter
Genentech Nonclinical Biostatisics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Gabor Grothendieck
Sent: Tuesday, August 25, 2009 2:03 PM
To:
On Mon, Aug 24, 2009 at 8:04 PM, Donald Boyd wrote:
> Hi. Am brand new to R and to mailing lists - have never posted anywhere
> before, so hope I do this right.
>
> Am using R 2.9.1 with lattice graphics (just installed, fully up to date).
>
> Am doing trellis xyplot with y (emp=employment), x (yea
Instead of using .* use [^)]* so that you only get up to the next ) and also
it seems that you want to trim spaces so add space star at the beginning
and end:
gsub(" *\\([^)]*\\) *", "", myvector)
On Tue, Aug 25, 2009 at 4:17 PM, Judith Flores wrote:
> Hello dear R-helpers,
>
> I haven't been
On Mon, Aug 24, 2009 at 4:27 PM, jebyrnes wrote:
>
> Hrm. I have to admit, I don't entirely understand how to use the scaling,
> and that seems like a lot of unneeded extra code. It is what it is, though.
That's true, and I think it would be easy enough to have a flag to
panel.3dscatter etc. tel
Hi Judith,
This probably isn't the only way to do it, but:
gsub("\\(.*?\\)", "", myvector, perl=TRUE)
seems to do the trick.
The problem is that regular expressions are greedy, so you were matching
everything between the first and last parens, as you noticed. Putting
the question mark ther
Judith:
I believe that this is, indeed, tough; it might require PERL regex's to do
entirely within the regular expression language. You might also wish to
check out the gsubfn package to see if it could help.
However, a reasonably simple alternative approach that I think will work is
to use strsp
Hi R users:
I will like to have a legend with math symbols and also with
the value of a variable.
But I cannot obtain both at the same time (symbol + value of a
variable):
Here is a reproducible example:
m1<-5
m2<-12
plot(1:5,1:5,type="n")
legend("topleft",legend=c(paste(expression(mu),"=",m1),
Hello dear R-helpers,
I haven't been able to figure out of find a solution in the R-help archives
about how to delete all the characters contained in groups of parenthesis. I
have a vector that looks more or less like this:
myvector<-c("something (80 km/h, sd) & more (6 kg/L,sd)", "something
On Aug 25, 2009, at 3:45 PM, David G. Tully wrote:
I am sure there is a simple way to do the following, but i haven't
been able to find it. I am hoping a merciful soul on R-help could
point me in the right direction.
I am doing a factor analysis on survey data with missing values. to
do
On Sun, Aug 23, 2009 at 5:41 PM, Ben Bolker wrote:
>
>
>
> Fredrik Karlsson wrote:
>>
>> Dear list,
>>
>> I have a two character vector with two different values in them (two
>> each, that is). Naturally, when I use these vectors as grouping
>> factors in a lattice plot, I get four panels.
>> Now,
You are required to provide specifications for all the activation functions,
the number of which is equal to the number of hidden layers plus 1. Thus in
your case actfns=c(2,2) would suffice.
BKMooney wrote:
>
> Hi,
>
> I am trying to build a VERY basic neural network as a practice before
> h
>From a somewhat ill-designed education experiment, I have categorical data
with two between subject variables and one within subject variable. Since it
is categorical (essentially counts of answer choices on multiple choice
questions), I'm looking for some chi-squared method.
I know how to gener
It estimates the model for every value of the L1 parameter. See
?predict.enet. When you predict, you have to specify the other
parameter (which you can do a variety of ways).
Max
On Tue, Aug 25, 2009 at 8:54 AM, Alex Roy wrote:
> Dear R users,
> I am using "enet" package in
I am sure there is a simple way to do the following, but i haven't been
able to find it. I am hoping a merciful soul on R-help could point me in
the right direction.
I am doing a factor analysis on survey data with missing values. to do
this, I run:
FA1<-factanal(na.omit(DATA), factors = X,
And of course I did not test this :). Within the data.frame argument list,
please change the <- operators to = signs. Then it should work.
Erik
-Original Message-
From: Erik Iverson
Sent: Tuesday, August 25, 2009 1:17 PM
To: 'w_poet'; r-help@r-project.org
Subject: RE: [R] table, xypl
With n = 100 I can see the two modes well separated in a density
plot.
However, I don't see why you cannot plot the density function, if
that's what you need:
curve(0.6 * dnorm(x, 0.4, 0.1) + 0.4 * dnorm(x, 0.8, 0.1), 0, 1.2)
Best,
Giovanni
> Date: Tue, 25 Aug 2009 13:55:26 +0100
> From: ri
>Example,
>data("GlaucomaM", package = "ipred") is accepted. Now instead of GlaucomaM,
I> need to give my own data. But the data format for GlaucomaM is not given.
>So how can I know that?
Not familliar with the packages at all but if you simply enter:
> data ("GlaucomaM", package="ipred")
> Glau
Assuming your starts and ends are the number of days since some
starting point (day 0), then following this example might do what you
want:
x <- 45678
class(x) <- 'Date'
x
[1] "2095-01-23"
You will have to study the documentation to find out if your starting
point is the same as R's, an
Hello all,
I got a reccomandation from a friend to try:
http://git-scm.com/
For verstion control of my R code.
I am currently using notepad++ for my IDE.
Does anyone got experience with this system?
Any tips or recommandation will be warmly welcomed.
Tal
--
Dear Admins:
Could you please add me to the r-h...@stat.math.ethz.ch mailing
list. I prefere to use my yahoo account: aboue...@yahoo.com
Thank you so much for your attention to this matter, and I look forward to
hearing from you soon.
My other contact information is:
Hello,
"I'm just starting out in R and have a basic question about xyplot and
tables. Suppose I had a table of data with the following names: Height,
Age_group, City. I'd like to plot mean Height vs Age_group for each City"
You did not provide a sample data.frame, so I generated one. This examp
And an alternative to nested if else's, in a case where they are
actually needed, is
switch
-Don
At 1:06 PM -0400 8/25/09, Ista Zahn wrote:
Nice one Gabor! It works great, and is really simple. Bill Dunlap's
solution of
GET$pass.var <- as.name(GET$pass.var)
eval(substitute(mean(x), list
Dear All,
is there a way to plot to a connection? I mean, I would like to do
something like
> tc <- textConnection("foo", "w")
> png(tc)
> plot(...)
> close(tc)
According to the documentation of the Cairo package, this should work
with CairoPNG() (among others), but all I get is
> tc <- textCon
That is what I wanted but grid(10,10) works just fine too
Thanks again.
--- On Mon, 8/24/09, David Winsemius wrote:
> From: David Winsemius
> Subject: Re: [R] plotting a grid with grid() ?
> To: "John Kane"
> Cc: "R R-help"
> Received: Monday, August 24, 2009, 4:37 PM
>
> On Aug 24, 2009, a
ARGGH! I knew I was missing something simple!! It was just too obvious.
Many thanks.
--- On Mon, 8/24/09, David Winsemius wrote:
> From: David Winsemius
> Subject: Re: [R] plotting a grid with grid() ?
> To: "John Kane"
> Cc: "R R-help"
> Received: Monday, August 24, 2009, 5:14 PM
>
> On
Thanks, Henrique.
I think you mean the following:
fConn <- file('test.txt', 'r+')
Lines <- readLines(fConn)
writeLines(c("Text at beginning of file\n", Lines), con = fConn)
close(fConn)
(With paste(), it does not work.)
Paul
On Tue, Aug 25, 2009 at 5:02 PM, Henrique Dallazuanna wrote:
> Try th
Just plotting the mean is not very informative (at least not more
informative than providing the same information in a table). If you want to
show some of the distributional properties of height for each Age_group x
City pair, try box-and-whisker plots ?bwplot
Imagine you have four age categories
Hi Bernardo,
I suggest you give a look at:
Dale MRT & Fortin MJ, 2009. Spatial Autocorrelation and Statistical Tests:
Some Solutions.
Journal of Agricultural, Biological and Environmental Statistics,
14(2):188-206.
Cheers
milton
On Tue, Aug 25, 2009 at 1:08 PM, B Garcia Carreras <
bernardo.garc
Is there a function in R to calculate the semi deviation of a data set? I
cannot seem to find a reference to one in the manual.
--
View this message in context:
http://www.nabble.com/Semi-Standard-Deviation-tp25138640p25138640.html
Sent from the R help mailing list archive at Nabble.com.
__
On Aug 25, 2009, at 12:15 PM, mirko86 wrote:
I want to draw lines using a matrix with the X-axe on the first
column, and
the the Y-axe on the second column.
These lines have to link the n points of a graph, so they are n-1,
but the
resout using these two commands:
X <- m
Another alternative is to use SSlogis which is very similar to the
model you're fitting except with one additional parameter:
Asym <- 3
xmid <- 0
scal <- 10
model <- nls(bod ~ SSlogis(time, Asym, xmid, scal), data = mydata)
summary(model)
plot(bod ~ time, mydata)
newdata <- data.frame(time = seq(1
Hi R community,
I'm just starting out in R and have a basic question about xyplot and
tables. Suppose I had a table of data with the following names: Height,
Age_group, City. I'd like to plot mean Height vs Age_group for each City.
When I try to do the following:
> library(lattice)
> xyplot(mea
Hi,
I have two sets of data for a given set of (non-lattice) locations. I would
like to know whether the two are significantly different. This would be
simple enough if it wasn't for the fact that the data is spatially
autocorrelated. I have come across several possible solutions (including
Cliff
Nice one Gabor! It works great, and is really simple. Bill Dunlap's solution of
GET$pass.var <- as.name(GET$pass.var)
eval(substitute(mean(x), list(x=GET$pass.var)))
also works.
Thanks!
On Tue, Aug 25, 2009 at 12:59 PM, Gabor
Grothendieck wrote:
> Try this:
>
> x <- get(GET$pass.var)
> mean(x)
Hi, Michael,
I think the SPSS answer is wrong. Your starting values are way off.
Look at this plot for verification:
con <- textConnection("time bod
11 0.47
22 0.74
33 1.17
44 1.42
55 1.60
67 1.84
79 2.19
8 11 2.17")
mydata <- read.table(con, header = TRUE)
close(co
Thank you for your suggestion.
Unfortunately, like most texts on the topic of nonlinear regression, examples
tend to be either a) a simple regression with one predictor and one response or
b) a continuous predictor and response with a categorical variable (i.e.
differentiating for example two cu
Try this:
x <- get(GET$pass.var)
mean(x)
On Tue, Aug 25, 2009 at 12:26 PM, Ista Zahn wrote:
> Hi everyone,
> I'm building a website (http://yourpsyche.org) using Jeffrey Horner's
> awesome Rapache module. I want to take user input, and pass it to an R
> script. At first I was simply using if el
Try this:
RSiteSearch("Weibull regression")
On Tue, Aug 25, 2009 at 12:02 PM, Lindsay Banin wrote:
> Dear R-users,
>
> I am trying to create a model using the NLS function, such that:
>
> Y = f(X) + q + e
>
> Where f is a nonlinear (Weibull: a*(1-exp(-b*X^c)) function of X and q is a
> covariat
I updated the previously posted function for Cramer's V so that it
automatically prints Cramer's V, chi-square, the degrees of freedom, and the
significance level of Cramer's V based on the chi-square value and the
degrees of freedom with desired (user-supplied) levels of precision. An
example is
Hi Ista -
I've had success using the do.call function with RApache
do.call(func, arglst)
where 'func' is the function you desire to use, 'arglst' is a list of arguments.
You may need to do some cleanup of the arguments using evalq()
function (so that numbers are treated as numbers, for example
Great! Thanks again!
Vivek
2009/8/25 Uwe Ligges :
> I finally found some time to look at it:
>
> Yes, it does not work properly for angles > 180 degrees. Will try to find a
> fix and make a new release soon.
>
> Best wishes,
> Uwe Ligges
>
>
> Vivek Ayer wrote:
>>
>> Hey guys,
>>
>> Not sure if I
Hi everyone,
I'm building a website (http://yourpsyche.org) using Jeffrey Horner's
awesome Rapache module. I want to take user input, and pass it to an R
script. At first I was simply using if else statements, but after a
while I had so many nested if else's in my code that my head was
spinning. S
I got hold of 'Modern Applied Statistics with S' by Venables and Ripley (p.
335-336) and I was able to answer my own question:
###
AA<-read.table("http://www.natursyn.dk/online/fingerprinting.txt",header=T)
aa.grp <- factor(c(rep('f',13),rep('b',10),rep('p',10)))
aa.lda<-lda(as.matrix(AA[3:9]),
I think you need to consult a local statistician, as you appear to be so far
out of your depth statistically that even the best-intentioned help from
this list may not suffice. For example, if I were that local statistician, I
would ask: what is the context? -- what is the underlying scientific iss
I finally found some time to look at it:
Yes, it does not work properly for angles > 180 degrees. Will try to
find a fix and make a new release soon.
Best wishes,
Uwe Ligges
Vivek Ayer wrote:
Hey guys,
Not sure if I encountered a bug with the scatterplot3d function.
Here's the calls I made
I want to draw lines using a matrix with the X-axe on the first column, and
the the Y-axe on the second column.
These lines have to link the n points of a graph, so they are n-1, but the
resout using these two commands:
X <- matrix(scan("MaxInterEdges.R",0), ncol=2);
Manuel,
Thanks for the reference. I printed it out and read through it this
morning.
I think I'm going to take a gls approach. I've spent the last couple weeks
reading about spatial autocorrelation, and found that the world of SAC is
large, complex, and requires more time than I currently have. U
Hi All,
I'm trying to run nls on the data from the study by Marske (Biochemical
Oxygen Demand Interpretation Using Sum of Squares Surface. M.S. thesis,
University of Wisconsin, Madison, 1967) and was reported in Bates and Watts
(1988).
Data is as follows, (stored as mydata)
time bod
11 0.
Try this;
fConn <- file('test.txt', 'r+')
Lines <- readLines(fConn)
writeLines(paste("Text at beginning of file", Lines, sep = "\n"),
con = fConn)
On Tue, Aug 25, 2009 at 12:54 PM, Paul Smith wrote:
> Dear All,
>
> I have a piece of text that I want to append to a text file at the
>
On Aug 25, 2009, at 8:45 AM, maram salem wrote:
Dear All,
I have a matrix m of the form
m
[,1] [,2]
[1,] 9072.302 0.0004462366
[2,] 9086.811 0.0005628169
[3,] 9101.320 0.0007126347
[4,] 9115.830 0.0008986976
[5,] 9130.339 0.001126
Dear R-users,
I am trying to create a model using the NLS function, such that:
Y = f(X) + q + e
Where f is a nonlinear (Weibull: a*(1-exp(-b*X^c)) function of X and q is a
covariate (continous variable) and e is an error term. I know that you can
create multiple nonlinear regressions where x i
Dear All,
I have a piece of text that I want to append to a text file at the
beginning of the text file.
I have thought about using cat() with the option 'append=T', but the
appending, in this case, is done at the bottom of the text file. Any
ideas?
Thanks in advance,
Paul
Sorry,
That worked.
However, is it possible to single out specific boxes and change their colour? I
want the four graph headers in
the top right of the trellice to be different colour to the rest.
Thanks very much
From: Sundar Dorai-Raj [sdorai...@gmail
(posting for my friend)
Hi all,
We are looking for a tutor who could teach me statistical learning and
data-mining using the book:
The Elements of Statistical Learning: Data Mining, Inference, and Prediction
Please drop me a line if you are interested. Thank you!
[[alternative HTML ver
It worked perfectly.
Thanks a lot.
On Mon, Aug 24, 2009 at 4:16 AM, Gabor Grothendieck wrote:
> 1. Just enter
> arima
> at the R console to see its source code (without comments). The source
> tar.gz for R is found by googling for R, clicking on CRAN in left column
> and choosing mirror. Or
Hello, List,
I am a new user of the R project, and I need some help in plotting a legend.
I am using the PBSmapping library to plot map of Ohio and heat color it with
the count of employees in each county. As a guide, I am using "Data Mashups
in R." I am able to plot the map with the colors; ho
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ronggui Huang
> Sent: Monday, August 24, 2009 9:26 PM
> To: Steven Kang
> Cc: r-help@r-project.org
> Subject: Re: [R] Removing objects from workspace
>
> It depends.
>
> If ther
Thanks Uwe. The wordnet is working now. My previous configuration that
causes the error was:
R - version 2.9.1
wordnet - version 0.1-4
rJava 0.6-2
After updating rJava to version 0.7-0 everything works now.
Uwe Ligges-3 wrote:
>
> Looks like nobody answered so far:
>
>
> Kelvin Lam wrote
I am trying to understand what the lda function in the MASS package calculates
when there are more dimensions than examples. It is my understanding that the
Fisher Linear Discriminant is not applicable in this case, because the inverse
of the covariance matrix cannot be calculated.
My question
Dear Users,
I would like to plot the log-likelihood (depending on two
parameters).
I wrote the following code:
library(mvtnorm)
sigma<-
matrix(c(4,2,2,3),ncol=2)
x<-rmvnorm(n=500,mean=c(1,2),sigma=sigma)
likel<-
function(param,data){
A reproducible example would be nice.
Try grid = FALSE for the first question, though I'm unaware which
lattice plot you are using where the default is TRUE. So I can't
guarantee that will even work.
For your second question, add
par.settings = list(strip.background = list(col = "white"))
to yo
To: silwood-r
Subject: Removing lattice graph gridlines and editing label box colour
Hi,
Is it possible to remove the background gridlines from a lattice graph (ie
graph made up of multiple individual graphs with annoying blue grid in the
backgroun)?
Also, Is it possible to change the colour o
Gabor: Wow, that seems to be exactly what I need! Does it matter that
"my" incidence matrices represent neighborhood relations between
vertices and faces rather than between vertices and edges?
Steve: Yep, I realize that this package is exactly what I'm searching
for. :)
Gabor Grothendieck s
Hi,
It looks like you're getting more good stuff, but just to follow up:
On Aug 24, 2009, at 4:01 PM, Michael Kogan wrote:
Steve: The two matrices I want to compare really are graph matrices,
just not adjacency but incidence matrices. There should be a way to
get an adjacency matrix of a gra
They can be regarded as incidence matrices rather than adjacency
matrices and in that case it follows:
library(igraph)
# incidence matrix to canonical edge list
inc2canel <- function(m) {
g <- graph.incidence(m)
cp <- canonical.permutation(g)
can <- permute.vertices(g, cp$
On Aug 25, 2009, at 6:25 AM, Inchallah Yarab wrote:
Thank you very much Peter it works!!
the second step is to calculate the sum of variable in each class!!!
how can i do this!!!
?table
?xtab
De : Petr PIKAL
Cc : r-help@r-project.org
Envoyé le : Mardi, 2
Vitalie S. wrote:
This is a BUGS error message that indeed tells you that BUGS cannot
truncate that density - not related to R at all.
Best,
Uwe Ligges
Yes, indeed, that does not work in OpenBugs as well. Means that
documentation of Open bugs is incorrect (they are using dnorm there)
> what do you mean by "%d-%b-%y". is it reading format or writing format.
"%d-%b-%y" is a date format - see the help page for strptime.
Example usage:
strptime("01-Jan-84", "%d-%b-%y")
strftime(Sys.time(), "%d-%b-%y")
Regards,
Richie.
Mathematical Sciences Unit
HSL
---
Dear R users,
I am using "enet" package in R for applying "elastic
net" method. In elastic net, two penalities are applied one is lambda1 for
LASSO and lambda2 for ridge ( zou, 2005) penalty. But while running the
analysis, I realised tht, I optimised only one lambda. ( eve
> I'm trying to draw the density function of a mixed normal distribution
> in the form of:
> .6*N(.4,.1)+ .4*N(.8,.1)
> At first I generate a random sample with size 200 by the below code:
> means = c(.4,.8)
> sds = sqrt(c(.1,.1))
> ind = sample(1:2, n, replace=TRUE, prob=c(.6,.4))
> x=rnorm(n,mean
Try this:
plot(0)
ix <- as.double(1:3)
legend("top", as.expression(lapply(ix, function(i) bquote(r^.(i)
On Tue, Aug 25, 2009 at 8:45 AM, Michael Cho wrote:
> Hi,
>
> Trying to do something fairly simple. I'm trying to get a legend that
> combines superscripts with a sequence, like this:
> r
Hi every one,
what do you mean by "%d-%b-%y". is it reading format or writing format.
Thanks in Advance.
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___
Dear All,
I have a matrix m of the form
m
[,1] [,2]
[1,] 9072.302 0.0004462366
[2,] 9086.811 0.0005628169
[3,] 9101.320 0.0007126347
[4,] 9115.830 0.0008986976
[5,] 9130.339 0.001126
[6,] 9144.848 0.0014018405
[7,] 9159.357 0
Hi,
Trying to do something fairly simple. I'm trying to get a legend that
combines superscripts with a sequence, like this:
r^2 = 1
r^2 = 2
r^2 = 3
...
Except that r^2 is properly formatted as a superscript.
I've been playing with "substitute" and "expression" and can get an
individual line to a
This is a BUGS error message that indeed tells you that BUGS cannot
truncate that density - not related to R at all.
Best,
Uwe Ligges
Yes, indeed, that does not work in OpenBugs as well. Means that
documentation of Open bugs is incorrect (they are using dnorm there).
Also "I" works
Hello everyone,
These are some questions about the 'pec' function in R. These questions deal
with prediction error curves and their derivation. Prediction error curves are
documented in, for example, "Efron-type measures of prediction error for
survival analysis" by Gerds and Schumacher.
I h
I'm trying to draw the density function of a mixed normal distribution
in the form of:
.6*N(.4,.1)+ .4*N(.8,.1)
At first I generate a random sample with size 200 by the below code:
means = c(.4,.8)
sds = sqrt(c(.1,.1))
ind = sample(1:2, n, replace=TRUE, prob=c(.6,.4))
x=rnorm(n,mean=means[ind],sd=s
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