Hi Berend,
>The question belongs on the R-devel mailinglist.
I try to find this mailing-list ...
>You are calling your Fortran routines directly from an R file with .Fortran, I
>presume?
Yes. Is there another solution (possibility ?)
>Declare them with (:,:) or (*,*), allocate and the return t
Hi Fabio,
You should write:
class(...)
where ... is the same as what you would type to have the variable
displayed on the console. Looking at your earlier message, it might
be:
x$trait3
so try:
class(x$trait3)
Jim
On Fri, Mar 4, 2016 at 11:30 AM, Fabio Monteiro
wrote:
> i just called trait
Thanks. I will try both the options 1) another mirror 2) upgrading R, and
revert in case of issues.
Br /
On Fri, Mar 4, 2016 at 10:56 AM, Jeff Newmiller
wrote:
> The usual thing to try in cases like this is another mirror.
>
> Another worthwhile step is upgrading your R software to the latest.
The usual thing to try in cases like this is another mirror.
Another worthwhile step is upgrading your R software to the latest... if only
to comply with the Posting Guide.
--
Sent from my phone. Please excuse my brevity.
On March 3, 2016 9:33:05 PM PST, Burhan ul haq wrote:
>Hi,
>
>I was pl
Hi,
I was planning to use GGally, which required me to upgrade ggplot2 but
despite trying multiple times, I have been unable to do so:
The ggplot2 downloads and installs, but when I load it, I get the following
message:
> library("ggplot2", lib.loc="/usr/local/lib/R/site-library")
Error in get(m
i just called trait3 to my variable.
Is this what i'm suppose to wright? class(trait3), or class
(my_trait3_variable?
both give error
2016-03-03 23:42 GMT+00:00 Jim Lemon :
> Hi Fabio,
> It is possible that your remaining "numeric" variable is a factor. What
> does:
>
> class(my_numeric_variabl
On 03/03/2016 02:13 PM, KMNanus wrote:
When I do that,
When you do what exactly?
It's impossible for anyone here to know what you're doing if you
don't show the code.
I get "Error in `$<-.data.frame`(`*tmp*`, "site", value
= integer(0)) :
replacement has 0 rows, data has 6”
The data fram
Hi Fabio,
It is possible that your remaining "numeric" variable is a factor. What does:
class(my_numeric_variable)
say? (where you substitute the name of your "numeric" variable)
Jim
On Fri, Mar 4, 2016 at 2:25 AM, Fabio Monteiro
wrote:
> Hello, my name is Fábio and I'm a Marine Ecology stude
> On Mar 3, 2016, at 12:33 PM, jake88 wrote:
>
> Data set attached … rename to mydata.csv .
>
>
>
>
>
> require ( RSNNS )
> mydata = read.csv("mydata.csv",header = TRUE)
# Needed to change to mydata.txt
>
> mydata.train = mydata[3000:1,]
>
> mydata.test = mydata[10005:10006,]
>
> m
When I did that, I got -
"Error in `$<-.data.frame`(`*tmp*`, "site", value = integer(0)) :
replacement has 0 rows, data has 6”
The data frame has 6 rows.
Ken
kmna...@gmail.com
914-450-0816 (tel)
347-730-4813 (fax)
> On Mar 3, 2016, at 4:14 PM, Ista Zahn wrote:
>
> Like this?
>
> x <-
When I do that, I get "Error in `$<-.data.frame`(`*tmp*`, "site", value =
integer(0)) :
replacement has 0 rows, data has 6”
The data frame has 6 rows.
Ken
kmna...@gmail.com
914-450-0816 (tel)
347-730-4813 (fax)
> On Mar 3, 2016, at 4:52 PM, Hervé Pagès wrote:
>
> Hi,
>
> On 03/03/2016 1
Hi,
On 03/03/2016 12:18 PM, KMNanus wrote:
I have a factor variable that is 6 digits and hyphenated. For example, 001-014.
I need to extract the first 3 digits to a new variable using mutate in dplyr -
in this case 001 - but can’t find a function to do it.
substr will do this for character s
Like this?
x <- factor("001-014")
y <- substr(as.character(x), 1, 3)
Best,
Ista
On Thu, Mar 3, 2016 at 3:18 PM, KMNanus wrote:
> I have a factor variable that is 6 digits and hyphenated. For example,
> 001-014.
>
> I need to extract the first 3 digits to a new variable using mutate in dply
I have a factor variable that is 6 digits and hyphenated. For example, 001-014.
I need to extract the first 3 digits to a new variable using mutate in dplyr -
in this case 001 - but can’t find a function to do it.
substr will do this for character strings, but I need the variable to remain as
> On Mar 3, 2016, at 10:30 AM, jake88 wrote:
>
> Data set attached …
Perhaps it was, and if you replied-all then it might be in the copy that was
sent directly to Charles, but it was not attached after the rhelp mail-server
scrubbed it. It wasn't labeled correctly by your email client as MIME
Not possible, because the hessian is singular. Recoded as follows (your
code should be executable before you put it in a help request).
# asindii2.R -- Is it possible to estimate the likelihood parameter
#and test for significant as follows:
x <- c(1.6, 1.7, 1.7, 1.7, 1.8, 1.8, 1.8, 1.8)
y <-
Data set attached …
require ( RSNNS )
mydata = read.csv("mydata.csv",header = TRUE)
mydata.train = mydata[3000:1,]
mydata.test = mydata[10005:10006,]
myfit <- elman ( mydata.train[,2:19],mydata.train[,1], size =100 ,
learnFuncParams =c (0.1) , maxit =1000)
pred <-predict (myfit ,
Hi all,
Is it possible to estimate the likelihood parameter and test for significant as
follows:
x <- c(1.6, 1.7, 1.7, 1.7, 1.8, 1.8, 1.8, 1.8)
y <- c( 6, 13, 18, 28, 52, 53, 61, 60)
n <- c(59, 60, 62, 56, 63, 59, 62, 60)
# note: there is no need to have the choose(n, y) term in the likelihood
I think the answer is in Venables and Ripley, Modern Applied Statistics with S.
4th Edition, 2002, page 332:
"Fisher (1936) introduced a linear discriminant analysis seeking a linear
combination
xa of the variables that has a maximal ratio of the separation of the class
means to the within-class
Ruofei,
Ben's suggestion is simple and gets you close:
require(MASS)
nsim <- 100
rho <- -.9
Z <- mvrnorm(nsim, mu=c(0,0),Sigma = cbind(c(1,rho),c(rho, 1)))
U <- pnorm(Z);
a <- Z[,1]
b <- qunif(U[,2])
cor(a,b)
Pearson correlation characterizes the linear relationship between normal
r.v.'s, bu
> On 3 Mar 2016, at 13:57, MAURICE Jean - externe
> wrote:
>
> Hi,
> I am 'translating' R functions in FORTRAN subroutines.
>
> Very often, an R function gives an 'array' as result and you don't have to
> bother with the dimension of the array : R creates automatically an array
> with the g
I'm looking for an R package to do Polytopic Vector Analysis (PVA) without
success. I've checked CRAN, then did a site searched there, scanned the
packages list, looked over old Help posting (there was a query about the same
back in 2005 or so - only one response that didn’t give much direction
Hello, my name is Fábio and I'm a Marine Ecology student in Portugal.
I'm currently using the FD package for my work and yesterday one message
appeared that I wasn't expecting and I really need your help to try to
figure out what's happening.
I'm using the dbFD function and the following message a
Mohsen,
Check at Bioconductor.
Andrés
> El 03/03/2016, a las 9:43, Mohsen Jafarikia escribió:
>
> Hello everyone:
>
> I have about a couple of thousands of samples each with about 100 SNP
> genotypes and I would like to do PCA using genotypes. I looked on the
> web and found different option
Hello everyone:
I have about a couple of thousands of samples each with about 100 SNP
genotypes and I would like to do PCA using genotypes. I looked on the
web and found different options available on R for PCA. I was
wondering if I could have advice about the program fits better what I
am trying
Dear David,
Thank you very much! I am sending this mail for the record.
I am on mac OSX 10.8.5. The "standard" binary installer packages worked and
fixed the problem (R-3.2.1-snowleopard.pkg, RStudio 0.99.891 - Mac OS X 10.6+
(64-bit)). So, a caution against fink (r-base and rstudio-desktop)
Thank you for your answer. Please let me provide additional information:
You have a pooled variance covariance matrix S (2x2). The matrix needs to be
inverted: S^-1.
The calculation of the coeffcients is done by S^-1 (average (x1) - average
(x2)). average (x1) is the vector of means (2x1) in
Unfortunately we can only provide so much help without a reproducible
example. Can you use a dataset that everyone would have access to to
reproduce the problem? Otherwise it is difficult for anyone to help you.
Regards,
Charles
On Tue, Mar 1, 2016 at 12:35 AM, jake88 wrote:
> I am new to R a
Le 03/03/2016 03:02 PM, Marc Schwartz a écrit :
On Mar 3, 2016, at 7:40 AM, Sebastien Moretti wrote:
Hi
I have issues with an R package developed in 2004.
It works perfectly in R < 3.
It can be installed in R > 3 but functions are not in the namespace.
Do you know a good - and simple - docu
> On Mar 3, 2016, at 7:40 AM, Sebastien Moretti
> wrote:
>
> Hi
>
> I have issues with an R package developed in 2004.
> It works perfectly in R < 3.
> It can be installed in R > 3 but functions are not in the namespace.
>
> Do you know a good - and simple - documentation to help me to solve
If the textbook provides the equations, you can work through them directly. But
without knowing more, it is hard to say. You could also contact the author of
the textbook.
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77
You searched, but did not tell us what you found, nor why it was unsuitable
for you undescribed use case. So all we can do is guess: my guess is
http://docs.rexamine.com/R-man/stringi/stringi-search-boundaries.html
Best,
Ista
On Mar 3, 2016 8:14 AM, "Sascha Wolfer" wrote:
> Hello list members,
>
Hi
I have issues with an R package developed in 2004.
It works perfectly in R < 3.
It can be installed in R > 3 but functions are not in the namespace.
Do you know a good - and simple - documentation to help me to solve that?
I have already fixed some problems with R CMD check but remaining one
Hello list members,
I am looking for an implementation of Unicode text segmentation (word boundary
detection) algorithms in R. You can find information about the algorithms here:
http://www.unicode.org/reports/tr29/#Word_Boundaries
The help page for the function ‚casefuns‘ from the excellent ‚U
Hi,
I am 'translating' R functions in FORTRAN subroutines.
Very often, an R function gives an 'array' as result and you don't have to
bother with the dimension of the array : R creates automatically an array with
the good length. It's not really the case with FORTRAN.
Until now, I create an ar
Dear Dennis
Thank you very much for your detail reply . It was really helpful to understand.
Tanvir Ahamed
Göteborg, Sweden | mashra...@yahoo.com
From: Dennis Murphy
Sent: Thursday, 3 March 2016, 4:38
Subject: Re: [R] Extract row form a dataframe by row
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