I have a situation when I need to save matrix with file names that are
programmatically created.
for (i in levels(mergeTrn$Continent)) {
matrix here
# I want to save this matrix with a file name that carries i from for
loop. The following does not work.
paste(plotroc_GBM_Trn_, i,
Hi all,
I am drawing some 3D surfaces using the Triangle tools (package misc3) and
drawScene.rgl. Do you know if it is possible to add axes and graduation on
the scene?
Christophe
--
View this message in context:
Hello,
?assign
assign(paste(plotroc_GBM_Trn_, i, sep=), matrix(rnorm(100),10,10))
HTH,
Pascal
Le 30/01/2013 17:04, Kumar Mainali a écrit :
I have a situation when I need to save matrix with file names that are
programmatically created.
for (i in levels(mergeTrn$Continent)) {
On Jan 30, 2013, at 04:58 , Bert Gunter wrote:
You almost never need dummy variables in R. R creates them
automatically from factors given model and possibly contrasts
specification.
?contrasts ## for some technical details.
If you have not read An Introduction to R do so now. Pay
On 01/29/2013 06:53 PM, Jean Véronis wrote:
Hello,
I am trying to use cluster.overplot from package plotrix and I get an error message when
I add the away parameter:
require(plotrix)
distance- read.table(distance.txt)
cmd- cmdscale(distance)
cp- cluster.overplot(cmd, away=2)
Error in if
Hi all,
I am very grateful to all those who write to me
1) how i can obtain relative risk (risk ratio) in logistic regression in R.
2) how to obtain the predicted risk for a certain individual using
fitted regression model in R.
Many thanks, in advance, for your help.
Amin.
Hello,
I would like to estimate Arima model by Maximum likelihood conditional on a
zero initial residual.
I am using:
model - arima(y[1:N], order=c(1, 0, 1), method=ML, include.mean=TRUE)
model
What I have to use to estimate arima model by Maximum likelihood
conditional on a zero initial
Hello together,
i have a question for substring.
I know i can filter a number like this one:
bill$No-substring(bill$Customer,2,4)
in this case i get the 2nd, 3rd and 4th number of my Customer ID.
But how can i do this, if i want the 2nd, 3rd and 4th number of a column.
Like this one.
I have:
great! many thanks!
Le 30 janv. 2013 à 09:52, Jim Lemon j...@bitwrit.com.au a écrit :
On 01/29/2013 06:53 PM, Jean Véronis wrote:
Hello,
I am trying to use cluster.overplot from package plotrix and I get an error
message when I add the away parameter:
require(plotrix)
distance-
Hi Mat,
The following should get you started:
s - Mercedes_02352
substr(s, nchar(s) - 3, nchar(s) - 1)
[1] 235
# defining a function
foo - function(x, a = 3, b = 1) substr(s, nchar(x) - a, nchar(x) - b)
foo(s)
[1] 235
HTH,
Jorge.-
On Wed, Jan 30, 2013 at 8:05 PM, Mat wrote:
Hello
Hello everybody!
I have again a rather simple question concerning recoding of variables:
I have a variable/data-frame column BIRTHPLACE containing abbreviations of
the 26 swiss counties (AG, AI, AR, BE, ZH, ... )
as well as international country codes (USA, GER, ESP, etc.) and another
variable
Hi David,
Check
?%in%
for a simpler approach.
Regards,
Jorge.-
On Wed, Jan 30, 2013 at 8:42 PM, David Studer wrote:
Hello everybody!
I have again a rather simple question concerning recoding of variables:
I have a variable/data-frame column BIRTHPLACE containing abbreviations of
the
http://www.consultadifesapdl.it/6cmpqy.php
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained,
Hi all,
I am very grateful to all those who write to me
1) how i can obtain relative risk (risk ratio) in logistic regression in R.
2) how to obtain the predicted risk for a certain individual using
fitted regression model in R.
Many thanks, in advance, for your help.
Amin.
Douglas M. Hultstrand dmhultst at metstat.com writes:
Hello R-Group,
I am new working with netcdf files and the raster package in R.I am
trying to read in a netcdf file using the package ncdf.I am able to
get the lat, lon and parameter I need and can plot using
fill.contour.
Please
Hi,
set.seed(125)
dat1-data.frame(BIRTHPLACE=sample(c(AG,AI,AR,BE,ZH,USA,GER,ESP),20,replace=TRUE),RES_STA=sample(LETTERS[c(1:3,24:25)],20,replace=TRUE))
dat1$VARNEW-ifelse(dat1$RES_STA==X
dat1$BIRTHPLACE%in%c(AG,AI,AR,BE,ZH),swiss,unknown)
A.K.
- Original Message -
From: David
Hi,
Your dataset had already some missing values. So, I need to subset only those
rows that are not missing.
!is.na(temp$ACTIVE_KWH)
# [1] TRUE TRUE TRUE FALSE TRUE FALSE FALSE TRUE TRUE TRUE TRUE FALSE
#[13] TRUE TRUE TRUE
temp$ACTIVE_KWH[!is.na(temp$ACTIVE_KWH)]
#[1] 1201.9 1202.2
Hi Dimitri,
Does this help?
k1-data.frame(item=sample(rep(letters),10,replace=T),a=c(1:10),b=11:20)
k2-data.frame(item=f,a=3,b=10)
merge-function(y,x)
{
if(y$amin(x$a))
{
x-rbind(x,y)
x-x[-which.min(x$a),]
}
return(x)
}
merge(k2,k1)
or much faster way would be to refer library(sqldf).
Hello,
You could do something like the following.
fun - function(x, mean, sd1, sd2, p)
dnorm(x, mean, sd1)*p + dnorm(x, mean, sd2)*(1 - p)
fun2 - function(x1, x2, mean, sd1, sd2, p){
p1 - pnorm(x2, mean, sd1) - pnorm(x1, mean, sd1)
p2 - pnorm(x2, mean, sd2) - pnorm(x1,
If you wanted this for all values in x that are smaller, i'd use
x[x$a y$a,] - y
for just the smallest:
x[intersect(which(x$a y$a),which.min(x$a)),] - y
On 29.01.2013, at 22:11, Dimitri Liakhovitski wrote:
Hello!
I have a large data frame x:
On 13-01-29 6:04 PM, Troy S wrote:
Ok, yes I realize it. So let me try to fix it:
I removed the sty files, and set TEXINPUTS to
'C:\R\R-2.15.2\share\texmf\tex\latex
I am back where I started: the tex file will not process.
Please let me know what I should do to fix the issue.
Follow my
Carol,
Actually, you have only five nodes, numbered 1, 2, 3, 6, and 7.
And all five nodes are included in your plot.
Nodes 1 and 3 are branching nodes; nodes 2, 6, and 7 are terminal nodes.
Try typing just the name of the rpart object for a very brief text version
of the tree.
rpart.res
Readers,
For a graph plot instruction:
plot(seq(10:50),type='h',yaxt='n',yaxs='i',lab=c(20,2,2),xlab='x axis
label',bty='l',main='graph title')
how to remove y-axis label and keep the x-axis label?
_
r2151
__
R-help@r-project.org mailing list
Hello,
Just use ylab = .
Hope this helps,
Rui Barradas
Em 30-01-2013 13:33, e-letter escreveu:
Readers,
For a graph plot instruction:
plot(seq(10:50),type='h',yaxt='n',yaxs='i',lab=c(20,2,2),xlab='x axis
label',bty='l',main='graph title')
how to remove y-axis label and keep the x-axis
dat1$VARNEW-rep(unknown,nrow(dat1))
dat1$VARNEW[dat1$RES_STA==X dat1$BIRTHPLACE %in%
c(AG,AI,AR,BE,ZH)]-swiss
--- On Wed, 30/1/13, arun smartpink...@yahoo.com wrote:
From: arun smartpink...@yahoo.com
Subject: Re: [R] recoding variables again :(
To: stude...@gmail.com stude...@gmail.com
Cc: R
On 1/29/2013 6:04 PM, Troy S wrote:
Ok, yes I realize it. So let me try to fix it:
I removed the sty files, and set TEXINPUTS to
'C:\R\R-2.15.2\share\texmf\tex\latex
I am back where I started: the tex file will not process.
Please let me know what I should do to fix the issue.
Troy
Don't do
Relative risk = exp(coef(model))
--- On Wed, 30/1/13, aminreza Aamini amin.r@gmail.com wrote:
From: aminreza Aamini amin.r@gmail.com
Subject: [R] Relative Risk in logistic regression
To: R-help R-help@r-project.org
Date: Wednesday, 30 January, 2013, 4:19 PM
Hi all,
I am very grateful
On 13-01-30 8:52 AM, Michael Friendly wrote:
On 1/29/2013 6:04 PM, Troy S wrote:
Ok, yes I realize it. So let me try to fix it:
I removed the sty files, and set TEXINPUTS to
'C:\R\R-2.15.2\share\texmf\tex\latex
I am back where I started: the tex file will not process.
Please let me know what
Hi,
I guess you could also use:
x[match(min(x$a),x$a[x$ay$a]),]- y
x
# item a b
#1 f 3 10
#2 b 2 12
#3 c 3 13
#4 d 4 14
#5 e 5 15
A.K.
- Original Message -
From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com
To: r-help r-help@r-project.org
Cc:
Sent: Tuesday,
Example from linear regression help (?lm)
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
lm.D9 - lm(weight ~ group)
lm.D90 - lm(weight ~ group - 1)
Hi,
You could use:
dat1-data.frame(col1=c(Mercedes_02352,Audi_03555))
dat1[,1]-as.numeric(gsub(.*_\\d{1}(.*)\\d{1}$,\\1,dat1[,1])) #if the
number of digits are the same
dat1
# col1
#1 235
#2 355
A.K.
- Original Message -
From: Mat matthias.we...@fnt.de
To: r-help@r-project.org
Cc:
Hi all,
Ive just started to learn working with R.
I wonder if it is possible to make a temporal correlogram in R and how to
do it. Actually what I have are movement tracks of birds, existing as
XY-positions every 5 minutes for a duration of at least 60 minutes (but up
to 8 hours). From these
Dear Team,
I am getting the following error message when try to run vb application
The program was running fine in 32 windows 7 machine.
When i moved the same program to 64 bit windows 8 machine i am getting the
following error
Error in inDL(x, as.logical(local), as.logical(now), ...) : unable
Dear R-users,
Though it's a silly thing to ask, but I'm not getting a way out. I wish to
find the percentage distribution for a data vector 'stop'. The coomand
below is giving the frequency distribution. May I know the option to see
the percentages instead of frequencies. Similarly, what option
It's possible the direct way in R:
customer - c(Mercedes_02352, Audi_03555)
substr(customer, nchar(customer) - 3, nchar(customer) - 1)
Isidro
-Mensaje original-
De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
En nombre de Jorge I Velez
Enviado el: miércoles,
Hello,
Try
barplot(table(stop)/sum(table(stop)))
Hope this helps,
Rui Barradas
Em 30-01-2013 11:34, Naser Jamil escreveu:
Dear R-users,
Though it's a silly thing to ask, but I'm not getting a way out. I wish to
find the percentage distribution for a data vector 'stop'. The coomand
below is
Hi,
I am interested in mapping 6 different images on the faces of a cube. I found
the full code for doing this here :
http://rwiki.sciviews.org/doku.php?id=graph_gallery:cube
But I am unable to adapt the code for my purpose. I would appreciate it if I
could get some help on the following points
Please reply with context to the list. Most R-help readers do not use nabble.
I don't quite understand your question but do you mean something like ?names?
John Kane
Kingston ON Canada
-Original Message-
From: matthias.we...@fnt.de
Sent: Tue, 29 Jan 2013 07:41:21 -0800 (PST)
To:
I don't see what is happening from your code but you have a typo in the emai if
not in your code. It should be library (pROC)
John Kane
Kingston ON Canada
-Original Message-
From: feth...@yahoo.fr
Sent: Mon, 28 Jan 2013 14:44:50 + (GMT)
To: r-help@r-project.org
Subject: [R]
Michael,
I added C:\R\R-2.15.2\share\texmf to the list of roots and hit apply and
that solved the problem. Thanks!
Duncan, I did try the R CMD Sweave you suggested and that did not solve the
issue. The problem was in running LaTeX so how was your suggestion help
me? It would have to copy the
HI,
Sorry, my previous solution doesn't work.
This should work for your dataset:
set.seed(1851)
x-
data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F)
y- data.frame(item=f,a=3,b=10,stringsAsFactors=F)
On 01/30/2013 09:02 AM, nalluri pratap wrote:
Relative risk = exp(coef(model))
Only if you fit using the log link. Using the logit link, this gives
odds ratios.
--- On Wed, 30/1/13, aminreza Aamini amin.r@gmail.com wrote:
From: aminreza Aamini amin.r@gmail.com
Subject: [R]
There is now a blog post that attempts to
answer the question in the subject line:
http://www.burns-stat.com/the-three-dots-construct-in-r/
Pat
On 17/01/2013 14:36, Ivan Calandra wrote:
Dear users,
I'm trying to learn how to use the
I have written a function (simplified here) that uses
I am not sure why one would want a relative risk from a logistic regression.
The measure of association from a logistic regression is the odds ratio, not
the relative risk.
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine
On 01/30/2013 11:17 AM, John Sorkin wrote:
I am not sure why one would want a relative risk from a logistic
regression. The measure of association from a logistic regression is the
odds ratio, not the relative risk.
John
Yes, the natural measure, when using the logit link, is the OR. I
If you use a log link, you are not, I believe, performing a logistic regression!
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD
This is FAQ 7.21. The most important part of that answer is at the end
where it says that it is better to use a list. Your code could be
something like:
plotroc - list()
for (i in levels(mergeTrn$Continent) {
# matrix defined here
plotroc[[ paste(plotroc_GBM_TRN_,i, sep=) ]] - matrix
}
now
On 01/30/2013 11:26 AM, John Sorkin wrote:
If you use a log link, you are not, I believe, performing a logistic
regression!
I guess strictly speaking, that is true. I was being a little sloppy in
terminology.
Kevin E. Thorpe kevin.tho...@utoronto.ca 1/30/2013 11:22 AM
On 01/30/2013
If you wish to remove missing values, you can use the option
na.action=na.omit.If you wish to Impute you can use rfImpute.
--- On Mon, 28/1/13, Lorenzo Isella lorenzo.ise...@gmail.com wrote:
From: Lorenzo Isella lorenzo.ise...@gmail.com
Subject: [R] RandomForest and Missing Values
To:
Because R can be interactive, I find that a little exploring through
the use of strategically placed browser() calls (?browser if you are
unfamiliar with this handy debugging tool) is often the fastest way to
solve little R puzzles like this.
For example, try this (in an R GUI):
f2 -
On 13-01-30 11:00 AM, Troy S wrote:
Michael,
I added C:\R\R-2.15.2\share\texmf to the list of roots and hit apply and
that solved the problem. Thanks!
Duncan, I did try the R CMD Sweave you suggested and that did not solve
the issue. The problem was in running LaTeX so how was your
Duncan,
Alas, I had left out the --pdf flag when I tried the solution.
I am glad I have an answer now, and a better understanding of the problem.
Troy
On Wed, Jan 30, 2013 at 9:50 AM, Duncan Murdoch murdoch.dun...@gmail.comwrote:
On 13-01-30 11:00 AM, Troy S wrote:
Michael,
I added
Thank you, everyone! I'll try to test those different approaches. Really
appreciate your help!
Dimitri
On Wed, Jan 30, 2013 at 11:03 AM, arun smartpink...@yahoo.com wrote:
HI,
Sorry, my previous solution doesn't work.
This should work for your dataset:
set.seed(1851)
x-
Sorry - I should have clarified:
My identifiers (in column item) will always be unique. In other words,
one entry in column item will never be repeated - neither in x nor in y.
Dimitri
On Wed, Jan 30, 2013 at 1:27 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Thank you,
Hi,
Any chance x$a to have the same number repeated?
If `Item` and `a` are unique, I guess both the solutions should work.
set.seed(1851)
x-
data.frame(item=sample(letters[1:20],20,replace=F),a=sample(1:45,20,replace=F),b=sample(20:50,20,replace=F),stringsAsFactors=F)
y-
Thanks for your replies! It seems, that I can fit my model now, when I can
provide the right starting values; however there remain warnings, such as:
1: In log(ifelse(y == 1, 1, (1 - y)/(1 - mu))) : NaNs wurden erzeugt
2: step size truncated due to divergence
3: step size truncated: out of
Hi, we're trying to install rJava on SLES 11.
With:
R CMD INSTALL rJava_0.9-3.tar.gz
It says:
configure: error: One or more Java configuration variables are not set.
Make sure R is configured with full Java support (including JDK). Run
R CMD javareconf
as root to add Java support to R.
So we
Try this:
Age_log_model = glm(Arthrose ~ Alter, data=x, start=c(-1, 0),
family=quasibinomial(link = log))
Ravi
Ravi Varadhan, Ph.D.
Assistant Professor
The Center on Aging and Health
Division of Geriatric Medicine Gerontology
Johns Hopkins University
Hello,
I would like to perform an NMDS on the following:
I have two independent variables, which are sites and treatments.
I have 6 sites which are peatlands. I collected 5 replicates (at the same time)
from each of the sites.
I used each of the replicates in a treatment.
There were 4
For a given linear regression, I wish to find the 2-tailed t-dist
probability that Y-hat = newly observed values. I generate prediction
intervals in predict() for plotting, but when I calculate my t-dist
probabilities, they don't agree. I have researched the issues with variance
of individual
I did not get any warnings when I ran your data/model example.
From: Fischer, Felix [mailto:felix.fisc...@charite.de]
Sent: Wednesday, January 30, 2013 11:19 AM
To: Ravi Varadhan
Cc: r-help@r-project.org
Subject: AW: [R] starting values in glm(..., family = binomial(link =log))
Thanks for your
On Jan 30, 2013, at 3:09 AM, cgenolin wrote:
Hi all,
I am drawing some 3D surfaces using the Triangle tools (package
misc3) and
drawScene.rgl. Do you know if it is possible to add axes and
graduation on
the scene?
You offer no code or data, so a specific answer is not called for.
In realy, values in a will be not integers, but numeric. They will never be
identical, but it could be that they are pretty close - I don't know after
how many points after the comma matter.
Dimitri
On Wed, Jan 30, 2013 at 2:06 PM, arun smartpink...@yahoo.com wrote:
Hi,
Any chance x$a to have
Why, in R, does (0.1 + 0.05) 0.15 evaluate to True? What am I missing
here? How can I ensure this (ostensibly incorrect) behavior doesn't
introduce bugs into my code? Thanks for your time.
Dave Mitchell
[[alternative HTML version deleted]]
R FAQ 7.31 (Note, this isn't R specific, rather it's a problem with
the finitude of computers)
MW
On Wed, Jan 30, 2013 at 8:32 PM, Dave Mitchell dmmtc...@gmail.com wrote:
Why, in R, does (0.1 + 0.05) 0.15 evaluate to True? What am I missing
here? How can I ensure this (ostensibly incorrect)
On Jan 30, 2013, at 5:49 AM, aminreza Aamini wrote:
Hi all,
I am very grateful to all those who write to me
1) how i can obtain relative risk (risk ratio) in logistic
regression in R.
2) how to obtain the predicted risk for a certain individual using
fitted regression model in R.
You
On 30-01-2013, at 21:32, Dave Mitchell dmmtc...@gmail.com wrote:
Why, in R, does (0.1 + 0.05) 0.15 evaluate to True? What am I missing
here? How can I ensure this (ostensibly incorrect) behavior doesn't
introduce bugs into my code? Thanks for your time.
R-FAQ 7.31:
Why, in R, does (0.1 + 0.05) 0.15 evaluate to True?
Because floating point arithmetic is done with a fixed
number of digits. If you are working in base 10 and have
2 digits to work with you would have
1/3 - .33
2/3 - .67
so that
1/3 + 1/3 2/3
How can I ensure this (ostensibly
Just look at the code of predict.lm(). It is reasonably perspicuous.
In particular look at res.var.
cheers,
Rolf Turner
On 01/31/2013 05:50 AM, Kurt Rinehart wrote:
For a given linear regression, I wish to find the 2-tailed t-dist
probability that Y-hat = newly observed values.
Hi
I have 25 samples in my dataset. I have written a multiple regression model
and I would like to test it.
I would like to train my model on 20 samples and then test it on 5
remaining. However I would like to test the model several times, each time
using different 5 samples out of 25 and check
Hello all,
When I tried to install fSeries in R, I got the following error messages:
install.packages(fSeries,dependencies=T)
Warning message:
package 'fSeries' is not available (for R version 2.15.2)
Is this package changing/merging to another package?
Thanks,
Rebecca
Works for me, so you will have to provide more information.
x11()
hist(rnorm(100))
dev.copy2pdf(file='mytest.pdf')
X11
2
list.files(patt='mytest')
[1] mytest.pdf
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On
Thank you Pascal and Greg for the suggestion. That is exactly what I needed!
- Kumar
On Wed, Jan 30, 2013 at 2:14 AM, Pascal Oettli kri...@ymail.com wrote:
Hello,
?assign
assign(paste(plotroc_GBM_Trn_**, i, sep=), matrix(rnorm(100),10,10))
HTH,
Pascal
Le 30/01/2013 17:04, Kumar
On 30-Jan-2013 20:39:34 Berend Hasselman wrote:
On 30-01-2013, at 21:32, Dave Mitchell dmmtc...@gmail.com wrote:
Why, in R, does (0.1 + 0.05) 0.15 evaluate to True? What am I missing
here? How can I ensure this (ostensibly incorrect) behavior doesn't
introduce bugs into my code? Thanks
Folks,
As the subject describes: I would like to know if there are packages that have
functionality tailored for standard Oil/Gas exploration and monetization.
Thanks,
KW
--
[[alternative HTML version deleted]]
__
R-help@r-project.org
Has anyone explored pulling XBRL formatted data into R?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
On Jan 30, 2013, at 4:19 AM, Johannes Radinger wrote:
Hi,
I already found a conversation on the integration of a normal
distribution and two
suggested solutions
(https://stat.ethz.ch/pipermail/r-help/2007-January/124008.html):
1) integrate(dnorm, 0,1, mean = 0, sd = 1.2)
and
2) pnorm(1,
Keith Weintraub kw1958 at gmail.com writes:
Folks,
As the subject describes: I would like to know if there are
packages that have functionality tailored for
standard Oil/Gas exploration and monetization.
Check out the sos package: it will help you answer the
question yourself,
Kurt: You missed including the term 1/n in your var.y.hat calculation. Do
that and pred and hand are the same.
Brian
Brian S. Cade, PhD
U. S. Geological Survey
Fort Collins Science Center
2150 Centre Ave., Bldg. C
Fort Collins, CO 80526-8818
email: brian_c...@usgs.gov
tel: 970 226-9326
Hello,
I tried to make a betadisper plot; however, it is quite messy at the moment
with lines and symbols.
I made two plots, one focusing on sites and the other on treatments.
This is the code that I used:
plot(betadisper(vegdist(y.nth,method=euclidean),site))
Hi.
I'm trying to export a .wav file using the writeWave function from tuneR
package in a different folder than the default getwd().
After reading through the manuals of some audio packages I couldn't figure
it out.
I'm picking one 3-hour .wav file and asking the function to take a sample
of 1
Hi.
I'm trying to export a .wav file using the writeWave function from tuneR
package in a different folder than the default getwd().
After reading through the manuals of some audio packages I couldn't figure
it out.
I'm picking one 3-hour .wav file and asking the function to take a
Hello,
I would like to perform a Box-Cox (âbcPowerâ) transformation on my data.
For this, I am determining lambda using the âpowerTransformâ function.
However, with one of my variables I get the following
Warning Message:
In estimateTransform(x, y, NULL, ...) :
Convergence failure:
Dear list,
Can I use a character to set the name of a R package? like this (-)
for example (sdp-R)
Thanks,
John
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PLEASE do read the posting guide
Put the directory name into the filename string. The syntax for doing that is
somewhat OS dependent, but for most cases you can use / as the separator. You
really should Google file path and your OS and learn how to do this, because
it is broadly applicable outside of R (not an R-help
Hi David,
Thanks for your answer.
Here is my (simplified) code :
8 -
library(misc3d)
#
### Fonction that draw a point A=(x,y,z), with radius r
misc3dPoint - function(A,r,color=black,alpha=1){
t1 -
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