On Thu, 27 Jan 2011, Julie Smith wrote:
Could someone please help me with instructions on running a
principal coordinate analysis using R?
?cmdscale
Julie Smith
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
U
Hi,
I have 25 normal and 25 tumor samples and generated 50 boxplots one for
each
Is there a possibility to alternate the colors for the boxplots
Green Red Green Red.
Example:
A B C D
10 23 23 34
20 24 24 30
20 2434 34
this would generate 4 boxplot one for each. I want the boxplot for
I need to import a large number of simple, space-delimited text files with a
few columns of data each. The one quirk is that some rows are missing data and
some contain junk text at the end of each line. A typical file might look like:
a b c d
1 2 3 x
4 5 6
7 8 9 x
1 2 3 x c c
4 5 6 x
7 8 9 x
Hi,
I am trying to convert a 2D correlation matrix to 3 columns for graphical
representation:
rdata = replicate(100, rnorm(15)) #construct a 2D matrix
c1 = cor(rdata) #outputs a correlation matrix
Now I want to convert the 2D c1 to
(row#, col#, correlation)
1 1 cor1
1 2 cor2
1 3 cor3
...
2 1
Could someone please help me with instructions on running a principal
coordinate analysis using R?
Julie Smith
Sent from my iPod
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PLEASE do read the posting guide http://w
Hi,
I am trying to work with the ape package, and there is one thing I am
struggling with. When calling the *read.GenBank()* function, I can get it
to work with an object created like this:
*>x <- c("AY395554","AY611035", ...)*
*>read.GenBank(x)*
However, I am trying to use the function to fetc
On Jan 27, 2011, at 6:29 PM, Peter Jaksons wrote:
Hello,
I want to create a flexible code for the following example: In stead
of
using a different code for each n (as in my example below), I want to
write a general code for every n (n from 1 to 10).
Have you done any thinking about the
On Jan 28, 2011, at 12:37 AM, Johannes Huesing wrote:
David Winsemius [Thu, Jan 27, 2011 at
10:08:00PM CET]:
You got a perfectly sensible reply from Thereau, the author of the
package, a day after your posting and then failed to respond to his
questions. I'm not sure what more you are expe
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of eric
> Sent: Thursday, January 27, 2011 7:07 PM
> To: r-help@r-project.org
> Subject: [R] There must be a smarter way
>
>
> Newbie and trying to learn the right way of doing thi
David Winsemius [Thu, Jan 27, 2011 at 10:08:00PM CET]:
> You got a perfectly sensible reply from Thereau, the author of the
> package, a day after your posting and then failed to respond to his
> questions. I'm not sure what more you are expecting.
>
My apologies. Next time I'll be checking Gm
On Thu, Jan 27, 2011 at 11:56 AM, John E. Kaprich wrote:
> When I try to convert the zoo object to a timeSeries object, which would
> allow me to utilize Rmetrics packages, I get an error message.
>
>
>
>> Data<-read.zoo("c:\\DOWUBSPRICING.txt,na.strings="NA",sep="\t",header=T)
>> is(Data)
> "zoo"
Newbie and trying to learn the right way of doing things in R. in this case,
I just have that feeling that my convoluted line of code is way more
complicated than it needs to be. Please help me in seeing the easier way.
I want to do something pretty simple. I have a dataframe called x that is
694
R works by going down the columns. If you make the rows into columns, it then
does what you want. You just have to make the columns back into rows to get the
original shape of your matrix.
So the code in one line is :
t(t(z) - means)
Original message
>Date: Fri, 28 Jan 2011 01:16:45
You could use the survey package to run the bootstrapping, if you mean
the Rao & Wu bootstrap that samples n-1 of n PSUs in each replicate.
Set up a survey design object with bootstrap replicate weights: use
svrepdesign() if you already have replicate weights, use svydesign()
and then as.svrepdesi
In addition to what has already been suggested you could use ..
mapply(function(x,y) x-y, z,means)
which returns
V1 V2
[1,] 0.333 -2.7142857
[2,] NA 7.2857143
[3,] -0.667 -3.7142857
[4,] -6.667 NA
[5,] NA -0.7142857
[6,] 1.333
It looks like the text didn't show assigning the results of factanal
to an object. Try:
pgdata<-read.table("pgfull.txt",header=T)
names(pgdata)
pgd<-pgdata[,1:54]
#missing line
model <- factanal(pgd,8)
par(mfrow=c(2,2))
plot(loadings(model)[,1],loadings(model)[,2],pch=16,xlab="Factor
1",ylab="Fac
On Jan 27, 2011, at 7:16 PM, Ernest Adrogué i Calveras wrote:
Hi,
I have this data.frame with two variables in it,
z
V1 V2
1 10 8
2 NA 18
3 9 7
4 3 NA
5 NA 10
6 11 12
7 13 9
8 12 11
and a vector of means,
means <- apply(z, 2, function (col) mean(na.omit(col)))
means
V1
The empirical statement on the proc mixed line gives you robust
standard errors, I don't think you get them in R.
In SAS you specify that the predictors are to be dummy coded using the
class . Are they factors in R? I can't tell from the SAS output,
because the formatting has been lost. However
sapply(z, function(row) ...) does not actually grab a row at a time out of
'z'. It grabs a column (because 'z' is a data.frame)
You may want:
t(apply(z, 1, function(row) row - means))
or:
t(t(z) - means)
Hope that helps,
-David Johnston
--
View this message in context:
http://r.789695.n4.n
Hi Harold:
I know the outputs are different between SAS and R, but the results that I
got have big difference.
Here is part of the result based on the SAS code I provided earlier:
Cov Parm SubjectEstimate Error
Value Pr > Z
Hi,
I have this data.frame with two variables in it,
> z
V1 V2
1 10 8
2 NA 18
3 9 7
4 3 NA
5 NA 10
6 11 12
7 13 9
8 12 11
and a vector of means,
> means <- apply(z, 2, function (col) mean(na.omit(col)))
> means
V1V2
9.67 10.714286
My intention was substracting mean
Hello,
I want to create a flexible code for the following example: In stead of
using a different code for each n (as in my example below), I want to
write a general code for every n (n from 1 to 10).
I tried a lot of things, but I always got stuck with the number of
subloops or witch closing t
May be this problem is OS specific (mac OS or windows...) i use ubuntu and
have no problem with this function
R> packageVersion("base")
[1] ‘2.12.1’
R> packageVersion('lme4')
[1] ‘0.999375.37’
R> sessionInfo()
R version 2.12.1 (2010-12-16)
Platform: x86_64-pc-linux-gnu (64-bit)
locale:
[1] LC_C
Hello,
I've drawn a black rectangle over the plotting area, and when I add an image()
heatmap, it doesn't take up all the area, but is set inward from the black
rectangle. Can anyone suggest how to make it stretch out to the entire area ?
Minimal example :
y <- matrix(runif(2000*20), nrow = 20
Hello,
Yes, that's right, it is a values matrix. Not a dissimilarity matrix.
i.e.
> str(iMatrix)
num [1:23371, 1:56] -0.407 0.198 NA -0.133 NA ...
- attr(*, "dimnames")=List of 2
..$ : NULL
..$ : chr [1:56] "-8100" "-7900" "-7700" "-7500" ...
For the snippet of checking for NAs, I get all
Belle:
Before I provide any more help, you'll need to follow some posting guide rules
and describe what you have done with some R code, describe what the problem is,
and describe what you are hoping to accomplish.
The fact that the results you get from R and SAS are different doesn't mean
ther
Hi Harold:
Yes, this was the R code that I tried, and got different result from SAS.
Is that mean I cannot actually use R to run unstructured covariance matrix?
How can I solve this problem if I need an unstructured covariance matrix
method?
Thanks for the help.
--
View this message in contex
Dear R community,
I am using the mlogit package to analyze discrete choice data. Apart
from a main effects model, I want to estimate interactions between the
attributes of the choice set (e.g. the existence of a certain attribute)
and some subject-specific data (like gender or income).
Studyin
You got a perfectly sensible reply from Thereau, the author of the
package, a day after your posting and then failed to respond to his
questions. I'm not sure what more you are expecting.
--
David
On Jan 27, 2011, at 3:43 PM, Johannes Huesing wrote:
Is there a special mailing list for th
On Jan 27, 2011, at 3:02 PM, Martin Hvidberg wrote:
Dear list
I have a large data frame with house prices. Prices have been
converted
to price per m2, to be able to compare. Prices are spread over
almost 10
years, each price is tagged with the year it relates to.
I would like to 'normali
Try:
library(TeachingDemos)
plot(Sepal.Length~Sepal.Width, data=iris)
tmp.y <- iris$Sepal.Length
for( i in unique(iris$Sepal.Width) ) {
tmp <- iris$Sepal.Width == i
tmp.y[ tmp ] <- spread.labs( tmp.y[tmp], .6*strheight('A'),
maxiter=1000 )
}
# optional
with(iris,
Is there a special mailing list for the survival package where my
question would have been more appropriate?
Johannes Huesing [Mon, Jan 10, 2011 at 09:26:37PM CET]:
> I am trying to model survival data with a Weibull distribution
>
On Jan 27, 2011, at 2:38 PM, mcitkowicz wrote:
Hello,
I have written the function I would like to integrate in two ways:
denfxn <- function(yy,vv,a2,b2,mu2) {
pp <- 1-pnorm(yy/sqrt(vv))
part1 <- pp^(a2-1)
part2 <- (1-pp)^(b2-1)
part3 <- dnorm(yy,mu2,sqrt(vv))
return(part1*part2*par
Dear list
I have a large data frame with house prices. Prices have been converted
to price per m2, to be able to compare. Prices are spread over almost 10
years, each price is tagged with the year it relates to.
I would like to 'normalize' all the prices to, eg. 2010 prices.
The data frame loo
I think it should be
fm <- lmer(score ~ trt + pair + grade + school + (1|team), test)
The unstructured covariance matrix you use in proc mixed is not available in Rs
function for mixed models
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Hi
I am trying to convert SAS codes to R, but some of the result are quite
different from SAS.
When I ran proc mixed, I have an option ddfm=bw followed by the model. How
can I show this method in R (I am thinking that this maybe the reason that I
can't get the similar results)
below is my SAS
Hello,
I have written the function I would like to integrate in two ways:
denfxn <- function(yy,vv,a2,b2,mu2) {
pp <- 1-pnorm(yy/sqrt(vv))
part1 <- pp^(a2-1)
part2 <- (1-pp)^(b2-1)
part3 <- dnorm(yy,mu2,sqrt(vv))
return(part1*part2*part3) }
denfxnorg <- function(yy,vv,a2,b2,mu2)
David,
I would do this with the HH package.
Rich
install.packages("HH") ## if you don't already have it
library(HH)
## the first call is to learn the number .8162 and see what the scaling is.
normal.and.t.dist(mu.H0=5.01, obs.mean=log(300), std.dev=.77,
Use.alpha.left=TRUE
Hi,
I received a bug report from a French user who told me there was
something wrong with the function packageVersion(), but
packageDescription() worked well. Below is the session info; I'm not
sure if other French users can reproduce the error message:
> packageVersion('base')
Erreur : spécifica
Hi all,
I have two basic questions, hope you should help me:
1. How do I save a log file in R with the results? For example, in Stata
it can be done by using "log using c:\...\test.txt"
2. How do I display the execution time of one function like the one below:
CalculaCorrelacao <- function(co
On Thu, Jan 27, 2011 at 11:30:37AM +0100, Serena Corezzola wrote:
> Hello everybody!
>
>
>
> I?m trying to define the optimal number of surveys to detect the highest
> number of species within a monitoring season/session.
>
> To do this I want to run all the possible combinations between a set
Hi,
I have some surprising trouble installing the gputools package.
This is a linux box running Fedora with an Nvidia GeForce GTX 260 card
and R 2.12.1
I had gputools running on it for the past several months. Wanting to
update to the newest version, I simply executed
install.packages("gputools"
Dear R users,
I'm using igraph package for clustering, just wondering is
leading.eigenvector.community function result deterministic or not?
I got slightly different results on different runs...
Many thanks
Yan
***
First for general info: Although I am listed as maintainer I am not
maintaining this version currently. Thanks to the developers who offer
this version and work on it. We will try to get a version on CRAN soon.
For your specific question:
Download the file and rename it to
BRugs_0.6-1.zip.
Th
Hello,
I wanted to ask if there is an R package to fit GLM (logistic for
example) via empirical likelihood.
--
-Tony
[[alternative HTML version deleted]]
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PLEASE
Hello, all. When I try to install the most recent BRugs package (the one
that uses version 3.1.2), found here:
http://www.openbugs.info/w/UserContributedCode
I get the following error:
> utils:::menuInstallLocal()
Error in gzfile(file, "r") : cannot open the connection
In addition: Wa
When I try to convert the zoo object to a timeSeries object, which would
allow me to utilize Rmetrics packages, I get an error message.
> Data<-read.zoo("c:\\DOWUBSPRICING.txt,na.strings="NA",sep="\t",header=T)
> is(Data)
"zoo"
> as.timeSeries.zoo(Data)
Error in .local (.Object, . )
Is
On Jan 27, 2011, at 8:45 AM, KATSCHKE, ADRIAN CIV DFAS wrote:
Dennis,
This is excellent. Thank you for the help. I knew I had a tangle
mess, but I didn't realize how much of a tangle until I used this
code. The coding definitely simplified the process and sped up the
execution time.
Ad
My colleagues that use one of the .Net languages/libraries can make
scatter plots that look better than R's because they have better
spreading of the labels.
If someone could spread this labels on the following graph, I would be
impressed.
plot(Sepal.Length~Sepal.Width, data=iris)
with(iris,text(
Hello Dear List members,
as you can see (and guess) from the code below
adaptIntegrate(f,lowerLimit=c(-1,-1),upperLimit=c(.,.))
$integral
[1] 9.997e-09
$error
[1] 1.665168e-16
$functionEvaluations
[1] 17
$returnCode
[1] 0
> adaptIntegrate(f,lowerLimit=c(-1,-1),upperLimit=c(1,1))
the
michalseneca gmail.com> writes:
> Hi,
> Basically what I am trying is to rewrite matlab code into R ...This is code
> for famous SABR model Calibration.
> I did most of the code up to optimalization issue.
> In the attachment you can find matlab and my unfinished R code..This is very
> helpful co
?system.time
On Thu, Jan 27, 2011 at 10:31 AM, Alaios wrote:
> Dear list members,
> I would like to measure how much time one function call makes from the time
> is call until the time it returns.
>
> Could you please tell me if that is possible in R?
>
>
> Best Regards
> Alex
>
> __
Dear list members,
I would like to measure how much time one function call makes from the time is
call until the time it returns.
Could you please tell me if that is possible in R?
Best Regards
Alex
__
R-help@r-project.org mailing list
https://stat.e
Hi,
Basically what I am trying is to rewrite matlab code into R ...This is code
for famous SABR model Calibration.
I did most of the code up to optimalization issue.
In the attachment you can find matlab and my unfinished R code..This is very
helpful code...Basically I am trying to achieve the sam
Thanks again for the pointer to spread.labs Greg.
I implemented it into the function and also extended it to deal with
formulas so it could behave just like boxplot.
Code and examples are available here:
http://www.r-statistics.com/2011/01/how-to-label-all-the-outliers-in-a-boxplot/
I'd be happy
Try this:
sweep(y, 2, x, "/")
On Thu, Jan 27, 2011 at 1:18 PM, adam_pgsql wrote:
>
> Hi,
>
> I'd like to divide each element of a matrix by a specific value per column.
> These specific values are stored in a list. For example:
>
> > x <- c(1,2,3,4,5)
> > y <- matrix(c(1:30), nrow = 6)
>
> Now
I sent this to Siri Bjoner so I dont know if that is allowed or even if it went
through.
Simulate 1000 observations from the student-t distribution with 3
> degrees of freedom and then calculate the truncated mean by
> excluding bottom 5% and top 5%.
This is the question I had and I poste
On Jan 27, 2011, at 9:18 AM, adam_pgsql wrote:
>
> Hi,
>
> I'd like to divide each element of a matrix by a specific value per column.
> These specific values are stored in a list. For example:
>
>> x <- c(1,2,3,4,5)
>> y <- matrix(c(1:30), nrow = 6)
>
> Now I want to divide each element in
Here's one way:
y/outer(rep(1, nrow(y)), x)
[,1] [,2] [,3] [,4] [,5]
[1,]1 3.5 4.33 4.75 5.0
[2,]2 4.0 4.67 5.00 5.2
[3,]3 4.5 5.00 5.25 5.4
[4,]4 5.0 5.33 5.50 5.6
[5,]5 5.5 5.67 5.75 5.8
[6,]6 6.0 6.00 6.00 6.0
Run the outer c
Hi:
Using your data below as the template,
u <- cut(x, breaks = c(0, seq(1000, 7000, by = 1000)), dig.lab = 4, right =
FALSE)
as.data.frame(table(u))
u Freq
1[0,1000)0
2 [1000,2000)5
3 [2000,3000)0
4 [3000,4000)0
5 [4000,5000)0
6 [5000,6000)0
7 [6000,7000)
On 11-01-27 14:58, Louise Mair wrote:
Hello,
I am working with a dataset which essentially has only one column - a list of
distances in metres, accurate to several decimal places. eg
distance
1000
6403.124
1000
1414.214
1414.214
1000
I want to organise this into a frequency table, grouping
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
[cc'ing back to r-help]
On 01/27/2011 10:06 AM, Ben Boyadjian wrote:
> Yes the it says the uniform distribution on the interval (-5,5).
If this is real-valued (which again is not precisely defined, but
that would be how I would interpret) then the
Hi,
I'd like to divide each element of a matrix by a specific value per column.
These specific values are stored in a list. For example:
> x <- c(1,2,3,4,5)
> y <- matrix(c(1:30), nrow = 6)
Now I want to divide each element in y[,1] by x[1], y[,2] by x[2] etc. I have
tried this
> my_function
On 2011-01-27 05:16, Ben Boyadjian wrote:
Hello I am trying to solve these problems and I am not allowed to use loops or
ifs.
1st Question
My first question is that I have generated 100 random numbers from the uniform
distribution then
A)add only the negative integers.
B)add elements until the
On 27.01.2011 14:08, michalseneca wrote:
Hi ,
can i ask if somebody know the syntax for optim function with multiple
variables:
I have function f (params,y,g,h,j) returnig z
y<-C(0.2,0.5,0.6)
where
params<-c(x1,x2)
I define g,h,j and then I define params, then use
optim to structure it ..ho
On 27.01.2011 11:16, Temel Canpolat wrote:
Hi,
I need to install "lmtest package".I have tried to several times to install
packages however I could not. Can anyone help?
Best regards
install.packages ()
--- Please select a CRAN mirror for use in this session ---
Warning: unable to access
Hello,
I am working with a dataset which essentially has only one column - a
list of distances in metres, accurate to several decimal places. eg
distance
1000
6403.124
1000
1414.214
1414.214
1000
I want to organise this into a frequency table, grouping into categories
of 0 - 999, 1000 - 199
On 2011-01-27 03:34, Juan Andres Hernandez wrote:
Hello I have the following data.frame and xyplot. I need this plot without
the borders. Does anybody know how to get this xyplot without borders o with
white borders?
Thank's in advance
Juan Hernández
my.label<- data.frame(quantil=rep(20,8),
my
For question 2,
TTT <- rt(1000,3)
mean(TTT[rank(TTT) <= 975 & rank(TTT) >25])
On Thu, Jan 27, 2011 at 8:16 AM, Ben Boyadjian wrote:
> Hello I am trying to solve these problems and I am not allowed to use loops
> or ifs.
>
> 1st Question
> My first question is that I have generated 100 random num
Siterer "Sascha Vieweg" :
On 11-01-27 14:24, Ben Bolker wrote:
Ben Boyadjian hotmail.com> writes:
Double post?
Yes, probably just hoped that if he changed the subject someone would
do his homework for him...
__
R-help@r-project.org mailing
On 11-01-27 14:24, Ben Bolker wrote:
Ben Boyadjian hotmail.com> writes:
Double post?
Hello I am trying to solve these problems and I am not allowed to use loops or
ifs.
1st Question
My first question is that I have generated 100 random numbers from the uniform
distribution then
A)add o
Thank you Terry,
I suspected that this was a mistake in the help file.
Who should be contacted regarding this?
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) |
Ben Boyadjian hotmail.com> writes:
>
> Hello I am trying to solve these problems and I am not allowed to use loops or
ifs.
>
> 1st Question
> My first question is that I have generated 100 random numbers from the uniform
distribution then
> A)add only the negative integers.
> B)add elements u
Dear list,
Indeed, there is no need adding error bars to boxplot. Points with error
bars as suggested will be quite better.
Many thanks to those who have advised me on this.
Regards
Ogbos
On 27 January 2011 12:29, Jim Lemon wrote:
> On 01/27/2011 05:04 AM, ogbos okike wrote:
>
>> Dear all,
>> I
On 11-01-27 15:12, Sascha Vieweg wrote:
On 11-01-27 13:16, Ben Boyadjian wrote:
Hello I am trying to solve these problems and I am not allowed to use
loops or ifs.
1st Question
My first question is that I have generated 100 random numbers from the
uniform distribution then
A)add only th
On 11-01-27 13:16, Ben Boyadjian wrote:
Hello I am trying to solve these problems and I am not allowed to use loops or
ifs.
1st Question
My first question is that I have generated 100 random numbers from the uniform
distribution then
A)add only the negative integers.
x <- c(1, 1.3)
x==round
Hello I am trying to solve these problems and I am not allowed to use loops or
ifs.
1st Question
My first question is that I have generated 100 random numbers from the uniform
distribution then
A)add only the negative integers.
B)add elements until the first appearance of a negative element.
Hello I am trying to solve these problems and I am not allowed to use loops or
ifs.
1st Question
My first question is that I have generated 100 random numbers from the uniform
distribution then
A)add only the negative integers.
B)add elements until the first appearance of a negative element.
Sorry about re-posting this, it never went out to the mailing list when I
posted this to r-help forum on Nabble and was pending for a few days, now
that I am subscribe to the mailing list I hope that this goes out:
I've been a viewer of this forum for a while and it has helped out a lot,
but this
Hi,
I have a few questions regarding machine learning packages in R, e.g.
earth package
and its forward pass and prunning pass, and hinge functions selection in the
example below.
1) How could I force the algorithm to generate knots in hinge function only
with values < 90 for e.g. variable doy?
2
Antje Niederlein yahoo.de> writes:
[snip]
> But what goes wrong if I want to display confidence intervals? I get a
> lot of warnings but I simply don't know why...
>
> confint(fit)
>
> Has it something to do with constraints for my parameters (lambda
> should be > than zero and prop should r
Dennis,
This is excellent. Thank you for the help. I knew I had a tangle mess, but I
didn't realize how much of a tangle until I used this code. The coding
definitely simplified the process and sped up the execution time.
Adrian
-Original Message-
From: Dennis Murphy [mailto:djmu...@gm
> splits, a matrix of split labels
This is a mistake in the help file.
> I'm also not sure I understand what the yval2 columns signify (even
that I read what it says in the help).
For the kyphosis data set y is a factor with two levels. For the
first split yval2 contains (1, 64, 17, .79012
Hi ,
can i ask if somebody know the syntax for optim function with multiple
variables:
I have function f (params,y,g,h,j) returnig z
y<-C(0.2,0.5,0.6)
where
params<-c(x1,x2)
I define g,h,j and then I define params, then use
optim to structure it ..however if i use syntax
optim(params , f,y=y,
Hi there.
Out of curiosity, why are you doing it that way instead of this:
<>=
plot(runif(100))
@
Kevin
On January 27, 2011 05:30:08 am Matthieu Stigler wrote:
> great, solved! Thanks a lot for your quick help!
>
> Le 27. 01. 11 11:22, Prof Brian Ripley a écrit :
> > Note that the 'problem' is
Hello everybody!
Im trying to define the optimal number of surveys to detect the highest
number of species within a monitoring season/session.
To do this I want to run all the possible combinations between a set of
samples and to calculate the total number of species for each combination of
2,
Hi,
I need to install "lmtest package".I have tried to several times to install
packages however I could not. Can anyone help?
Best regards
install.packages ()
--- Please select a CRAN mirror for use in this session ---
Warning: unable to access index for repository
http://cran.cict.fr/bin/wind
Hi R-helpers,
I am using MONMLP package for Neural network.I am able to do regression
using that.Now, i want to use it for classification.Say, for classifying
Iris dataset.I have not seen any example regarding that. Can you please let
me know if classification is possible using monmlp NN? If so,
Have you tried using the arrows function with angle = 90? you will have to
create a mean and standard error variable
-
- AK
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Thank you for the clarification, this makes sense now! dose.p from MASS also
does the job perfectly, in returning x values for specified proportions. I'm
curious, if I use the other parameter in predict.glm type="link" (instead of
type="response"), in the case of a logistic binomial, what does this
On Thu, 2011-01-27 at 10:45 +0100, Uwe Ligges wrote:
>
> On 27.01.2011 05:00, Dario Strbenac wrote:
> > Hello,
> >
> > In the documentation for agnes in the package 'cluster', it says that NAs
> > are allowed, and sure enough it works for a small example like :
> >
> >> m<- matrix(c(
> > 1, 1, 1,
Eduardo,
On 01/27/2011 12:53 PM, Philipp Pagel wrote:
caused by a problem with font substitution in some version of
the poppler library which is uses by many LINUX PDF viewers. Try to
view the file in acrobat reader and possibly other viewers.
I'm running Ubuntu, and uninstalling package ttf-s
On Thu, 2011-01-27 at 00:10 -1000, Ahnate Lim wrote:
> Thank you for the clarification, this makes sense now! dose.p from
> MASS also does the job perfectly, in returning x values for specified
> proportions. I'm curious, if I use the other parameter in predict.glm
> type="link" (instead of type="r
On Thu, Jan 27, 2011 at 01:29:31AM -0200, Eduardo de Oliveira Horta wrote:
> Hi there,
>
> yet on the topic of greek letters and pdf plotting: when I run the
> following code
>
> pdf(file="temp.pdf")
> mu=seq(from=-pi, to=pi, length=100)
> plot(mu, sin(mu^2),
> type="l",
> xlab=expressi
Dear R-helpers and Dr. Hijmans,
I can not invoke maxent() in GNU Linux. Could you give me any directions
on that? There is maxent.jar in version 3.3.3e in the right folder.
R:
> jar <- paste(system.file(package="dismo"), "/java/maxent.jar", sep='')
> jar
[
Hello I have the following data.frame and xyplot. I need this plot without
the borders. Does anybody know how to get this xyplot without borders o with
white borders?
Thank's in advance
Juan Hernández
my.label <- data.frame(quantil=rep(20,8),
my.factor=factor(c('FA','FB','FC','FD','FE','FG','FH',
On Thu, Jan 27, 2011 at 1:41 AM, gaja wrote:
>
> Hey Josh. Thanx for an explicit post, but I still can't get it. Would be nice
> if someone would just do that maths instead of posting "forum isn't for
> that". I know its for R help and that, but if I can't do it, u could, huh?
The whole point of
Hi all,
Is there any way to plot multiple linear models(lm) on the same
window. I have data in the following format where I would like get a
log-log plot between 3rd column and 4th column and fit a straight line
through the points. The problem is that I have multiple files with such a
f
great, solved! Thanks a lot for your quick help!
Le 27. 01. 11 11:22, Prof Brian Ripley a écrit :
Note that the 'problem' is with dev.off(), not pdf(), and the answer
is ?invisible.
pdf()
invisible(dev.off())
On Thu, 27 Jan 2011, Matthieu Stigler wrote:
Hi
I am sorry, my question seems sim
On 01/27/2011 05:04 AM, ogbos okike wrote:
Dear all,
I am trying to add error bars on a boxplot but have encountered an error as
indicated below. Is there a package I need to install or a library I have to
load before this goes please.
Thanks for any idea.
Ogbos
x<-replicate(20,rnorm(50))
boxp
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