Thank you all! This is a great course on Lattice par settings (though
not answering the question)!
And a truly must have: Lattice: Multivariate Data Visualization with R
(Use R).
Thank you Dr. Sarkar!
Here's it all summed up.
#Looking at lattice par settings:
library(lattice)
str(trellis.par.get
Bars are corresponding to bins, and the bin-width for lines is 0;
please tell me what is the "frequency" at a fixed point (rather than
over an interval)?
Regards,
Yihui
--
Yihui Xie <[EMAIL PROTECTED]>
Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086
Mobile: +86-15810805877
Homepage: http://www.y
All -
When I plot something like
a<-rnorm(5)
b<-rnorm(5)
plot(a,b,col = "red")
points(10,-10)
The last point is missing because it is out of
range of the first plot.
I just try to switch from Matlab to R. In Matlab,
it always can automatic adjust the xlim and ylim
for such case.
Is it possi
Dear Himanshu,
Try ?sink() :
data_file.out <- krukal.test(data_file)
sink("~/DATA/results/data_file_out.txt")
data_file.out
sink()
HTH,
Jorge
On Thu, Oct 16, 2008 at 4:31 PM, Himanshu Ardawatia
<[EMAIL PROTECTED]>wrote:
> Hello,
>
> I am running Kruskal-Walis test in R. When I try to save r
Em Qui, 2008-10-16 às 22:31 +0200, Himanshu Ardawatia escreveu:
> Hello,
>
> I am running Kruskal-Walis test in R. When I try to save results using
> write.table it gives me the following error :
>
> Error in as.data.frame.default(x[[i]], optional = TRUE, stringsAsFactors =
> stringsAsFactors) :
On 10/16/08, Paul Boutros <[EMAIL PROTECTED]> wrote:
> Hello,
>
> I have encountered some unexpected behaviour with levelplot that may simply
> be a misunderstanding on my part.
>
> If I create a levelplot from a matrix with named columns, some "padding"
> space appears at the top and bottom of t
Hello,
I have encountered some unexpected behaviour with levelplot that may
simply be a misunderstanding on my part.
If I create a levelplot from a matrix with named columns, some
"padding" space appears at the top and bottom of the heatmap. Here is
an example, and I've used panel.fill t
?try
On Thu, Oct 16, 2008 at 4:46 PM, hgreatrex <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> Is there any way that I can supress error messages so that they don't stop
> for loops running?
>
> I'm using the gstat package and have created a variogram model for the
> double exponential model:
>
>"dexp
Hi,
Is there any way that I can supress error messages so that they don't stop
for loops running?
I'm using the gstat package and have created a variogram model for the
double exponential model:
"dexp <- vgm( .5, "Exp", 10,.1, add.to = vgm(.5, "Exp", 100))"
and fitted it using
"dexp
repkakala Gazeta.pl wrote:
Rigth, sorry. R is great for many usages even when it did not got
modernised as many realy modern languages did.
This is a nice read if you have some time about some possibly surprising
truths about 'modern' languages.
http://www.paulgraham.com/icad.html
Rigth, sorry. R is great for many usages even when it did not got
modernised as many realy modern languages did.
Appologies for developers, I were too rude.
\misiek
2008/10/16 <[EMAIL PROTECTED]>
>
> that's extremely rude , especially to all the people who made and make R
> what it is. If you
Hi everyone,
How to constrain the residual variance to be 0 when using the lmer
function?
Wen Luo
Wen Luo, Ph.D.
Assistant Professor
Department of Educational Psychology
University of Wisconsin -- Milwaukee
Enderis Hall 759
P.O. Box 413
Milwaukee, WI 53201-0413
Office: 414 229 4998
Fax: 414 229
Dears,
Any help?
Thanks,
LFRC
LFRC wrote:
>
> Dears,
>
> I'm trying to find the parameters (a,b, ... l) that optimize the function
> (Model)
> described below.
>
> 1) How can I set some constraints with MLE2 function? I want to set p1>0,
> p2>0,
> p3>0, p1>p3.
>
> 2) The code is giving
that's extremely rude , especially to all the people who made and make
R what it is. If you don't like R, noone is forcing you to use it.
On Thu, Oct 16, 2008 at 5:50 PM, repkakala Gazeta.pl wrote:
On 10/16/2008 10:50 AM, culpritNr1 wrote:
Now, modern high level languages like the cont
On 10/16/2008 10:50 AM, culpritNr1 wrote:
>
>>
>> Now, modern high level languages like the continually improving R,
>
>
"
References:
Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) _The New S
Language_. Wadsworth & Brooks/Cole.
"
This is most important reference in many help pa
2008/10/16 Erin Hodgess <[EMAIL PROTECTED]>:
> Hi R People!
>
> Is there a setup for Roman numerals similar to that of LETTERS and
> letters, please?
>
> I was putting together a randomized block design and thought that it
> might be nice for factors.
>
> Thanks,
> Erin
>
as.roman() in utils:
>
Hi R People!
Is there a setup for Roman numerals similar to that of LETTERS and
letters, please?
I was putting together a randomized block design and thought that it
might be nice for factors.
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
Un
This idea is very wrong, Have a look at the function cnormal1 in the
VGAM package
John frain
2008/10/15 drbn <[EMAIL PROTECTED]>:
>
> I'm using stableFit from the package fBasics to estimate the parameters of a
> truncated normal distribution (I'm interested in the parameters of the
> underlying
On Tue, Oct 14, 2008 at 7:15 PM, john verzani <[EMAIL PROTECTED]> wrote:
> Tim Smith yahoo.com> writes:
>
> >
> > Thanks Charlie - I just tried it, but still get the same error:
> >
> > ---
> > > install.packages("gWidgets",dependencies=TRUE)
>
> ...
>
> > > library(gW
XLSolutions-corp has 2 day training courses; one for beginners and one
for advanced.
Here is the website:
www.xlsolutions-corp.com
Hope this helps!
Sincerely,
Erin
On Thu, Oct 16, 2008 at 1:50 AM, timpanister <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Can anyone suggest some short term statistical c
Dear R users,
I have two uniform random variables and I need to estimate the joint
density. I like to know whether
there is any package which estimates bivariate uniform densities.
Thanks,
Lavan
--
View this message in context:
http://www.nabble.com/bivariate-uniform-density-estimation-tp2
If you want to avoid using a loop, rather than coding a loop, then someting
like:
DV <- rnorm(sum(n))
DV <- DV * rep(sigma, times=n) + rep(mu, times=n)
will achieve the same as the loop you specified, but generalised to
non-constant n.
HTH
Ray Brownrigg
On Fri, 17 Oct 2008, Bert Gunter wrote:
Hello,
I am running Kruskal-Walis test in R. When I try to save results using
write.table it gives me the following error :
Error in as.data.frame.default(x[[i]], optional = TRUE, stringsAsFactors =
stringsAsFactors) :
cannot coerce class "htest" into a data.frame
The overall code is as follow
Yes, but the difference is that the looping in mapply is done in C.
There are no interpreted loops in mapply,as far as I can see.
-Christos
> -Original Message-
> From: Bert Gunter [mailto:[EMAIL PROTECTED]
> Sent: Thursday, October 16, 2008 4:13 PM
> To: [EMAIL PROTECTED]; 'David Afsha
mapply is still a (disguised) loop (at the interpreted level). So other than
improving code readability (always a good thing!), it shouldn't make much of
an efficiency difference.
A longer answer is: if all you're doing is a location-scale family of
distributions, then creating a matrix of standar
Have a look at mapply.
-Christos
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of David Afshartous
> Sent: Thursday, October 16, 2008 3:47 PM
> To: r-help@r-project.org
> Subject: [R] Loop avoidance in simulating a vector
>
>
>
> All,
>
> I'd
All,
I'd like to simulate a vector that is formed from many distinct
distributions and avoid a loop if possible. E.g, consider:
mu = c(1, 2, 3)
sigma = c(1, 2, 3)
n = c(10, 10, 10)
And we simulate a vector of length 30 that consists of N(mu[i], sigma[i])
distributed data, each of length n[i]
Try this (to evaluate them at 10):
polynomial.values(l, 10)
On Thu, Oct 16, 2008 at 2:40 PM, Chernomoretz Ariel <[EMAIL PROTECTED]> wrote:
> Hi All,
> I need to evaluate a series expansion using Legendre polynomials.
> Using the 'orthopolinom' package I can get a list of the first n
> Legendre p
try this:
library(orthopolynom)
l <- legendre.polynomials(4)
fun.l <- lapply(l, as.function)
fun.l[[3]](1:3)
I hope it helps.
Best,
Dimitris
Chernomoretz Ariel wrote:
Hi All,
I need to evaluate a series expansion using Legendre polynomials.
Using the 'orthopolinom' package I can get a list
On 17/10/2008, at 8:22 AM, useR wrote:
Let's try to bring this discussion back again after Frank made
very funny remark!
Frank's remark was *serious*. Take it seriously.
cheers,
Rolf Turner
###
On 17/10/2008, at 7:40 AM, Chernomoretz Ariel wrote:
Hi All,
I need to evaluate a series expansion using Legendre polynomials.
Using the 'orthopolinom' package I can get a list of the first n
Legendre polynomials as character strings.
library(orthopolynom)
l<-legendre.polynomials(4)
l
[[1]]
Let's try to bring this discussion back again after Frank made
very funny remark!
What I'm doing at the moment is:
1. I split dataset in two (development and holdout)
2. I fit single predictor logistic model for every variable and
collect following stats:
DMaxDeriv=modelD$stats[2]
DModelLR=m
Hi All,
I need to evaluate a series expansion using Legendre polynomials.
Using the 'orthopolinom' package I can get a list of the first n
Legendre polynomials as character strings.
> library(orthopolynom)
> l<-legendre.polynomials(4)
> l
[[1]]
1
[[2]]
x
[[3]]
-0.5 + 1.5*x^2
[[4]]
-1.5*x + 2.5*
On Thu, 16 Oct 2008, Duncan Murdoch wrote:
On 10/16/2008 12:19 PM, Felipe Carrillo wrote:
This is my sessionInfo() and Sys.getenv("PATH")
library(tools)
testfile <- system.file("Sweave", "Sweave-test-1.Rnw", package = "utils")
## enforce par(ask=FALSE)
options(device.ask.default=FALSE)
## cre
Thank- you all for you suggestions and help. I have now resolved my
issue. This help forum is one of the things that makes R a great platform
for statistics.
Stephen Cole
Marine Ecology Lab
Saint Francis Xavier University
On Wed, Oct 8, 2008 at 8:07 AM, Erin Hodgess <[EMAIL PROTECTED]>wrote:
Duncan:
I installed R in C:\R\R-2.7.2 and worked beautifully. I hope this will help
others with the same problem. Thanks a lot for your help.
--- On Thu, 10/16/08, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 10/16/2008 12:19 PM, Felipe Carrillo wrote:
> > This is my sessionInfo() and
> Sys.
On 10/16/2008 11:43 AM, Greg Snow wrote:
I wonder if including the p-values for the normality test is the best approach in you
animation? The clt does not say that the distribution of the means will be normal, just
that it approaches normality (and therefore may be a decent approximation). Th
On 10/16/2008 12:19 PM, Felipe Carrillo wrote:
This is my sessionInfo() and Sys.getenv("PATH")
library(tools)
testfile <- system.file("Sweave", "Sweave-test-1.Rnw", package = "utils")
## enforce par(ask=FALSE)
options(device.ask.default=FALSE)
## create a LaTeX file
Sweave(testfile)
Writing t
On Oct 16, 2008, at 12:55 PM, Jorge Ivan Velez wrote:
Dear Jörg,
See ?cut and ?table. Is this what you want?
x <- c(1, 3, 1.2, 5, 5.9)
table(cut(x,breaks=c(0,2,4,6)))
(0,2] (2,4] (4,6]
2 1 2
Perhaps even greater future efficiency could be had by also adding
?seq
table(cut(x, bre
Thanks for your great help!
I have now what I wanted.
For sure it's not well written and you can realize it with much less
lines.
But it works and it's like how I wanted it to look like;
z1 <- rnorm(4500, mean=20, sd=5)
z2 <- rnorm(3600, mean=28, sd=5)
z3 <- rnorm(1300, mean=40, sd=7)
z4 <
On Thu, 16 Oct 2008, Paulo Cortez wrote:
Hi,
I want to save a RWeka model into a file, in order to retrive it latter with
a load function.
See this example:
library(RWeka)
NB <- make_Weka_classifier("weka/classifiers/bayes/NaiveBayes")
model<-NB(formula,data=data,...) # does not run but yo
Hi,
I want to save a RWeka model into a file, in order to retrive it latter
with a load function.
See this example:
library(RWeka)
NB <- make_Weka_classifier("weka/classifiers/bayes/NaiveBayes")
model<-NB(formula,data=data,...) # does not run but you get the idea
save(model,file="model.dat"
Hi,
is there a way to replace long splits with some short naming
conventions?
For example if my resulting tree from rpart has 50-100 values can I
display some summary text like group1 instead of group num = 1 or 2 or
3100?
is there a way to specify different display text for nod
AFAICS, you have a 2GB address space: please do read the rw-FAQ and fix
that, in which case R will automatically make use of the increased space.
Hint: what is /3GB for?
On Thu, 16 Oct 2008, liujb wrote:
Dear R users,
I am running lmer() and having memory problem: "reached total allocation
Does this do what you want?
colhist <- function(x,col){
tmp <- hist(x,plot=F)
br <- tmp$breaks
w <- as.numeric(cut(x,br,include.lowest=TRUE))
sy <- unlist(lapply(tmp$counts,function(x)seq(length=x)))
sy <- sy[order(order(x))]
plot( range(br)
On Thu, Oct 16, 2008 at 11:42 AM, x0rr0x <[EMAIL PROTECTED]> wrote:
>
> Hi all,
>
> I'm trying to create a histogram which shows the frequency of variables
> within a certain timeframe.
>
> I've been using SPSS before, but I didn't quite like it...
>
> To describe my problem further here are some e
I would personally use grid for this.
library(grid)
grid.rect(width = 5, height = 3, default.units = "cm")
Barry Rowlingson wrote:
2008/10/16 Thomas Steiner <[EMAIL PROTECTED]>:
Hi I want to draw sth in a pdf file with a predefined defined size.
Say a 5cm x 3cm rectangle (a ruler):
pdf("rect
Dear R users,
I am running lmer() and having memory problem: "reached total allocation of
1535Mb".
The lmer() function was successful (no errors or warnings). However when I
do "summary(fit.lmer), I got this "Error: cannot allocate vector of size
19.5 Mb). I used memory.size() and got 1880. I us
Dear Jörg,
See ?cut and ?table. Is this what you want?
x <- c(1, 3, 1.2, 5, 5.9)
table(cut(x,breaks=c(0,2,4,6)))
(0,2] (2,4] (4,6]
2 1 2
HTH,
Jorge
On Thu, Oct 16, 2008 at 12:46 PM, Jörg Groß <[EMAIL PROTECTED]> wrote:
> Hi,
>
>
> Is there a function which counts the frequencies o
?cut or ?findInterval and ?table
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> project.org] On Behalf Of Jörg Groß
> Sent: Thursday, October 16, 20
Hi all,
I'm trying to create a histogram which shows the frequency of variables
within a certain timeframe.
I've been using SPSS before, but I didn't quite like it...
To describe my problem further here are some example variables:
the "event" is actually a string which I recoded using:
[code]
2008/10/16 Jörg Groß <[EMAIL PROTECTED]>:
> Hi,
>
>
> Is there a function which counts the frequencies of the occurence of a
> number within an interval?
>
> for example I have this vector:
>
> x <- c(1, 3, 1.2, 5, 5.9)
>
> and I want a vector that gives me the frequencies within an interval of 2,
Add to that list the functions grconvertX and grconvertY which can take
coordinates in many different units and convert them to others (including
inches, then just multiply/divide by 2.54 for the conversion from/to cm).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Health
Hi,
Is there a function which counts the frequencies of the occurence of a
number within an interval?
for example I have this vector:
x <- c(1, 3, 1.2, 5, 5.9)
and I want a vector that gives me the frequencies within an interval
of 2, beginning at 0
(so the intervals are 0-2, 2-4, 4-6 a
Since water quality is ordered, you could use cor.test with method of kendall
or spearman. This will basically give a test of whether the rank based
correlation is significantly different from 0. Often more interesting
information can come from the data if you do some more work. Start by grap
2008/10/16 Thomas Steiner <[EMAIL PROTECTED]>:
> Hi I want to draw sth in a pdf file with a predefined defined size.
> Say a 5cm x 3cm rectangle (a ruler):
>
> pdf("rect.pdf", paper="a4")
> plot(c(1,5,5,1,1),c(1,1,3,3,1),asp=1,axes="n")
> dev.off()
>
> but how do I fix that one unit is 1cm?
I jus
Dear John,
Perhaps is possible to do the job using apply or any of its friends. For
now, here is a solution using for() :
# Data sets
set.seed(123)
sample1<-matrix(data=c(1:20),byrow=TRUE,ncol=5,nrow=4)
sample2<-sample1+rnorm(20)
# t-test p-values
k=nrow(sample1)
res=NULL
for(i in 1:k) res[i]=t.t
This is my sessionInfo() and Sys.getenv("PATH")
library(tools)
testfile <- system.file("Sweave", "Sweave-test-1.Rnw", package = "utils")
## enforce par(ask=FALSE)
options(device.ask.default=FALSE)
## create a LaTeX file
Sweave(testfile)
Writing to file Sweave-test-1.tex
Processing code chunks .
How about the following?
sapply(1:nrow(sample1), function(x) t.test(sample1[x,], sample2[x,])$p.value)
Gabor
On Thu, Oct 16, 2008 at 6:07 PM, John Sorkin
<[EMAIL PROTECTED]> wrote:
> Although I am grateful to Jorge and David for their suggestions, their
> solution will not solve my problem as s
I don't think that your question was bad form, there are some things you can do
in the future to make it better form. Your questions was certainly better form
than others who just ask for an answer to their homework without even admitting
that it is homework. You were confused about some very
Although I am grateful to Jorge and David for their suggestions, their solution
will not solve my problem as sample2 is not fixed. Let me demonstrate what I
want to do:
# Define two matricies
sample1<-matrix(data=c(1:20),byrow=TRUE,ncol=5,nrow=4)
sample2<-sample1+rnorm(20)
These are the comput
Am 16.10.2008 um 17:12 schrieb Yihui Xie:
plot(density(x))
Regards,
Yihui
--
Thanks!
Is there also a way to plot the frequency-bars (not the density) as
lines instead of bars?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/l
I wonder if including the p-values for the normality test is the best approach
in you animation? The clt does not say that the distribution of the means will
be normal, just that it approaches normality (and therefore may be a decent
approximation). The normality test can just reject the null
Jorge, David,
Thank you for a brilliant solution to my problem!
Thanks
John
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 21201-1
On 10/16/2008 10:18 AM, Felipe Carrillo wrote:
Hi:
I'm still trying to figure out how use Sweave. Trying the example below I get
the error message when texi2dvi is executed. Any ideas about how to make
texi2dvi work?
Try running it as texi2dvi("Sweave-test-1.tex", pdf=TRUE, quiet=FALSE),
an
On 10/16/2008 10:50 AM, culpritNr1 wrote:
Hello Gabor,
First of all, thanks for your reply.
Indeed, your strategy solves the problem of speed=). My question was,
however, more of an R design question.
Your suggestion of pre-allocating a big and continuous portion of memory is
probably the firs
On Thu, Oct 16, 2008 at 8:05 AM, kdebusk <[EMAIL PROTECTED]> wrote:
> What test do I use to determine if there is a correlation between a
> discrete variable and a continuous variable?
>
> For example - I have water quality ratings for streams (excellent,
> good, fair, poor) and a corresponding nit
Hi I want to draw sth in a pdf file with a predefined defined size.
Say a 5cm x 3cm rectangle (a ruler):
pdf("rect.pdf", paper="a4")
plot(c(1,5,5,1,1),c(1,1,3,3,1),asp=1,axes="n")
dev.off()
but how do I fix that one unit is 1cm?
Thanks
thomas
__
R-help
Thanks a lot Stephen. It worked for me. also Jorge had a solution, it's a
short and easy code. U could have a look in case U want to use it in future.
Thanks again for all your help
stephen sefick wrote:
>
> a <- c(1:10)
> b <- c(.5, .6, .9, 10, .4, 3, 4, 9, 0, 11)
> d <- c(21:30)
>
> z <- dat
plot(density(x))
Regards,
Yihui
--
Yihui Xie <[EMAIL PROTECTED]>
Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086
Mobile: +86-15810805877
Homepage: http://www.yihui.name
School of Statistics, Room 1037, Mingde Main Building,
Renmin University of China, Beijing, 100872, China
On Thu, Oct 16, 20
Jorge you are awesome,
This code is very efficient. it's exactly what I wanted to do and it's very
short and not complicated.
Thanks a lot
Jorge Ivan Velez wrote:
>
> Dear Alex,
> Is this what you want?
>
> my=read.table(textConnection("
> X8 X9 X10 X102 X110 X177 X283 X284 X286 X292 X297 X
Lawrence Hanser wrote:
Dear Colleagues,
I run this model:
mod1 <- lmer(x~category+subcomp+category*subcomp+(1|id),data=impchiefsrm)
obtain this summary result:
Linear mixed-effects model fit by REML
Formula: x ~ category + subcomp + category * subcomp + (1 | id)
Data: impchiefsrm
AIC BIC
The real problem is that running a loop to build up an object bit by bit
is not the preferred way to write an R program. The preferred way
is to use lapply or other apply command in which case the object
gets built for you.
On Thu, Oct 16, 2008 at 10:50 AM, culpritNr1 <[EMAIL PROTECTED]> wrote:
>
Hi,
I know how to plot an histogram and how to add a density line.
But how can I plot only the density line without the bars?
Is there a way to say the hist() function not to plot bars, but a
density line instead?
__
R-help@r-project.org mailing li
Hello Gabor,
First of all, thanks for your reply.
Indeed, your strategy solves the problem of speed=). My question was,
however, more of an R design question.
Your suggestion of pre-allocating a big and continuous portion of memory is
probably the first manual work around to try. And it will wo
Dear Colleagues,
I run this model:
mod1 <- lmer(x~category+subcomp+category*subcomp+(1|id),data=impchiefsrm)
obtain this summary result:
Linear mixed-effects model fit by REML
Formula: x ~ category + subcomp + category * subcomp + (1 | id)
Data: impchiefsrm
AIC BIC logLik MLdeviance REMLdev
Hi:
I'm still trying to figure out how use Sweave. Trying the example below I get
the error message when texi2dvi is executed. Any ideas about how to make
texi2dvi work?
library(tools)
> testfile <- system.file("Sweave", "Sweave-test-1.Rnw", package = "utils")
> options(device.ask.default=FAL
Thanks for the correction David! You're absolutely right! A typo :- )
...
Cheers,
Jorge
On Thu, Oct 16, 2008 at 10:00 AM, David Winsemius <[EMAIL PROTECTED]>wrote:
>
> On Oct 16, 2008, at 8:11 AM, Jorge Ivan Velez wrote:
>
> Dear John,
>> Yes. Assuming that 'sample2' is fixed, something li
On Oct 16, 2008, at 8:11 AM, Jorge Ivan Velez wrote:
Dear John,
Yes. Assuming that 'sample2' is fixed, something like this should do
what
you want:
# Data sets
set.seed(123)
sample1=matrix(rnorm(100),ncol=10)
sample2=rnorm(10,5,4)
# t-test p-values
apply(X,1,function(x) t.test(x,sample2)$p
Dear Alex,
Is this what you want?
my=read.table(textConnection("
X8 X9 X10 X102 X110 X177 X283 X284 X286 X292 X297 X306 X308 X314
0 1 000100000000
0 0 001000000010
0 1 0000000
Hello,
When i use tune svm with sigmoid oy polynomial kernels to optimize
Cost/gamma/coef0 (sigmoid) or cost/gamma/coef0/degree (polynomial)
parameters, i can't get a summary of the best parameters. instead I got this
result:
dummyparameter
1 0
I use the following command line (7 fo
a <- c(1:10)
b <- c(.5, .6, .9, 10, .4, 3, 4, 9, 0, 11)
d <- c(21:30)
z <- data.frame(a,b,d)
library(fields)
results <- c()
for(i in 1:(length(rownames(z))-1)){
results[i] <- rdist(z[i,], z[(i+1),])
}
results.1 <- data.frame(results)
f <- rownames(z)
r <- f[-1]
rownames(results.1) <- r
coln
This may be incorrect, but couldn't you just assign a number to the
ratings 1:4 and then scatterplot them agianst N (x-axis). Or you may
want to plot boxplots of N by water quality ratings. If these are
water quality ratings derived from aquatic macroinvertebrates and you
have access to the actua
Hi everyone,
I have dataset which I take random samples of it 5 times. each time I get
the mean for rows for each sample.
at the end I need to calculate the Average of all means for each sample and
each row. to clear it up I give an example:
say this is my dataset.
X8 X9X10X12 X13 X14 X15 X1
What test would you use to determine if a correlation exists between a
discrete variable and a continuous variable?
For example, I have a rating for stream water quality (excellent,
good, fair, poor) and a corresponding nitrogen concentration. I want
to see if there is a correlation between the wa
Thanks for your responses. I know it is bad form asking about these things
but i was really having trouble getting my head around it, and i wanted to
make sure that the cause wasn't due to the commands i was entering into R.
At least now i know it's a conceptual error i am making rather than a
tec
Excellent - now it works. Thank you, Christian!
(I still seem to be unable to replicate the results as reported in the
paper, but that is unrelated to R-help. Thomas, do you have any idea
what I may have entered wrongly?)
Best,
Christian
Christian Ritz wrote:
Hi Christian,
I believe that the
What test do I use to determine if there is a correlation between a
discrete variable and a continuous variable?
For example - I have water quality ratings for streams (excellent,
good, fair, poor) and a corresponding nitrogen concentration for each
rating. I want to know if the the ratings correl
Duncan Murdoch wrote:
This isn't really about R, and maybe it is homework, but now that we
got you in the appropriate frame of mind:
(a) p values should look at "this or more unfavourable" events. You
have arranged things so that that translates to -6.13 or _lower_. I.e.
you're looking a
Dear users
I am fitting a Generalized Additive Mixed Models (gamm) model to
establish possible relationship between explanatory variables (water
temperature, dissolved oxygen and chlorophyll) and zooplankton data
collected in the inner and outer estuarine waters. I am using monthly
time-se
On 10/16/2008 7:35 AM, Peter Dalgaard wrote:
rr400 wrote:
Hi, i am doing a statistics course and am having trouble with an exercise
where i need to determine whether my success rate at something is higher
than 80%.
I was successful in 29 out of 60 trials, so these were the commands i
entered i
Dear John,
Yes. Assuming that 'sample2' is fixed, something like this should do what
you want:
# Data sets
set.seed(123)
sample1=matrix(rnorm(100),ncol=10)
sample2=rnorm(10,5,4)
# t-test p-values
apply(X,1,function(x) t.test(x,sample2)$p.value)
[1] 0.002348970 0.004733230 0.004810952 0.004907549
R 2.7.2
Windows XP
I am using apply to compute a series of Student's t-test from two matrices,
sample1 and sample2.
boo<-apply(sample1,1,t.test,sample2)
I want to pick of the p-values from the tests, but can't seem to get it to
work. I have tried several methods to get the values including:
bo
Gad Abraham wrote:
Frank E Harrell Jr wrote:
Gad Abraham wrote:
This approach leaves much to be desired. I hope that its
practitioners start gauging it by the mean squared error of
predicted probabilities.
Is the logic here is that low MSE of predicted probabilities equals a
better calibra
try this:
> x <- read.table(textConnection(" user_id site_id name
+ 11 11 februari
+ 21 11 redbook
+ 31 11 tips
+ 73 6 sleep
+ 83 6monitoring
+ 93 6 alarm"), header=T
rr400 wrote:
Hi, i am doing a statistics course and am having trouble with an exercise
where i need to determine whether my success rate at something is higher
than 80%.
I was successful in 29 out of 60 trials, so these were the commands i
entered into R:
n=60
p.hat=29/n
p.0=0.8
se.0=sqrt(p.0*
bg708 wrote:
Hi,
I am trying to plot a large matrix by using color2D.matplot. A substantial
part of the matrix has no data (NAs). However, when trying to plot it, I get
an error. It is only when I change all NAs to zeros that it works, but that
in some way introduces 'wrong' data. Is there any w
Hi, i am doing a statistics course and am having trouble with an exercise
where i need to determine whether my success rate at something is higher
than 80%.
I was successful in 29 out of 60 trials, so these were the commands i
entered into R:
>n=60
>p.hat=29/n
>p.0=0.8
>se.0=sqrt(p.0*(1-p.0)/n)
>
Dear all
Recently the "granova" package was launched. I installed but after when I
invoked it in R it requested for other libraries. They were downloaded and
install automatically.
I tried to run the example syntax of granova.1w and granova.2w but two
things happened: i) either a file called
Please read the last line of every message to r-help and provide
reproducible code.
On Thu, Oct 16, 2008 at 12:44 AM, Ted Byers <[EMAIL PROTECTED]> wrote:
> Thanks Gabor,
>
> To be clear, would something like testframe$est[[i]] <- fp$estimate be
> valid within my loop, as in (assuming I created te
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