On Thu, Mar 06, 2008 at 06:54:41PM -0500, Charilaos Skiadas wrote:
> On Mar 6, 2008, at 1:49 PM, Mag. Ferri Leberl wrote:
>
> > Dear everybody!
> > Is there a command in \LaTeX to display the R-Logo or has anybody
> > made it up?
> > Thank you in advance.
>
> Isn't it just an image? Hence you w
If you are calling libR.so, it is an option set when you initialize R.
See Rf_initEmbeddedR in 'Writing R Extensions' and the examples in the
tests/Embedded directory in the sources.
BTW, discussion of embedded R is definitely off topic for R-help: use
the R-devel list.
On Thu, 6 Mar 2008, Scot
Whoa! If you want to optimise in R, you need to write an R function to
define your objective. If that function is already written in C, as
appears to be the case, you need to write an interfact to make it
available to R. This is not exactly something novices would take on
lightly.
If you wish t
Hi,
I can not comprehend the linear fitting results of polynoms. For
example, given the following data (representing y = x^2):
> x <- 1:3
> y <- c(1, 4, 9)
performing a linear fit
> f <- lm(y ~ poly(x, 2))
gives weird coefficients:
> coefficients(f)
(Intercept) poly(x, 2)1 poly(x, 2)2
4.66
poly() computes by default orthogonal polynomials; check the online
help file for poly() for more info. Probably you want to use the 'raw'
argument in this example, i.e.,
x <- 1:3
y <- c(1, 4, 9)
lm(y ~ poly(x, 2, raw = TRUE))
I hope this helps.
Best,
Dimitris
Dimitris Rizopoulos
Biosta
It does help if you read the help information for poly.
> ?poly
> x <- 1:3
> y <- c(1, 4, 9)
> f <- lm(y ~ poly(x, 2, raw = TRUE)) ## note raw = TRUE
> coef(f)
(Intercept) poly(x, 2, raw = TRUE)1 poly(x, 2, raw = TRUE)2
0 0 1
>
You were
Dear,
I need to analyse reaction time.
The general idea might be described as this: There are 8x8 circles. My
program light a circle at one time. The subject then click the lighted
circled as fast as possible. After the correct circle is clicked, then the
next circle will be lighted for the subje
On 07-Mar-08 08:16:06, Firas Swidan, PhD wrote:
> Hi,
> I can not comprehend the linear fitting results of polynoms.
> For example, given the following data (representing y = x^2):
>
>> x <- 1:3
>> y <- c(1, 4, 9)
>
> performing a linear fit
>
>> f <- lm(y ~ poly(x, 2))
>
> gives weird coeffici
Thanks for the clarifications. It seems the confusion resulted from
making one assumption more than necessary regarding the behavior of
poly().
Best wishes,
Firas.
On Fri, 2008-03-07 at 18:33 +1000, [EMAIL PROTECTED] wrote:
> It does help if you read the help information for poly.
>
> > ?poly
>
Hi,
I can not comprehend the linear fitting results of polynoms. For
example, given the following data (representing y = x^2):
> x <- 1:3
> y <- c(1, 4, 9)
performing a linear fit
> f <- lm(y ~ poly(x, 2))
gives weird coefficients:
> coefficients(f)
(Intercept) poly(x, 2)1 poly(x, 2)2
4.66
Nathanael,
On 7 Mar 2008, at 09:40, nathan3073 wrote:
>
> Dear,
> I need to analyse reaction time.
>
> The general idea might be described as this: There are 8x8 circles. My
> program light a circle at one time. The subject then click the lighted
> circled as fast as possible. After the correct c
I believe that the procedure you describe at the end (resampling the
cases) is the original interpretation of bagging, and that using weighting
is equivalent when a procedure uses case weights.
If you are getting different results when replicating cases and when using
weights then rpart is not
Hi, yes it's true Mathematica is too expensive (I think one of the most
expensive out there), but yacas has a lng way to go before becoming
mathematica. I agree mathematica is not really for data analysis but I think
it's to help researchers and inventors invent new ideas, understand concepts
Hi,
I'm no expert programmer at all;
I'm running an R script ("mariam1_2.R"). This scripts calls another
script, which contains an R function, which .Call some C code.
It runs several times without any problem, but sometimes I get the error:
---
*** caught segfault ***
address 0x1c404ec8, caus
Dear All,
I am using xyplot() with many groups like this:
statselect <- levels(dat$stat)
xyplot(relmse~T|lambda, groups=stat, data=dat,
panel = panel.superpose,
key=simpleKey(statselect, lines=T))
However, I want lines not scatterplots and if I set
panel.groups=panel.linejoin that
Dear Javier,
>From your description, it seems you are not the author of the C
function rrfunc, which is where the problem is probably located (the
segmentation fault is most likely a pointer related problem where the
C code is trying to do something with memory it should not be trying
to do). Depe
Thanks a lot, John, Gavin; Hadley and Greg, for your helpful comments
and suggestions. I finally achieved what I wanted using the suggested
method from Gavin with corrections from Greg.
Out of curiosity (and interest to learn): Hadley, how would you
simplify that code using lattice or ggplot and h
Today HTML help stopped working. The menu command "Help > Html help" usually
brings up my default web browser (Opera 9.26), but now R gives the error
Error: file association for 'doc\html\index.html' not available or invalid
If I try the same menu command a second time, R crashes with the message
Dear R Users:
We have splitted up the micEcon package into three packages:
a) Package "maxLik" provides tools for maximum likelihood estimations
(see http://www.maxLik.org).
b) Package "sampleSelection" provides tools for estimating Heckman-type sample
selection/generalized tobit models (see h
sms.ed.ac.uk> writes:
> I am using xyplot() with many groups like this:
>
> statselect <- levels(dat$stat)
> xyplot(relmse~T|lambda, groups=stat, data=dat,
>panel = panel.superpose,
>key=simpleKey(statselect, lines=T))
>
Add lty="l". And remove the panel=panel.superpose, it's
Hi,
Was wondering if it is possible to pass function name as a parameter, smth
along this line
param.to.pass<-c(1,'max','h')
dd<-function(dfd, param=param.to.pass,...){
ttime.int <- format(ttime,fmt)
data.frame(
param[3] = tapply(dfd[,param[1]],ttime.int,param[3]),
...)
}
1. I have used gdata::read.xls() with much happiness. But every now
and then it breaks. I have not, as yet, been able to construct a
mental model about the class of .xls files for which it works. Does
someone have a simple rule for predicting the circumstances under
which it will work?
On Fri, 7 Mar 2008, Jon Olav Vik wrote:
> Today HTML help stopped working. The menu command "Help > Html help" usually
> brings up my default web browser (Opera 9.26), but now R gives the error
> Error: file association for 'doc\html\index.html' not available or invalid
> If I try the same menu co
Duncan Murdoch wrote:
> On 06/03/2008 6:45 PM, Thomas Manke wrote:
>> Hi,
>>
>> I have a character-valued vector (old_names) and want to translate
>> its entries whenever possible, using a dictionary (dict=data.frame).
>> The translation direction is dict$V3 --> dict$V2, but
>> some values may be
Yes,
tapply(rnorm(100), gl(5,20), "max")
On 07/03/2008, Yuri Volchik <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> Was wondering if it is possible to pass function name as a parameter, smth
> along this line
>
> param.to.pass<-c(1,'max','h')
>
> dd<-function(dfd, param=param.to.pass,...){
> ttime
Dear Ramon,
I'm afraid I'm the author of the C function. Although I'm not a 'real'
programmer I need to do some programming in my research work.
As you say, I've used MAKEFLAGS="CFLAGS=-O1", and valgrind, with the
expression you've said.
The content of the log file contains three blocks of the typ
Hi
I created a complex layout with using layout() and it looks exactly as
I need it. But I don't want to print in the order in which the
subfigure are numbered, but in a different order.
How can I navigate in the layout so that I can specify the subfigure
in which to plot?
At the moment I am usi
Hi
I have a raster which I would like to plot in a greyscale instead of
colour. Is this possible, and how?
Thanks
Rainer
--
Rainer M. Krug, Dipl. Phys. (Germany), MSc Conservation Biology (UCT)
Plant Conservation Unit Department of Botany
University of Cape Town
Rondebosch 7701
South Africa
Hello!
I need some help, because I don't know how this error means: Error:
variables ‘Output1’, ‘Output2’, ‘Output3’, ‘Output4’, ‘Output5’ were
specified with different types from the fit
Execution halted
Can you help me?
Thank You
__
R-help@r-projec
Dear Javier,
On Fri, Mar 7, 2008 at 1:09 PM, <[EMAIL PROTECTED]> wrote:
> Dear Ramon,
> I'm afraid I'm the author of the C function. Although I'm not a 'real'
Oh, oh... that's too bad. There is not anyone else to blame, then :-).
> programmer I need to do some programming in my research wo
On Mar 7, 2008, at 7:18 AM, Rainer M Krug wrote:
> Hi
>
> I created a complex layout with using layout() and it looks exactly as
> I need it. But I don't want to print in the order in which the
> subfigure are numbered, but in a different order.
>
> How can I navigate in the layout so that I can s
On 07/03/2008, Charilaos Skiadas <[EMAIL PROTECTED]> wrote:
> On Mar 7, 2008, at 7:18 AM, Rainer M Krug wrote:
>
> > Hi
> >
> > I created a complex layout with using layout() and it looks exactly as
> > I need it. But I don't want to print in the order in which the
> > subfigure are numbered,
On Mar 7, 2008, at 2:17 AM, Oldrich Kruza wrote:
> Hello Soumyadeep,
>
> if you store the data in a tabular file, then I suggest using standard
> text-editing tools like cut (say your file is called data.csv, fields
> are separated with commas and you want to get rid of the third and
> sixth colum
> I need some help, because I don't know how this error means: Error:
> variables ?Output1?, ?Output2?, ?Output3?, ?Output4?, ?Output5? were
> specified with different types from the fit
> Execution halted
>
> Can you help me?
No!
Please read the posting guide at
http://www.R-project.org/posti
On Mar 7, 2008, at 8:02 AM, Rainer M Krug wrote:
> On 07/03/2008, Charilaos Skiadas <[EMAIL PROTECTED]> wrote:
>> On Mar 7, 2008, at 7:18 AM, Rainer M Krug wrote:
>>
>>> Hi
>>>
>>> I created a complex layout with using layout() and it looks
>>> exactly as
>>> I need it. But I don't want to print
On 3/7/08, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
> > Today HTML help stopped working. The menu command "Help > Html help"
> usually
> > brings up my default web browser (Opera 9.26), but now R gives the error
> > Error: file association for 'doc\html\index.html' not available or invalid
On Fri, 7 Mar 2008, Prof Brian Ripley wrote:
> I believe that the procedure you describe at the end (resampling the cases)
> is the original interpretation of bagging, and that using weighting is
> equivalent when a procedure uses case weights.
>
> If you are getting different results when repl
> Was wondering if it is possible to pass function name as a parameter
Yes. This isn't exactly what you wanted, but it demonstrates the
principle.
x = rnorm(5)
[1] -0.6510448 0.4591730 1.3225205 1.2314391 -0.0888139
myfun <- function(fname, x) eval(parse(text=paste(fname,"(x)",sep="")))
myfu
On 07/03/2008, Charilaos Skiadas <[EMAIL PROTECTED]> wrote:
> On Mar 7, 2008, at 8:02 AM, Rainer M Krug wrote:
>
> > On 07/03/2008, Charilaos Skiadas <[EMAIL PROTECTED]> wrote:
> >> On Mar 7, 2008, at 7:18 AM, Rainer M Krug wrote:
> >>
> >>> Hi
> >>>
> >>> I created a complex layout with usin
Dear Rers,
I understand that I can include R-code in LaTeX using Sweave. Is there a way
to do it the other way round? Particularly, I need some TeX symbols in the
legend of an R-plot. This can be done in matlab easily, so I am optimistic
with R. Any suggestions for a command or package?
Best,
Mari
On Mar 7, 2008, at 8:41 AM, Rainer M Krug wrote:
>>
>> I'm not sure I understand it, why don't you want to just number the
>> subfigures in the order in which you will draw them?
>
> Because i thought it would be easier the other way round? Thanks
> anyway
Yes, I agree it should not be as har
Hi
Martin Kaffanke <[EMAIL PROTECTED]> napsal dne 06.03.2008 18:05:03:
>
> Am Donnerstag, den 06.03.2008, 17:47 +0100 schrieb Petr PIKAL:
> > Hi
> >
> > [EMAIL PROTECTED] napsal dne 06.03.2008 17:41:06:
> >
> > > Hi there,
> > >
> > > When i do
> > >
> > > mean(fl[1:20], na.rm=T)
> >
> > e
> >> I need some help, because I don't know how this error means: Error:
> >> variables ?Output1?, ?Output2?, ?Output3?, ?Output4?, ?Output5? were
> >> specified with different types from the fit
> >> Execution halted
> I have two tables and I join them like this: tabela <-
> data.frame(dados.i
Mario Maiworm wrote:
> Dear Rers,
> I understand that I can include R-code in LaTeX using Sweave. Is there a way
> to do it the other way round? Particularly, I need some TeX symbols in the
> legend of an R-plot. This can be done in matlab easily, so I am optimistic
> with R. Any suggestions for
Carla Rebelo wrote:
> Hello!
>
> I need some help, because I don't know how this error means: Error:
> variables ‘Output1’, ‘Output2’, ‘Output3’, ‘Output4’, ‘Output5’ were
> specified with different types from the fit
Please read the posting guide!
Please tell us at least what produced the er
Rainer M Krug wrote:
> Hi
>
> I have a raster which I would like to plot in a greyscale instead of
> colour. Is this possible, and how?
Probably, and we may tell you how given you tell us what "raster" means:
> ?raster
No documentation for 'raster' in specified packages and libraries:
you cou
Andrew,
You haven't defined the problem clearly, but I will try to guess what
you mean...
> i.e. what properties are within 3 km from properties infected on
> "2008-01-01" over the last 14 days.
I'm guessing that each row of your sample data represents one
infection event, and you want to find
Andrew McFadden maf.govt.nz> writes:
>
>
> Hi all
>
> The code of trying to write relates to selecting properties (given by x
> and y co-ordinates) spatially (distance "X" from "infected" properties
> identified by date) over a certain time period.
>
> i.e. what properties are within 3 km fro
Dear Experts,
I am trying to do a time-stratified case-crossover analysis on air
pollution data and number of myocardial infarctions. In order to avoid
model selection bias, I started with a simple simulation.
I'm still not sure if my simulation is right. But the results I get from
the "ts
Dear Mag. Ferri Leberl,
I'm using something like:
--- tex.tex ---
\documentclass{article}
\usepackage{graphicx}
\usepackage{fancyvrb}
\newcommand{\Rlogo}{\protect\includegraphics[height=1.8ex,keepaspectratio]{Rlogo.pdf}}
\newcommand{\myinput}[1] {\begin
Dear Experts,
I am trying to do a time-stratified case-crossover analysis on air pollution
data and number of myocardial infarctions. In order to avoid model selection
bias, I started with a simple simulation.
I'm still not sure if my simulation is right. But the results I get from the
"ts-ca
Thank you, uwe and jeremy. I was actually looking exactly for that! But
something still doesn't work:
I want to plot a symbol in a legend of a plot, lets say "\sigma = 2". 2
should be the value of a variable. So, when I try
mySigma=2;plot(1:10,dnorm(1:10,sd=mySigma),type='l')
legend(x="topright",
Mario Maiworm wrote:
> Thank you, uwe and jeremy. I was actually looking exactly for that! But
> something still doesn't work:
> I want to plot a symbol in a legend of a plot, lets say "\sigma = 2". 2
> should be the value of a variable. So, when I try
>
> mySigma=2;plot(1:10,dnorm(1:10,sd=mySi
Jean,
this is nice, but 1) the logo is a bitmap, it is ugly if you
resize it, 2) you don't need a pdf version for pdflatex, it
handles jpg (and maybe also png as well), so you can
just use the logos at the R developer site.
It would be really nice to have a non-bitmap version, though.
If it e
Hi,
Thakns all for your help
I am doing the next in my dataframe tabla, column pend1, because the
Lilliefors (Kolmogorov-Smirnov) test give me a pvalue < alfa. (data no
normal distribution). I need do a transformation with box-cox or
other:
> bc <- boxcox.fit(tabla$pend1)
R send to me:
Error in
Hi, netters,
This is probably a rookie question but I couldn't find the answer after hours
of searching and trying.
Suppose there'a a dataframe M:
x y
10 A
13 B
8 A
11 A
I want to locate the rows where x >=10 and y="A". I know how to do it to
vectors by using
which, but how t
On Mar 7, 2008, at 10:50 AM, zhihuali wrote:
>
> Hi, netters,
>
> This is probably a rookie question but I couldn't find the answer
> after hours of searching and trying.
>
> Suppose there'a a dataframe M:
>
> x y
> 10 A
> 13 B
> 8 A
> 11 A
>
> I want to locate the rows where x >=
Finally, this should work for an array of sigmas. I just realized that the
substitute()-command is not evaluated within a c()-environment :(
mySigma[1] <- 2; mySigma[2] <- 3;
plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l') ;
lines(dnorm(1:10,sd = mySigma[2]),lty = 2);
legend(x = "topright", l
Hi Zhihua,
M <- data.frame (x=c(10, 13, 8, 11), y=c('A', 'B', 'A', 'A'))
which (M$x >= 10 & M$y == 'A')
# [1] 1 4
Hope it helps,
Nael
2008/3/7 N. Lapidus <[EMAIL PROTECTED]>:
> Hi Zhihua,
>
> M <- data.frame (x=c(10, 13, 8, 11), y=c('A', 'B', 'A', 'A'))
> which (M$x >= 10 & M$y == 'A')
> # [1]
Try:
with(m, which((x >= 10 & y == "A")))
On 07/03/2008, zhihuali <[EMAIL PROTECTED]> wrote:
>
> Hi, netters,
>
> This is probably a rookie question but I couldn't find the answer after
> hours of searching and trying.
>
> Suppose there'a a dataframe M:
>
> x y
> 10 A
> 13 B
> 8
You might want to read
Ligges, U. (2002): R Help Desk: Automation of Mathematical Annotation
in Plots. R News 2 (3), 32-34.
with an example at the end that meets your requirements:
(please note that I removed those ugly ";"
mySigma[1] <- 2
mySigma[2] <- 3
plot(1:10, dnorm(1:10, sd = mySigma[1
Hello,
I am looking for advice on a task I am trying to complete.
I have a 4 column dataframe defining the start and end coordinates of
line edges (from a CGAL alpha shapes function to define concave hulls
from point clusters). I would like to create polygon shapefiles from
these line edges,
Or my personal favorite if the length of mySigma is variable:
mySigma <- 2:3
plot(1:10, dnorm(1:10, sd = mySigma[1]), type = 'l')
lines(dnorm(1:10,sd = mySigma[2]),lty = 2)
leg <- as.expression(lapply(mySigma, function(x) bquote(sigma == .(x
legend(x = "topright", lty = c(1,2),legend = leg)
T
Hi,
Can anyone explain the following error??
Error in FUN(newX[, i], ...) : missing observations in cov/cor
In addition: Warning message:
In FUN(newX[, i], ...) : NAs introduced by coercion
svm_modelSAheart1 <- svm(x_training, y_training) is the command i am
using.my x/y training are working fi
I would like to thank Brian Ripley and Torsten Hothorn for their quick and
thoughtful responses.
I rerun the example given by Professor Ripley by just starting R and
sourcing the code below and I got slightly different results. Then I ran
it again setting the random seed before the sample command
Gabor,
> this is nice, but
> 1) the logo is a bitmap, it is ugly if you resize it
Sure, it's a bitmap, but the naked eye resolution is only
100 $\mu$m so that a vectorized solution is overkilling
in most common situations IMHO. I have to zoom by a factor
1200% to see some pixellization problems
I finally got everything plotted as I had planned. So, thank you again.
Sundars suggestion looks more flexible when it comes to large numbers of
categories (i.e. sigmas). In the plot I will be using, there are 3 sigmas...
I am keen to learn about that bquote and substitute stuff. I will definitely
After some struggling with the data format, non-standard in
BioConductor, I have gotten cor.balance in package CORREP to work. My
desire was to obtain maximum-likelihood p-values from the same data
object using cor.LRtest, but it appears that this function wants
something different, which I can
I have a set of data that consists of a number of biological measurements.
The columns are Time that runs from 01/01/2005 to 01/5/2007, Group which has
23 levels and postcode which is nested within group. This is a balanced panel
but the number of postcodes differs within groups, from 15 to
Jean,
On Fri, Mar 07, 2008 at 06:09:35PM +0100, Jean lobry wrote:
> Gabor,
>
> > this is nice, but
> > 1) the logo is a bitmap, it is ugly if you resize it
>
> Sure, it's a bitmap, but the naked eye resolution is only
> 100 $\mu$m so that a vectorized solution is overkilling
> in most common sit
Hi
I have been trawling the web, FAQs, and R manuals for help on the following
issue, but have failed and was wondering if anyone has a solution to the
following problem:
After having installed R 2.6.2 for Windows (binary), I tried to install various
packages. Every time I try loading a packa
Friends,
I changed one line of a package at the source level and then rebuilt it.
When I run R CMD check, I get an error:
installing R.css in C:/polsplineRS.Rcheck
-- Making package polsplineRS
adding build stamp to DESCRIPTION
making DLL ...
making hareall.d from harea
On 3/7/2008 1:21 PM, Shewcraft, Ryan wrote:
> Friends,
>
> I changed one line of a package at the source level and then rebuilt it.
> When I run R CMD check, I get an error:
You're getting lots of warnings from the compiler; presumably those have
nothing to do with you. Then at the end you're g
Dear UseRs,
I'm curious about the derivative of n!.
We know that Gamma(n+1)=n! So when on takes the derivative of
Gamma(n+1) we get Int(ln(x)*exp(-x)*x^n,x=0..Inf).
I've tried code like
> integrand<-function(x) {log(x)*exp(x)*x^n}
> integrate(integrand,lower=0,upper=Inf)
It seems that R doesn
On Fri, 7 Mar 2008, Max wrote:
> Dear UseRs,
>
> I'm curious about the derivative of n!.
>
> We know that Gamma(n+1)=n! So when on takes the derivative of
> Gamma(n+1) we get Int(ln(x)*exp(-x)*x^n,x=0..Inf).
>
> I've tried code like
>
>> integrand<-function(x) {log(x)*exp(x)*x^n}
>> integrate(inte
Thanks! That's seemed to partially work, but now when I try to load it
in R I get:
library(polsplineRS)
Error in library.dynam("polspline", pkg, lib) :
shared library 'polspline' not found
Error in library(polsplineRS) : .First.lib failed for 'polsplineRS'
-Original Message-
Fr
DateTimeRM61
11/30/2006 12:31NA
11/30/2006 12:46NA
11/30/2006 13:012646784125
11/30/2006 13:16NA
11/30/2006 13:31NA
11/30/2006 13:46NA
11/30/2006 14:012666435177
11/30/2006 14:16NA
11/30/2006 14:31NA
11/30/2006 14:46
Hi Max,
The analytic integral \int _0 ^\Inf exp(-t) t^n log(t) might not converge
because the integrand tends to -Inf as t -> 0.
So, here is a numerical approach to estimating the derivative of the gamma
function:
library(numDeriv)
fx <- function(x, n) exp(-x) * x^n
gf <- function(n) {integrat
Prof Brian Ripley formulated on Friday :
> On Fri, 7 Mar 2008, Max wrote:
>
>> Dear UseRs,
>>
>> I'm curious about the derivative of n!.
>>
>> We know that Gamma(n+1)=n! So when on takes the derivative of
>> Gamma(n+1) we get Int(ln(x)*exp(-x)*x^n,x=0..Inf).
>>
>> I've tried code like
>>
>>> in
On 3/7/2008 1:53 PM, Shewcraft, Ryan wrote:
> Thanks! That's seemed to partially work, but now when I try to load it
> in R I get:
>
> library(polsplineRS)
> Error in library.dynam("polspline", pkg, lib) :
> shared library 'polspline' not found
> Error in library(polsplineRS) : .First.li
Hi max,
Prof. Ripley is right. Your problem is that you missed a (-) sign in the
exponential. Here is a demonstration showing the agreement between
numerical and analytical results:
gx <- function(x, n) exp(-x) * x^n * log(x)
df <- function(n) {integrate(gx, lower=0, upper=Inf, n=n)$val}
lib
Thank you, thats really good and gives me very good information.
Thanks,
Martin
Am Donnerstag, den 06.03.2008, 14:35 -0500 schrieb Chuck Cleland:
> On 3/6/2008 2:07 PM, Martin Kaffanke wrote:
> > Am Mittwoch, den 05.03.2008, 14:38 -0300 schrieb Henrique Dallazuanna:
> >> Try this:
> >>
> >> On 05
Hi,
I get the following error when I try to predict the probabilities of a
test sample:
Error in predict.randomForest(fit.EBA.OM.rf.50, x.OM, type = "prob") :
New factor levels not present in the training data
I have about 630 predictor variables in the dataset x.OM (25 factor
variables and th
1. This is not reproducible.
Lines was not provided in reproducible form.
Please look at my prior emails and use that
form so that one can copy from your post and
paste it directly into R and observe the error.
2. What do you mean by does not plot? Do you get
an error or does nothing appear? If
bayesm version 2.2-0 is now available on CRAN.
Major changes include:
1. general density estimation using a Dirichlet Process Prior and a
normal base
2. linear instrumental variable models with unknown error distributions
(the Bayesian analogue of IV methods). Achieved via DP priors.
peter r
Dear All,
I have a function f(x,y,z)=exp(x^3+y^4+x^2*y+x*z^2+y/z) over D, where is D={
(x,y,z)| 0 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, reproducible code.
Also, see the nearZeroVar function in the caret package.
MAx
On Fri, Mar 7, 2008 at 7:41 AM, Charilaos Skiadas <[EMAIL PROTECTED]> wrote:
>
> On Mar 7, 2008, at 2:17 AM, Oldrich Kruza wrote:
>
> > Hello Soumyadeep,
> >
> > if you store the data in a tabular file, then I suggest using standard
On Fri, 7 Mar 2008, Max wrote:
> Prof Brian Ripley formulated on Friday :
>> On Fri, 7 Mar 2008, Max wrote:
>>
>>> Dear UseRs,
>>>
>>> I'm curious about the derivative of n!.
>>>
>>> We know that Gamma(n+1)=n! So when on takes the derivative of
>>> Gamma(n+1) we get Int(ln(x)*exp(-x)*x^n,x=0..Inf)
Is there a way to combine two columns within a data frame?
Example data:
id snp AL1 AL2
150030 A B
151030 A A
152030 A B
This is what I would like:
indvsnp AL1AL2
150030 AB
151030 AA
152030 AB
Any help
Try:
transform(x, AL1Al2 = paste(AL1, AL2, sep=''))[-c(3:4)]
On 07/03/2008, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> Is there a way to combine two columns within a data frame?
>
> Example data:
>
> id snp AL1 AL2
> 150030 A B
> 151030 A A
> 15
Or perhaps:
myfun <- function(fname, ...)match.fun(fname)(...)
On 07/03/2008, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> > Was wondering if it is possible to pass function name as a parameter
>
> Yes. This isn't exactly what you wanted, but it demonstrates the
> principle.
>
> x = rnorm(5)
Depending on your purpose you might want to look
at ?interaction or its synonym : (i.e. a colon)
On Fri, Mar 7, 2008 at 4:09 PM, <[EMAIL PROTECTED]> wrote:
> Is there a way to combine two columns within a data frame?
>
> Example data:
>
> id snp AL1 AL2
> 150030 A B
>
Dale,
Other than the first SAS contrast, does the following demonstrate what
your asking for?
> summary(twoway)
material temp voltage
1:12 50:12 Min. : 20
2:12 65:12 1st Qu.: 70
3:12 80:12 Median :108
Mean :106
3rd Qu.
Dear list,
I just made a very simple mistake, but it was hard to spot. And I
think that I should warn other people, because it is probably so
simple to make...
=== R code ===
# Let us create a matrix:
(a <- cbind(c(0,1,1), rep(1,3)))
# [,1] [,2]
# [1,]01
# [2,]11
# [3,]
Hello,
I have an irregular time series of the form :
Time Data Time1 Data1
1 b1 e
7 g 4i
NA NA 5 k
NA NA NA NA ...
(the columns have varying length of NAs after a certain point)
Converting this to regular time se
and you might want to check ?prop.table
> prop.table(a, 2)
[,1] [,2]
[1,] 0.0 0.333
[2,] 0.5 0.333
[3,] 0.5 0.333
or even ?sweep (which will be useful for more complex situations)
> sweep(a, 2, colSums(a), "/")
[,1] [,2]
[1,] 0.0 0.333
[2,] 0.5 0.333
Hi all:
I have always used SPSS to estimate weekly
covariance based on a linear regression model
but have to hard code the model Std. Error and the
Mean-Square and then execute
one week a the time. I was wondering if someone
could give me an idea on how to estimate
weekly(WK) covariance usi
R is working exactly as documented. If you look at how the matrix is
stored (column-wise) and then what you are dividing by, you will see
that it is doing what you asked (when recycling values):
> as.vector(a)
[1] 0 1 1 1 1 1
> as.vector(a)/c(2,3)
[1] 0.000 0.333 0.500 0.333 0.50
Alexey Shipunov wrote:
> Dear list,
>
> I just made a very simple mistake, but it was hard to spot. And I
> think that I should warn other people, because it is probably so
> simple to make...
>
> === R code ===
>
> # Let us create a matrix:
> (a <- cbind(c(0,1,1), rep(1,3)))
>
> # [,1] [
Dear all,
Thank you very much for all your valuable suggestions!
Best wishes,
Alexey Shipunov
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