It would, I consider, be useful to note, on the help pages for Sweave
and RweaveLatex, that the path to Sweave.sty in the R installation can
be extracted as paste(R.home(),"share/texmf/", sep="/") or its
equivalent.
John Maindonald email: john.maindon...@anu.edu.au
phone : +61 2
Thanks Hadley: I had already gone to your website and stat="identity" is what I
needed.
--- On Fri, 12/26/08, hadley wickham wrote:
> From: hadley wickham
> Subject: Re: [R] ggplot2 Xlim
> To: mazatlanmex...@yahoo.com
> Cc: r-help@r-project.org, "Wayne F"
> Date: Friday, December 26, 2008, 3
Hi Felipe,
It sounds like ForkLength is a factor - what deos str(FL) tell you?
You might also need geom_bar(..., stat = "identity") since your data
are pretabulated.
Hadley
On Fri, Dec 26, 2008 at 2:44 PM, Felipe Carrillo
wrote:
> Wayne:
> What's crowded are my x axis labels. The bars look fine
Wayne:
What's crowded are my x axis labels. The bars look fine on my graph but the
labels are being displayed from 29 to 170 one by one. I need something like a
seq(29,170 by:10) or something like that. I don't want to treat my FL as factor
because I don't want a bar per each FL value, I only wa
>In my limited experience (we have some insurance projets), 100% can occur,
>but otherwise a beta distbribution may suit, which suggests a mixture
>distribution. But start with an empirical examination (histogram, ecdf,
>density plot) of the distribution, since it may reveal other features.
G
Try variations of this:
library(leaps)
b<-regsubsets(Fertility~.,data=swiss)
w <- summary(b)$which
lapply(1:nrow(w), function(i) coef(lm(Fertility ~., swiss[w[i, ]])))
On Fri, Dec 26, 2008 at 1:57 PM, Murtaza Das wrote:
> Thanks for replying Gabor.
>
> I checked the leaps() function and i think
Hello again,
thank you very much for your help so far.
To be more specific, I generate a simplified data set that is similar to
my real world data:
set.seed( 123 )
data <- data.frame( x = runif( 200 ), y = NA )
for( i in 1:200 ){
data$y[ i ] <- rweibull( 1, 1, 70 + 10 * data$x[ i ] ) - 30
}
d
Fantastic - thanks for all the helpful replies.
Best
Rory
On Fri, Dec 26, 2008 at 6:51 PM, Matthias Kohl wrote:
> indeed one could use package distr ...
>
> library(distr)
> X <- Unif(Min = 0, Max = 1)
> Y <- convpow(X, 10)
> p(Y)(6) - p(Y)(4)
>
> Best
> Matthias
>
> Prof Brian Ripley wrote:
>
>
Hi All,
I am currently using Rserve to call R as a service and execute R
scripts. I find problems cropping up because I write my scripts in R
(2.8.0) on Windows and communicated with Rserve on Ubuntu to execute
them.
Rserve running on the Ubuntu machine has R-base 2.6.0
Many a times, when i encoun
Dear R community and Kurt,
I am using consensus.r from clue to compute hard consensus partitions
(method: "hard/euclidean"; same considerations apply for "hard/manhattan").
The function gives in output a hard consensus partition that minimizes a
criterion measure Crit, namely the sum of the w
Thanks for replying Gabor.
I checked the leaps() function and i think it is intended to find the
best combination of predictors in the linear model.
Does leaps have a way to combine different factor columns in my data
frame as follows :
I have the regression model fixed. The combination of predic
indeed one could use package distr ...
library(distr)
X <- Unif(Min = 0, Max = 1)
Y <- convpow(X, 10)
p(Y)(6) - p(Y)(4)
Best
Matthias
Prof Brian Ripley wrote:
Look at packages distr* : they can do your example and might do what
your real applications.
On Fri, 26 Dec 2008, Rory Winston wrote:
See the leaps package.
On Fri, Dec 26, 2008 at 12:37 PM, Murtaza Das wrote:
> Hi,
>
> I am trying to find an efficient way of applying a linear regression
> model to different factor combinations in a data frame.
> I want to obtain the output with minimal or no use of loops if
> possible. Please
Hi,
I am trying to find an efficient way of applying a linear regression
model to different factor combinations in a data frame.
I want to obtain the output with minimal or no use of loops if
possible. Please let me know if this query is unclear.
Thanks,
Murtaza
*
Nidhi,
Presumably, you are trying to simulate 20 items all sharing one
general factor but having some error.
The model as you specified it has no error. Thus all the
correlations will be 1 and the factors will not make any sense.
Most items have loadings on a general factor of the orde
On Fri, 26 Dec 2008, Bert Gunter wrote:
Thankyou for the clarification, Brian. This is very helpful (as usual).
However, I think the important point, which I misstated, is that whether it
be for() or, e.g. lapply(), the "loop" contents must be evaluated at the
interpreted R level, and this is w
Thankyou for the clarification, Brian. This is very helpful (as usual).
However, I think the important point, which I misstated, is that whether it
be for() or, e.g. lapply(), the "loop" contents must be evaluated at the
interpreted R level, and this is where most time is typically spent. To get
t
Look at packages distr* : they can do your example and might do what your
real applications.
On Fri, 26 Dec 2008, Rory Winston wrote:
Hi
Firstly , happy Christmas to R-Help! Secondly, I wonder if anyone can help
me with the following query: I am trying to reproduce some explicit
probability c
Actually the last line could be simplified to just:
> s > 4 & s < 6
mean sd sims
[1] 0.72 0.45 2500
On Fri, Dec 26, 2008 at 8:33 AM, Gabor Grothendieck
wrote:
> Try a simulation approach. vignette("rv") for more info.
>
>> set.seed(1)
>> library(rv)
>> x <- rvunif(10)
>> s <- simapply(x,
Try a simulation approach. vignette("rv") for more info.
> set.seed(1)
> library(rv)
> x <- rvunif(10)
> s <- simapply(x, sum)
> mean(s > 4 & s < 6)
mean sd 1% 2.5% 25% 50% 75% 97.5% 99% sims
[1] 0.72 0.45 00 0 1 1 1 1 2500
On Fri, Dec 26, 2008 at 7:42 AM, Rory Winston
I am trying to simulate a dataset using Parallel Latent CTT model and this is
what i have done so far:
(START)
#Importing psych library for all the simulation related functions
library(psych)
# Settting the working directory path to C:/NCME
path="C:/NCME"
setwd(path)
#Using the function to g
Hi
Firstly , happy Christmas to R-Help! Secondly, I wonder if anyone can help
me with the following query: I am trying to reproduce some explicit
probability calculations performed in APPL (a Maple extension for
computational probability). For instance, in APPL, to compute the
probability that the
austernkommunikat...@googlemail.com wrote:
hello,
i am using the hist function with classified values. The class breaks are >1, so
histogram$density is != 1.
How to plot the histogram with freq=FALSE and the real class density values.
I used:
h2 = hist(value, breaks = breaks_vector)
h2$den
hello,
i am using the hist function with classified values. The class breaks are >1,
so
histogram$density is != 1.
How to plot the histogram with freq=FALSE and the real class density values.
I used:
> h2 = hist(value, breaks = breaks_vector)
> h2$density = round(h2$counts/sum(h2$counts), 2)
Prof Brian Ripley wrote:
On Thu, 25 Dec 2008, Oliver Bandel wrote:
The apply-functions do bring speed-advantages.
This is not only what I read about it,
I have used the apply-functions and really got
results faster.
The reason is simple: an apply-function does
make in C, what otherwise
I recently upgraded from 2.8.0 to 2.8.1 by first installing the 2.8.1 version
then copying the binaries and the library to the 2.8.0 folder. Now Tinn-R will
not start up. I just see that start up splash screen for a long period of time.
It seems fozen to me. Any guesses on what I did wrong in th
Not an R question as yet .
In my limited experience (we have some insurance projets), 100% can occur,
but otherwise a beta distbribution may suit, which suggests a mixture
distribution. But start with an empirical examination (histogram, ecdf,
density plot) of the distribution, since it m
On Thu, 25 Dec 2008, Oliver Bandel wrote:
Bert Gunter gene.com> writes:
FWIW:
Good advice below! -- after all, the first rule of optimizing code is:
Don't!
For the record (yet again), the apply() family of functions (and their
packaged derivatives, of course) are "merely" vary carefully wr
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