Hi,
we noticed there was a error in the "arules" package.
After reading the source code, we saw that the Dice similarity index was
"miscalculated" in "dissimilarity" function : an copy-paste from Jaccard
Index was not corrected (2* a_b_c, ie 2*(a+b+c) in the code instead of
2*a +b + c !!!).
A
Am Montag, 22. Januar 2007 12:33 schrieb Matthieu Mourroux:
> Hello,
>
> I'd like to know if the D'Agostino test of normality is reliable,
The test is not consistent. The test statistic
can be used for testing the hypothesis of uniformity.
See the paper
Baringhaus, L.; Henze, N.
A test for u
On Mon, 22 Jan 2007, Louisell, Paul wrote:
> Hello,
>
> Does anyone know of an R version of loess that allows more than 4
> predictors and/or allows the specification of offsets? For that matter,
> does anyone know of _any_ version of loess that does either of the
> things I mention?
Why would yo
On Mon, 22 Jan 2007, Paul Smith wrote:
> Dear All
>
> I would like to use rpart to obtain a regression tree for a dataset
> like the following:
>
> Y X1 X2 X3 X4
> 5.500033 B A 3 2
> 0.35625148D B 6 5
> 0.8062546 E C 4
> TeamInfo
TEAMNAME LEVEL WORKTIME BONUS
1 batch sunan B 135 9,818
2 batch Chenqi E 121 6,050
3 batch jiangxu F 97 4,189
4 online zhouxi F 63 2,720
5 online chenhe H 36 1,064
## try:
> factor(TeamInfo$TEAM)
[1] batch batch batch o
Usually (that is, not limited in R language), when error occurs in
try, stacks are rollbacked, so the variables defined in try no longer
exists after calling try.
One non-elegant solution is:
fit<-NULL
try ( (fit = lm(y~x, data = data_fitting)), silent =T)
if(!is.null(fit)){
coeffs = as.list(coef
I have been using the wonderful xtable package lately, in combination
with Sweave, and I have a couple of general questions along with a
more particular one.
I'll start with the particular question. I basically have a 1x3 array
with column names but no row names. I want to create a latex tab
Hello,
Does anyone know of an R version of loess that allows more than 4
predictors and/or allows the specification of offsets? For that matter,
does anyone know of _any_ version of loess that does either of the
things I mention?
Thanks,
Paul Louisell
650-833-6254
[EMAIL PROTECTED]
Research Asso
why not use lda{MASS} and it has cv=T option; it does "loo", though.
Or use randomForest.
if you have to use lrm, then the following code might help:
n.fold <- 5 # 5-fold cv
n.sample <- 50 # assumed 50 samples
s <- sample(1:n.fold, size=n.sample, replace=T)
for (i in 1:n.fold){
# create your tr
Benjamin Tyner said the following on 1/22/2007 3:18 PM:
> Hi,
>
> Say I have
>
> z<-data.frame(y=runif(190),
>x=runif(190),
>f=gl(5,38),
>g=gl(19,10))
>
> plot<-xyplot(y~x|g,
> data=z,
>
So, I take it, given that the use of a pipe is suggested for
sequential reading, that the standard approach to processing a data
frame is to load the entire file? Please correct if wrong.
BTW, I am not interested in finding direct translations of SAS data
step statements to R, but instead in f
Dear All
I would like to use rpart to obtain a regression tree for a dataset
like the following:
Y X1 X2 X3 X4
5.500033B A 3 2
0.35625148 D B 6 5
0.8062546 E C 4 3
5.100014C A 3
Hi,
Say I have
z<-data.frame(y=runif(190),
x=runif(190),
f=gl(5,38),
g=gl(19,10))
plot<-xyplot(y~x|g,
data=z,
layout=c(5,4),
groups=f,
auto.key
One option for processing very large files with R is split:
## split a large file into pieces
#--parameters: the folder, file and number of parts
FLD=/home/user/data
F=very_large_file.dat
parts=50
#---split
cd $FLD
fn=`echo $F | awk -F\. '{print $1}'` #file name without extension
Hi Lalitha,
Use
try()
or
tryCatch()
Cheers
Andrew
On Mon, Jan 22, 2007 at 12:43:28PM -0800, lalitha viswanath wrote:
> Hi
> I am a newbie to R and am using the lm function to
> fit my data.
> This optimization is to be performed for around 45000
> files not all of which lend themselves to
Hi
Thanks for your response.
However I seem to be doing something wrong regarding
the try block resulting in yet another error described
below.
I have a function that takes in a file name and
does the fit for the data in that file.
Hence based on your input, I tried
try ( (fit = lm(y~x, data = da
This is answered in the FAQ list.
-thomas
On Mon, 22 Jan 2007, lalitha viswanath wrote:
> Hi
> I am a newbie to R and am using the lm function to
> fit my data.
> This optimization is to be performed for around 45000
> files not all of which lend themselves to
> optimization. Some of t
I am trying to implement a simple r-svm example using the iris data (only two
of the classes are taken and data is within the code). I am running into some
errors. I am not an expert on svm's. If any one has used it, I would appreciate
their help. I am appending the code below.
Thanks../Murli
Hi
I am a newbie to R and am using the lm function to
fit my data.
This optimization is to be performed for around 45000
files not all of which lend themselves to
optimization. Some of these will and do crash.
However, How do I ensure that the program simply goes
to the next file in line without
In the two solutions for the repeated measures problem given in the
original reply below, the F and p values given by aov() with the error
strata defined by Error() are different from those given by lme().
However, when one does the problem "by hand" using the standard split
plot model, the results
Hi,
I am new to R (and not really a stats expert) and am having trouble
interpreting its output. I am running a human learning experiment, with
6 scores per subject in both the pretest and the posttest. I believe I
have fitted the correct model for my data- a mixed-effects design, with
subject
Nils Hoeller wrote:
>
> for example.
>
> BUT what can I do for dynamic m and sd?
> I want something like integrate(dnorm(,0.6,0.15),0,1), with the first
> dnorm parameter open for the
> integration but fixed m and sd.
integrate(function(x)dnorm(x,0.1,1.2), 0, 1)
is a way of "fixing" additional
On Mon, 22 Jan 2007, Lynette wrote:
> Dear all, especially to Thomas,
>
> I have figured out the problem. For the step function, something wrong with
> my C codes. I should use the expression ((x>=0.25)&&(x<=0.75)) ? 2:1 instead
> of ((x>=1/4)&&(x<=3/4)) ? 2:1 ). Have no idea why 0.25 makes diff
Dear all, especially to Thomas,
I have figured out the problem. For the step function, something wrong with
my C codes. I should use the expression ((x>=0.25)&&(x<=0.75)) ? 2:1 instead
of ((x>=1/4)&&(x<=3/4)) ? 2:1 ). Have no idea why 0.25 makes difference from
1/4 in C. But now I can go ahead
This is to submit a commented example function for use in the data
argument to the bigglm(biglm) function, when you want to read the data
from a file (instead of a URL), or rescale or modify the data before
fitting the model. In the hope that this may be of help to someone out
there.
make.data <
On Mon, 22 Jan 2007, Lynette wrote:
> Well,
>
> I have no idea either. I can get correct answers for continous functions but
> incorrect for step functions.
>
I have just tried using Rdqags from C for the function x>0 and it worked
fine (once I had declared all the arguments correctly). The code
On Mon, 22 Jan 2007, Markus Schmidberger wrote:
> Hello,
> thanks for help and code.
> We did a lot of work to speedup our function in R. We have a nested loop,
> vectorizing is the fastest way. But there we got a very big matrix and
> problems with memory. So we want to stay by loops an speedup
On Jan 22, 2007, at 10:39 AM, John Fox wrote:
> One thing that seems particularly
> striking in your results is the large difference between elapsed
> time and
> user CPU time, making me wonder what else was going on when you ran
> these
> examples.
Yes, indeed there were a lot of other thin
Dear R-user,
I am trying to use the R "nlme" function to fit a non linear mixed
effects model. The method has some problem to reach the convergence.
I am trying to understand causes of the problem by following step by
step evolution of the iterative algorithm (verbose=TRUE command).
However, I
Well,
I have no idea either. I can get correct answers for continous functions but
incorrect for step functions.
Sign, I have been trying to realize the integration in C for long time.
Thank you for your answering.
Best,
Lynette
- Original Message -
From: "Thomas Lumley" <[EMAIL PROT
Hello,
thanks for help and code.
We did a lot of work to speedup our function in R. We have a nested
loop, vectorizing is the fastest way. But there we got a very big matrix
and problems with memory. So we want to stay by loops an speedup with C.
My code is similar to this. (my_c is code from Br
Dear Brian,
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Prof
> Brian Ripley
> Sent: Monday, January 22, 2007 11:06 AM
> To: Charilaos Skiadas
> Cc: John Fox; r-help@stat.math.ethz.ch
> Subject: Re: [R] efficient code. how to reduce running tim
Thank you Alain and Max for your swift responses.
It might be that I'm misunderstanding your responses, but aren't
you testing if there is a difference between the two full models?
What I want to know, os whether the effect of a specific predictor
(x) differs between model1 and model2. I'm n
Dear R experts
I am looking for a package which gives me latin hyper cube samples
from the grid of values produced from the command "expand.grid". Any
pointers to this issue might be very useful. Basically, I am doing the
following:
> a<-(1:10)
> b<-(20:30)
> dataGrid<-expand.grid(a,b)
Now, is t
Dear Haris,
My timings were on a 3 GHz Pentium 4 system with 1 GB of memory running Win
XP SP2 and R 2.4.1.
I'm no expert on these matters, and I wouldn't have been surprised by
qualitatively different results on different systems, but this difference is
larger than I would have expected. One thi
On Mon, 22 Jan 2007, Charilaos Skiadas wrote:
> On Jan 21, 2007, at 8:11 PM, John Fox wrote:
>
>> Dear Haris,
>>
>> Using lapply() et al. may produce cleaner code, but it won't
>> necessarily
>> speed up a computation. For example:
>>
>>> X <- data.frame(matrix(rnorm(1000*1000), 1000, 1000))
>>> y
?box.cox
?boxcox
On Jan 22, 2007, at 2:25 AM, Arun Kumar Saha wrote:
> I have a dataset 'data' and I want to see the effect of Box-Cox
> transformation on it Interactively for different lambda values. I
> already
> got a look on function "vis.boxcoxu" in package "TeachingDemos". But I
> didn't
It can't be done with the current code.
In a nutshell, you are trying to use a feature that I never got around to
coding. It's been on my "to do" list, but may never make it to the top.
Terry
__
R-help@stat.math.ethz.ch mailing list
http
On Sun, 21 Jan 2007, Lynette wrote:
> Dear all,
>
> I am using Rdqags in C to realize the integration. It seems for the
> continous C function I can get correct results. However, for step functions,
> the results are not correct. For example, the following one, when integrated
> from 0 to 1 gives
You can use the anova function a la:
> anova(model1, model2)
Analysis of Variance Table
Model 1: y ~ x
Model 2: y ~ x + z
Res.DfRSS Df Sum of Sq F Pr(>F)
1 13 4.4947
2 12 4.4228 10.0720 0.1952 0.6665
I would suggest getti
Aimin:
The problem is that the columns you choose for training (only 4
variables) do not match the ones used for prediction (all except y).
David
I try to use naiveBayes
> p.nb.90<-naiveBayes(y~aa_three+bas+bcu+aa_ss,data=training)
>
pr.nb.90<-table(predict(p.nb.90,training[,-13],ty
Dear helpeRs,
I'm estimating a series of linear models (using lm) in which in every
new model variables are added. I want to test to what degree the new
variables can explain the effects of the variables already present in
the models. In order to do that, I simply observe wether these
effe
On Jan 21, 2007, at 8:11 PM, John Fox wrote:
> Dear Haris,
>
> Using lapply() et al. may produce cleaner code, but it won't
> necessarily
> speed up a computation. For example:
>
>> X <- data.frame(matrix(rnorm(1000*1000), 1000, 1000))
>> y <- rnorm(1000)
>>
>> mods <- as.list(1:1000)
>> system.
Dear R useres,
I'm interested in getting a series of time-varying correlation, simply
between two random variables.
Could you please introduce a package to do this task?
Thank you so much for any help.
Amir
-
Don't pick lemons.
You might find the following code useful. It's part of a package I'm
developing for interactive model exploration.
Hadley
# Generate all models
# Fit all combinations of x variables ($2^p$)
#
# This technique generalises \code{\link{fitbest}}. While it is much
# slower it will work for any type
You don't say what model you want to do. It isn't necessary to store
each combination of predictors in a separate matrix unless you really
need to do this for some other reason, in which case I imagine you
could adopt this idea. I dare say there are better ways, but this
should work (assuming you
Dear Prof. Ripley and Christoph,
thank you very much for your comments. You have helped me a lot.
Thanks,
Tomas Goicoa
>Dear Prof. Ripley
>
>Thank you for your email. Yes, this is of course the correct
>syntax to save us the extra calculation. And I forgot the
>"lower.tail = FALSE" fo
Hello,
I'd like to know if the D'Agostino test of normality is reliable,
because somme of our results are not really coherent.
This test seems to be very sensitive. Even compared to a normal
distribution generated by R, the results are not very clear.
thanks for any help
Matthieu.
__
Hi,
List , i have 6 predictor variables and i want to make possible combinations
of these 6 predictors ,all the data is in matrix form ,
if i am having 6 predictors than possible combination of sets are 64 2 power 6,
or 63 ,whatever it may be i want to store the result in another variable to
Jerry Pressnell wrote:
> I have been running this script regularly for some time. This morning the
> following error message appeared.
> EWL<-get.hist.quote("EWL",start=(today <- Sys.Date())-350,quote="Cl")
> Error in if (dat[n] != start) cat(format(dat[n], "time series starts
> %Y-%m-%d\n")) :
On 1/22/07, Dirk Eddelbuettel <[EMAIL PROTECTED]> wrote:
>
> On 22 January 2007 at 00:05, Ramon Diaz-Uriarte wrote:
> | On 1/20/07, Dirk Eddelbuettel <[EMAIL PROTECTED]> wrote:
> | > Just confirms my suspicion that even after all these years, I barely
> | > scratched the surface of ess. That '2+ y
I have been running this script regularly for some time. This morning the
following error message appeared.
Any suggestions to correct this change would be appreciated.
EWL<-get.hist.quote("EWL",start=(today <- Sys.Date())-350,quote="Cl")
trying URL 'http://chart.yahoo.com/table.csv?s=EWL&a=1
play below after your code and look at tk window:
tkentryconfigure(editMenu,0,state="disable")
tkentryconfigure(editMenu,0,state="active")
tkentryconfigure(topMenu,1,state="disable")
tkentryconfigure(topMenu,1,state="active")
HTH
On 1/22/07, Jarno Tuimala <[EMAIL PROTECTED]> wrote:
> Hi!
>
> I'
The following three lines will do what you want. You will probably
want to change some of the default behaviour; just look at the
relevant help pages.
plot(x,y)
text(x,y,ID)
grid(2)
On 21/01/07, gnv shqp <[EMAIL PROTECTED]> wrote:
> Hi my friends,
>
> I'm trying to make a scatterplot like this.
Dear Tomas
You can produce the results in Montgomery Montgomery with
lme. Please remark that you should indicate the nesting with the
levels in your nested factor. Therefore I recreated your data,
but used 1,...,12 for the levels of batch instead of 1,...,4.
purity<-c(1,-2,-2,1,-1,-3,0,4, 0,-4, 1
Hi!
I've constructed a small menu-driven interface to a couple of R functions
using the possibilities offered by the tcltk package. When user runs some
specific analyses, I would then like to disable some of the menus (or menu
choises) that are not applicable after the performed analysis. I tri
Hi,
List , i have 6 predictor variables and i want to make possible combinations
of these 6 predictors ,all the data is in matrix form ,
if i am having 6 predictors than possible combination of sets are 64 2 power 6,
or 63 ,whatever it may be i want to store the result in another variable to
the reason is that is more natural to use pnorm(), which also should a
more efficient approximation of the Normal integral than intgrate(),
you may even use
diff(pnorm(0:1, mean = 0.5, sd = 1.2))
however, the point I meant to make was to use the '...' argument that
can found in many R function
see:
?commandArgs
or more detail for R startup mechanisms:
?Startup
On 1/22/07, Deepak Chandra <[EMAIL PROTECTED]> wrote:
> Hi All,
>
> A simple and naive question from a newbie. How can one access command-line
> arguments in R i.e. equivalent of argv in C?
> Have spent a lot of time on findin
Hi Nils,
I would say, pnorm is faster and has a higher precision.
Best,
Matthias
- original message
Subject: Re: [R] Integration + Normal Distribution + Directory Browsing
Processing Questions
Sent: Mon, 22 Jan 2007
From: Nils Hoeller<[EMAIL PROTECTED]>
> Thank you,
>
> both wor
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