Dear R Users,
Is it possible to make file names dependent on a changing variable?
For instance. I generate random numbers in a loop and at each iteration I
want data to write to file (I do not want to write everything in one file
using 'append'):
for (i in 1:50){
x-matrix(runif(100,
Use paste(Data_,i, sep=)
From: Robert Mcfadden [EMAIL PROTECTED]
Date: Mon, 24 Jul 2006 11:06:38 +0200
To: r-help@stat.math.ethz.ch
Subject: [R] change the name of file
Dear R Users,
Is it possible to make file names dependent on a changing variable?
For instance. I generate random
Dear all
The following is a very basic and beginner's question on loops.
Suppose you have data (say, 1000 cases) with two variables named
answer (string) and class (numeric). The latter runs from 1 to 78
and categorizes the data. I need to create a file answer_1.txt for
the cases with
Do you mean write.table instead of Write() ? Try
fn - paste(Data_, i, .txt, sep=)
write.table( t(x), file=fn, sep=\t )
Regards, Adai
On Mon, 2006-07-24 at 11:06 +0200, Robert Mcfadden wrote:
Dear R Users,
Is it possible to make file names dependent on a changing variable?
For instance. I
Of course file name Data_i.txt will be the same for changing i,
unfortunately.
?paste, e.g. paste(Data_, i, .txt, sep=)
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R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide
Great, little change and it works. Thank you.
Look
for (i in 1:50){
x-matrix(runif(100, min=0,max=1),nrow=5,ncol=20)
write(t(x),file=paste(data_,i,.txt,sep=),ncolumns=5,sep=\t)
}
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: Monday, July 24, 2006
Dear Robert,
maybe (!) the following might do the job:
for (i in 1:50){
x-matrix(runif(100, min=0,max=1),nrow=5,ncol=20)
Write(t(x),file=paste(Data_ i .txt, sep=),ncolumns=5,sep=\t)
}
Hope it really does...
Thilo
On Monday 24 July 2006 11:06, Robert Mcfadden wrote:
Dear R Users,
Is it
for (i in 1:78){
write.table(answer[class == i], paste(answer_,i, .txt, sep=''))
# if you want the variable
assign(paste(k_, i, sep=''), answer[class == i])
}
On 7/24/06, Chris Kopp [EMAIL PROTECTED] wrote:
Dear all
The following is a very basic and beginner's question on
Gabor == Gabor Grothendieck [EMAIL PROTECTED]
on Sun, 23 Jul 2006 09:02:35 -0400 writes:
Gabor Moving this to r-devel.
[would have been a good idea ... but you didn't;
I think it's too late now; rather keep the msg thread together]
Gabor Looking at the diff.POSIXt
Gabor code
I'm aware that S N Krishna asked the same
question. However, I have failed to implement the
posted solution for running rank order
correlations on multiple subsets of data using the by() function.
Here is my problem:
Take a set of data from two subjects, who
provided numerical infant
Thank you for your suggestion Phil, but they are already installed... Any
other idea, please ?
Thanks in advance
Bernat
A Dijous 20 Juliol 2006 20:22, Phil Spector va escriure:
Bernat -
Just a guess, but maybe you need to install the packages
xfonts-100dpi
and
After loading the package Matrix (version 0.995-12), using the summary
function with an lme (package nlme version 3.1-75) object results in an
error message saying
Fehler in dim(x) : kein Slot des Namens Dim für dieses Objekt der
Klasse correlation
(translated: 'Error in dim(x) : no slot of
On 7/24/06, Peter J. Lee [EMAIL PROTECTED] wrote:
I'm aware that S N Krishna asked the same
question. However, I have failed to implement the
posted solution for running rank order
correlations on multiple subsets of data using the by() function.
Here is my problem:
Take a set of data from
Many thanks,
your solutions solve a number of my problems (not least weening myself off SAS)
Best wishes,
Peter
_
Peter James Lee
Assistant Professor
Psikoloji Bölümü
Bilkent University
Bilkent
Ankara
Turkey
06800
e-mail: [EMAIL PROTECTED]
office: (90) 312 290 1807
Dear list,
I was wondering if there is a function to plot league tables, sometimes
also known as caterpillar plots?
A league table is conceptually very similar to a box plot. One difference
is that the inter-quartile ranges are not shown. If there isn't such a
function a first attempt for a
Dirk == Dirk Enzmann [EMAIL PROTECTED]
on Mon, 24 Jul 2006 13:55:44 +0200 writes:
Dirk After loading the package Matrix (version 0.995-12),
Dirk using the summary function with an lme (package nlme
Dirk version 3.1-75) object results in an error message
Dirk saying
Dirk
Hi all,
I am looking for a function in R to trim the last two characters of an
8 character string in a vector. For example, I have the codes
37-079-2, 370079-3,37-079-8 and want to trim them to 37-079 by
removing the last two characters. Is sub the correct function to use,
and if so how can I
See ?substr and ?nchar
Try:
substr(Hello World, 0, nchar(Hello World)-2)
Regards
Markus Gesmann
FPMA
Lloyd's Market Analysis
Lloyd's * One Lime Street * London * EC3M 7HA
Telephone +44 (0)20 7327 6472
Facsimile +44 (0)20 7327 5718
http://www.lloyds.com
-Original Message-
From: [EMAIL
Eran Tauber (PhD)
Lecturer in Molecular Evolution
Dept. of Genetics
University of Leicester
University Rd, Leicester LE1 7RH
England
Phone: 44 (0)116 252-3455, 252-3421 (lab) Fax: 44 (0)116 252-3378
www.le.ac.uk/genetics/et22
A . (dot) matches any character and $ matches the end of string so
this replaces the last two characters with the empty string:
sub(..$, , x)
On 7/24/06, Wade Wall [EMAIL PROTECTED] wrote:
Hi all,
I am looking for a function in R to trim the last two characters of an
8 character string in
Dear R-users,
We are monitoring the activity of animals during a few days period. The
data from each animal (crossing of infra-red beam) are collected as a
time series (in 30 min bins). An example is attached below.
y -
c(0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,3,28,27,46,76,77,60,19,35,55,59,48
Hi all,
I have a matrix of 474 rows (samples) with 565 columns (variables).
each of the 474 samples belong to one of 120 groups, with the
groupings as a column in the above matrix. For example, the group
column would be:
1
1
1
2
2
2
.
.
.
120
120
I want to randomly select one from each group.
Try:
RSiteSearch(finding peaks)
On 7/24/06, Tauber, Dr E. [EMAIL PROTECTED] wrote:
Dear R-users,
We are monitoring the activity of animals during a few days period. The
data from each animal (crossing of infra-red beam) are collected as a
time series (in 30 min bins). An example is
Hello R-helpers!
I have a question concerning extracting sequence information from a
vector. I have a vector (representing the bins of a time series where
the frequency of occurrences is greater than some threshold) where I
would like to extract the min, median and max of each group of
Look at index vectors in the R intro.
Best
Niels
On Mon, 24 Jul 2006, Kevin J Emerson wrote:
Hello R-helpers!
I have a question concerning extracting sequence information from a
vector. I have a vector (representing the bins of a time series where
the frequency of occurrences is
Dear Wade,
Say that your groups are
groups - sort(sample(1:10, 100, replace = TRUE))
Create a dummy
rows - 1:length(groups)
Then
tapply( rows, groups, function(x) sample(x, 1))
does the trick to select the row numbers you need for your sampling.
Sincerely,
Carlos J. Gil Bellosta
Well, how you do it might be a matter of taste with respect to how you
want the results.
You could try using by with sample
by(x,x[,3],function(y){y[sample(nrow(y),1),]})
This will return a list with one list element for each sample group.
You can the combine the list back into a matrix.
Dear Kevin,
Try something like
groups - cut( tmp, c(-Inf, which(diff(tmp) 1 ) + 0.5, Inf) )
Sincerely,
Carlos J. Gil Bellosta
http://www.datanalytics.com
http://www.data-mining-blog.com
Quoting Kevin J Emerson [EMAIL PROTECTED]:
Hello R-helpers!
I have a question concerning extracting
Let me clarify one thing that I dont think I made clear in my posting.
I am looking for the max, min and median of the indicies, not of the
time series frequency counts. I am looking to find the max, min, and
median time of peaks in a time series, so i am looking for the
information concerning
?unique says
Value:
An object of the same type of 'x'. but if an element is equal to
one with a smaller index, it is removed.
However, I need to keep the one with the LARGEST index.
Can someone please show me the light?
I thought about reversing the row order twice, but I couldn't
Eik Vettorazzi sent the following at 24/07/2006 13:04:
Dear list,
I was wondering if there is a function to plot league tables, sometimes
also known as caterpillar plots?
A league table is conceptually very similar to a box plot. One difference
is that the inter-quartile ranges are not
On Mon, 2006-07-24 at 12:00 -0500, [EMAIL PROTECTED] wrote:
?unique says
Value:
An object of the same type of 'x'. but if an element is equal to
one with a smaller index, it is removed.
However, I need to keep the one with the LARGEST index.
Can someone please show me the
I explicitly did is this way
isoforms-as.vector(rownames(mult.comp$estimate))
estimate-as.vector(mult.comp$estimate)
lower-as.vector(mult.comp$conf.int[,1])
upper-as.vector(mult.comp$conf.int[,2])
p.val.raw-as.vector(mult.comp$p.value.raw)
p.val.bon-as.vector(mult.comp$p.value.bon)
As you do not seem to have received what you consider to be satisfactory
reply, here is a function that I **think** does what you want:
sequences - function(x,incr = 1)
{
ix - which(abs(diff(c(FALSE,diff(x) == 1))) ==incr)
if(length(ix)%%2)c(ix,length(x))
else ix
}
This
Try:
largestDF - DF[nrow(DF)- which(!duplicated(rev(DF$t)))+1,]
You can then sort this however you like in the usual way. Row names will be
preserved.
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
The business of the statistician is to catalyze the scientific
Hi R-users
I'm working with a data.frame of 4 x 10, for which I need to apply the
split function. The result is very long and cannot be stored in a variable
due to memory exceeding. I've tried to send the result directly to a file
through sink(filename) function, but the problem still occurs.
This might work:
tmp - c(24,25,29,35,36,37,38,39,40,41,42,43,44,45,46,47,68,69,70,71)
# generate breaks
group - c(0, cumsum(diff(tmp) != 1))
tapply(tmp, group, summary)
$`0`
Min. 1st Qu. MedianMean 3rd Qu.Max.
24.00 24.25 24.50 24.50 24.75 25.00
$`1`
Min. 1st Qu.
Hi
This is a question regarding classification performance using different methods.
So far I've tried NaiveBayes (klaR package), svm (e1071) package and
randomForest (randomForest). What has puzzled me is that randomForest seems to
perform far better (32% classification error) than svm and
Thank you, Bert and Mark.
I believe Mark's solution works, but it was taking a very long time.
Bert's is very fast.
My day is saved!
David L. Reiner
Rho Trading Securities, LLC
Chicago IL 60605
312-362-4963
-Original Message-
From: Berton Gunter [mailto:[EMAIL PROTECTED]
Sent:
Dear all,
First of all I apologize if you received this twice: I was checking the
archive and I noticed that the text was scrubbed from the message,
probably due to some setting in my e-mail program.
I am unsure about how to specify a model in R and I thought of asking
some advice to the
Sorry, all. My previous post was mixed up. Here's the corrected version:
sequences - function(x,incr = 1)
{
ix - which(abs(diff(c(FALSE,diff(x) == incr))) ==1)
if(length(ix)%%2)c(ix,length(x))
else ix
}
-- Bert Gunter
Genentech Non-Clinical Statistics
South San
Here is something that will mark the beginning/end of your sequence. You
will have to adjust the low limits and the lengths to fit your requirements.
y -
c(0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,3,28,27,46,76,77,60,19,35,55,59,48
out.data.mat can be created compactly using with:
out.data.mat - with(mult.comp,
cbind(estimate, conf.int, p.value.raw = c(p.value.raw),
p.value.bon, p.value.adj)
)
On 7/24/06, Nair, Murlidharan T [EMAIL PROTECTED] wrote:
I explicitly did is this way
On Mon, 24 Jul 2006 11:18:10 -0400,
Wade Wall [EMAIL PROTECTED] wrote:
Hi all, I have a matrix of 474 rows (samples) with 565 columns
(variables). each of the 474 samples belong to one of 120 groups, with
the groupings as a column in the above matrix. For example, the group
column would be:
Can I run jobs in the background and the check the status of it from
time to time in Windows version of R?
Thanks ../Murli
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
I have a question about deparse function in R
What is the reason that deparse use an argument like width.cutoff ?
Why the maximum cutoff is 500?
I was manipulating an R formula and used deparse. Since the length of user's
formula was greater then 500, my code didnt work.
thanks
Johan
I have a question about deparse function in R
What is the reason that deparse use an argument like width.cutoff ?
Why the maximum cutoff is 500?
I was manipulating an R formula and used deparse. Since the length of user's
formula was greater then 500, my code didnt work.
thanks
Johan
johan
Using ranef() (package nlme, version 3.1-75) with an 'lme' object I can
obtain random effects for intercept and slope of a certain level (say:
1) - this corresponds to (say level 1) residuals in MLWin. Maybe I'm
mistaken here, but the results are identical.
However, if I try to get the
I can't add much to your question, being a complete novice at
classification, but I have tried both randomForest and SVM and I get better
results from randomForest than SVM (even after tuning). randomForest is
also much, much faster. I just thought randomForest was a much better
algorithm,
Here is an example of creating the set of indices (instead of the
dataframes) and then using them to compute on some of the values.
# create some test data
n - 1000
x - data.frame(a=sample(letters[1:5], n, TRUE), b=sample(LETTERS[1:5], n,
TRUE),
c=sample(1:10, n, 1000))
x.split -
Hi Spencer,
Thanks for the direction. Yes, I am after the sigma for the estimated variance
of the whitened residuals. I'll give your code a try!
Cheers,
Melissa
On Sun, 23 Jul 2006, Spencer Graves wrote:
Confidence intervals for the fracdiff parameter estimates shouldn't
be too
Hi WizaRds,
I'd like to overplot UK fuel consumption per quarter over the course of five
years.
Sounds simple enough?
Unless I'm missing something, the following seems very involved for what I'm
trying to do. Any suggestions on simplifications?
The way I did it is awkward mainly because of
Try:
matplot(levels(data$Quarter), matrix(data$Consumption, 4), type = o)
On 7/24/06, John McHenry [EMAIL PROTECTED] wrote:
Hi WizaRds,
I'd like to overplot UK fuel consumption per quarter over the course of five
years.
Sounds simple enough?
Unless I'm missing something, the following
And if lattice is ok then try this:
library(lattice)
xyplot(Consumption ~ Quarter, group = Year, data, type = o)
On 7/24/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Try:
matplot(levels(data$Quarter), matrix(data$Consumption, 4), type = o)
On 7/24/06, John McHenry [EMAIL PROTECTED]
Am I correct to understand from the previous discussions on this topic (a
few years back) that if I have a matrix with missing values my PCA options
seem dismal if:
(1) I dont want to impute the missing values.
(2) I dont want to completely remove cases with missing values.
(3)
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