Dear expeRts,
I use foreach to do parallel computations. Is it possible to have some progress
output written while the computations are done? In the minimal example below, I
just print a number (n) to check the progress. If you run this example with
%do% instead of %dopar%, then the
On Thu, Dec 23, 2010 at 09:12:41AM -0500, Data Analytics Corp. wrote:
Hi,
Does anyone know of a package for or any implementation of a Bayesian
Belief Network in R?
Different types of graphical models in R including Bayesian networks are
described in CRAN Task View gR
I run a batch file with the following command in Windows XP:
C:\R\R-2.12.1\bin\Rterm.exe --no-save --no-restore C:\users\me\file.R
C:\users\me\file.out 21
Is there any way to get only the output of R in file.out, without getting all
the code from file.R too?
Any help greatly appreciated,
Dear Eu,
On Wed, Dec 22, 2010 at 12:00 AM, EU JIN LOK ejl...@hotmail.com wrote:
Dear R users
I'm a novice user of R and have absolutely no prior knowledge of social
network analysis, so apologies if my question is trivial. I've spent alot of
time trying to solve this on my own but I
On Dec 28, 2010, at 8:09 AM, Mikkel Grum wrote:
I run a batch file with the following command in Windows XP:
C:\R\R-2.12.1\bin\Rterm.exe --no-save --no-restore C:\users\me
\file.R C:\users\me\file.out 21
Is there any way to get only the output of R in file.out, without
getting all the
On Dec 28, 2010, at 8:27 AM, David Winsemius wrote:
On Dec 28, 2010, at 8:09 AM, Mikkel Grum wrote:
I run a batch file with the following command in Windows XP:
C:\R\R-2.12.1\bin\Rterm.exe --no-save --no-restore C:\users\me
\file.R C:\users\me\file.out 21
Is there any way to get only
On Tue, Dec 28, 2010 at 8:09 AM, Mikkel Grum mi2kelg...@yahoo.com wrote:
I run a batch file with the following command in Windows XP:
C:\R\R-2.12.1\bin\Rterm.exe --no-save --no-restore C:\users\me\file.R
C:\users\me\file.out 21
Is there any way to get only the output of R in file.out,
At 20:08 27/12/2010, Louisa wrote:
Dear,
I'm very new to R Gui and I have to make an assignment on Gamma Regressions.
Surfing on the web doesn't help me very much so i hope this forum may be a
step forward.
Well since you are so honest about it being homework try Googling for
lognormal gamma
Hi
I have a large dataset, containing a wide range of binary variables.
I would like first of all to compute a jaccard matrix, then do a PCA on this
matrix, so that I finally can do a hierarchical clustering on the principal
components.
My problem is, that I don't know how to compute the jaccard
Whoops - thought I was replying to google medstats instead of r-help.
Frank
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
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Sent from the R help mailing
Flabbergaster jlunding at gmail.com writes:
My problem is, that I don't know how to compute the jaccard dissimilarity
matrix in R? Which package to use, and so on...
http://rss.acs.unt.edu/Rdoc/library/arules/html/dissimilarity.html
http://cc.oulu.fi/~jarioksa/softhelp/vegan/html/vegdist.html
Jacob,
You might have a look at the vegan package. It might compute the Jaccard
distance and it might have some other toolsa that you might be interested
in.
Dave
From:
Flabbergaster jlund...@gmail.com
To:
r-help@r-project.org
Date:
12/28/2010 08:26 AM
Subject:
[R] Jaccard dissimilarity
Thanks. The way I run it, I can determine what version of R to run with which
script. Don't know how to do that with R CMD BATCH.
Placing options(echo = FALSE) in the infile solves my problem. I got that from
the page you linked to.
Mikkel
--- On Tue, 12/28/10, David Winsemius
jaccard in package prabclus computes a Jaccard matrix for you.
By the way, if you want to do hierarchical clustering, it doesn't seem to
be a good idea to me to run PCA first. Why
not cluster the dissimilarity matrix directly without information loss by
PCA? (I should not make too general
On Dec 28, 2010, at 10:38 AM, Mikkel Grum wrote:
Thanks. The way I run it, I can determine what version of R to run
with which script. Don't know how to do that with R CMD BATCH.
Seems as though something like this (using absolute path to the
instance of R.exe) should work:
Hi,
I have a simple task, but I am looking for a clever and fast way to do it:
I have a vector x with 0,1 or 2 and I want to create another vector y with
the same length following the rules:
If the element in x is equal to 0, the element in y is equal to 0
If the element in x is equal to 2, the
Try this:
replace(replace(x, x == 1, sample(0:1, 1)), x == 2, 1)
On Tue, Dec 28, 2010 at 2:43 PM, M.Ribeiro mresende...@yahoo.com.br wrote:
Hi,
I have a simple task, but I am looking for a clever and fast way to do it:
I have a vector x with 0,1 or 2 and I want to create another vector y
Hi friends,
I get different values for McNemar's test in R and SPSS. Which one should i
rely on when the p values differ.
I came across this problem when i started learning R and seriously give up
on SPSS or any other proprietary software.
Thank u in advance
Output in SPSS follows
*Crosstab*
On Dec 28, 2010, at 11:05 AM, Manoj Aravind wrote:
Hi friends,
I get different values for McNemar's test in R and SPSS. Which one should i
rely on when the p values differ.
I came across this problem when i started learning R and seriously give up
on SPSS or any other proprietary software.
Hi Henrique,
Thanks for the fast answer,
The only problem in your code, which I think I didn't mention in my message
is that I would like one different random sampling procedure for each 1 in
my vector
The way it was written, it samples only once and replace by every 1:
x =
This sounds like something I could use..
I'm kind of new with R, meaning I've having some minor troubles all the
time...
Say I have a range of binary(0,1) variables X1 to Xn, with missing data for
different cases.
At the moment my data is a binary indicator matrix; rows representing the i
Marc Schwartz marc_schwa...@me.com [Tue, Dec 28, 2010 at 06:30:59PM CET]:
On Dec 28, 2010, at 11:05 AM, Manoj Aravind wrote:
Hi friends,
I get different values for McNemar's test in R and SPSS. Which one should i
rely on when the p values differ.
[...]
The SPSS test appears to be
Try this indeed
replace(replace(x, x == 1, sample(0:1, sum(x == 1), rep = TRUE)), x == 2, 1)
On Tue, Dec 28, 2010 at 3:14 PM, M.Ribeiro mresende...@yahoo.com.br wrote:
Hi Henrique,
Thanks for the fast answer,
The only problem in your code, which I think I didn't mention in my message
is
I don't know if it's any faster, but it is also possible this way:
y - ifelse(x ==1, round(runif(x)), sign(x))
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
Is the room still a room when
Hello,
I am trying to filter a data set like below so that the peaks in the Phase
value are more obvious and can be identified by a peak finding function
following the useful advise of Carl Witthoft. I have written the following
for(i in length(data$Phase)){
On 28/12/2010 1:08 PM, Nathan Miller wrote:
Hello,
I am trying to filter a data set like below so that the peaks in the Phase
value are more obvious and can be identified by a peak finding function
following the useful advise of Carl Witthoft. I have written the following
for(i in
On 28.12.2010 19:08, Nathan Miller wrote:
Hello,
I am trying to filter a data set like below so that the peaks in the Phase
value are more obvious and can be identified by a peak finding function
following the useful advise of Carl Witthoft. I have written the following
for(i in
On Dec 28, 2010, at 11:47 AM, Johannes Huesing wrote:
Marc Schwartz marc_schwa...@me.com [Tue, Dec 28, 2010 at 06:30:59PM CET]:
On Dec 28, 2010, at 11:05 AM, Manoj Aravind wrote:
Hi friends,
I get different values for McNemar's test in R and SPSS. Which one should i
rely on when the p
On 28/12/2010 1:13 PM, Uwe Ligges wrote:
On 28.12.2010 19:08, Nathan Miller wrote:
Hello,
I am trying to filter a data set like below so that the peaks in the Phase
value are more obvious and can be identified by a peak finding function
following the useful advise of Carl Witthoft. I have
On Dec 28, 2010, at 1:08 PM, Nathan Miller wrote:
Hello,
I am trying to filter a data set like below so that the peaks in the
Phase
value are more obvious and can be identified by a peak finding
function
following the useful advise of Carl Witthoft. I have written the
following
for(i
Thank you Michael!
--
View this message in context:
http://r.789695.n4.nabble.com/Gamma-Lognormal-Model-tp3165408p3166318.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Part of the reason I was having difficulty is that I'm trying to add a
legend with more than one element:
plot(1,1)
obv = 5
txt = Pop mean
# this works
legend(topleft, legend=bquote(.(txt) == .(obv)*degree))
# but this doesn't
legend(topleft, legend=c(bquote(.(txt) == .(obv)*degree), Von Mises
Hi all,
I haven't solved the problem of filtering the data, but I have managed to
find all the peaks in the data despite their relatively flat nature using
peaks() in the IDPmisc package. It works really well for my data and the
ability to set a lower threshold for peaks to report is convenient
Hi,
this seems to work,
plot.new()
legend(topleft, legend=as.expression(c(bquote(.(txt) ==
.(obv)*degree), Von Mises distribution)))
HTH,
baptiste
On 28 December 2010 20:17, Tyler Dean Rudolph
tylerdeanrudo...@gmail.com wrote:
legend(topleft, legend=c(bquote(.(txt) == .(obv)*degree), Von
On Tue, Dec 28, 2010 at 5:09 AM, Mikkel Grum mi2kelg...@yahoo.com wrote:
I run a batch file with the following command in Windows XP:
C:\R\R-2.12.1\bin\Rterm.exe --no-save --no-restore C:\users\me\file.R
C:\users\me\file.out 21
I'm a bit surprised this worked for you...did you customize your
Here is a basic example:
tmp.df - expand.grid( x= 1:100, y=1:100 )
tmp.df$z - with(tmp.df, x+2*y)
library(lattice)
levelplot( z ~ x + y, data=tmp.df )
tx2 - with(tmp.df, cut(x, seq(0.5, 100.5, 10) ) )
ty2 - with(tmp.df, cut(y, seq(0.5, 100.5, 20) ) )
tmp.df2 - aggregate(tmp.df, list( tx2, ty2
Marc Schwartz marc_schwa...@me.com [Tue, Dec 28, 2010 at 07:14:49PM CET]:
[...]
An old question of mine: Is there any reason not to use binom.test()
other than historical reasons?
(I meant in lieu of the McNemar approximation, sorry if some
misunderstanding ensued).
I may be missing the
Hi,
I would like to be able to plot data from each of the sp.id on individual
plots. At the moment I can plot all the data on one graph with the following
commands but I cannot figure out how to get individual graph for each sp.id.
i- function(df)plot(lnbm,ln.o2con,data=df)
j-
The data= argument to plot only makes sense if the first
argument is a formula. So if you change the plot command
in your function to
plot(ln.o2con~lnbm,data=df)
you might get what you want. But I would suggest you take a
look at the plot produced by
library(lattice)
I have a data frame with three columns
client ID | date | value
For each cilent ID I want to determine Min date and Max date and for
any dates in between that are missing I want to insert a row
Client ID | date| NA
Any help would be appreciated.
__
Hi,
I have been examining large data and need to do simple linear regression
with the data which is grouped based on the values of a particular
attribute. For instance, consider three columns : ID, x, y, and I need to
regress x on y for each distinct value of ID. Specifically, for the set of
On Dec 28, 2010, at 9:23 PM, Entropi ntrp wrote:
Hi,
I have been examining large data and need to do simple linear
regression
with the data which is grouped based on the values of a particular
attribute. For instance, consider three columns : ID, x, y, and I
need to
regress x on y for
Dear 'analyst41' (it would be a courtesy to know who you are)
Here is a low-level way to do it.
First create some dummy data
allDates - seq(as.Date(2010-01-01), by = 1, length.out = 50)
client_ID - sample(LETTERS[1:5], 50, rep = TRUE)
value - 1:50
date - sample(allDates)
clientData -
library(nlme)
lmList(y ~ x | factor(ID), myData)
This gives a list of fitted model objects.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Entropi ntrp
Sent: Wednesday, 29 December 2010 12:24 PM
To: r-help@r-project.org
Subject:
Hi:
There are some advantages to taking a plyr approach to this type of problem.
The basic idea is to fit a linear model to each subgroup and save the
results in a list, from which you can extract what you want piece by piece.
library(plyr)
# One of those SAS style data sets...
df -
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