On Jun 3, 2011, at 17:42 , Carl Witthoft wrote:
Hi,
I feel dumb even asking, but isn't there an R function somewhere that I can
use to reduce the resolution of a vector (or matrix) by summing terms in
uniform blocks? That is, for a vector X, reduce it to some X.short as
X.short[1]-
Is there a way to prepare a questionnaire in R like html forms whose data
can be directly populated into R?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
Hi all,
I am currently doing a consistency analysis using an lmer model and
trying to use parametric bootstrapping for the confidence intervals.
My model is like this:
model-lmer(y~A+B+(1|C/D)+(1|E),binomial)
where E is the individual level for consistency analysis, A-D are
other fixed and
here's the code to generate the matrix
set.seed(1)
types = c(1,2,3)
n=5
p=1
pop.props = c(0.6,0.2,0.2)
x=matrix(pop.props,nrow=n,ncol=length(pop.props), byrow=T)
b=10
habs = rMultinom((x),b)
print(habs)
--
View this message in context:
Hi,
I have a matrix
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]131131123 2
[2,]131211122 1
[3,]122321112 1
[4,]321113121
Thank you all guys!!
I have some other questionsFor garch model, for example, we have 10 time
periods, and the function use MLE to get the parameters based on these 10
time periods. Then, the function calculates covariance matrix at each time
period based on the estimated parameters. Is this
Hi,
Try apply function:
apply(matrix,1,table)
Regards
M
Le 08/06/11 08:23, the_big_kowalski a écrit :
Hi,
I have a matrix
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]131131123 2
[2,]1312111
Dear Jenni,
In the newest version of the doBy package there is a function called PBrefdist
(and PBrefdist.mer) for calculating the reference distribution of the
likelihood ratio statistic for comparing nested models. Looking into this
function may help you. Perhaps the functions PBmodcomp and
Thanks Sarah for the response; with the command
str(echoknn.train)
the coloumn class is a logi value (i think without any immagination that
is a logical value ). So, how can I handle this type of data?
Thanks a lot.
P.S. Yes, is a course assignment and i was hoping to solve this problem
Vijayan,
I cannot find an error in your code, but I had a look at the code of
triangle.test -- unless I'm missing something, it contains a bug. If you study
the way in which the matrix pref is updated, you find that the vector
preference is compared to 1, 2 and 3 instead of X, Y and Z as it
Hello everybody,
I am trying different ways of interpolating a surface.
Now, I am trying to interpolate these values using the MBA package. In order to
do it I am using the surface option of the package.
mba.surf(xyz, 20, 20, n = 1, m = 1, h = 8, extend=FALSE,
sp=FALSE, ...)
The function
Hello,
this question is a bit out of the blue.
I am a big R fan and user and in my new job I do some decision
modeling (mostly health economics). For that decision trees are often
used (I guess the most classic example is the investment decision A,
B, and C with different probabilities, what is
Dear Paco,
I tried dbscan on my computer with method=hybrid and a 155000*3
data matrix and it works. Needs some time though.
(You can track the progress using something like
countmode=c(1:10,100,1000,1,10).)
Note that for some reason I don't exactly understand, it takes *much*
longer
Hi all,
I have a huge dataset of values and I have precalculated outside of R the
frequency/count distribution (as for example counts returned by hist
function or the output of table function). For example,
x-hist(c(1,2,2,2,1,3), breaks=0:3, plot=F)
x
$breaks
[1] 0 1 2 3
$counts
[1] 2 3 1
Is
Hi, I just installed the 2.13 version for Mac (without uninstalling the
previous versions). I transferred the folder library containing the packages I
normally use to the 2.13 library folder. When I require any package this is
what I get:
require(ape)
Loading required package: ape
Error in
Date: Wed, 8 Jun 2011 12:37:33 +0530
From: ammasamri...@gmail.com
To: r-help@r-project.org
Subject: [R] Can we prepare a questionaire in R
Is there a way to prepare a questionnaire in R like html forms whose data
can be directly populated into
If the result is limited to just the counts of 1,2 and 3, you can do
the following:
habs - matrix(sample(1:3, 50, TRUE, prob=c(.6,.2,.2)),5,10)
habs
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]132113111 2
[2,]123131
Date: Tue, 7 Jun 2011 09:50:10 -0700
From: egregory2...@yahoo.com
To: r-help@r-project.org
Subject: [R] RgoogleMaps Axes
R Help,
I posted a question on StackOverflow yesterday regarding an issue I've been
having with the RgoogleMaps
How can we create HTML forms in R
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and
See packages rpart, randomForest, party.
Also, typing R Decision Trees produced good google results.
http://www.google.com/search?aq=fsourceid=chromeie=UTF-8q=R+Decision+Trees
On Wed, Jun 8, 2011 at 7:02 AM, stefan.d...@gmail.com
stefan.d...@gmail.com wrote:
Hello,
this question is a bit out
On Wed, 8 Jun 2011, Mariana Varela wrote:
Hi, I just installed the 2.13 version for Mac (without uninstalling
As the posting guide says, there is no such version of R
Also, I think you did uninstall a previous version, for it is missing
and the default behaviour of the Apple installer
Thank you so much for reply. But I am looking for the exact opposite.
I do not have a data set which I want to partition. But already a
sequence/tree-like set of decision rules and with which I want to
simulate what is my expected outcome/pay-off given a particular
scenario.
As far as I
Unless I missed it, neither the OP nor the list was CC'd on this, so for anyone
interested, I forward this solution (untested from my side) from the package
maintainer. Not sure whether the file comes through, so I include the updated
code in the message's body below.
Cheers, Michael
So TreeAge fits models but won't predict from them? That seems like
bizarre behavior. I suppose I would recommend, then, looking at the
source code from the aforementioned packages for how they store their
split data. It sounds like you would have to write code to hack
TreeAge outputs into another
Colin Wahl wrote:
I would rather use cleveland dot plots than bar charts to display my study
results. I have not been able to find (or figure out) an R package that is
capable of producing the publication quality dot charts Im looking for. I
have either not been able to get error bars
As Mike had written, there are frameworks for web-development with R.
RApache http://www.rapache.net is one. Also, see the R package Rook:
http://cran.r-project.org/web/packages/Rook/index.html .
On Wed, 2011-06-08 at 17:26 +0530, amrita gs wrote:
How can we create HTML forms in R
Wouldn't
Dear list,
i'm checking the residuals plots of a gam model after a processus of model
selection. I found the best model, all my terms are significant, the
r-square and the deviance explained are good, but I have strange residuals
plots:
http://dl.dropbox.com/u/1169100/gam.check.png
On Wed, Jun 8, 2011 at 12:56 PM, amrita gs ammasamri...@gmail.com wrote:
How can we create HTML forms in R
HTML is Just Plain Text, so you can create them using R's text output
'cat' function. E.g.
cat('form First name: input type=text name=firstname /br
/Last name: input type=text
Hi,
I am trying to install the impute package in unix. but I get the
following error message. I followed the following steps. Do you know what is
causing this and how I can solve this problem?
source(http://www.bioconductor.org/biocLite.R;)
biocLite(impute)
Using R version 2.11.1,
Hello,
say I have a 2D-matrix (indexed by x and y), which contains
z values, which I want to plot over x-y.
Either dotted, or if possible as a landscape.
I tried around with persp and plot3d (from rgl)
and persp3d (from rgl).
I sometimes get something that looks good and a while later, when
This is a consequence of predict.ols calling predictrms which relies on
model.frame which re-issues the expression of x.knots. You would have the
same problem if using update(fit object) in another session. For that
reason you have to keep an external knots vector available in your
environment.
I manage to install the package. But I cant load it now.
It says:
library(CGHcall)
Loading required package: impute
Error in dyn.load(file, DLLpath = DLLpath, ...) :
unable to load shared library
'/nfs/users/nfs_s/st5/R-modules/impute/libs/impute.so':
libgfortran.so.1: cannot open
Dear all,
What is the simplest way of producing a bar graph using ggplot but
avoiding calling qplot?
That is, given:
d - data.frame(x=seq(1,5), y=seq(1,5))
Why does the following line return an error?
ggplot(d, aes(x=x, y=y)) + stat_identity() + geom_bar(bindwidth=1)
Thanks in advance,
jcb!
Dear all,
I receive the very same error message on a Debian computer (testing)
with R 2.13.0 also.
install.packages('emu')
Installing package(s) into
‘/home/olivier/R/i486-pc-linux-gnu-library/2.13’ (as ‘lib’ is
unspecified)
Error in ret[i, ] - c(pkgs[i], lib, desc) :
number of items to
Hi Juan,
Each geom can have it's own stat, so stat_identity() doesn't change
the stat used by geom_bar(). You need
ggplot(d, aes(x=x, y=y)) + geom_bar(stat=identity)
Best,
Ista
2011/6/8 Juan Carlos Borrás jcbor...@gmail.com:
Dear all,
What is the simplest way of producing a bar graph using
Hi,
i am trying to use the normalization function from CGHcall library.
The command i use is
normal.fullData - normalize(fullData, method = median, cellularity = 1)
and i get the following error message. How can i solve this issue.
Error in function (classes, fdef, mtable) :
unable to find
Hello again.
Thank you for the comments. I have written these codes.
iy=function(x)
{
res=NULL
ress=0
for (i in (1:2))
{
for (xx in x[i])
{
fy=function(y)
(exp(-exp(y+log(xx)))*(-exp(y+log(xx)))^2)/(1-exp(-exp(y+log(xx
res=c(res,integrate(fy,-6.907,-1.246)$value)
ress=ress+res
}
}
Jon,
So TreeAge fits models but won't predict from them? That seems like
bizarre behavior.
Nothing bizarre about TreeAge, just a different tool in a different
disicpline.
http://en.wikipedia.org/wiki/Decision_tree
Graham
[[alternative HTML version deleted]]
TreeAge works just fine. But its commercial, thats all...
On Wed, Jun 8, 2011 at 6:17 PM, Graham Smith myotis...@gmail.com wrote:
Jon,
So TreeAge fits models but won't predict from them? That seems like
bizarre behavior.
Nothing bizarre about TreeAge, just a different tool in a different
Dear Prof Ripley,
Apologies - I've re-sent that to Duncan Temple Lang, along with your note about
lib versions.
Version info was included in my original post - I gave full sessionInfo(). It's
XML_3.4-0.
I only have a very sketchy understanding of libraries and systems
administration, but it
Hi
I am attempting to do a loadings plot from a plsr object. I have managed to do
this using the gasoline data that comes with the pls package. However when I
conduct this on my dataset i get the following error message.
plot(BHPLS1, loadings, comps = 1:2, legendpos = topleft, labels =
I am using ggplot2 to make a boxplot that overlays a scatterplot:
pp = qplot(time, error, data=times, size=I(1), geom=jitter, main=title,
ylab=Error (min), xlab=Time before ON (min), alpha=I(1/10),
color=times$runway,
ylim=c(-30,40))
pp2 = pp + with(times, facet_wrap(~ runway,
This new version includes a plot method for plotting
a particular instance along the path.
Trevor Hastie has...@stanford.edu
Professor, Department of Statistics,
Hi Jim,
See ?scale_x_discrete
Best,
Ista
On Wed, Jun 8, 2011 at 3:26 PM, James Rome jamesr...@gmail.com wrote:
I am using ggplot2 to make a boxplot that overlays a scatterplot:
pp = qplot(time, error, data=times, size=I(1), geom=jitter, main=title,
ylab=Error (min), xlab=Time before
Hi,
I'm having trouble with getting the png function to properly produce multiple
graphs. RIght now I have:
for (z in data) {
png(file=z,bg=white)
thisdf-data[[z]]
plot(thisdf$rc,thisdf$psi)
dev.off()
}
Which should take the data object, a list of data sets and produce a graph of
each
Hello ,
I am trying to create a histogram in order to compare between two groups and
would like it to be similar to the figure attached. How can I generate this
using R ?
Thank you,
Nandini http://r.789695.n4.nabble.com/file/n3582448/5634-15977-1-PB.gif
--
View this message in context:
Hello ,
I am trying to create a histogram in order to compare between two groups and
would like it to be similar to the figure attached. How can I generate this
using R ?
Thank you,
Nandini
__
Hello.
I'm a visually impaired statistician, working at the National Institute of
Public Health in Denmark.
I would like to use R for some analysis and have succesfully installed version
2.13.0 on my Windows XP labtop.
I then would like to run R interactively but unfortunately the textfont of
I'm trying to develop this function so that I can efficiently generate
all possibile combinations of the strings.
So I have certain roots, prefixes, and sufixes. I also have different
combinations of the data, some with two strings (roots, prefix) and others
with three strings (roots, prefix,
st1=c(5,3)
st2=c(-1,2)
st3=c(1,3)
st4=c(2,6)
st = rbind(st1,st2,st3,st4)
st
dx=round(dist(st), digits=2)
dx #À¯Å¬¸®µå °Å¸®Çà·Ä
D1=dist(st, method=quot;euclideanquot;)
D1 # À¯Å¬¸®µå°Å¸® ±¸Çϱâ
D2=dist(st, method=quot;manhattanquot;)
D2 # ¸ÇÇÏź °Å¸®Çà·Ä
and I want know
How can i use
Hi again, I was using package.skeleton() function to create the
skeleton of my package in windows. Here is my attempt:
rm(list = ls())
setwd(F:/R_PackageBuild)
package.skeleton(trial1, namespace = TRUE, code_files =
F:/R_PackageBuild/trial.r)
In the trial.r file, there are 2 objects, one is a
Thanks for your reply, Frank. I've noticed that the x.knots object doesn't
actually have to be the vector of knots. Just having x.knots - 0 or even
x.knots - a will allow predict to work.
Mark Seeto
Frank Harrell wrote:
This is a consequence of predict.ols calling predictrms which relies on
It is difficult for someone from a statistical frame of mind to understand
what this is about --- you need to think a bit differently. It is mostly a
simulation and decision analysis, with some use of statistical functions to
draw random samples to simulate the fact that outcome of interest can
I've just found the lavaan package, and I really appreciate it, as it
seems to succeed with models that were failing in sem::sem. I need
some clarification, however, in the output, and I was hoping the list
could help me.
I'll go with the standard example from the help documentation, as my
What do you mean by latent estimate?
The table of variances has variances for each factors.
Is there something different in the sem output that you don't see here?
Yes, this looks normal.
Jeremy
On 8 June 2011 13:14, R Help rhelp.st...@gmail.com wrote:
I've just found the lavaan package,
Hello R users,
I have difficulties when trying to make R packages. I tried to read many
tutorials but still could not find out the right way. Could any one help me
out please? (I'm using Windows xp.)
After running package.skeleton() and edit those RD files, I don't know how
to use Rtools (or CMD
Dear Sam,
In each case, the first observed variable is treated as a reference
indicator with its coefficient fixed to 1 to establish the metric of the
corresponding factor and therefore to identify the model. If you didn't do
the same thing (or something equivalent, such as fixing the factor
Hi,
I am trying to learn time series, and I am attending a colleague's
course on Econometrics. However, he uses e-views, and I use R. I am
trying to reproduce his examples in R, but I am having problems
specifying a AR(1) model. Would anyone help me with my code?
Thanks in advance!
Reproducible
see inline below.
On Wed, Jun 8, 2011 at 12:37 PM, Anupam anupa...@gmail.com wrote:
It is difficult for someone from a statistical frame of mind to understand
what this is about --- you need to think a bit differently. It is mostly a
simulation and decision analysis, with some use of
Hi,
Our sysadmin updated libxml2 to version 2.7.8, and now xmlTreeParse works fine
with no segfault.
Thank you very much - that was very helpful,
Janet
On Jun 8, 2011, at 11:59 AM, Janet Young wrote:
Dear Prof Ripley,
Apologies - I've re-sent that to Duncan Temple Lang, along with your
Hi,
There must be an easy way to do this, but I'm not finding it..
I'd just like to know the syntax to move up and down folder levels, without
necessarily entering a full file path. Also, how to construct file and folder
paths using variables.
For example 1, if I wanted to print to the
I think the issue is that the x axis is a factor. How would ggplot2 know
which ones to drop? So it labels them all.
If I do the following, the labels get better:
pp = qplot(time, error, data=times, size=I(1), geom=jitter,
main=title,
ylab=Error (min), xlab=Time before ON (min),
On Wed, Jun 8, 2011 at 5:02 PM, James Rome jamesr...@passur.com wrote:
I think the issue is that the x axis is a factor.
Rather the opposite I think. In the data you sent, time is numeric,
not a factor. This works for me:
qplot(factor(time), error, data=times, size=I(1), geom=boxplot) +
On 11-06-08 5:35 PM, J wrote:
Hi,
There must be an easy way to do this, but I'm not finding it..
I'd just like to know the syntax to move up and down folder levels, without
necessarily entering a full file path. Also, how to construct file and folder
paths using variables.
For example
Yes, that is the difference. For the last SEM I built I fixed the
factor variances to 1, and I think that's what I want to do for the
CFA I'm doing now. Does that make sense for a CFA?
I'll try figuring out how to do that with lavaan later, but my model
takes so long to fit that I can't try it
On 11-06-08 4:17 PM, Xia.Li wrote:
Hello R users,
I have difficulties when trying to make R packages. I tried to read many
tutorials but still could not find out the right way. Could any one help me
out please? (I'm using Windows xp.)
Read the Writing R Extensions manual, not the many
Try this:
lapply(dir(myPathDir, recursive = TRUE, pattern = myFile.txt$,
full.names = TRUE), readLines, warn = FALSE)
On Wed, Jun 8, 2011 at 6:35 PM, J jonsle...@gmail.com wrote:
Hi,
There must be an easy way to do this, but I'm not finding it..
I'd just like to know the syntax to move up
Hi,
I am given the following table:
head(hsa_refseq)
chr genome regionstart stop nu strand nu.1nu.2
gene_id
1 chr1 hg19_refGeneCDS 6742 6751 0 +0 gene_id
NM_032291
2 chr1 hg19_refGene exon 66999825 6751 0 +. gene_id
NM_032291
3 chr1
On 11-06-08 9:28 AM, Michael Davidsen wrote:
Hello.
I'm a visually impaired statistician, working at the National Institute of
Public Health in Denmark.
I would like to use R for some analysis and have succesfully installed version
2.13.0 on my Windows XP labtop.
I then would like to run R
I think the command you want is barplot
x = rbinom(10,15,0.65)
y = rbinom(10,15,0.25)
barplot(rbind(x,y),beside=TRUE)
Sam
On Wed, Jun 8, 2011 at 10:14 AM, nandini_bn nandini...@hotmail.com wrote:
Hello ,
I am trying to create a histogram in order to compare between two groups and
would
Have a look at:
http://addictedtor.free.fr/graphiques/thumbs.php
One of the graph examples they have is exactly what you are after.
On Wed, Jun 8, 2011 at 11:14 PM, nandini_bn nandini...@hotmail.com wrote:
Hello ,
I am trying to create a histogram in order to compare between two groups and
SNV Krishna krishna at primps.com.sg writes:
Hi All,
I am unable to get the plot geomap in googleVis package. data is as follows
head(index.ret)
countryytd
1 Argentina -10.18
2 Australia -3.42
3 Austria -2.70
4 Belgium 1.94
5Brazil -7.16
6Canada 0.56
On 06/08/2011 03:06 PM, ads pit wrote:
Hi,
I am given the following table:
head(hsa_refseq)
chr genome regionstart stop nu strand nu.1nu.2
gene_id
1 chr1 hg19_refGeneCDS 6742 6751 0 +0 gene_id
NM_032291
2 chr1 hg19_refGene exon 66999825 6751 0
Dear Sam,
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of R Help
Sent: June-08-11 5:57 PM
To: John Fox
Cc: r-help
Subject: Re: [R] Results of CFA with Lavaan
Yes, that is the difference. For the last SEM I built I fixed
On Wed, 8 Jun 2011, Iuri Gavronski wrote:
Hi,
I am trying to learn time series, and I am attending a colleague's
course on Econometrics. However, he uses e-views, and I use R. I am
trying to reproduce his examples in R, but I am having problems
specifying a AR(1) model. Would anyone help me
I have a series of strings and I am trying to find all combinations and then
assign 1 or 0 to them based
on whether they contain the words car or budged. I want the data to look
like:
car budget
cheap car insurance quote10
budget car insurance quote 11
I'm writing a function and keep getting the following error message.
myfunc - function(lst) {
lst - list(roots = c(car insurance, auto insurance),
roots2 = c(insurance), prefix = c(cheap, budget),
prefix2 = c(low cost), suffix = c(quote, quotes),
suffix2 = c(rate, rates), suffix3 = c(comparison))
I'd like to conduct one-way ANOVA's on multiple IV's. Is there a function for
aov (y~all of my IV's)? Thank you!
--
View this message in context:
http://r.789695.n4.nabble.com/ANOVA-with-many-IV-s-tp3583788p3583788.html
Sent from the R help mailing list archive at Nabble.com.
I'm trying to run a function inside a function but get an error message.
lst - list(roots = c(car insurance, auto insurance),
roots2 = c(insurance), prefix = c(cheap, budget),
prefix2 = c(low cost), suffix = c(quote, quotes),
suffix2 = c(rate, rates), suffix3 = c(comparison))
myfunc -
Hello,
I know of some various methods out there to utilize multiple processors but
am not sure what the best solution would be. First some things to note:
I'm running dependent simulations, so direct parallel coding is out
(multicore, doSnow, etc).
I'm on Windows, and don't know C. I don't plan
Hi:
You can try something like this: assuming the factor variables of
interest and the response variable are in a data frame named df,
ivset - c(comma separated vector of factor names)
myaovs - lapply(ivset, function(x) {
form - as.formula(substitute(yvar ~ foo, list(foo = as.name(x
Hi Abraham,
Your example runs fine for me. I get this as the newdf object (you
should be able to copy and paste into your console):
newdf - structure(list(keyword = structure(c(7L, 3L, 5L, 1L, 8L, 4L,
6L, 2L, 15L, 11L, 13L, 9L, 16L, 12L, 14L, 10L, 20L, 18L, 19L,
17L, 23L, 21L, 24L, 22L, 27L,
Hi Abraham,
mylist - list(roots = car, prefix = cheap)
myfoo - function(x) {
print(mylist$x)
}
myfoo(roots) ## fails, but in a sneaky way
## you actually extract variable x from mylist
## but there is no variable x (it is just NULL)
## so while no error is thrown, you get nothing
myfoo -
Hi Allen and list,
See the code below. I've tried it on R2.13 and R2.8.0 using either
heatmap.plus 1.3 or the latest. All gave the same results. The problem is in
the last line: when I tried to plot two different color bars, the one
corresponding to cm.colors(10) is not correct (it starts
I'm using package.skeleton() windows 7, 64 bit.
When I try to specify the code_files
package_skeleton(code_files = some directory)
I get a warning that that the connection cannot be opened and
I get a Permissions denied error.
I'm running R as admin and I've given everybody full
here i wrote a step by step tutorial.
http://stevemosher.wordpress.com/2011/06/09/making-simple-packages-in-r-on-windows/
On Wed, Jun 8, 2011 at 1:17 PM, Xia.Li oddity...@gmail.com wrote:
Hello R users,
I have difficulties when trying to make R packages. I tried to read many
tutorials but
It is not nearly as complicated as Dennis Murphy makes out.
Just do
aov(y ~ ., data=X)
where X is your data frame with one column (the response) name ``y''
and then any number of other columns which will then form the
predictors (which may be either numeric predictors or factors).
Hi Sam,This is exactly what I wanted. Could you please explain the code ? what
does 15, 0.65 and 0.25 stand for ?Nandini
Date: Wed, 8 Jun 2011 15:16:45 -0700
From: ml-node+3583766-897200094-233...@n4.nabble.com
To: nandini...@hotmail.com
Subject: Re: Histogram
I think the command
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