David Winsemius wrote:
On Jul 26, 2010, at 2:36 PM, xin wei wrote:
hi, this is more a statistical question than a R question. but I do
want to
know how to implement this in R.
I have 10,000 data points. Is there any way to generate a empirical
probablity distribution from it (the problem
Michael Lachmann wrote:
Hi,
R seems to have a feature that isn't used much, which I don't really
now how to call. But, the dimnames function, can in addition to giving
names to rows/columns/dim 3 rows/dim 4 rows... can also give labels to
the dimensions themselves.
Actually, this is
Hi:
Another approach might be to use the melt() function in package reshape
before creating the plot with xyplot, something along the lines of the
following:
library(reshape)
mdat - melt(data, id = 'X')
This should create a data frame with three columns: X, variable (all the D*
names as factor
If I run proc.time() function, I would get following:
proc.time()
user system elapsed
2.82 4.18 792.39
However I am struggling the meaning of the object what it returned. In
help file it says that:
user time is the time required to execute the calling process. Here
what is the
Hi R,
I was using 'as.dendrogram' with the DIST coefficient, where the smaller
values of the DIST coefficient, say that the objects are closer to each
other, while the larger values of the coefficient say that the objects
are far from each other.
But now, I have my coefficient as the DICE
Hi:
On Mon, Jul 26, 2010 at 11:36 AM, xin wei xin...@stat.psu.edu wrote:
hi, this is more a statistical question than a R question. but I do want to
know how to implement this in R.
I have 10,000 data points. Is there any way to generate a empirical
probablity distribution from it (the
Hi Dennis,
you should take a look at the CRAN task view for distributions
http://cran.r-project.org/web/views/Distributions.html
Beside that our distr-family of packages might be useful, see also
http://www.jstatsoft.org/v35/i10/
Dear all,
Sorry to bother you with a question that must be a FAQ, but I'm not
clever enough to have Google give me the answer.
I want to filter out some bytes that I read in a file, before applying
rawToChar (to prevent null character problems)
So, I tried things like match(some_byte,
On Tue, Jul 27, 2010 at 5:12 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
The following code should return 1, but it returns 0:
source(http://r-bc.googlecode.com/svn/trunk/R/bc.R;)
bc(9 % 2)
See FAQ 2 on the r-bc package home page:
http://r-bc.googlecode.com
Thanks to all
Thank you! As an R newbie I sometimes find it confusing that in R everything
can be done in at least four different ways. My question about S4/OOP was
motivated by the fact that those two class systems have the same designer
(Chambers). It is as if Chambers has designed an S5 system that
You failed to supply a raw value to match, and if you do it works:
match(as.raw(2), a)
[1] 9
for the ninth value is *not* '2, indeed': it is as.raw(2).
I think you are confusing R objects with their printed representation.
(In this case values are coerced to strings, and a[9] is thus coerced
Hi
r-help-boun...@r-project.org napsal dne 27.07.2010 00:48:52:
On Jul 26, 2010, at 10:56 AM, Steffen Uhlig wrote:
Dear David, Petr, and Alain,
thank you very much for your fast responses. It's a typical
handbook-not-read-error at my side. I will dig deeper into the
This is reporting on the accumulated values of the user CPU, system
CPU and elapsed time. The 'user + system' values indicate how much
CPU has been used to process whatever has occurred in the R session so
far. To get the time to execute a statement, or block of code, you
can use 'system.time'
Hi
you can look at package cluster or maybe mvpart or tree. You could also
look to CRAN search facilities where you can find other possible packages.
BTW there is no attached sample data, the list has strict policy for
allowed attachments. See Posting Guide
Regards
Petr
On Jul 26, 2010, at 11:54 PM, Murray Jorgensen wrote:
I am introducing the scan() function to my class. Consider the
following file (Scanexamp.Rnw )
\documentclass[12pt]{article}
\begin{document}
=
height = scan()
64 62 66 65 62
69 72 72 70
Have you considered adding an empty line or
Which mgcv version are you using, and which package (and version number) are
you using for the `contrast' function? (I assume you mean `gamm' below, btw)
On Thursday 22 July 2010 20:13, Xia Li wrote:
Dear All,
I met problems when doing contrast and now really need some help in the
model
Dear Prof Ripley,
You wrote :
You failed to supply a raw value to match, and if you do it works:
match(as.raw(2), a)
[1] 9
for the ninth value is *not* '2, indeed': it is as.raw(2).
Thanks for the clarification.
I think you are confusing R objects with their printed representation.
Dear list members
I am calculating kernel home-ranges and would like to visualize them with the
function image(). The xy-values consist of longitude and latitude data.
example:
x y
36154.97 355143.0
R-code:
xy-hares[,c(x,y)]
id-hares[,date]
(ud1 - kernelUD(xy, id))
ud2 -
Both suggestions generate similar errors to those of the original code.
I would also be worried if the results would not puzzle my students.
But thanks! Murray
David Winsemius wrote:
On Jul 26, 2010, at 11:54 PM, Murray Jorgensen wrote:
I am introducing the scan() function to my class.
On Tue, 27 Jul 2010, bruno Piguet wrote:
Dear Prof Ripley,
You wrote :
You failed to supply a raw value to match, and if you do it
works:
match(as.raw(2), a)
[1] 9
for the ninth value is *not* '2, indeed': it is as.raw(2).
Thanks for the clarification.
I only recently discovered options(available_packages_filters =
list(add = TRUE, license/FOSS)) [cf. help(available.packages,
package=utils) in R 2.10.0 or later] which goes nicely with my
options(checkPackageLicense = TRUE) [new in R 2.11].
But now I want to purge my library of packages I
on 2010/7/27, Prof Brian Ripley wrote :
OK. Had I tried 02, it would have worked, but only by chance.
No, not 'by chance': that is what the help page for match() tells you
happens.
Well, i mean that *I* would have typed 02 by chance.
I read, but didn't fully understand the sentence
Hello Shubha,
To my understanding,
Once you get a dist matrix, you would wish to put it into a clustering
algorithm, who's output you would then put into a plot (which will produce a
dendrogram).
You might want to have a look at the steps given here:
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of xin wei
Sent: Monday, July 26, 2010 11:36 AM
To: r-help@r-project.org
Subject: [R] how to generate a random data from a empirical
distribition
hi, this is more a
Allan Engelhardt allane at cybaea.com writes:
I only recently discovered options(available_packages_filters =
list(add = TRUE, license/FOSS)) [cf. help(available.packages,
package=utils) in R 2.10.0 or later] which goes nicely with my
options(checkPackageLicense = TRUE) [new in R
Hi,
Recently I started working with netcdf IPCC climate model data and I am
using R to analyze these data. Some problems occur while doing this, as
I am using the output of different climate models which are different in
for example time calendar and grid coordinates.
To analyze monthly
On Jul 27, 2010, at 7:01 AM, Murray Jorgensen wrote:
Both suggestions generate similar errors to those of the original
code. I would also be worried if the results would not puzzle my
students.
You are teaching them about R or about Sweave? You are setting up code
that is designed to
Hi!
I have a list of 24 elements, all of the same type (dataframe, for example).
I am looking for an alternative to mylist[[1]] + mylist[[2]] + ... +
mylist[[24]] to obtain the sum.
Anyone can help me?
Thanks in advance.
Nicola S.
[[alternative HTML version deleted]]
Have a look at Reduce(), e.g.,
Reduce(+, your.list)
I hope it helps.
Best,
Dimitris
On 7/27/2010 2:36 PM, Nicola Sturaro Sommacal wrote:
Hi!
I have a list of 24 elements, all of the same type (dataframe, for example).
I am looking for an alternative to mylist[[1]] + mylist[[2]] + ... +
Hi,
do.call(sum, mylist)
?do.call
baptiste
On 27 July 2010 14:36, Nicola Sturaro Sommacal
mailingl...@nicolasturaro.com wrote:
Hi!
I have a list of 24 elements, all of the same type (dataframe, for example).
I am looking for an alternative to mylist[[1]] + mylist[[2]] + ... +
Hi,
Is there any way to estimate a DEPENDENT variable through a GLM/LM model?
Suppose I have the linear model: y=a0+a1*x1+a2*x2 (a0=1, a1=0.6, a2=0.8,
x1~N(1,1), x2~N(0,1)).
The alphas and the auxiliary variables are given and I have to estimate y.
The point is if I estimate it, let¹s say
On 27/07/10 13:20, Ben Bolker wrote:
Allan Engelhardtallaneat cybaea.com writes:
I only recently discovered options(available_packages_filters =
list(add = TRUE, license/FOSS)) [cf. help(available.packages,
package=utils) in R 2.10.0 or later] which goes nicely with my
One of the nice features of R's formula syntax is that
you can create a character string containing a formula,
and pass it to the formula() function. For example:
xyplot(formula(paste(paste(paste('D',1:10,sep=''),collapse='+'),'X',sep='~')),data)
will do what you want.
Hi,
I'm trying to plot a graph with error bars using xYplot in the Hmisc
package. My data looks like this.
mortstand sitetype
0.042512776 0.017854525 Plot A ST
0.010459803 0.005573305 PF ST
0.005188321 0.006842107MSFST
0.004276068
Hi,
On Tue, Jul 27, 2010 at 4:14 AM, Albert-Jan Roskam fo...@yahoo.com wrote:
Thank you! As an R newbie I sometimes find it confusing that in R everything
can be done in at least four different ways. My question about S4/OOP was
motivated by the fact that those two class systems have the
On Tue, Jul 27, 2010 at 7:17 AM, Allan Engelhardt all...@cybaea.com wrote:
I only recently discovered options(available_packages_filters = list(add =
TRUE, license/FOSS)) [cf. help(available.packages, package=utils) in R
2.10.0 or later] which goes nicely with my options(checkPackageLicense =
If the x-axis variable is really a factor, xYplot will not handle it.
You probably need a dot chart instead (see Hmisc's Dotplot).
Note that it is unlikely that the confidence intervals are really
symmetric.
Frank
On Tue, 27 Jul 2010, Kang Min wrote:
Hi,
I'm trying to plot a graph with
Pablo, we've had success using
http://mephisto.unige.ch/traminer/preview.shtml to look at marketing paths.
Question would be how many distinct case step discriptions are there?
HTH, Jim
On Jul 26, 2010 9:44 AM, Pablo Cerdeira pablo.cerde...@gmail.com wrote:
Hi all,
I have no idea if this
Hi Ya'll
I'm trying to install the zoo library... I've downloaded about 5 or 6 of
their releases (inc. the latest one), all of them giving me the same syntax
error when trying to install...
Anyone seen this problem before? its not the lib argument that is the
problem (ive been installing
On Tue, Jul 27, 2010 at 10:02 AM, Rnoobie j...@stellaconcepts.com wrote:
Hi Ya'll
I'm trying to install the zoo library... I've downloaded about 5 or 6 of
their releases (inc. the latest one), all of them giving me the same syntax
error when trying to install...
Anyone seen this problem
On Tue, Jul 27, 2010 at 9:48 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Tue, Jul 27, 2010 at 7:17 AM, Allan Engelhardt all...@cybaea.com wrote:
I only recently discovered options(available_packages_filters = list(add =
TRUE, license/FOSS)) [cf. help(available.packages,
Hi,
Could anyone help me understand how the mincriterion threshold works in
ctree and cforest of the party package? I've seen examples which state that
to satisfy the p 0.05 condition before splitting I should use mincriterion
= 0.95 while the documentation suggests I should use mincriterion =
Hi Charles,
On Fri, 2010-07-23 at 14:40 -0700, Charles C. Berry wrote:
On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
Sorry. I should have included some data. I've attached a subset of my
data (50/192) cases in a Rdata file and have pasted it below.
Running anova I get the
If I understand you correctly, set the filter and use packageStatus().
Its summary() method tells you which packages you have installed which
are 'unavailable'. E.g. my Mac (with pkgType = source) shows in
R-devel
summary(packageStatus(.libPaths()[1]))$Libs[[1]]$unavailable
[1] BayesX
On Tue, 27 Jul 2010, Matthew OKane wrote:
Hi,
Could anyone help me understand how the mincriterion threshold works in
ctree and cforest of the party package? I've seen examples which state that
to satisfy the p 0.05 condition before splitting I should use mincriterion
= 0.95 while the
Hi Ana,
Does the predict function do what you want? Type in ?predict.lm
--Gray
On 7/27/10, Ana De Barros belindadebar...@gmail.com wrote:
Hi,
Is there any way to estimate a DEPENDENT variable through a GLM/LM model?
Suppose I have the linear model: y=a0+a1*x1+a2*x2 (a0=1, a1=0.6, a2=0.8,
Dear all,
I am struggling with the calculation of standard error of the coefficient in
Binary logistic regression analysis.
I built a binary logsitic regression model as follows and got confused since
the calculation of standard error of coefficients of X1, X2 and X3 are not
the same as the
Great, thanks. I couldn't quite get your syntax to work, but
z - packageStatus(.libPaths()[1])[[1]]
unname( z$Package[z$Status == unavailable] )
seems to do the trick for me.
Thanks again.
Allan
On 27/07/10 16:31, Prof Brian Ripley wrote:
If I understand you correctly, set the filter and
On Tue, 27 Jul 2010, Allan Engelhardt wrote:
Great, thanks. I couldn't quite get your syntax to work, but
Did you use R-devel? The syntax has changed ... and that's why I said
'e.g.'.
z - packageStatus(.libPaths()[1])[[1]]
unname( z$Package[z$Status == unavailable] )
seems to do the
On Tue, 27 Jul 2010, Christopher David Desjardins wrote:
Hi Charles,
On Fri, 2010-07-23 at 14:40 -0700, Charles C. Berry wrote:
On Fri, 23 Jul 2010, Christopher David Desjardins wrote:
Sorry. I should have included some data. I've attached a subset of my
data (50/192) cases in a Rdata file
Thanks for the quick reply. That makes sense, although I am still confused
why the functions prevent mincriterion from being lower than 0 as to run
cforests with testtype as Teststatistic and grow maximum depth trees you
need to put in qnorm(0) (= -inf) which is impossible? I will stick to other
On 7/27/2010 6:00 AM, r-help-requ...@r-project.org wrote:
Date: Mon, 26 Jul 2010 11:36:29 -0700 (PDT)
From: xin weixin...@stat.psu.edu
To:r-help@r-project.org
Subject: [R] how to generate a random data from a empirical
distribition
Message-ID:1280169389379-2302716.p...@n4.nabble.com
Not to beat a dead horse...
I've found that I like the useR conferences more than most statistics
conferences. This isn't due to the difference in content, but the
difference in the audience and the environment.
For example, everyone is at useR because of their appreciation of R.
At most other
On Mon, 26 Jul 2010, Alon Friedman wrote:
Hi
I am looking for R templates to introduce the R to my students at
Seton hall university. The templates are predefined scripts in R that
will retain its primary intent when individually customized with their
own variable data or text.
Well the help
Do not worry about the SE. The SE listed on the output is the SE of the log
odds. You can use the estimate (beta) and SE from the listing to compute a
confidence interval (CI)as follows:
CI exp(beta-1.96*SE) to exp(beta-1.96*SE)
John
John Sorkin
Chief Biostatistics and Informatics
Univ. of
Hi Allan,
It helps a lot. I´ll try to read more about it.
But, as you asked me, here goes a brief explanation about the necessary
columns of the sample date paste at the end:
id_processo: identify a legal case, it is its primary key.
ordem_andamento: is the step number inside a legal case
On Tue, 27 Jul 2010, John Sorkin wrote:
Do not worry about the SE. The SE listed on the output is the SE of the log
odds. You can use the estimate (beta) and SE from the listing to compute a
confidence interval (CI)as follows:
CI exp(beta-1.96*SE) to exp(beta-1.96*SE)
The standard errors
On Tue, 27 Jul 2010, Murray Jorgensen wrote:
I am introducing the scan() function to my class. Consider the following file
(Scanexamp.Rnw )
\documentclass[12pt]{article}
\begin{document}
=
height = scan()
64 62 66 65 62
69 72 72 70
part = scan(what = character(0))
Soprano Soprano Soprano
Hi Jim,
Ow! Very nice job at http://mephisto.unige.ch/traminer/preview.shtml I´m
going to read more about it.
I have a lot of different steps, in a sequence. Actually, 586 different
possible steps, but I have 4269 legal cases, with a maximum of 379 steps
each one.
If you want, I can send this
Hi,
Just to extend the excellent suggestions, if you are interested in the
odds ratio, you can just use exp():
#Odds Ratio
exp(fit4$coefficients)
#Confidence interval around OR
exp(confint(fit4))
To give you an idea graphically of the log odds (or logit) look at:
p - seq(0, 1, by = .001)
Hi,
I am running the following model:
fit1.full - coxme(Surv(age_sym1, sym1) ~ sex + lifedxm*sex + (1|famid),
data=bip.surv)
I would like to extract the AIC from that object to calculate the AICC.
However, when I look at str(fit1.full) and summary(fit1.full) (pasted
below) I don't see anything
When I ask R to compute: predict(data, int = c), following a linear
regression, what is it computing exactly? How are these lower and upper
prediction limits different than what I would get for confidence limits?
Thanks,
Matt.
[[alternative HTML version deleted]]
I am writing a function where the arguments are names of objects or variable
names in a data frame. To convert the strings to the objects I am using
eval(parse(text=name)):
f.graph.two.vbs-function(dataname,v1){
val-paste(dataname,v1,sep=$)
val-eval(parse(text=val))
val
}
Dear Jorge,
It might help if you included some sample data that replicates your
problem, or, perhaps, the results of sessionInfo(). I cannot
replicate it. Here are my results:
data - data.frame(A = 1:10, RECORD = 1:10)
f.graph.two.vbs-function(dataname,v1){
+ val-paste(dataname,v1,sep=$)
Hi Jose,
What's the problem?
# some data
set.seed(123)
mydata - data.frame(x = rnorm(10), y = rpois(10, 10))
mydata
f.graph.two.vbs-function(dataname,v1){
val-paste(dataname,v1,sep=$)
val-eval(parse(text=val))
val
}
f.graph.two.vbs('mydata', 'x')
[1] -0.56047565
Well, the data set I am using is quite large, but RECORD is also a numeric
variable and data is a data frame.
Here is my sessionInfo():
sessionInfo()
R version 2.11.0 (2010-04-22)
i386-apple-darwin9.8.0
locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1]
On 27/07/2010 2:34 PM, Jorge A. Ahumada wrote:
I am writing a function where the arguments are names of objects or variable
names in a data frame. To convert the strings to the objects I am using
eval(parse(text=name)):
f.graph.two.vbs-function(dataname,v1){
Thanks for all the help.
I had tried using the index in caret to try to dictate which rows of the
sample would be used in each of the tree building in RF. (e.g. use all data
from A B site for training, hold out all data from C site for testing etc)
However after running, when I cross-checked
I want to plot points with different colors to show different
selections of points in a 3d plot.
My problem is that if I plot a red point at a location where already a
blue point was plotted, than the point is still blue.
Is there a parameter or so which can be used to draw over a existing
Hello,
I'm using the package survey, version 3.22-2 in R 2.11.1, on windows
XP, but this isn't a platform issue.
I am trying to calculate some regression and probability proportional
to size (pps) estimates for a dataset I have. The data has 2 strata,
weights, and cluster IDs. However, I can't
Yes, this solved the problem. Thanks.
Jorge
On 7/27/10 3:13 PM, Duncan Murdoch murdoch.dun...@gmail.com wrote:
On 27/07/2010 2:34 PM, Jorge A. Ahumada wrote:
I am writing a function where the arguments are names of objects or
variable
names in a data frame. To convert the strings to the
Hello,
I see some people including myself confused by the different object-oriented
approaches in R (S3, S4, OOP, R.oo etc.).
Would it be ok to collect examples and solutions for the different OO-packages
in one package and add a vignette for documentation?
(assuming I find time for this
I've found that opening a connection, and scanning (in a loop)
line-by-line, is far faster than either read.table or read.fwf. E.g,
here's a file (temp2) that has 1500 rows and 550K columns:
showConnections(all=TRUE)
con - file(temp2,open='r')
system.time({
for (i in 0:(num.samp-1)){
The index indicates which samples should go into the training set.
However, you are using out of bag sampling, so it would use the whole
training set and return the OOB error (instead of the error estimates
that would be produced by resampling via the index).
Which do you want? OOB estimates or
Hi,
I ran R (version 2.9.0) CMD build under root in Fedora (9). When it
tried to remove junk files it removed EVERYTHING in my local
account! (See below).
Can anyone tell me what happened, and even more importantly if I can I
restore what was lost.
Panickingly,
Jarrod
[jar...@localhost
On Tue, Jul 27, 2010 at 4:17 PM, schuster m...@friedrich-schuster.de wrote:
Hello,
I see some people including myself confused by the different object-oriented
approaches in R (S3, S4, OOP, R.oo etc.).
S3 is the primary OO approach in R and if you are new you should
initially focus on that
On 27/07/2010 3:20 PM, rrich...@fh-lausitz.de wrote:
I want to plot points with different colors to show different
selections of points in a 3d plot.
My problem is that if I plot a red point at a location where already a
blue point was plotted, than the point is still blue.
Is there a
This is a frequently asked/answered question (7.21 in the FAQ). What searching
did you do and why did it not find this FAQ or previous discussion of it? How
could the documentation/search/etc. be improved so that you (and the next n
people with this question) will find the answer easier?
The
myDF:
d1 d2 d3 d4
d5
-0.1669103510.022304377 -0.00825924 0.008330689 -0.000925938
-0.1669103510.022304377 -0.00825924 0.008330689 -0.000925938
-0.1669103510.022304377 -0.00825924
Hopefully the right now will be a forever. The fact that trellis graphics do
not support hatching is a conscious decision or feature, not a bug or missing
feature to be corrected some day (at least I for one will be disappointed if
Deepayan (or someone else) gives in and implements it).
Have
It all depends on what you are going with the data. First in your
scan example, I would not read in a line at a time, but probably
several thousand and then process the data. Most of your time is
probably spent in reading. I assume that you are not reading it all
in at once (but then maybe you
Write a function that incorporates doit and the column shuffle.
Let's call it doitbetter
replicate(1, doitbetter())
You'll probably want to read the help for replicate to make sure the
defaults are what you want.
--Gray
On Tue, Jul 27, 2010 at 4:43 PM, jd6688 jdsignat...@gmail.com wrote:
On 27/07/2010 4:39 PM, Jarrod Hadfield wrote:
Hi,
I ran R (version 2.9.0) CMD build under root in Fedora (9). When it
tried to remove junk files it removed EVERYTHING in my local
account! (See below).
Can anyone tell me what happened, and even more importantly if I can I
restore what was lost.
On Jul 27, 2010, at 4:53 PM, Duncan Murdoch wrote:
On 27/07/2010 4:39 PM, Jarrod Hadfield wrote:
Hi,
I ran R (version 2.9.0) CMD build under root in Fedora (9). When it
tried to remove junk files it removed EVERYTHING in my local
account! (See below).
Can anyone tell me what happened, and
Hi all,
I am dealing with very large numbers but I am only interested in their last
digits.
244^244
[1] Inf
As far as I can tell, the largest number R can take has 308 digits (1+E308).
Is there a way I could see the last digits only of 244^244?
Many thanks for your help.
Vincent Deluard,
Another option for fitting a smooth distribution to data (and generating future
observations from the smooth distribution) is to use the logspline package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original
I am trying to do the following to accomplish the tasks, can anybody to
simplify the solutions.
Thanks,
for (i in 1:1){
d-apply(s,2,sample)
pos_neg_tem-t(apply(d,1,doit))
if (i1){
pos_neg_pool-rbind(pos_neg_pool,pos_neg_tem)
}else{
pos_neg_pool- pos_neg_tem
}}
--
View this
You can get the 'bc' package for R, or just use 'bc' itself:
bc 1.05
Copyright 1991, 1992, 1993, 1994, 1997, 1998 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'.
244^244
Easiest thing is to sample with replacement from the original data.
This is the idea behind the bootstrap, which is sampling from the
empirical CDF.
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
It looks to me like you keep sampling from some dataset 's' 10,000
times. Since you can sample() with replacement, I wonder if you could
just take a sample of the size you want, rather than using a loop with
sample. Perhaps along these lines:
d - apply(s, 2, sample, size = 1*nrow(s),
Colleagues
I am trying to replicate some analyses performing in SAS using a rank ANCOVA
model. I assume that this is a non-parametric ANOVA with covariates but I
have not been able to get confirmation from the people who did the original
analyses. Can anyone direct me to the comparable
Hello,
I am trying to use copula-GARCH to model a multivariate time series. So far, I
have:
-Estimated the GARCH(1,1) model
-Obtained and standardized the residuals
-Applied the pdf (Student's t in this case) to obtain (pseudo) uniform variables
-Estimated the copula
-Obtained a random sample
Look at the HSAUR package (and the book that it goes with). This may give you
(your students) enough to start on several topics. If you use these examples
then you should probably include the book as at least an optional text for the
class.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data
As far as I can tell, the largest number R can take has 308 digits
(1+E308).
This is not a correct statement. The magnitude of the largest 64-bit double
precision floating
point number in R is approximately 1E308. The internal storage in R (and
in most computers today)
contains only the
This has been shown to yield unreliable analyses. Use the more formal
proportional odds ordinal logistic model. This is a generalization of
the Wiloxon-Mann-Whitney-Kruskal-Wallis statistic. This is
implemented in the rms package and elsewhere.
Frank E Harrell Jr Professor and Chairman
Hi I want to plot an x,y plot something like an scatter plot.
I always have the same doubt, were is the last point of my file?
Imagine it is a time series so I want the last point to indicate with an
arrow (but athomatically if posible).
Someone knows if it is posible?
Could it be posible to
It is probably easier to do this without eval and parse (see fortune(106)).
Something like:
val - get(dataname)[[v1]]
hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From:
Hi,
Here are two options. I would colour the last point differently, or
make it a different shape rather than using an arrow, but its your
choice.
x - 1:10
y - 1:10
#With an arrow
plot(x, y)
arrows(x0 = x[length(x)], y0 = y[length(y)] - 1,
x1 = x[length(x)], y1 = y[length(y)] - .5)
It won't happen automatically unless you make it do so yourself.
Fortunately this is not hard. See,
?arrows
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Trying To learn again
Sent: Wednesday, 28 July 2010 7:07 AM
To:
I'm somewhat a new user and have been trying to figure out how to repeat
rows a certain number of time based on a variable. Currently, the number of
rows is not reflective of the number of observations. To get the number of
observations (n=22 in this case), I have to multiply by the variable
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