On 06/26/2010 11:20 PM, Karl Brand wrote:
Dear List,
I'd really appreciate tip's or code demonstrating how i can achieve some
common axis labels integrated into a multiple plot.
In my example (below), i'm trying to achieve:
-a single "Results 1 (Int)" centered & btwn row 1 and row 2;
-a single
On Sun, 27 Jun 2010, Sebastian Gibb wrote:
Hello,
is there a possibilty in R to convert numbers (double precision, 64bit) into
single precision ones (32bit).
I need that for compatibility reasons. Until now I call a C application which
casts a double to a float.
float precision32(double value)
Hello,
is there a possibilty in R to convert numbers (double precision, 64bit) into
single precision ones (32bit).
I need that for compatibility reasons. Until now I call a C application which
casts a double to a float.
float precision32(double value) {
return (float)value;
}
But I want to u
-- Forwarded message --
From: John-Paul Bogers
Date: Sat, Jun 26, 2010 at 10:14 PM
Subject: Re: [R] Recoding dates to session id in a longitudinal dataset
To: jim holtman
Dear Jim,
he data concerns HPV screening data.
The data looks as follows
pat1 sampledate1 HPV16 0.3
pat2 sa
A slightly better scaling is the following:
par.scale <- c(1.e06, 1.e06, 1.e-05, 1) # "q" is scaled differently
> SPoptim <- optim(pars, SPsse, B=d$catch, CPE=d$cpe, control=list(maxit=1500,
> parscale=par.scale))
> SPoptim
$par
B0Kqr
7.320899e+05
2010/6/27 顾小波 :
> Hi,
>
> I post this message to the general r-help list hoping anyone within a wider
> range have suggestions:
>
>
>
> There are three ways to integration R and postgres, especially on 64bit
> Microsoft windows Platform,
>
>
>
> 1. via RODBC package, which has 32 bit and 64 bit v
Derek,
The problem is that your function is poorly scaled. You can see that the
parameters vary over 10 orders of magnitude (from 1e-04 to 1e06). You can get
good convergence once you properly scale your function. Here is how you do it:
par.scale <- c(1.e06, 1.e06, 1.e-06, 1.0)
SPoptim <
Hi,
I post this message to the general r-help list hoping anyone within a wider
range have suggestions:
There are three ways to integration R and postgres, especially on 64bit
Microsoft windows Platform,
1. via RODBC package, which has 32 bit and 64 bit version for windows
2. via RPostg
Atte, note the similarity between what Greg described and a bootstrap. The
difference to a true bootstrap is that in Greg's version you subsample the
population (or in other instances the data). This is known as subsampling
bootstrap and discussed in Politis, Romano, and Wolf (1999).
HTH,
Daniel
Why do you use *double* square brackets on the left side of the replacement?
>From the help info for "[[":
"The most important distinction between [, [[ and $ is that the [ can select
more than one element whereas the other two select a single element."
You seem to be selecting 20 elements.
--
Thanks! Works like a charm. -Seth
--
View this message in context:
http://r.789695.n4.nabble.com/use-a-data-frame-whose-name-is-stored-as-a-string-variable-tp2269095p2269732.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-p
If I understand correctly from their website, "discrete choice" models
are mostly generalized linear models with the common link functions
for discrete data? Apart from a few names I didn't recognize, all
analyses seem quite "standard" to me. So I wonder why you would write
the log-likelihood yours
Dear R-help users,
I'd like to use the R-package "pls" and want to extract the explained
Y-variance to identify the important (PLS-) principal components in my
model, related to the y-data. For explained X-variance there is a
function: "explvar()". If I understand it right, the summary()
It does work, thank you, but the literal 5 do.call(subset, list(dat, expression(5mailto:bill.venab...@csiro.au]
Sent: Saturday, June 26, 2010 1:08 AM
To: Vadim Ogranovich; r-help@r-project.org
Subject: RE: [R] subset arg in subset(). was: converting result of substitute
to 'ordidnary' expression
Thanks! The results were similar to the t.test p-values show (I have
four samples).
Thank you also for using that replicate-function which i didn't know.
Till now I have just used for-loops that are not so beautiful... i
don't know about the speed. Have to test that.
Atte
Greg Snow kirjoitt
Your function is very irregular, so the optim is likely to return
local minima rather than global minima.
Try different methods (SANN, CG, BFGS) and see if you get the result
you need. As with all numerical optimsation, I would check the
sensitivity of the results to starting values.
Ni
On Sat, 26 Jun 2010, Bryan Hanson wrote:
Thanks Chuck, I understand much better what is going on with your example.
But I'm still uncertain why the b2$t array does not have the dimensions of R
x no. of strata.
Because the test statistic returned by mm() is a scalar. It has nothing to
do with
No I mean something like this, assuming that the iris dataset contains the full
population and we want to see if Setaso have a different mean than the
population (the null would be that there is no difference in sepal width
between species, or that species tells nothing about sepal width):
out
I am trying to use optim() to minimize a sum-of-squared deviations function
based upon four parameters. The basic function is defined as ...
SPsse <- function(par,B,CPE,SSE.only=TRUE) {
n <- length(B) # get number of years of data
B0 <- par["B0"]
Hi,
I want to estimate a dynamic paneldata model with the following code, but
unfortenately I received the error message below.
form<-PB~Activity+Solvency+Cap_Int
dynpanel<-pgmm(dynformula(form,list(1,1,1,1)),data=panel[1:2185,1:37],effect="twoways",model="onestep",index=c("Aktie","Datum"),g
Look at the summary.formula function inside package Hmisc
Christos
> Date: Sat, 26 Jun 2010 05:17:34 -0700
> From: raoul.t.dso...@gmail.com
> To: r-help@r-project.org
> Subject: [R] Calculating Summaries for each level of a Categorical variable
>
>
> Hi,
>
> I have a dataset which has a categ
Thanks Chuck, I understand much better what is going on with your example.
But I'm still uncertain why the b2$t array does not have the dimensions of R
x no. of strata.
Any further insight would be appreciated. Bryan
*
Bryan Hanson
Acting Chair
Professor of Chemistry & Biochemistry
De
On Sat, Jun 26, 2010 at 7:42 AM, Tal Galili wrote:
> And there are also the "brew", and "Sweave" packages (as Henrique
> mentioned).
>
Also, odfWeave and Sweave via LyX. I believe that this is FAQed.
Liviu
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https://st
Hi Martin,
Thanks a lot for your advice.
I tried the process you suggested as below, it worked, but in a
different way that I planned.
library(Biostrings)
x <- c("ACTCCCGCCGTTCGCGCGCAGCATGATCCTG",
"ACTCCCGCCGTTCGCGCGC",
"CAGGATCATGCTGCGCGCGAACGGCGGGAGT",
"CAGGATCATG
On 2010-06-26 3:52, (Ted Harding) wrote:
On 25-Jun-10 21:46:13, Albert Lee, Ph.D. wrote:
How do I become a member of R user community?
Albert Lee,
Ph.D. statistician
1. By using R
2. By subscribing to the R-help mailing list and keeping in touch
with the rest of us!
To subscribe your em
On 26 Giu, 17:19, Allan Engelhardt wrote:
> On 26/06/10 16:07, Muenchen, Robert A (Bob) wrote:
>
> > I've been trying to make sense of Google Scholar searches. I'm obviously
> > missing something basic. Here are two searches onwww.google.com:
>
> > sas - gets 68M hits
> > sas OR spss - gets 74.3M
Thanks Bill, that worked great!!
> You ask:
# How can I use
> predict here, 'newdata'
>
> crashes
predict(m1,newdata=wolf$predicted);wolf # it doesn't work
To
>
> use predict() you need to give a fitted model object (here m1) and a
> *data
> frame* to specify the values of the predicto
On Sat, 26 Jun 2010, Bryan Hanson wrote:
Hello All. I must be missing the really obvious here:
mm <- function(d, i) median(d[i])
b1 <- boot(gravity$g, mm, R = 1000)
b1
b2 <- boot(gravity$g, mm, R = 1000, strata = gravity$series)
b2
Both b1 and b2 seem to have done (almost) the same thing, but
Bob, i'm confused.
You try a search with "Google Scholar" or with "Google Insights for search"?
---
useful references for "Google Insights for search":
* matching terms:
http://www.google.com/support/insights/bin/answer.py?hl=en&answer=94777
* interpreting search volumes:
http://www.google.com/s
On 26/06/10 16:07, Muenchen, Robert A (Bob) wrote:
I've been trying to make sense of Google Scholar searches. I'm obviously
missing something basic. Here are two searches on www.google.com:
sas - gets 68M hits
sas OR spss - gets 74.3M hits. A bigger number as "OR" would imply.
But when I do t
>-Original Message-
>From: Joris Meys [mailto:jorism...@gmail.com]
>Sent: Friday, June 25, 2010 10:10 PM
>To: Muenchen, Robert A (Bob)
>Cc: Dario Solari; r-help@r-project.org
>Subject: Re: [R] Popularity of R, SAS, SPSS, Stata...
>
>>>I had taken the opposite tack with Google Trends by su
Hi,
I have a dataset which has a categorical variable "R",a count variable C
(integer) and 4 or more numeric variables (A,T,W,H - integers) containing
measures for "R". I would like to summarize each level of the variable R by
the average for A,T,W and H.
I have written a function to calculate
Thank you Henrik for your answer.
I hope now I am inline with the posting huide and perhaps I get an
answer, thank you.
> sessionInfo()
R version 2.9.0 alpha (2009-03-23 r48200)
i386-pc-mingw32
locale:
LC_COLLATE=German_Austria.1252;LC_CTYPE=German_Austria.1252;LC_MONETARY=German_Austria.1252;LC
I think the hardest thing about true EDMA (meaning the Richtsmeier and Lele
version) is the bootstrapping to get significance. Have you tried their
software?
http://www.getahead.psu.edu/resource_new.html
-
Corey Sparks, PhD
Assistant Professor
Department of Demography and Organization Studi
Did you try tapply?
?tapply
tapply(RT, RT$R, fun=WA)
or something like that
-
Corey Sparks, PhD
Assistant Professor
Department of Demography and Organization Studies
University of Texas at San Antonio
501 West Durango Blvd
Monterey Building 2.270C
San Antonio, TX 78207
210-458-3166
corey.sp
Hello All. I must be missing the really obvious here:
mm <- function(d, i) median(d[i])
b1 <- boot(gravity$g, mm, R = 1000)
b1
b2 <- boot(gravity$g, mm, R = 1000, strata = gravity$series)
b2
Both b1 and b2 seem to have done (almost) the same thing, but it looks like
the strata argument in b2 has
t(e$values * t(e$vectors))
Uwe Ligges
On 25.06.2010 20:42, rkevinbur...@charter.net wrote:
I am trying to check the results from an Eigen decomposition and I need to force a
scalar multiplication. The fundamental equation is: Ax = lx. Where 'l' is the
eigen value and x is the eigen vector co
On Fri, 2010-06-25 at 23:28 -0400, Carrie Li wrote:
> Hello everyone,
>
> I have a question about integration of two density function
> Intuitively, I think the value after integration should be 1, but they are
> not. Am I missing something here ?
>
> > t=function(y){dnorm(y, mean=3)*dnorm(y/2, m
I sent you a "doubble integration" solution a couple of days ago, that answered
your question. You did not have the coureteousy to acknowledge that. Now, you
are asking a different question that is "incorrectly" formulated. What you are
doing is not "multivariate" integration. You are just i
Dear List,
I'd really appreciate tip's or code demonstrating how i can achieve some
common axis labels integrated into a multiple plot.
In my example (below), i'm trying to achieve:
-a single "Results 1 (Int)" centered & btwn row 1 and row 2;
-a single "Results 2 (Int)" centered & btwn row 2
On 06/26/2010 02:07 PM, Duncan Murdoch wrote:
Jim Hargreaves wrote:
Hi Duncan, list,
Thanks for the advice, but unfortunately that wasn't what was causing
my problem. I'm still getting the "Recursive indexing failed at level
2" message even after replacing my unlist(pulse[i]) with pulse[[i]].
Jim Hargreaves wrote:
Hi Duncan, list,
Thanks for the advice, but unfortunately that wasn't what was causing my
problem. I'm still getting the "Recursive indexing failed at level 2"
message even after replacing my unlist(pulse[i]) with pulse[[i]].
Read the second part of my first message,
Hi Duncan, list,
Thanks for the advice, but unfortunately that wasn't what was causing my
problem. I'm still getting the "Recursive indexing failed at level 2"
message even after replacing my unlist(pulse[i]) with pulse[[i]].
Error:
>
pulse_subset[[1:as.numeric(length(pulse[[i]][as.numeric(
Hi,
I'm fairly new to R but I have a large dataset (30 obs) containing
patient material. Some patients came 2-9 times during the three year
observation period. The patients are identified by a unique idnr, the
sessions can be distinguished using the session date. How can I recode the
date of t
On 26/06/2010 8:29 AM, Jim Hargreaves wrote:
On 06/26/2010 01:20 PM, Duncan Murdoch wrote:
On 26/06/2010 7:53 AM, Jim Hargreaves wrote:
Dear fellow R users,
I am replacing elements of a list like so:
pulse_subset[[1:20]]=unlist(pulse[i])[1:20]
If pulse is a list, then pulse[i]
Hi,
I'm fairly new to R but I have a large dataset (30 obs) containing
patient material. Some patients came 2-9 times during the three year
observation period. The patients are identified by a unique idnr, the
sessions can be distinguished using the session date. How can I recode the
date of t
On 06/26/2010 01:20 PM, Duncan Murdoch wrote:
On 26/06/2010 7:53 AM, Jim Hargreaves wrote:
Dear fellow R users,
I am replacing elements of a list like so:
pulse_subset[[1:20]]=unlist(pulse[i])[1:20]
If pulse is a list, then pulse[i] is also a list, with one element. I
think you want pulse[
On 26/06/2010 7:53 AM, Jim Hargreaves wrote:
Dear fellow R users,
I am replacing elements of a list like so:
pulse_subset[[1:20]]=unlist(pulse[i])[1:20]
If pulse is a list, then pulse[i] is also a list, with one element. I
think you want pulse[[i]], which extracts element i.
If pulse_s
Dear fellow R users,
I am replacing elements of a list like so:
pulse_subset[[1:20]]=unlist(pulse[i])[1:20]
where pulse is a list of lists, and pulse [i] has >20 values.
This gives the error "Recursive Indexing failed at level 2". But,
interestingly this instruction is part of a loop which ha
first of all take a look at the object you created:
> df.list <- vector("list", 3)
>
>
> for(i in 1:3){
+ assign(paste("s",i, sep=""),matrix(0, nrow = 20, ncol = 5, byrow
+ = FALSE, dimnames = NULL))
+ }
>
> # and then insert them with a loop like this
>
> # put matrices names in a vector
> matri
b <- paste("C:\\rphp\\",arg, sep='')
On Sat, Jun 26, 2010 at 12:55 AM, Maulik Shah wrote:
> I am fitting 3 parameter model to my response matrix and want to generate
> item characterstic curve.
> I want to specify file name to save item characterstic curve by passing it
> as external parameter to
Hi I can install RWinEdt if I start R with administrator rigths, but it
does not paste my code to the console. I found advice in the link below
how to manage the problem, but it did not work, any other idea?
http://yusung.blogspot.com/2009/01/rwinedt-and-windows-vistawindow-7.html
Thanks a lot,Jo
On 25-Jun-10 21:46:13, Albert Lee, Ph.D. wrote:
> How do I become a member of R user community?
>
> Albert Lee,
> Ph.D. statistician
1. By using R
2. By subscribing to the R-help mailing list and keeping in touch
with the rest of us!
To subscribe your email address to the list, visit the R-h
Atte Tenkanen kirjoitti 26.6.2010 kello 5.15:
>
> Greg Snow kirjoitti 25.6.2010 kello 21.55:
>
>> Let me see if I understand. You actually have the data for the
>> whole population (the entire piece) but you have some pre-defined
>> sections that you want to see if they differ from the popul
I am using EDMA for comparing the interlandmark distances of two forms.
Actually there is a software called EDMA of course developed by Lele and
Richstmeier but main point is I am trying to solve quite different problem
on the same data set using R and EDMA so data entry format of EDMA software
is
Greg Snow kirjoitti 25.6.2010 kello 21.55:
> Let me see if I understand. You actually have the data for the
> whole population (the entire piece) but you have some pre-defined
> sections that you want to see if they differ from the population,
> or more meaningfully they are different from
I am fitting 3 parameter model to my response matrix and want to generate
item characterstic curve.
I want to specify file name to save item characterstic curve by passing it
as external parameter to the R batch script. The following is the code I
have written for this.
*R Script:*
library(ltm)
c
Hi there,
I cannot seem to figure out how to access the elements of a list if those
elements are a matrix.
For example I have a the following list
df.list <- vector("list", 3)
and I have made each of the elements a matrix as follows
for(i in 1:3){
assign(paste("s",i, sep=""),matrix(0, nrow =
Hi all,
Sorry to bother you. I'm estimating a discrete choice model in R using
the maxBFGS command. Since I wrote the log-likelihood myself, in order to
double check, I run the same model in Limdep. It turns out that the
coefficient estimates are quite close; however, the standard errors are v
How do I become a member of R user community?
Albert Lee,
Ph.D. statistician
Confidentiality Notice: This communication, and any file...{{dropped:12}}
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read th
Greg Snow kirjoitti 25.6.2010 kello 21.55:
> Let me see if I understand. You actually have the data for the
> whole population (the entire piece) but you have some pre-defined
> sections that you want to see if they differ from the population,
> or more meaningfully they are different from
Dear useRs, and expeRts,
tahanks
I have found idea how to add to oryginal data a column with markers to know
with all data in wchich period in c2 they are, suimply in the code I could add:
stacked_idx<-stack(idx)
merge(stacked_idx,C.df,by.x=c('values'),by.y=c('c0'), all=T)
thanks for su
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