How is the pointsize set in trellis.devices?
>From my reading of the trellis.device help file, I understood that the
pointsize arg would be referenced to the call to the pdf function.
So I set up a trellis pdf device as so:
trellis.device(device = pdf, file = "Singers.pdf", height = 160/25.4,
Please don't post HTML email.. we don't necessarily see what you see when you
do that. Read the Posting Guide for more list etiquette.
Have you played with the "usr" coordinates? Have you read the help for the par
function?
The par(usr=...) call is telling base graphics what the x and y coordin
On Jun 8, 2014, at 6:27 PM, Alexsandro Cândido de Oliveira Silva wrote:
> Hello,
>
> I am using the bnlearn package in R to handle large amounts of data in
> Bayesian networks. The variables are discrete and have more than 3 million
> observations.
> With bn.fit function I could easily get the
(1) you hit a memory error, because you are including too many variable
levels. even if you narrowed your 16 variables down to these four, the
`rake` function would need room for matricies containing as much data as a
6-million record table:
nrow( expand.grid( data473t[ , c( 'm11' , 'm12c' , 'm13
This is in reference to the post here by Ray Brownrigg:
http://r.789695.n4.nabble.com/inset-one-map-on-top-of-another-map-td3848752.html
using this code with the maps package:
map("state", region= "ohio", xlim=c(-85, -80), ylim=c(38, 42))
par(usr=c(-216, -66, 24, 144)) # you should be able to '
Which formula for standard deviation are you using?
If you know the population mean then you should divide by n (3 in this
case), but if you don't know the population mean and use the mean
calculated from the sample then it is more usual to use n-1 as the
denominator (this makes the variance an un
Hi,
If you have library(qdap) installed:
library(qdap)
as.vector(bracketXtract(nw.str,"square",with=T))
#[1] "[D]" "[A|D]" "[T|A:D]" "[C|T]"
A.K.
On Sunday, June 8, 2014 4:31 PM, Alexsandro Cândido de Oliveira Silva
wrote:
Hi,
I have a string something like that:
nw.str <- "[D][A|D
Hello,
I am using the bnlearn package in R to handle large amounts of data in
Bayesian networks. The variables are discrete and have more than 3
million observations.
With bn.fit function I could easily get the conditional probability
distribution. However, some variables have unobserved va
Compare 7059 with nrow(x). Looks like they do not match, it is best not to
hardcode such things.
Cheers, Mike
On 9 Jun 2014 05:40, "lijiaming" wrote:
> Hi All,
> I try to empoly Monte Carlo to find out the distribution of my data.
> The code is following
> library(sp)
> library(rgdal)
> libr
Hello everybody. I am using ltm package, an following this sequence:
library(ltm)fit3 <- ltm(LSAT ~ z1)factor.scores(fit2, resp.patterns =
rbind(c(0,1,1,0,0), c(0,1,0,1,0)))
And I obtain the following result:
Call:rasch(data = LSAT)
Scoring Method: Empirical Bayes
Factor-Scores for specified respo
On Jun 8, 2014, at 1:46 PM, Duncan Murdoch wrote:
> On 08/06/2014, 4:30 PM, Alexsandro Cândido de Oliveira Silva wrote:
>> Hi,
>>
>> I have a string something like that:
>>
>> nw.str <- "[D][A|D][T|A:D][C|T]"
>>
>> And I need to split it in this way:
>>
>> "[D]" "[A|D]" "[T|A:D]" "[C|T]"
>
>
On 08/06/2014, 4:30 PM, Alexsandro Cândido de Oliveira Silva wrote:
> Hi,
>
> I have a string something like that:
>
> nw.str <- "[D][A|D][T|A:D][C|T]"
>
> And I need to split it in this way:
>
> "[D]" "[A|D]" "[T|A:D]" "[C|T]"
You could probably use lookahead and lookbehind Perl regular
expre
try this:
> nw.str <- "[D][A|D][T|A:D][C|T]"
> x <- strsplit(nw.str, "]")
> paste0(x[[1]], ']')
[1] "[D]" "[A|D]" "[T|A:D]" "[C|T]"
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Sun, Jun 8, 2014
Hi,
I have a string something like that:
nw.str <- "[D][A|D][T|A:D][C|T]"
And I need to split it in this way:
"[D]" "[A|D]" "[T|A:D]" "[C|T]"
Thanks!!
Regards.
Alexsandro Cândido
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/li
Here is some code with a subset of 50 cases that produces a similar set
of errors to what I got with the full data set (31,690 cases):
##
# Load "survey" package
#
library(survey)
##
# Create raking margins
##
pop.m01 <- data.frame(m01=1:14,
Freq=c(1013620, 136
Hi All,
I try to empoly Monte Carlo to find out the distribution of my data. The
code is following
library(sp)
library(rgdal)
library(raster)
x<-readOGR(".", "finance") #get one spatial file
N.list<-list()
for(i in 1:1000){
Sample.x<-sample(1:7059,600) #the shp file includes 7059 points£¬
Thanks Max and Duncan for the replies.
To Max in particular I would say that ready code written in possibly non-optimal
style (though I did not read enough of caret to have an opinion on that code
in particular) would be good practice too.
When I was a professor and I got complaints about other p
Hi VP,
Not sure what you wanted.
Perhaps:
library(XML)
URL <-
"http://money.securebank.in/index.php?option=com_dashboard&view=history&Itemid=56&startdate=01/01/2013&enddate=6/9/2014&exchange=MCX&sid=1";
doc <- htmlParse(URL)
tableNodes <- getNodeSet(doc, "//table")
dat1 <- readHTMLTable(tableN
Hi,
You may try:
fun1 <- function(len, low, high, mean, sd) {
x <- rnorm(len * 2, mean, sd)
x <- x[x < high & x > low]
x[1:len]
}
##slow
fun2 <- function(len, low, high, mean, sd) {
i <- 1
while (i < len) {
x <- rnorm(1, mean, sd)
if (x < high & x > low) {
Thanks so much!
The 'truncnorm' package did the trick for me! Such variables are uncommon in
my field (psychology), so this is the first time I stumbled across this kind
of distribution.
--
View this message in context:
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Dear William,
Thanks to your answer, i no longer waste time trying to get a solution for
this question. I'm waiting your next release.
Tham
--
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Sent from the R
Dear list-members,
I discover recently bbmle package from an answer in r-list. It makes
some analyses more easily to solve. However I have one problem using
fixed parameter of mle2 and one question about parametrization of mle2.
I send the question directly to the maintainer of the package but
Dear All,
seeking input (or assurance) that I am using the optFederov in AlgDesign
apropriatelly...
I have:
require(AlgDesign)
c <-seq(1,64,by=0.1)
econ=9.16
ic=4.796
hill=1.217
x=5.618
eff <-econ-(ic*(c^hill/(x^hill+c^hill)))
cand.list <-data.frame(c=c,eff=eff,econ=econ,ic=ic,hill=hill,x=x)
Yet another way which returns the row and column of the items you want
rc <- which(t(x[,-c(1,ncol(x))]),arr.ind=TRUE) #this identifies the rows and
columns but is one column off
rc[,1] <- rc[,1] +1 #this adjusts the columns
colnames(rc) <- c("col","row")
rc #show them
Bill
On Jun 6, 2014,
On 07.06.2014 23:34, Milan Bimali wrote:
Dear R Community,
I am in process of developing an R package which in turn depends on a
package that is not available in CRAN but has to be downloaded from a web
source (as follows). Could someone guide me on how to include the package
listed in "Depe
Firstly, you both need to subscribe to the mailing list. Please go to
https://stat.ethz.ch/mailman/listinfo/r-help and subscribe. In this way
you will also get emails from people asking questions and may benefit or
even contribute help to another. There are several other specialty help
lists tha
could you provide a reproducible example ?dput is your friend
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
On Sat, Jun 7, 2014 at 11:22 AM, Michael Willmorth <
mwillmo...@clearwater-research.com> wrote:
> I'm teaching myself how to use rake() in the R "
Hi,
Please check this link:
http://stats.stackexchange.com/questions/25956/what-formula-is-used-for-standard-deviation-in-r
A.K.
It is my understanding that the R function SD finds the standard deviation of a
random variable or a list. Please consider the following list: { 1, 2, 3 }. I
claim t
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