Hi
r-help-boun...@r-project.org napsal dne 08.03.2010 11:39:27:
> Hi everyone,
>
> BEGINNER question:
> I get the error below when running hier.part. Probably i´m doing
> something wrong.
>
> Error in glm.fit(x = X, y = Y, weights = weights, start = start,
etastart =
> etastart, :
> object
Hi
r-help-boun...@r-project.org napsal dne 09.03.2010 01:44:38:
>
> On 9/03/2010, at 11:17 AM, Mike Prager wrote:
>
> > Rolf Turner wrote:
> >>
> >> I solved the problem by putting in a colClasses argument in my
> >> call to read.csv(). But I really think that the read functions
> >> are bei
Your description of your data isn't clear to me.
What are the values in column 2, for example?
Are you trying to construct regression or classification trees?
A reproducible example would really help.
-Peter Ehlers
On 2010-03-08 20:40, valeriano.parravic...@unige.it wrote:
Hi,
I have a probl
Hi
r-help-boun...@r-project.org napsal dne 08.03.2010 18:55:10:
> Here is the example.
>
> > age=18:29
> > height=c(76.1,77,78.1,78.2,78.8,79.7,79.9,81.1,81.2,81.8,82.8,83.5)
> > type=c("A", "B", "C", "D","A", "B", "C", "D","A", "B", "C", "D")
> > typec=c("0","4","2","9","0","7","2","3","0","1",
It's probably binary data - which implies that you can only "read" it with
the application that created it. What is the filename extension?
Ravi
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On 2010-03-08 18:00, Guy Green wrote:
Hi Peter& others,
Thanks (Peter) - that gets me really close to what I was hoping for.
The one problem I have is that the "cut" approach breaks the data into
intervals based on the absolute value of the "Target" data, rather than
their frequency. In othe
R does provide support for basic FTP requests. Not for DELETE
requests. And not for communication on the same connection.
I think your best approach is to use the RCurl package
(http://www.omegahat.org/RCurl).
D.
Orvalho Augusto wrote:
> Dears I need to make some very basic FTP operations wit
Hi Peter & others,
Thanks (Peter) - that gets me really close to what I was hoping for.
The one problem I have is that the "cut" approach breaks the data into
intervals based on the absolute value of the "Target" data, rather than
their frequency. In other words, if the data ranged from 0 to 50
Hi,
I have a problem with ctree of party package.
I have data on distribution of more than one species (about 50 species) and I
would like identify the relation of this multivariate object (species
distribution) with a number of explanatory variables.
rs is the name of my dataframe containing the
Dear Michael, thanks. I have installed gtk+ 2-12.9 revision 2 from
http://gladewin32.sourceforge.net/, it still doesn't work.
I tried gtk2-runtime-2.16.6-2010-02-24-ash.exe from
http://gtk-win.sourceforge.net/home/index.php/en/Downloads, it did not
solve the issue neither.
Could you please give me
try
as.numeric(read_data$DEC)
this should turn it into a numeric variable that you can work with
hth
David Freedman
CDC, Atlanta
Guy Green wrote:
>
> Hi Peter & others,
>
> Thanks (Peter) - that gets me really close to what I was hoping for.
>
> The one problem I have is that the "cut" app
Ditching T/F for TRUE/FALSE would get my vote, too.
-Peter Ehlers
On 2010-03-08 17:44, Rolf Turner wrote:
On 9/03/2010, at 11:17 AM, Mike Prager wrote:
Rolf Turner wrote:
I solved the problem by putting in a colClasses argument in my
call to read.csv(). But I really think that the read
On 2010-03-08 14:45, Marco Bressan wrote:
Hi,
my name is Marco Bressan i'm working to improve ADati package. I study
psicology ad Padua University (Italy). I have this problem: why bartlett.test
function running good and my anova.welch function no?
Ciao,
il mio nome � Marco Bressan e sto lavor
On 9/03/2010, at 11:17 AM, Mike Prager wrote:
> Rolf Turner wrote:
>>
>> I solved the problem by putting in a colClasses argument in my
>> call to read.csv(). But I really think that the read functions
>> are being too clever by half here. If field entries are surrounded
>> by quotes, shouldn
Rolf Turner wrote:
>
> I solved the problem by putting in a colClasses argument in my
> call to read.csv(). But I really think that the read functions
> are being too clever by half here. If field entries are surrounded
> by quotes, shouldn't they be left as character? Even if they are
> all F
Dear R-help members,
I have a question regarding how to use varComb function to specify a
variance function for the "weights" in the gls. I need to fit a
linear model with heteroscedasticity. The variance function is
exp(c0+nu0*W +nu1*W^2) where W is a covariate. Initially I want to use
va
Alternatively
library(latticeExtra)
xyplot(y ~ x | site, d) +
xyplot(y ~ x | site, q, col = "red")
(which is a shortcut for:)
xyplot(y ~ x | site, d) +
as.layer(xyplot(y ~ x | site, q, col = "red"))
On 9 March 2010 11:17, Seth W Bigelow wrote:
> Ah, wonderful, thank you for the code De
Ah, wonderful, thank you for the code Deepayan. To recap for posterity: I
have two datafiles, d and q: each has x-y coordinates that are conditioned
by site (The actual data, for me, is maps of parent trees and their
seedlings). I wanted to superimpose the xy plots of d and q, by site,
without
On 9/03/2010, at 3:48 AM, Liaw, Andy wrote:
(in response to a question on the meaning of the sentence:
>> "Independent variables whose correlation with the response
>> variable was not significant at 5% level were removed")
> If your ultimate interest is in real scientific progress, I'd sugge
You can just access the data from the list:
result <- lapply(output, function(.data){
lettermatch(creator, .data)
})
You can then take the "result" and possibly 'cbind' back into the matrix you
want.
On Mon, Mar 8, 2010 at 10:59 AM, Laetitia Schmid wrote:
> Hi!
> I need some help to finish
On Mon, Mar 8, 2010 at 3:44 PM, casperyc wrote:
>
> Hi Rolf Turner ,
>
> God, it directed to the wrong page.
>
> I firstly find the formula in wiki, than tried to verify the answer in R,
> now, given that 143/12 ((n^2-1)/12 ) is the correct answer for a discrete
> uniform random variable,
> I am s
Hi Achim Zeileis-4,
That's very helpful.
Thanks!
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Hi Rolf Turner ,
God, it directed to the wrong page.
I firstly find the formula in wiki, than tried to verify the answer in R,
now, given that 143/12 ((n^2-1)/12 ) is the correct answer for a discrete
uniform random variable,
I am still not sure what R is calculating there?
why it gives me 13?
On 9/03/2010, at 12:13 PM, casperyc wrote:
>
> Hi all,
>
> I am REALLY confused with the variance right now.
You need to learn the difference
(a) Between sample variance (*estimate* of population variance)
and
population variance.
and
Hi all,
I am REALLY confused with the variance right now.
for a discrete uniform distribution on [1,12]
the mean is (1+12)/2=6.5
which is ok.
y=1:12
mean(y)
then var(y)
gives me 13
1- on http://en.wikipedia.org/wiki/Uniform_distribution_%28discrete%29 wiki
the variance is (12^2-1)/12=
Hello, I have a simple class that looks like:
setClass("statisticInfo",
representation( max = "numeric",
min = "numeric",
beg = "numeric",
current = "numeric",
avg = "numeric",
Or for real power and flexibility, see the 'convert()' function in package
tis.
Jeff
Gabor Grothendieck writes:
> See ?aggregate.zoo, e.g.
>
> library(zoo)
> z <- zoo(1:1000, as.Date("2000-01-01") + 0:999)
> aggregate(z, as.yearmon, mean)
>
> or replace mean with whatever summarization you want.
Hello,
Warning, I'm guessing only those who have used the Hmisc package's
summary.formula function with LaTeX will be able to offer much help here.
I am using the Hmisc package's summary.formula function to produce
tables for a LaTeX report. The "latex" function in the same package
supports
I've found the "surveillance" package useful for monitoring walk-in
clinic visits in our county as the influenza pandemic evolved. It might
serve your needs for monitoring IFI (invasive fungal infections?) in
your hospital.
--Chris
Christopher W. Ryan, MD
SUNY Upstate Medical University Clinic
On 03/08/2010 07:18 AM, Albert-Jan Roskam wrote:
> Sorry: there was an error in the last sentence:
> And, inside those validity checks, is most of the checking done with 'if'
> 'else' computations, or is it also common to use try()?
For me it's a matter of taste, and usual to use if... (because y
Hi,
my name is Marco Bressan i'm working to improve ADati package. I study
psicology ad Padua University (Italy). I have this problem: why bartlett.test
function running good and my anova.welch function no?
Ciao,
il mio nome è Marco Bressan e sto lavorando per migliorare il pacchetto ADati.
St
Dears I need to make some very basic FTP operations with R.
I need to do a lot of "get" and issue a respective "delete" command
too on the same connection.
How can I do that?
Thanks in advance
Caveman
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https://stat
Hi,
I met with this error message with the following data set. Do you know how
to resolve it? Thanks.
> data<-read.table("c://temp3//abc.csv", sep = ",", header=T)
> classwt<-c( 0.5806452, 0.4193548)
> y<-data[,1]
> x<-data[,2:ncol(data)]
> print(y)
[1] 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 1 1 1 1 1
Hi all,
Does any one know how to write the likelihood function for Poisson distribution
in R when P(x=0).
For normal case, it an be written as follows,
n * log(lambda) - lambda * n * mean(dat)
Any help is highly appreciated
Ashta
Ma Ismail - NewYork-MEAG-NY meag-ny.com> writes:
>
> Hi,
>
> A few of the developers on our Quant team are using R for data
calculation andto generate a resulting CSV file. They have R
installed on their workstations. We are interested in having this
deployed to user workstations where the u
On Mon, Mar 8, 2010 at 8:46 PM, Barry Rowlingson
wrote:
> On Mon, Mar 8, 2010 at 6:44 PM, Ma Ismail - NewYork-MEAG-NY
> wrote:
>> Hi,
>>
>> A few of the developers on our Quant team are using R for data calculation
>> and to generate a
[snip]
> I've noticed a lot of financial corporates getti
I always use paste()
i <- 1
sqlcmd_ScaffLen <- paste("SELECT scaffold.length
FROM scaffold, scaffold2contig, contig2read
WHERE scaffold.scaffold_id=scaffold2contig.scaffold_id AND
scaffold2contig.contig_id=contig2read.contig_id AND
contig2read.read_id LIKE '%MG", i ,"%'", sep='')
That should cre
On Mon, Mar 8, 2010 at 6:44 PM, Ma Ismail - NewYork-MEAG-NY
wrote:
> Hi,
>
> A few of the developers on our Quant team are using R for data calculation
> and to generate a resulting CSV file. They have R installed on their
> workstations. We are interested in having this deployed to user works
Hi,
A few of the developers on our Quant team are using R for data calculation and
to generate a resulting CSV file. They have R installed on their workstations.
We are interested in having this deployed to user workstations where the users
will not have R installed on their workstations. Is
I had a bug in my last solution.
This one should work.
next.combin <- function(oldcomb,n){
lcomb <- length(oldcomb)
hole.pos <- last.hole.pos(oldcomb,n)
if ((hole.pos == lcomb) & oldcomb[lcomb]==n) {
return(NA)
}
newcomb<-oldcomb
newcomb[hole.pos:lcomb]<-oldcomb[hole.pos]+(1:(lcomb
On 2010-03-08 8:47, Guy Green wrote:
Hello,
I have a set of data with two columns: "Target" and "Actual". A
http://n4.nabble.com/file/n1584647/Sample_table.txt Sample_table.txt is
attached but the data looks like this:
Actual Target
-0.125 0.016124906
0.135 0.120799865
...
Thanks, it works!
Xumin
milton ruser
03/08/2010 01:02 PM
To
Xumin Zeng
cc
r-help
Subject
Re: [R] how to convert character variables into numeric variables directly
Try: as.numeric(as.character( typec))
milton
On Mon, Mar 8, 2010 at 12:55 PM, Xumin Zeng wrote:
Here is the example.
Try: as.numeric(as.character( typec))
milton
On Mon, Mar 8, 2010 at 12:55 PM, Xumin Zeng wrote:
> Here is the example.
>
> > age=18:29
> > height=c(76.1,77,78.1,78.2,78.8,79.7,79.9,81.1,81.2,81.8,82.8,83.5)
> > type=c("A", "B", "C", "D","A", "B", "C", "D","A", "B", "C", "D")
> > typec=c("0","4",
Can somebody tell me how can I calculate derivatives of some functions
through r citing some examples??
by Taylor's expansion formulae how can I handle complecated functions which
can not be handled properly manually by r-script?
--
View this message in context:
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Here is the example.
> age=18:29
> height=c(76.1,77,78.1,78.2,78.8,79.7,79.9,81.1,81.2,81.8,82.8,83.5)
> type=c("A", "B", "C", "D","A", "B", "C", "D","A", "B", "C", "D")
> typec=c("0","4","2","9","0","7","2","3","0","1","2","3")
> typen=c(0,1,2,3,0,1,2,3,0,1,2,3)
> data1=data.frame(age=age,height=
Dear all,
I have the following dataset:
t <- structure(list(pUrb = c(20.160307, 51.965649, 26.009581, 3.141484,
64.296826
), pUrb_class = structure(c(1L, 1L, 1L, 1L, 1L), .Label = c("0",
"1"), class = "factor"), pAgri = c(79.921386, 46.657713,
40.269204, 0, 0.440691), pAgri_class =
On 2010-03-08 8:24, vincent laperriere wrote:
Hi all,
I would like to fit a gamma pdf to my data using the method of RSS (Residual
Sum-of-Squares). Here are the data:
x<- c(86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138,
142, 146, 150, 154, 158, 162, 166, 170, 174)
y<
Thanks Liaw!
I just implemented it using tapply:
tapply(fit$splits[, "improve"], rownames(fit$splits), sum)
If you can reference me to any other source / example and so on - it would
be great. but either way - you helped me a lot, thank you !
Tal
Contact
Details:--
Hello,
The function bwlabel() in the Bioconductor package EBImage, extracts the
connected components of an image. Denoting your binary matrix by x, the
following code gives you the first 10 largest clusters (in size).
> library(EBImage)
> y = bwlabel(x)
> sort(table(y), dec=TRUE)[1:10]
See h
Thanks for making it quickly reproducible - I was able to see that message
in English within a few seconds.
The start has x=86, but the data is also called x. Remove x=86 from start
and you get a different error.
P.S. - please do include the R version information. It saves time for us,
and we l
Here's a variation on the theme:
boxplot(x[,i])
title(main =
bquote(.(mainlabel1)~~italic(.(predictor[i]))~~.(mainlabel2))
)
-Peter Ehlers
On 2010-03-08 8:46, Miguel Porto wrote:
Hello,
Try this way (not sure if it's the best way, but it works):
boxplot(x[,i],
main=substitute(expressi
Hi!
I need some help to finish my script.
I have two tables that I combine randomly to produce a third table.
This I do for hundreds of iterations. In the output file I get all the
simulated tables after each other. It looks like this (in this case 3
iterations):
output file:
[[1]]
Hello.
I have just started learning how to work with R program but I have
encountered a problem.
I can't think up how to remove the rows which contain two (2) or more NA or
Zero (0).
I would be glad if you could help me because I just have some basic
knowledge so far and I even haven't mastered
Hello,
I have a set of data with two columns: "Target" and "Actual". A
http://n4.nabble.com/file/n1584647/Sample_table.txt Sample_table.txt is
attached but the data looks like this:
Actual Target
-0.125 0.016124906
0.135 0.120799865
... ...
... ...
I want t
Another possibility is to use fn$ in the gsubfn package. Just preface
any command with fn$ to enable a quasi-perl-like string interpolation.
In this example $i is replaced with 1:
> library(gsubfn)
> library(sqldf)
> i <- 1
> fn$sqldf("select count(*) from CO2 where Plant like '%n$i%'")
count(*)
Try this:
i<-1
sqlcmd_ScaffLen<-sprintf('SELECT scaffold.length
FROM scaffold,scaffold2contig,contig2read
WHERE scaffold.scaffold_id=scaffold2contig.scaffold_id AND
scaffold2contig.contig_id=contig2read.contig_id AND contig2read.read_id LIKE
\'%%MG%d%%\'' ,i)
sqlcmd_ScaffLen
Your problem:
1. Need
D'oh -- thanks! I'm always forgetting the double-bracket extractor...
-Original Message-
From: Erik Iverson [mailto:er...@ccbr.umn.edu]
Sent: Monday, March 08, 2010 10:50 AM
To: Steve Jaffe
Cc: r-help@r-project.org
Subject: Re: [R] quickest way convert 1-col df to vector?
sjaffe wrote:
sjaffe wrote:
anything shorter than as.vector(as.matrix( df ) )?
df[[1]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide comm
Hello,
Try this way (not sure if it's the best way, but it works):
boxplot(x[,i],
main=substitute(expression(paste(a," ",italic(b),"
",c)),list(a=mainlabel1,b=predictor[i],c=mainlabel2)),
ylab=paste(ylabel),cex.lab=cexalabel,cex.main=cexmlabel,cex.axis=1.5)
Best,
Miguel
On Mon, Mar 8, 2010 at
anything shorter than as.vector(as.matrix( df ) )?
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If the data is a dataframe or matrix 'd':
d <- d[apply(d, 1, function(v) sum( is.na(v) ) <= 2 & sum(v==0, na.rm=T) <=
2 ), ]
which can be deconstructed as follows:
i1 <- apply(d, 1, function(v) sum(is.na(v)) <= 2 ) ## true for rows with 2
or fewer na's
i2 <- apply(d, 1, function(v) sum( v == 0,
Hi,
I cannot really test since your dataframe is completely distorted, but
what happens if you try: format(QuoteBID, nsmall=5)?
I think it's just a matter of printing, which uses the number of digits
from options(digits=).
See: ?options, ?format, etc.
But I'm not an expert and cannot really
I doubt it. Guess why?
Dr. Rubén Roa-Ureta
AZTI - Tecnalia / Marine Research Unit
Txatxarramendi Ugartea z/g
48395 Sukarrieta (Bizkaia)
SPAIN
-Mensaje original-
De: r-help-boun...@r-project.org [mailto
Dear all,
I have a table like this:
a <- read.csv("test.csv", header = TRUE, sep = ";")
a
UTM pUrb pUrb_class pAgri
pAgri_class pNatFor pNatFor_class
1 NF188520.160307 NA 79.921386NA
Try google.
On Mon, Mar 8, 2010 at 8:09 AM, ManInMoon wrote:
>
> What is "R News 4/1"?
> --
> View this message in context:
> http://n4.nabble.com/POSIXct-type-lost-tp1584379p1584464.html
> Sent from the R help mailing list archive at Nabble.com.
>
> _
Hi,
I'm trying to install tcltk in R-2.10.1, however I get error.
someone can help?
thanks
Vasco
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.
Dear R users,
in the example below the name of the genus will be displayed in the main
titles using the variable predictor[i] and paste.
I would like to have the genus name in italic. However all my attempts
using expression and substitute failed.
Does anybody know a solution?
Thanks a lot in a
Hi all,
I would like to fit a gamma pdf to my data using the method of RSS (Residual
Sum-of-Squares). Here are the data:
x <- c(86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138,
142, 146, 150, 154, 158, 162, 166, 170, 174)
y <- c(2, 5, 10, 17, 26, 60, 94, 128, 137, 128, 77,
What is "R News 4/1"?
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Hi:
I want to obtain a particular value from a data.frame. Following is my
dataframe:
> Quotes
BID ASK
Name
CT2 GOVT99.9296999.9375 CT2
TUM0 COMDTY 108.53125 108.5469TUM0
CT5 GOVT
Hello,
I am using RmySQL and would like to iterate through a few queries.
I would like to use sprintf but I think I'm having problems mixing and
matching the sprintf syntax and the SQL regex.
I have checked my sqlcmd and it works when I wan to match %MG1% but how
do I iterate for i 1-72? Escape
Sorry: there was an error in the last sentence:
And, inside those validity checks, is most of the checking done with 'if'
'else' computations, or is it also common to use try()?
Cheers!!
Albert-Jan
~~
In the face of ambiguity, r
Hi,
I'm reading up on S4 classes *). There seem to be at least two ways of input
validation:
setClass() (using the 'validity' argument) and setValidity(). Is it a matter
of taste which function is used? Or should more complex validation code better
be put in a setValiditity call?
*) A (Not
Thanks
I would never deduct it out from the help page of approx.
Regards
Petr
r-help-boun...@r-project.org napsal dne 08.03.2010 15:47:37:
> Perhaps approx:
>
>approx(range(x), c(0.5, 1), xout = x)$y
>
> A one-linear, but longer, is also possible based on lm:
>
> predict(lm(c(0.5, 1)
One way to do it (no p-values) is explained in the original CART book.
You basically add up all the "improvement" (in fit$split[, "improve"])
due to each splitting variable.
Andy
From: Tal Galili
>
> Simple example:
>
> # Classification Tree with rpart
>
> library(rpart)
>
> # grow tree
>
>
thanks Kieth. I wanted something generic code to check column data
type and loop through and create the interaction columns automatically
as I want to test this out as a new algorithm for data mining.
Traditional regression may give misleading results with
multi-collinearity and thus I wanted t
Perhaps approx:
approx(range(x), c(0.5, 1), xout = x)$y
A one-linear, but longer, is also possible based on lm:
predict(lm(c(0.5, 1) ~ x, data.frame(x = range(x))), data.frame(x))
On Mon, Mar 8, 2010 at 9:37 AM, Petr PIKAL wrote:
> Hi all
>
> I know I probably reinvented wheel but it was
If your ultimate interest is in real scientific progress, I'd suggest that you
ignore that sentence (and any conclusion drawn subsequent to it).
Cheers,
Andy
From: bbslover
>
> This topic refer to independent variables reduction, as we
> know ,a lot of
> method can do with it,however, for pre
That sounds like a particular form of permutation test. If the
"scrambling" is replaced by sampling with replacement (i.e., some data
points can be sampled more than once while others can be left out),
that's the simple (or nonparametric) bootstrap. The goal is to generate
the distribution of the
Hi all
I know I probably reinvented wheel but it was maybe simpler then search in
docs or ask help before I did my part.
I made a simple function which can scale a vector between chosen values.
Do anybody know simpler/better approach?
myscale<-function(x, miny=0.5, maxy=1) {
rx <- diff(range(
Hi,
on a lattice, I have binary 0/1 data. 1s are rare and may form clusters.
I would like
to know the size/length of largest cluster. Any help warmly welcome,
Sylvain.
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Nice shot of cumsum(). Just improve it a little:
> x <- c(0,0,1,2,3,0,0,4,5,6)
> x.groups <- split(x, (x != 0) * cumsum(x == 0))[-1]
> x.groups
$`2`
[1] 1 2 3
$`4`
[1] 4 5 6
> lapply(x.groups, mean)
$`2`
[1] 2
$`4`
[1] 5
On Mon, Mar 8, 2010 at 11:02 AM, jim holtman wrote:
> Try this:
>
>> x <
Dear R,
I have three matrices like this
K: pp-by-pp commutation matrix,
I: p-by-p diagonal matrix,
X: p-by-q dense matrix,
and I wish to calculate
K(IoX)
where `o' denotes Kronecker product.
Can you give me any suggestion to speed it up when `p' and `q' are large?
Thanks in adv
What I was looking for was the string of non-zero values and where they
'broke' at. I could have used 'rle', but I sometime find this approach just
as easy. Every place there is a zero will be TRUE which has the value 1.
'cumsum' will generate a running sum of these values. When there is a
non-z
Add the number to the POSIXct origin. See table at end of R News 4/1.
On Mon, Mar 8, 2010 at 7:11 AM, ManInMoon wrote:
>
> It appears that I am creating a matrix where als columns are of type number,
> so my Date column has been converted to a number.
>
> Is there a way to show or display this n
I am generating a column of dates using POSIXct, but when I try to assign it
to an existing dataframe - it gets "stored" as numbers instead of as
POSIXct.
Is there a way to force a column to be a specific type (POSIXct)?
Thanks,
Moon
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It appears that I am creating a matrix where als columns are of type number,
so my Date column has been converted to a number.
Is there a way to show or display this number column as a Date again?
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Sen
Duncan Murdoch wrote:
On 07/03/2010 5:26 PM, rkevinbur...@charter.net wrote:
Thatnk you.
The documentation indicates as you indicated that if there is not an exact
match then the next element is chosen. But it does not indicate the case that
contains an exact match but there is not value t
On Sun, 7 Mar 2010 20:23:25 -0800 (PST) thedoctor81877 wrote:
> Hello. I am new to R, and am typing some homework for my undergrad Analysis
> class. I am trying to graph the following function in R: f(x) = x^2 for x
> >=0, and f(x) = 0 for x <0. How do I do this in R?
f=function(x) ifelse(x >= 0,
Johannes wrote:
>
>
> How can I, given the code snippet below, draw the error bars in the center
> of each grouped bar rather than in the center of the group?
>
http://markmail.org/message/oljgimkav2qcdyre
Dieter
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Maybe you can create a helper vector first:
> helper <- structure(names = unlist(j), rep(names(j), sapply(j, length)))
> helper
acbd
"j1" "j1" "j2" "j2"
> helper[i]
aabbbccd
"j1" "j1" "j2" "j2" "j2" "j1" "j1" "j2"
On Sat, Mar 6, 2010 at 1:42 AM, Carlos
Hi everyone,
BEGINNER question:
I get the error below when running hier.part. Probably i´m doing
something wrong.
Error in glm.fit(x = X, y = Y, weights = weights, start = start, etastart =
etastart, :
object 'fit' not found
In addition: Warning messages:
1: In glm.fit(x = X, y = Y, weights =
On 4 March 2010 23:47, Jim Lemon wrote:
> On 03/05/2010 04:11 AM, Bert Gunter wrote:
>>
>> Folks:
>>
>> Rolf's (appropriate, in my view) response below seems symptomatic of an
>> increasing tendency of posters to hide their identities with pseudonyms
>> and
>> fake headers. While some of this may
Hi Patrick,
I've read it quickly and it seems to be a good resource for beginners
that have just downloaded R and have no idea what to do.
My guess is that it should be a good introduction to other documents
such as the "An introduction to R".
I'll test it with the next students in my team th
Thank you for answers.
My code is very slow compared with yours ;-)
#my code
system.time(r0<-f0(iBig,jBig))
user system elapsed
82.489 15.060 97.544
#Holtman's code
system.time(r1<-f1(iBig,jBig))
user system elapsed
0.100 0.012 0.113
#Dunlap's code
system.time(r2<-f2(iBig,jBig))
u
Dear friends,
I have problems to plot a matrix from an already distance-matrix. I Just want
to plot it, the distance calculation were already done, I don' have the
original data to re-calculate the distance in R that will be easy. If some one
can prove, here goes the data.
Thank you
Rosa._
Marc Schwartz wrote:
On Mar 5, 2010, at 3:45 PM, Ryan Garner wrote:
I just installed R on Redhat Linux at work for the first time and have two
questions.
1. I tried to install R to have png and cairo capabilities and was
unsuccessful. Before running make, I ran ./configure --with-libpng=yes
Hi,
How can I, given the code snippet below, draw the error bars in the center
of each grouped bar rather than in the center of the group?
Thanks for any hints,
Joh
library(lattice)
barley[["SD"]] <- 5
barchart(
yield ~ variety | site,
data = barley,
groups=year,
origin=0,
lowDev=b
Thanks Gabor - sprintf did the trick
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On 03/08/2010 04:48 AM, casperyc wrote:
http://n4.nabble.com/file/n1583733/100307070476876317b486a941.jpg
I want to get a histogram by factors.
Hi casperyc,
Have a look at the third example for the barp function in the plotrix
package.
Jim
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