On Jan 29, 2011, at 7:45 PM, Simon Goodman wrote:
How can I extract only the time component from an POSIXlt object?
For example if I try the following it still returns both the date and
time...
as.POSIXlt(tr.date[1])
[1] "2010-10-18 21:46:53"
as.POSIXlt(tr.date[1],"%H:%M:%S")
[1] "2010-1
Some code I have been playing with to do this follows ...
get.best.arima <- function(x.ts, minord=c(0,0,0,0,0,0),
maxord=c(2,1,1,2,1,1))
{
# function based on 'Introductory Time Series with R'
best.aic <- 1e8 # a big number
n <- length(x.ts)
for(p in minord[1]:maxo
How can I extract only the time component from an POSIXlt object?
For example if I try the following it still returns both the date and
time...
>as.POSIXlt(tr.date[1])
[1] "2010-10-18 21:46:53"
>as.POSIXlt(tr.date[1],"%H:%M:%S")
[1] "2010-10-18 21:46:53"
round and trunc don't help... is there
Hello,
Thankyou for the clarification about the NAs. For your interest, thankfully my
end goal was not to plot a dendrogram with 23371 elements, but just to use the
output of the clustering to re-order the rows of a matrix before plotting it
with image(). Since clara() and pam() are partitionin
Em 28/1/2011 20:15, Peter Francis escreveu:
Dear List,
I am not sure if A) this is possible or B) the correct place to ask.
I am looking to find the mean - i have n, and the two-tailed confidence intervals
0.95& 0.25 with a p-value of 0.05.
Can i find the mean from this data ?
Thanks
Pet
On Jan 29, 2011, at 5:21 PM, peterfran...@me.com wrote:
>
> Sorry for not being more explicit in my emails, the code I am using
> was not generated by me therefore I was not in the position to share
> it.
>
> Sorry if the information i have provided is insufficient or if I
> have wasted peo
Hi,
Just wondering for the SMA and EMA in package TTR, is it possible to me to
code it so that, say if I need to calculate SMA (x, n=100), when the sample
size is less than 100, it will give me the SMA (x, k) where k is the sample
size of the data? Right now it only gives me an invalid n error.
T
On 11-01-29 8:19 PM, Duncan Murdoch wrote:
On 11-01-29 1:11 PM, MM wrote:
Hello
I have a 3.5 million elements numeric vector x. I'm trying to calculate the
rolling std dev of the previous 144 elements.
rsd144<-runsd(x, 144, center=0, endrule="NA")
this crashes R (ie on the console disappe
On 11-01-29 1:11 PM, MM wrote:
Hello
I have a 3.5 million elements numeric vector x. I'm trying to calculate the
rolling std dev of the previous 144 elements.
rsd144<-runsd(x, 144, center=0, endrule="NA")
this crashes R (ie on the console disappears and the Rgui.exe process is not
there an
Sorry for not being more explicit in my emails, the code I am using was not
generated by me therefore I was not in the position to share it.
Sorry if the information i have provided is insufficient or if I have wasted
peoples time.
Basically I have the information listed below I.e n and the q
> "FL" == Feng Li
> on Sat, 29 Jan 2011 19:46:48 +0100 writes:
FL> I meant "sparse matrix", sorry for the typo.
aha.. :-)
FL> On Sat, Jan 29, 2011 at 7:02 PM, Feng Li wrote:
>> Dear R,
>>
>> I have a simple question concerning with a special case of
>> spa
> Prof Brian Ripley
> on Sat, 29 Jan 2011 21:00:19 + (GMT) writes:
> On Sat, 29 Jan 2011, David Winsemius wrote:
>>
>> On Jan 29, 2011, at 12:17 PM, Prof Brian Ripley wrote:
>>
>>> On Sat, 29 Jan 2011, David Winsemius wrote:
>>>
On Jan
On Sat, Jan 29, 2011 at 2:03 PM, john nicholas wrote:
> All,
>
> I have been just recently working with zoo objects for trading systems.
>
> Can someone please help with these basic questions?
>
> Given a daily time series downloaded using get.hist.quote() from the tseries
> package, ie..
>
>
On Jan 29, 2011, at 9:39 AM, Peter Francis wrote:
Dear all,
The data is generated from 1000 random samples of a phylogenetic
tree to calculate phylogenetic diversity. I sampled the tree 1000
times at with various species communities (600) to get a random PD
per community. I then want to
On Sat, Jan 29, 2011 at 1:23 PM, Joshua Wiley wrote:
>> xxAB <- dat$xxA * dat$xxB
>> meanAB <- meanA * meanB
>> sdAB <- sqrt(sdA^2 + sdB^2 + meanA^2 * sdB^2 + meanB^2 * sdA^2)
don't know why I did that the hard way (besides I think I messed up
the formula slightly), but the point is there's still
All,
I have been just recently working with zoo objects for trading systems.
Can someone please help with these basic questions?
Given a daily time series downloaded using get.hist.quote() from the tseries
package, ie..
startDate= as.Date("2000-01-01")
endDate= as.Date("2011-01-29")
freque
Hello all,
I came across a behavior of R with environments that I'm not sure what is
causing it.
It involves changing variables that are found through using model.frame on a
formula inside a function.
I wonder if it's a "bug" or a "feature". And in either case, how it might
be managed.
Here is
Thank you for the links and the information.You have been very helpful
On Sat, Jan 29, 2011 at 2:45 PM, David Winsemius wrote:
>
> On Jan 29, 2011, at 7:58 AM, David Winsemius wrote:
>
>
>> On Jan 29, 2011, at 7:22 AM, Alex Smith wrote:
>>
>> Hello I am trying to determine wether a given matrix
I am running this in a cluster,
I am trying to calculate this in a cluster,
cdata = replicate(2000,rnorm(6))
c1 = cor(cdata)
m1 = as.matrix(cdata)
cl = makeCluster(64)
out = clusterApplyLB(cl,m1,cor)
Error in checkForRemoteErrors(val) :
12000 nodes produced errors; first error: supply both '
Dear all:
I am going to explain better the problem I have. I would greatly appreciate
an answer because I cannot advance further in my project while I do not
solve this problem.
I have a series of mean sea surface temperatures measured by satellites.
These data are reported in degrees longitude
Dear all,
The data is generated from 1000 random samples of a phylogenetic tree to
calculate phylogenetic diversity. I sampled the tree 1000 times at with various
species communities (600) to get a random PD per community. I then want to test
my observed PD with that of a random sample to test
Hello
I have a 3.5 million elements numeric vector x. I'm trying to calculate the
rolling std dev of the previous 144 elements.
rsd144<-runsd(x, 144, center=0, endrule="NA")
this crashes R (ie on the console disappears and the Rgui.exe process is not
there anymore)
with smaller vectors
On Sat, Jan 29, 2011 at 1:05 PM, David Winsemius wrote:
>
> On Jan 29, 2011, at 3:49 PM, Joshua Wiley wrote:
>
>> On Sat, Jan 29, 2011 at 12:31 PM, David Winsemius
>> wrote:
>>>
>>> Huh?. With the same model and data, they should be the same:
>>
>> I must be missing something, because that is wha
On Jan 29, 2011, at 3:49 PM, Joshua Wiley wrote:
On Sat, Jan 29, 2011 at 12:31 PM, David Winsemius
wrote:
Huh?. With the same model and data, they should be the same:
I must be missing something, because that is what I would have
expected too, but it is not what I get (at least when I run t
On Sat, 29 Jan 2011, David Winsemius wrote:
On Jan 29, 2011, at 12:17 PM, Prof Brian Ripley wrote:
On Sat, 29 Jan 2011, David Winsemius wrote:
On Jan 29, 2011, at 10:11 AM, David Winsemius wrote:
On Jan 29, 2011, at 9:59 AM, John Fox wrote:
Dear David and Alex,
I'd be a little careful a
On Sat, Jan 29, 2011 at 12:31 PM, David Winsemius
wrote:
> Huh?. With the same model and data, they should be the same:
I must be missing something, because that is what I would have
expected too, but it is not what I get (at least when I run the code
as shown below).
>> lm.mod2 <- lm(out ~ scal
On Jan 29, 2011, at 2:36 PM, Joshua Wiley wrote:
On Sat, Jan 29, 2011 at 11:10 AM, David Winsemius
wrote:
On Jan 29, 2011, at 12:12 PM, Russell Pierce wrote:
Just in case someone else stumbles onto this thread and is facing a
similar issue: The quick solution for me turned out to be using
David,
Thanks for your feedback. However, I think if you look at the mean of
your output (which, as a minor point, doesn't provide the same data to
predict as expand.grid(X1=c(-1,0,1),X2=c(-1,0,1)), you'll see it is
very different from the mean of dat$out. Like Ista's previous answer,
I think in
Another option is
set.seed(10)
dat <- data.frame(xxA = rnorm(20,10), xxB = rnorm(20,20))
dat$out <- with(dat,xxA+xxB+xxA*xxB+rnorm(20,20))
lm.mod <- lm(out ~ I(scale(xxA))*I(scale(xxB)), data=dat)
newdata <- data.frame(xxA=c(-1,0,1),xxB=c(-1,0,1))
preds <- predict(lm.mod, newdata)
Best,
Ista
On Sat, Jan 29, 2011 at 11:10 AM, David Winsemius
wrote:
>
> On Jan 29, 2011, at 12:12 PM, Russell Pierce wrote:
>
>> Just in case someone else stumbles onto this thread and is facing a
>> similar issue: The quick solution for me turned out to be using Glm
>> and Predict in the rms package. Than
On Jan 29, 2011, at 12:12 PM, Russell Pierce wrote:
Just in case someone else stumbles onto this thread and is facing a
similar issue: The quick solution for me turned out to be using Glm
and Predict in the rms package. Thanks go to Joshua and Ista for
helping me out with this issue. Double
On Jan 29, 2011, at 12:17 PM, Prof Brian Ripley wrote:
On Sat, 29 Jan 2011, David Winsemius wrote:
On Jan 29, 2011, at 10:11 AM, David Winsemius wrote:
On Jan 29, 2011, at 9:59 AM, John Fox wrote:
Dear David and Alex,
I'd be a little careful about testing exact equality as in all(M
== t
I meant "sparse matrix", sorry for the typo.
Feng
On Sat, Jan 29, 2011 at 7:02 PM, Feng Li wrote:
> Dear R,
>
> I have a simple question concerning with a special case of sparse matrix
> multiplications. Say A is a 200-by-1 dense matrix. B is a 1-by-1
> block- diagonal matrix, and e
Uwe Ligges statistik.tu-dortmund.de> writes:
> On 29.01.2011 15:06, martanair wrote:
[snip]
> > I write:
> > radon.data<- list ("n", "J", "x", "y", "county")
> > radon.inits<- function (){
> >list (a=rnorm(J), b=rnorm(1), mu.a=rnorm(1),
> > sigma.y=runif(1), sigma.a=runif(1))
> >
Dear R,
I have a simple question concerning with a special case of spare matrix
multiplications. Say A is a 200-by-1 dense matrix. B is a 1-by-1
block- diagonal matrix, and each diagonal block B_i is 100-by-100. The usual
way I did A%*%B will take about 30 seconds which is to time cons
On Wed, Jan 26, 2011 at 4:24 PM, Markus Damian wrote:
> Dear all,
>
> I need to generate plots in which the points of the plot are replaced
> by text labels, such as "dog" and "cat". The usual way of specifying
> the plotting symbol with pch works only if the labels are single
> characters, as far
Hello,
Using the data and code below I've been using R to compare output from two
different solute transport codes, the red contours are from verified code
while the blue contours come from some modified code I've been working on.
The goal is for the contours to match, and although there is curr
Can you please
1. reply to me personally, not only to the list (which I won't read all
the time) and
2. cite what you wrote before. I tend to delete messages that are not of
importance for me and rely on the fact that old information is cited.
Now I do not have the data nor the code anymore
I have R2WinBUGS
and this is my radon1.bug
# Bugs code for multilevel model for radon
# with bsmt as an individual predictor
# varying-intercept model
model {
for (i in 1:n){
y[i] ~ dnorm (y.hat[i], tau.y)
y.hat[i] <- a[county[i]] + b*x[i]
}
b ~ dnorm (0, .0001)
tau.y <- pow(si
On Sat, 29 Jan 2011, David Winsemius wrote:
On Jan 29, 2011, at 10:11 AM, David Winsemius wrote:
On Jan 29, 2011, at 9:59 AM, John Fox wrote:
Dear David and Alex,
I'd be a little careful about testing exact equality as in all(M == t(M)
and
careful as well about a test such as all(eigen(
Just in case someone else stumbles onto this thread and is facing a
similar issue: The quick solution for me turned out to be using Glm
and Predict in the rms package. Thanks go to Joshua and Ista for
helping me out with this issue. Double thanks go to Joshua for
suggesting I take a closer look
On Jan 29, 2011, at 10:11 AM, David Winsemius wrote:
On Jan 29, 2011, at 9:59 AM, John Fox wrote:
Dear David and Alex,
I'd be a little careful about testing exact equality as in all(M ==
t(M) and
careful as well about a test such as all(eigen(M)$values > 0) since
real
arithmetic on a c
On Jan 29, 2011, at 9:59 AM, John Fox wrote:
Dear David and Alex,
I'd be a little careful about testing exact equality as in all(M ==
t(M) and
careful as well about a test such as all(eigen(M)$values > 0) since
real
arithmetic on a computer can't be counted on to be exact.
Which was why
Dear David and Alex,
I'd be a little careful about testing exact equality as in all(M == t(M) and
careful as well about a test such as all(eigen(M)$values > 0) since real
arithmetic on a computer can't be counted on to be exact.
Best,
John
John Fox
Senator Willi
On Jan 29, 2011, at 7:58 AM, David Winsemius wrote:
On Jan 29, 2011, at 7:22 AM, Alex Smith wrote:
Hello I am trying to determine wether a given matrix is symmetric and
positive matrix. The matrix has real valued elements.
I have been reading about the cholesky method and another method is
On Sat, 29 Jan 2011, Kang Min wrote:
Thanks Prof Ripley, the condition worked!
Btw I tried to search ?repl but I don't have documentation for it. Is
it in a non-basic package?
I meant grepl: the edit messed up (but not on my screen, as sometimes
happens when working remotely). The point is t
On 29.01.2011 15:06, martanair wrote:
Why this?
I write:
radon.data<- list ("n", "J", "x", "y", "county")
radon.inits<- function (){
list (a=rnorm(J), b=rnorm(1), mu.a=rnorm(1),
sigma.y=runif(1), sigma.a=runif(1))
}
radon.parameters<- c ("a", "b", "mu.a", "sigma.y", "sigma.a")
Why this?
I write:
radon.data <- list ("n", "J", "x", "y", "county")
radon.inits <- function (){
list (a=rnorm(J), b=rnorm(1), mu.a=rnorm(1),
sigma.y=runif(1), sigma.a=runif(1))
}
radon.parameters <- c ("a", "b", "mu.a", "sigma.y", "sigma.a")
# with 10 iterations
radon.1 <- bugs (rad
Sascha,
I guess the software you wan't is RKWard:
http://rkward.sourceforge.net/
I have tested a few R front-ends and to me this one seemed the most
adequate. I suppose as an ex-matlab user my interface preferences are
biased, but still RKWard has a good editor with syntax highlighting,
code fol
On Jan 29, 2011, at 7:22 AM, Alex Smith wrote:
Hello I am trying to determine wether a given matrix is symmetric and
positive matrix. The matrix has real valued elements.
I have been reading about the cholesky method and another method is
to find
the eigenvalues. I cant understand how to imp
On Sat, Jan 29, 2011 at 4:08 AM, Liviu Andronic wrote:
> Hello
>
> On Fri, Jan 28, 2011 at 8:03 PM, Jonathan P Daily wrote:
>> Geany also handles a ton of filetypes, but lacks a direct interface to R
>> in the windows version - I wrote an AutoHotKey script that did this for me
>> in about 5 minut
Thanks Prof Ripley, the condition worked!
Btw I tried to search ?repl but I don't have documentation for it. Is
it in a non-basic package?
On Jan 29, 6:54 pm, Prof Brian Ripley wrote:
> The grep comdition is "[A-J]"
>
> BTW, why there are lots of unnecessary steps here, including using
> cbind()
Many thanks for the answers. They helped me resolve my .RData problem.
--
View this message in context:
http://r.789695.n4.nabble.com/Problem-with-R-Installation-RData-tp3244121p3246036.html
Sent from the R help mailing list archive at Nabble.com.
__
Hello again,
many thanks for your help! after upgrading the lattice package to the
latest version, your solution works perfectly.
Deepayan Sarkar wrote:
On Fri, Jan 28, 2011 at 5:21 PM, E Hofstadler wrote:
Hello Deepayan,
many thanks for your reply and help. Does this solution with
"ylab.r
Hello I am trying to determine wether a given matrix is symmetric and
positive matrix. The matrix has real valued elements.
I have been reading about the cholesky method and another method is to find
the eigenvalues. I cant understand how to implement either of the two. Can
someone point me to the
The grep comdition is "[A-J]"
BTW, why there are lots of unnecessary steps here, including using
cbind() and subset():
x <- rep(LETTERS[1:20],3)
y <- rep(1:3, 20)
z <- paste(x,y, sep="")
random.data <- rnorm(60)
data <- data.frame(z, random.data)
data[grepl("[A-J]", z), ]
Now (for the paranoi
Hi:
This 'works':
> di <- read.csv(textConnection("
+ 10,2000
+ 12,2001
+ 13,2002
+ 15,2003
+ 17,2004"), header = FALSE)
> names(di) <- c('y', 'year')
> m <- lm(y ~ year, data = di)
> diextra <- data.frame(year = c(1990, 1991, 1993))
> predict(m, new = diextra)
123
-7.0 -5.3 -1.9
You
On 11-01-28 10:46 PM, Fernando Lima wrote:
Dear all,
I have a problem that has been driving me nuts. I have searched everywhere
but could not find a comprehensive answer. I only get (sometimes
contradictory) bits of information.
I have a series of measurements with associated standard deviation
On Sat, Jan 29, 2011 at 08:59:44AM +0100, Marius Hofert wrote:
> Dear expeRts,
>
> I somehow can't manage to do the following:
> Given two vectors x and y of lengths 3 and 2, respectively, I would like to
> apply
> a function to each combination of the entries of x and y and receive a 3-by-2
>
Hi all,
I would like to subset a dataframe by using part of the level name.
x <- rep(LETTERS[1:20],3)
y <- rep(1:3, 20)
z <- paste(x,y, sep="")
random.data <- rnorm(60)
data <- as.data.frame(cbind(z, random.data))
I need rows that contain the letters A to J, so I tried:
subset(data, grepl(LETTE
Readers,
Data was imported using the read csv command:
dataimport<-read.csv("/path/to/dataimport.csv")
10,2000
12,2001
13,2002
15,2003
17,2004
Using the help contents for 'predict.lm' (i.e. ?predict.lm) a new data
frame was created
dataimportextra<-data.frame(x=seq(1990,2010,1))
predict(lm(dat
Hello
On Fri, Jan 28, 2011 at 8:03 PM, Jonathan P Daily wrote:
> Geany also handles a ton of filetypes, but lacks a direct interface to R
> in the windows version - I wrote an AutoHotKey script that did this for me
> in about 5 minutes.
>
Could you please expand on that? I prefer to use Geany whe
Dear expeRts,
I somehow can't manage to do the following:
Given two vectors x and y of lengths 3 and 2, respectively, I would like to
apply
a function to each combination of the entries of x and y and receive a 3-by-2
matrix
containing the results. Now outer() seems to be the way to go, but out
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