On 06/17/2012 08:50 AM, @ngel wrote:
Hello again and thank you all so much for the help. Well, I tried and did it
myself, here's what I wrote:
library(tnet)
net<- read.table("data.txt")
net<- as.tnet(net, type="longitudinal tnet")
loop<- range(net[,1])
net.static<- vector(length=as.integer(loop
ozzi wrote
>
> Dear Berend,
>
> the coding is not working for more than 6. Try to plug 10 or other number.
> Would you please correct ?
>
My mistake.
mkupper <- function(A) {
n <- length(A)
m <- (1+sqrt(1+8*n))/2
stopifnot(m==as.integer(m))
print(m)
L <- matrix(0,nrow=m,nc
Hello,
It's stringsAsFactors = FALSE, just one '='.
Your data structure, a list, seems to have elements, vectors, of the
same lengths, all of them are of length 34773. You can check it with
sapply(dat, length)
It should return 4 times the value 34773. If so, there's no reason for
David's su
Hi
is there a easy way to get something like
http://addictedtor.free.fr/graphiques/graphcode.php?graph=137
pairs(USJudgeRatings[,c(2:3,6,1,7)],
lower.panel=panel.smooth, upper.panel=panel.cor)
but without the "lower.panel" (and without numbers and ticks at the border.)
thx
Christof
_
Hello,
The answer is yes, there is an easy way of getting that graphic without
the lower.panel and axis tickmarks.
op <- par(xaxt="n", yaxt="n")
pairs(USJudgeRatings[,c(2:3,6,1,7)],
upper.panel=panel.cor)
par(op)
where panel.cor() is the function in the link.
Hope this helps,
Rui Barra
Dear John,
Thank you very much for your help! It was very important for getting
along further.
I found out some additional things which I want to give back to you as
feedback (maybe other persons will have the same problems).
It's also possible to leave away the individual=T argument.
Instead
Thank you Arun for your time!
Your idea is maybe only the first step to what I want but it was
nevertheless a new tool for me and interessing to learn.
I added a "week"-column to your data set:
dattrial<-data.frame(a=c(1,NA,rnorm(4,10)), Week=c(3,3,3,4,4,4))
I am looking for a way to count the
Hello,
I've seen your reply to arun's reply and gave it a try.
Since arun's code included more than one column, I've added another in
one of the examples.
# Example 1
dattrial1 <- data.frame(a=c(1,NA,rnorm(4,10)), Week=c(3,3,3,4,4,4))
d1 <- split(dattrial1, dattrial1$Week)
count <- sapply(d1,
Dear Berend,
the coding is not working for more than 6. Try to plug 10 or other number.
Would you please correct ?
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Sent from the R help mailing list archive at Nabble.
Great! That works!
Thank you Rui!
I would have spent days (which I don't have left before handing my report
in) getting there by myself!
Have a great rest-weekend!
--
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you are right Jeff and sorry for this
I will try to explain what I want.
I have the following dataset
dat <- data.frame("country" = c(rep(1,4)),
"date" = c("23/11/08","28/12/08","25/01/09","22/02/09"),
"price" = c(2,3,4,5))
Normally, prices are observed every 4
Rui Barradas wrote
>
> It's stringsAsFactors = FALSE, just one '='.
>
> sapply(dat, length)
>
> It should return 4 times the value 34773.
>
> use dput()
>
it worked!
> dat2 <- data.frame(dat, stringsAsFactors = FALSE)
> sapply(dat2, length)
device_info_serial hour
or put differently, I want to change the data frequency of the time
series to monthly
thanks
On 6/17/12, stef salvez wrote:
> you are right Jeff and sorry for this
>
> I will try to explain what I want.
>
> I have the following dataset
>
> dat <- data.frame("country" = c(rep(1,4)),
>
Hi everyone, i'm not very good with R, and i know that my question is quite
simple.
I generated this vector for a sample of 100 and i want to know how to save
these as
ez1, ez1, ez3...ez100.
I guess i have to do it by "for", but i didn't succeed..
i put here my output
ezt
[1] -0.05470228 -1.76
Short answer: don't. R is built around vectors and lists. Trying to
put individual variates in their own "scalar" variables is almost
always a bad idea.
Long answer: you can do it with assign() but I won't tell you how in
light of my short answer.
Michael
On Sun, Jun 17, 2012 at 8:30 AM, Cassie
It is not clear exactly what you want. Are you saying that you want to create
an object out of each number in that vector or do you just want to name each
element in the vector?
It is best to supply the code that you used so that readers can see what you
are doing.
Also when you are providi
On Sun, Jun 17, 2012 at 6:30 AM, Cassie wrote:
> Hi everyone, i'm not very good with R, and i know that my question is quite
> simple.
>
So make an effort to educate yourself instead of badgering this list.
Please read "An Intro to R" -- which ships with R (help.start() or Help
Menu --> Manuals
It's requested to cc the list on our back-and-forths so that i) you
can get help faster (I was moving all yesterday and it's a Hallmark
holiday in the US today) since others may respond; ii) others can
benefit by reading the archives.
On Sun, Jun 17, 2012 at 4:34 AM, Giulia Motta wrote:
>
>
> t
Hi,
In my case, that was due to the existence of highly correlated
variables. see loggedEvents in the help file of mice.
Weidong Gu
On Sat, Jun 16, 2012 at 3:33 PM, b04877054 wrote:
> I have the same problem. How did you end up solving yours?
>
> --
> View this message in context:
> http://r.7
Dear R users,
I'm wonder if there is a easy way to make R use multi-CPUs on my computer.
My computer has four CPUs but R uses only one. Thanks.
Gary
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch
Hello,
Try
names(departures) <- NULL
dep <- data.frame(departures[ !sapply(departures, is.null) ])
You now have data.frame 'dep'.
Rui Barradas
Em 17-06-2012 12:19, capital_P escreveu:
Rui Barradas wrote
It's stringsAsFactors = FALSE, just one '='.
sapply(dat, length)
It should return 4
Take a look at the parallel package which ships with all current versions of R.
Michael
On Jun 17, 2012, at 11:39 AM, Gary Dong wrote:
> Dear R users,
>
> I'm wonder if there is a easy way to make R use multi-CPUs on my computer.
> My computer has four CPUs but R uses only one. Thanks.
>
> G
On 17.06.2012 19:04, R. Michael Weylandt wrote:
Take a look at the parallel package which ships with all current versions of R.
which is the right answer for the body of the message. For the subject
line: Take a look at multi-threaded BLAS which can be used with R.
Uwe Ligges
Michael
O
> I'm working on analyzing a large data set, lets asume that
> dim(Data)=c(1000,8700). I want to calculate the canberra distance
> between the columns of this matrix, and using a toy example ('test' is
> a matrix filled with random numbers 0-1):
>
>> system.time(d<-as.matrix(dist(t(test), method
On Jun 17, 2012, at 7:19 AM, capital_P wrote:
Rui Barradas wrote
It's stringsAsFactors = FALSE, just one '='.
sapply(dat, length)
It should return 4 times the value 34773.
use dput()
it worked!
dat2 <- data.frame(dat, stringsAsFactors = FALSE)
sapply(dat2, length)
device_info_ser
> a<-c(1,4)
> a
[1] 1 4
> b<-a*5
> b
[1] 5 20
a is a very long vector , how can i get c(1:5,4:20)? i do not want to
use a loop.
thanks very much!
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PLEASE do read the posti
"a" looks pretty short to me, just two elements.
You should spend some (more) time reading the "Introduction to R" document that
is supplied with the software... particularly the parts on indexing.
---
Jeff Newmiller
On Mon, Jun 18, 2012 at 01:24:21AM +0800, 李红旺 wrote:
> > a<-c(1,4)
> > a
> [1] 1 4
> > b<-a*5
> > b
> [1] 5 20
>
> a is a very long vector , how can i get c(1:5,4:20)? i do not want to
> use a loop.
Hi.
How large are the numbers in a? If max(a) is not too large, try
the following.
a <- c(1, 4
Hello,
i am trying to apply k-nn classification for my time-series, however the
euclidean distance is not the best choice as the features i use are not all
normalized (others have values form 0-1 others are negative etc.) and also
it doesn't do any feature evaluation and give different weights to
Thanks for all replied.
I read the introduction of R parallel. Is it for loops only?
Gary
On Sun, Jun 17, 2012 at 10:04 AM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> Take a look at the parallel package which ships with all current versions
> of R.
>
> Michael
>
> On Jun 17, 2
On 17-Jun-2012 17:24:21 ÀîºìÍú wrote:
>> a<-c(1,4)
>> a
> [1] 1 4
>> b<-a*5
>> b
> [1] 5 20
>
> a is a very long vector , how can i get c(1:5,4:20)?
> i do not want to use a loop.
>
> thanks very much!
Your question is not quite clear, but I think you mean:
a is a very long vector, and you wa
Hi,
Sorry, I didn't understand your question in the first post. I saw Rui's reply
and your reply that it is solved.
I have another solution if it helps you.
dattrial<-data.frame(a=c(1,NA,rnorm(4,10)), Week=c(3,3,3,4,4,4))
dattrial_wk3<-subset(dattrial,Week==3)
dattrial_wk4<-subset(dattrial,We
You can also try the corrgram package.
Kevin
On Sun, Jun 17, 2012 at 4:06 AM, Christof Kluß wrote:
> Hi
>
> is there a easy way to get something like
> http://addictedtor.free.fr/graphiques/graphcode.php?graph=137
>
> pairs(USJudgeRatings[,c(2:3,6,1,7)],
> lower.panel=panel.smooth, upper.panel
I would argue (somewhat emphatically) that the parallel facilities you
are looking at are absolutely not for `for` loops. `for` loops are a
control structure native to imperative programming and, as such, are
inherently stateful. This provides many advantages, but easy
parallelization is absolutely
I think multicore is one of answer if you can write your function in to lapply.
On Mon, Jun 18, 2012 at 12:14 PM, R. Michael Weylandt
wrote:
> I would argue (somewhat emphatically) that the parallel facilities you
> are looking at are absolutely not for `for` loops. `for` loops are a
> control st
Thanks Kjetil for your detailed code for log-Cholesky but my situation is like
optimization of the variables inside the matrix.
Let me change your own example matrix to explain the problem.
A martrix is [,1] [,2][,3]
[1,]3*Vs+Vn1*Vs 1*Vs
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