I thought mailing lists are very professional and filled with boring
people. But hey, R mailing list is pure fun and has lots of humour elements
;) So yeah, making it clear, logs as in the text dump spit out by your
software applications, servers and network elements.
Let me add to my query. This
On 01/05/2013 10:15 AM, Andrea Goijman wrote:
...
Hi Andrea,
This may be what you want:
occ.data<-data.frame(year=rep(2003,20),Ruta=rep(202,20),
+ Point=c(3,4,1,1,2,3,4,1,4,5,1,3,4,5,1,4,5,10,7,8),
+ Site=c(rep(2021,17),rep(2022,3)),
+ specie=c("MICH","MICH","MISA","MOBO","MOBO","MOBO","MOBO",
On 01/04/2013 10:13 PM, mary wrote:
Hi,
I'm Marianna
I'm trying to apply the command "repeat" to my matrix but the repeat process
doesn't work as I would.
In particular I would like to apply the function robustm () _that I have
created_ to my two matrices, if the difference between the two matr
Suzen, Mehmet gmail.com> writes:
>
> Hello List,
>
> Are there any R package that can process MT940/942?
>
> Thanks
>
> mem
I find it hard to say because I have no idea what those formats are
(although googling suggests they're banking codes).
library("sos") and using findFn() on "MT940
On Jan 4, 2013, at 10:46 AM, Christofer Bogaso wrote:
> Thanks all for your help. However I was looking for following type of
> operation:
>
> Mat1 <- Mat2 <- matrix(1:20, 4, 5); Mat2[1,5] <- 200
> Mat = rbind(Mat1, Mat2)
> apply(Mat, 2, function(x) return(all(x[1:4] == x[5:8])))
>
> However I
On Fri, Jan 4, 2013 at 5:53 PM, Matthijs Daelman
wrote:
> Hi
>
> I have to time series with a different time base.
>
> The first has only sporadic datapoints:
> 2011-02-01 15.29130
> 2011-02-08 17.60278
> 2011-02-15 17.99737
> 2011-02-22 25.43690
>
> The other has a daily datapoint:
> 2011-02-01 3
Hi Matthijs,
Look at the code in the attached file and try it out.
Best,
Mario
On Fri, Jan 4, 2013 at 5:53 PM, Matthijs Daelman wrote:
> Hi
>
> I have to time series with a different time base.
>
> The first has only sporadic datapoints:
> 2011-02-01 15.29130
> 2011-02-08 17.60278
> 2011-02-15
Hi
I have to time series with a different time base.
The first has only sporadic datapoints:
2011-02-01 15.29130
2011-02-08 17.60278
2011-02-15 17.99737
2011-02-22 25.43690
The other has a daily datapoint:
2011-02-01 342.34
2011-02-02 68.45
2011-02-03 130.47
2011-02-04 129.86
2011-02-05 81.98
20
List,
I want to reshape my data, but I'm not sure how to do it... it might be a
simple task, but don't know which package does this.
"occ.data" (see below) is how my original data are arranged, and I know
that with melt() I can reshape it like "y" (see below). However, I just
want to build a matri
Hi Mary,
Some things are clear from your code fragment:
- S_X is a variable external to your function, and is possibly a variable
accessible to robustm.
- Unless calling robustm changes S_X or abs(b1[i,i]-b)[i,i])<=0.001 is
always true for all S_X and all i, your repeat statement, and so your
func
On Jan 4, 2013, at 7:30 AM, Sam Steingold wrote:
Hi,
to count vector elements with some property, the standard idiom seems to
be length(which):
--8<---cut here---start->8---
x <- c(1,1,0,0,0)
count.0 <- length(which(x == 0)
Erin Hodgess gmail.com> writes:
>
> Dear R People:
>
> I'm not sure if I should be asking here or R-devel, but here goes:
>
> Will there be a new version of R coming out in the next few weeks, please?
>
> I'm setting up for the spring semester and want the students to have
> the latest versio
Erin,
We recently received an email message saying version 3.0 was coming soon. As I
remember it, there will be a number of important improvements, but no major
changes in functionality. Version 2.xx will probably be fine for your class.
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics
On Fri, Jan 4, 2013 at 12:45 PM, Saptarshi Guha wrote:
> Hello,
>
> How would i use bquote to create names for a list.
>
You don't.
-- Bert
(If you wish to ask how to change names in a list, or some other such
thing, then ask that.)
> e.g. to create list(a=10)
>
> bquote(list(.(X)=10), list(
Dear R People:
I'm not sure if I should be asking here or R-devel, but here goes:
Will there be a new version of R coming out in the next few weeks, please?
I'm setting up for the spring semester and want the students to have
the latest version.
Thanks very much,
Sincerely,
Erin
--
Erin Hod
I think you mean between column 1 and 2 of A? Why is 36003918 not
included? It is clearly between 35838396 and 36151202 in the first row of A.
My earlier solution should work fine. Just create a new matrix AX that has
the columns switched so that the start is always column 1 and use that to
identi
Hello,
How would i use bquote to create names for a list.
e.g. to create list(a=10)
bquote(list(.(X)=10), list(X="a"))
does not work.
The best i could come up with is
bquote({ a=list(10);names(a)=.(X); a}, list(X="a"))
which is quite ugly.
Is there an elegant way to solve this?
Regards
Sa
those were definitely different 'logs' than I was thinking about.
On Fri, Jan 4, 2013 at 1:50 PM, peter dalgaard wrote:
>
> On Jan 4, 2013, at 18:41 , jim holtman wrote:
>
>> what type of logs are you trying to process?
>
> Oregon pine?
>
>> what is their format?
>
> Cylindrical?
>
>> what inform
In Canada we use sawmills to process logs. Or we carve
them into totem poles. Or we carve dugout canoes.
Of course, sometimes we burn them in order to keep warm.
Peter Ehlers
On 2013-01-04 10:50, peter dalgaard wrote:
On Jan 4, 2013, at 18:41 , jim holtman wrote:
what type of logs are you tr
HI,
If you want to extract the month as in the format you mentioned in the original
post:
Month<-format(dat1[,1],format="%b")
head(Month)
#[1] "Jan" "Jan" "Jan" "Jan" "Jan" "Jan"
library(ISOweek)
Week<-date2ISOweek(dat1[,1])
Week1<-gsub(".*\\-(.*)\\-.*","\\1",Week)
head(Week1)
#[1] "W01" "W01
Thanks Giovanni,
but I already tried with the others "random.method" and it doesn't work anyway.
Do you think is it possible change the if statement in the plm formula, to set
the value to 0 if sigma<0 ?
Because if I insert a dummy variabile in the formula R gives me this error:
> Error in if (
HI Fares,
Sorry, that I misunderstand your question.
Probably, this works for you.
date1<-
seq.Date(as.Date("1jan2003",format="%d%b%Y"),as.Date("1jan2013",format="%d%b%Y"),by="day")
length(date1)
#[1] 3654
set.seed(51)
donation<-sample(1000:300,3654,replace=FALSE)
dat1<-data.frame(date1
On Jan 4, 2013, at 18:41 , jim holtman wrote:
> what type of logs are you trying to process?
Oregon pine?
> what is their format?
Cylindrical?
> what information do you want from them?
Tensile strength?
;-)
(Or: Base-10, 5-digit mantissa, geometric mean)
> On Fri, Jan 4, 2013 at 6:33 A
Thanks all for your help. However I was looking for following type of operation:
Mat1 <- Mat2 <- matrix(1:20, 4, 5); Mat2[1,5] <- 200
Mat = rbind(Mat1, Mat2)
apply(Mat, 2, function(x) return(all(x[1:4] == x[5:8])))
However I believe there must be some smarter method using
***mapply()*** or someth
I am having trouble predicting new data with a model created from package
mboost:
> mb1<-glmboost(as.formula(formula1),data=data_train,control=boost_control(mstop=400,nu=.1))
> f.predict<-predict(mb1,newdata=data_train)
Error in scale.default(X, center = cm, scale = FALSE) :
length of 'center'
Thomas,
thanks for your reply. I will switch to desolve. Unfortunatelly, the
differential equation can not be re-written to an analytical solution in this
case (to the best of my knowledge) because it is a non-linear process, which
may not have an exact analytical solution. Thank you
Andras
HI Fares,
You could try this:
dat1<- read.table(text="
date donation
3jan2003 20235
4jan2003 25655
5jan2003 225860
6jan2003 289658
7jan2003 243889
8jan2003 244338
9jan2003 243889
",sep="",header=TRUE,stringsAsFactors=FALSE)
The post is not very specific as to what you need.
what type of logs are you trying to process? what is their format?
what information do you want from them?
On Fri, Jan 4, 2013 at 6:33 AM, Ramprakash Ramamoorthy
wrote:
> Hello all,
>
> Need some suggestions on interesting use cases with R in the
> field of log processing. Any help wou
Hi,
I would like to do coloring of map regions based on the region values
"weight". The approach I am taking is first to break regions into equal
intervals,
classIntervals(spdf$weight,4)$brks #4 intervals in this case
and coloring all regions within the interval with the same color
col = brewer
On 4 January 2013 17:47, Bert Gunter wrote:
> Inline.
>
> On Fri, Jan 4, 2013 at 7:44 AM, Suzen, Mehmet wrote:
>>
>> I am always reserved about types and not sure how R auto casting works
>> internally.
>> In a large code using many different packages, I think being reserved
>> about
>> this woul
Inline.
On Fri, Jan 4, 2013 at 7:44 AM, Suzen, Mehmet wrote:
> I am always reserved about types and not sure how R auto casting works
> internally.
> In a large code using many different packages, I think being reserved about
> this would not hurt.
>
> Also, are there anyway to force R to be "st
On 4 January 2013 at 16:57, Suzen, Mehmet wrote:
| On 4 January 2013 11:36, Royden Fernandes wrote:
| > Hi,
| >
| > I am able to integrate C++ and R through RInside library. However when I
|
| Questions regarding RInside should go to the rcpp-devel mailing list.
| http://lists.r-forge.r-project.
On 4 January 2013 16:53, jim holtman wrote:
> Is performance a concern? How often are you going to do it and what
> other parts of your script also take longer? Why are you concerned
> about allocating/discarding two vectors?
I think Sam's question was about additional memory introduced by whi
Dear R users,
I want to group the d values in classes. If I use this script I have a
problem.
classes <- function(x, n){
s <- seq(0, ceiling(max(x)), by = n)
factor(n*findInterval(x, s), levels = s)
}
z<-sapply(tapply(t$d,t$plot,function(x) classes(t$d, 4)),table)
z<-cbind(z)
Thank you!
Initia
On 4 January 2013 11:36, Royden Fernandes wrote:
> Hi,
>
> I am able to integrate C++ and R through RInside library. However when I
Questions regarding RInside should go to the rcpp-devel mailing list.
http://lists.r-forge.r-project.org/mailman/listinfo/rcpp-devel
___
Hi Ista,
I was doing about something similar when I saw your post, so I took your
code and did some timings:
> system.time(replicate(100, count.0 <- length(which(x == 0
user system elapsed
5.590 0.173 5.834
> system.time(replicate(100, count.1 <- sum(x == 0)))
user syste
What is the concern if it works? you can also do
sum(x==0)
Is performance a concern? How often are you going to do it and what
other parts of your script also take longer? Why are you concerned
about allocating/discarding two vectors?
On Fri, Jan 4, 2013 at 10:30 AM, Sam Steingold wrote:
> H
Hi Sam,
Here is one alternative, which is at least faster:
system.time(count.0 <- length(which(x == 0)))
system.time(count.1 <- sum(x == 0))
all.equal(count.0, count.1)
Best,
Ista
On Fri, Jan 4, 2013 at 10:30 AM, Sam Steingold wrote:
> Hi,
> to count vector elements with some property, the sta
My 2 cents:
AFAIK both which and length are from C compiled code:
http://cran.r-project.org/doc/manuals/r-release/R-ints.html#g_t_002eInternal-vs-_002ePrimitive
so they must be quite efficient ie .Primitive and .Internal. Probably
combination
of this with a pattern in C would be more memory effi
Hi,
I am using the nls function and it stops because the number of iterations
exceeded 50, but i used the nls.control argument to allow for 500
iterations. Do you have any idea why it's not working?
fm1 <- nls(npe ~ SSgompertz(npo, Asym, b2, b3),
data=f,control=nls.control(maxiter=500))
Error i
The default shrinkage (learning rate) for mboost is much higher than for gbm.
--
View this message in context:
http://r.789695.n4.nabble.com/mboost-vs-gbm-tp4637518p4654637.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-pro
As long as there are no negative numbers, method2 and 3 works:
test[1]<- -1
length(which(test==0))
#[1] 6
length(which(test<1))
#[1] 7
length(which(test < .Machine$double.xmin))
#[1] 7
length(which(abs(test)<1))
#[1] 6
length(which(abs(test) < .Machine$double.xmin))
#[1] 6
A.K.
- Orig
Hello.
This is my dataset.
id order(t) A Previous A 1 1 0 NA 1 2 1 0 1 3 1 1 1 4 0 1 1 5 0 0 2 1
1 NA 2 2 0 1 3 1 1 NA 3 2 1 1 3 3 0 1 3 4 0 0
Id is individual. (ex.1=Mike, 2=Sandy)
Order(t) is order of individual's behavior.
A is idividual's behavior(ex. buying is 1 otherwise 0) at
Hi,
You could try this:
dat1<-read.table(text="
id,age,weight,height,gender
1,22,180,72,m
2,13,100,67,f
3,5,40,40,f
4,6,42,,f
5,12,98,66,
6,50,255,60,m
",sep=",",header=TRUE,stringsAsFactors=FALSE,na.strings="")
list1<-by(dat1[c("weight","height")],dat1[c("age","gender")],colMeans,na.rm=TR
Hi,
I'm Marianna
I'm trying to apply the command "repeat" to my matrix but the repeat process
doesn't work as I would.
In particular I would like to apply the function robustm () _that I have
created_ to my two matrices, if the difference between the two matrices is
less than 0.001, R give me ba
Hello all,
Need some suggestions on interesting use cases with R in the
field of log processing. Any help would be greatly appreciated.
--
With Thanks and Regards,
Ramprakash Ramamoorthy,
India,
+91 9626975420
[[alternative HTML version deleted]]
_
Hi,
I am able to integrate C++ and R through RInside library. However when I
use a jni call as my UI is through java it crashes at C++ side. My files
are
JNICallingClass.C:
JNIEXPORT void JNICALL Java_CallR_run (JNIEnv* env, jobject callr){
Thanks.
At 2013-01-04 17:41:23,"Jorge I Velez" wrote:
Dear meng,
Check I understood correctly, the Ryacas package might do what you want. Check
its vignette at
http://cran.r-project.org/web/packages/Ryacas/vignettes/Ryacas.pdf
Best,
Jorge.-
On Fri, Jan 4, 2013 at 5:59 PM, meng <>
No,I have no x1 and x2 values.
At 2013-01-04 17:32:24,"PIKAL Petr" wrote:
>Hi
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
>> project.org] On Behalf Of meng
>> Sent: Friday, January 04, 2013 8:00 AM
>> To: R help
>> Subject: [R] a question
I am always reserved about types and not sure how R auto casting works
internally.
In a large code using many different packages, I think being reserved about
this would not hurt.
Also, are there anyway to force R to be "strongly typed" similar to
Occaml etc...
mem
On 4 January 2013 16:35, arun
Hello Rui/Jorge,
This is shorter, and probably needs less memory for large matrices as
you create
an other copy by defining nas:
matrixOp <- function(m1, m2, op=`+`) {
rows <- min(nrow(m1), nrow(m2))
cols <- ncol(m1)
op(m1[1:rows, 1:cols], m2[1:rows, 1:cols])
}
Best,
mem
On 4 January 2013 1
Hi,
to count vector elements with some property, the standard idiom seems to
be length(which):
--8<---cut here---start->8---
x <- c(1,1,0,0,0)
count.0 <- length(which(x == 0))
--8<---cut here---end--->8---
however, this approac
Matteo,
I fully agree with David: please read the posting guide.
Anyway, the error message says it all: "the estimated variance of the
individual effect is negative". See e.g. the "basic panel" chapter (10
or 11) in Wooldridge's "Econometric Analysis of XS and Panel Data" to
understand why this m
So given B
> cbind(B, apply(B, 1, diff))
[,1] [,2][,3]
[1,] 35838396 36151202 312806
[2,] 35838674 35838584 -90
[3,] 36003908 35838674 -165234
[4,] 36004090 36003908-182
[5,] 36150188 36003992 -146196
Row 1 is start/end and rows 2 through 5 are end/start
so you only want
Muito obrigado, Rui. Cleaner and simpler than my approach.
Regards,
Jorge.-
On Sat, Jan 5, 2013 at 12:08 AM, Rui Barradas <> wrote:
> Hello,
>
> Using part of your code, it's possible to do without Reduce, have foo
> (fun, below) do the job.
>
> fun <- function(x, y, FUN = `+`){
> if(nrow(x)
Hello,
Using part of your code, it's possible to do without Reduce, have foo
(fun, below) do the job.
fun <- function(x, y, FUN = `+`){
if(nrow(x) < nrow(y)){
nas <- matrix(NA, ncol = ncol(x), nrow = nrow(y) - nrow(x))
x <- rbind(x, nas)
}else{
nas <- matrix(NA,
At Fri, 4 Jan 2013 17:17:40 +0530,
Christofer Bogaso wrote:
>
> Hello again,
>
> Let say I have 2 matrices which equal number of columns but different
> number of rows like:
>
> Mat1 <- matrix(1:20, 4, 5)
> Mat2 <- matrix(1:25, 5, 5)
>
> Now for each column 1-to-5 I need to fetch the correspond
You're right David, I'm really sorry. I deactivated html, hope now it works.
> R 2.15.2 plm() function on Windows 7.
I have a problem with the variance estimation in a random effect model
I used this formula to get my result, and it works:
> reg <- deltaF ~ L1.deltaF + L2.deltaF + deltaCDS + L
Dear Christofer,
You can try the following:
# proccess the matrices
foo <- function(m1, m2){
if(ncol(m1) != ncol(m2)) stop('number of columns should be equal')
if(nrow(m1) < nrow(m2)){
nas <- matrix(NA, ncol = ncol(m1), nrow = nrow(m2) - nrow(m1))
m1 <- rbind(m1, nas)
}
else{
nas <- matrix(NA
Hello again,
Let say I have 2 matrices which equal number of columns but different
number of rows like:
Mat1 <- matrix(1:20, 4, 5)
Mat2 <- matrix(1:25, 5, 5)
Now for each column 1-to-5 I need to fetch the corresponding columns
of these 2 matrices and add the corresponding elements (ignoring NA
v
Hi!
Thanks a lot for your help!
I now do have exactly the plot that I wanted!
(I also realized that I have to use the 'text' command outside of the plot
function; beginners mistake :-)
Cheers,
Nina
__
Dr. Nina Hubner
scientist quantitative proteomics
http://marionniles.com/myxgcvc.php
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible
Dear meng,
Check I understood correctly, the Ryacas package might do what you want.
Check its vignette at
http://cran.r-project.org/web/packages/Ryacas/vignettes/Ryacas.pdf
Best,
Jorge.-
On Fri, Jan 4, 2013 at 5:59 PM, meng <> wrote:
> Hi all:
> I have a question about the computation of expr
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of meng
> Sent: Friday, January 04, 2013 8:00 AM
> To: R help
> Subject: [R] a question about the computation of expression
>
> Hi all:
> I have a question about the computation
Hi all:
I have a question about the computation of expression:
y = 1+(2x1+2)-3*(5x1-1)+(3x2+3)-2*(x2-1)
The result of y is:
y = 11-13x1+x2
How can I compute the result of y via R function?
Many thanks!
My best.
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