On 20/02/2014 15:54, Matías Guzmán Naranjo wrote:
I'm trying to install various packages like lme4, ndl and RcppEigen but
they all fail with similar error messages:
install.packages("RcppEigen")
Installing package into '/home/matias/R/x86_64-pc-linux-gnu-library/3.0'
(as 'lib' is unspecified)
pr
On Feb 20, 2014, at 12:03 PM, Michael Friendly wrote:
>
> The vector, vec, below represents a time series of the number of some events
> recorded
> for 208 two-week intervals from 2005/01/01 up to the end of 2012. I want to
> represent this together with the date information. I tried ts(), bu
Thank you so much Arun:)
On Wednesday, February 19, 2014 6:06 PM, arun wrote:
Hi Farnoosh,
Try:
library(plyr)
res <- join(`DATA-A`,`DATA-B`,by="Var1",type="right")[,c(3,1:2,4)]
head(res)
# ID Var1 Var2 var3
#1 1 AETNA CARDINAL CARE TIER 2 10
#2 2
Please read the Posting Guide, which offers several applicable tips, such as:
Don't post in HTML format... it tends to corrupt your code samples.
Please provide a hand-generated example result that should be what the solution
should transform your sample data into.
Please show the code that did no
I have a data frame of vehicle trajectories. Here's a snapshot:
>dput(head(df))structure(list(vehicle = c(2L, 2L, 2L, 2L, 2L, 2L), frame =
>43:48, globalx = c(6451214.156, 6451216.824, 6451219.616, 6451222.548,
>6451225.462, 6451228.376), class = c(2L, 2L, 2L, 2L, 2L, 2L), velocity =
... and yet another approach (written for generalization)
> names(A)[-1][as.matrix(A[,-1])%*%(seq_len(ncol(A)-1))]
[1] "D" "B" "B" "B" "C" "C" "D"
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
Also,
rownames(which(t(!!A[,-1]),arr.ind=TRUE))
A.K.
On Thursday, February 20, 2014 6:48 PM, arun wrote:
Hi,
May be this helps:
A$Variable <- rep(colnames(A[,-1]),nrow(A))[t(!!A[,-1])]
A.K.
On Thursday, February 20, 2014 5:55 PM, ioanna ioannou wrote:
Hello,
Assuming that I have a da
Dear all,
I would like to fit some data (few samples and a much larger number of
features), using linear regression with lasso regularization.
So far I have used the packages enet and lars, that perform very well and
of which I really appreciated the easy interface.
Unfortunately, while on the maj
Hi,
May be this helps:
A$Variable <- rep(colnames(A[,-1]),nrow(A))[t(!!A[,-1])]
A.K.
On Thursday, February 20, 2014 5:55 PM, ioanna ioannou wrote:
Hello,
Assuming that I have a data frame
A<-data.frame(A=c(10,100,1000,30,50,60,300),
B=c(0,1,1,1,0,0,0),
Hello,
Assuming that I have a data frame
A<-data.frame(A=c(10,100,1000,30,50,60,300),
B=c(0,1,1,1,0,0,0),
C=c(0,0,0,0,1,1,0),
D=c(1,0,0,0,0,0,1))
What I would like is to introduce a new column Variable such that:
We are happy to inform you that abstract submission for
useR! 2014 is now available online, see
http://user2014.stat.ucla.edu/
The R User Conference, useR! 2014 is scheduled for July 1-3, 2014 at
the University of California, Los Angeles. Before the official
program, half-day tutorials will
Hi Michael
What about using zoo you can convert to ts if required
library(zoo)
tv <- seq(as.Date('2005/01/01'),as.Date('2013/01/01'),by=14)
zt <- zoo(vec,tv)
head(zt)
2005-01-01 2005-01-15 2005-01-29 2005-02-12 2005-02-26 2005-03-12
1 0 0 0 1
Depending what you really want to achieve, the following may be useful or
educational:
dat$ID2x <- with( dat, ave( rep( 1, nrow( dat ) ), ID, USE, FUN=cumsum ) )
dat$ID2y <- dat$ID2x
dat$ID2y[ dat$USE != "001" ] <- NA
On Thu, 20 Feb 2014, arun wrote:
Hi,
Try:
dat$ID2 <- with(dat,ave(seq_along
If the data are two week intervals, the frequency is 52/2 = 26
and that matches the amount of data 8*26=208:
> vec.ts <- ts(vec, start=c(2005, 1), frequency=26)
> str(vec.ts)
Time-Series [1:208] from 2005 to 2013: 1 0 0 0 1 1 1 0 2 0 ...
> print(vec.ts, calendar=TRUE)
p1 p2 p3 p4 p5 p6 p7 p8
Hi,
Try:
dat$ID2 <- with(dat,ave(seq_along(USE),ID,FUN=function(x){x1 <- USE[x] =='001';
ifelse(!x1,'',cumsum(x1))}))
A.K.
On Thursday, February 20, 2014 3:31 PM, Pedro Mardones
wrote:
Dear R community;
I'm kind of stuck with the following situation and would appreciate any
hint. Let's assu
Dear R community;
I'm kind of stuck with the following situation and would appreciate any
hint. Let's assume I have the following data frame:
dat <- data.frame(ID = c(rep("01",18), rep("02",16)), USE = c(c("001","004",
"005","007","001","004","005","007","012","001","004","005","007","001","004",
The vector, vec, below represents a time series of the number of some
events recorded
for 208 two-week intervals from 2005/01/01 up to the end of 2012. I
want to
represent this together with the date information. I tried ts(), but
don't quite understand
how to use the start=, end= and frequ
On 20.02.2014 14:45, Simone Gill-Alvarez wrote:
To Whom It May Concern:
I am subscribed to this listserv, but I would like to switch my settings so
that I receive a daily batched email rather than individual messages. Is it
possible to do that, or do I need to unsubscribe and then resubscribe?
Hi,
Try:
library(stringr)
res <-
transform(dat[rep(1:nrow(dat),str_count(dat$Result,"-")+1),],Result=unlist(strsplit(as.character(dat$Result),"
- ")),Batch=ave(Batch,Batch,FUN=function(x) paste0(x,letters[seq_along(x)])))
row.names(res) <- 1:nrow(res)
# Based on the expected results, it coul
I believe this is what you should do:
* In the lbfgsb.cpp from base R, include the GPL blurb and R
copyright holders. Also add your own names. (see e.g.
https://github.com/hadley/pryr/blob/master/src/typename.cpp)
* In the DESCRIPTION, add the R core team as a contributor,
https://github.com/had
I'm trying to install various packages like lme4, ndl and RcppEigen but
they all fail with similar error messages:
install.packages("RcppEigen")
Installing package into '/home/matias/R/x86_64-pc-linux-gnu-library/3.0'
(as 'lib' is unspecified)
probando la URL '
http://cran.at.r-project.org/src/con
Dear All,
my problem case: I want to use the apriori-function to analyze my test data.
Take a look (the data is attached).
WarenGruppe is a hierarchical higher class for Item. I analyzed with apriori()
Customer and Item but I want also to start a cross-rule-mining. That means for
example that
Hi,
I'm analyzing data from an unbalanced (non-orthogonal) factorial experiment
(counts as being dependent on two nominal variables and one continuous
variable) using GLM model with quasi-Poisson distribution.
Which function (library) accounting for non-orthogonality in the data
(something like A
Dear R users,
we try to submit a package to CRAN. One of our function pbcOptim.R uses
lbfgsb.cpp which is a slight modification of optim.c. In the documentation for
pbcOptim.R we said
"The code for pbcOptim is based on that of \code{\link{optim}}. In particular,
pbcOptim calls /src/lbfgsb.cpp
To Whom It May Concern:
I am subscribed to this listserv, but I would like to switch my settings so
that I receive a daily batched email rather than individual messages. Is it
possible to do that, or do I need to unsubscribe and then resubscribe?
Best,
Simone Gill
[[alternative HTML ver
Hello Bart,
I solved your problem.
Please find the code given below,
dat <-structure(list(Batch = c(1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 4L, 4L,
4L, 4L), Test =
I am baffled why you have gone so far down this road, Ted. Considerable effort
has gone into making prediction of value N of the RNG sequence unrelated to
value N-1 of the sequence as long as you don't know the internal state of the
RNG. This is true for both the R internal RNG and the platform-
Hi
You probably will get more answers when you post your question to
Crossvalidated or Stackoverflow as it is more about statistics than about R.
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of PQuery
> Sent: Wednesda
Your code produces four line graphs, one above the other. What
do you mean by drawing a contour line for the entire panel? A
contour map requires three vectors, two for the horizontal
position and one for the elevation or height. You have four
vectors of y variables with the same x. Since you do no
On 20/02/2014 12:00, (Ted Harding) wrote:> [see at end]
>
> It seems clear that Izhak seeks to detach the random generation of y
> from the random generation of x after using set.seed(). On my reading of
>?RNG
> once set.seed has been used, as RUI says, it affects all subsequent
> calls to the
[see at end]
On 20-Feb-2014 10:47:50 Rui Barradas wrote:
> Hello,
>
> I'm not sure I understand the question. When you use set.seed, it will
> have effect in all calls to the random number generator following it. So
> the value for y is also fixed.
> As for your code, you don't need the second s
Hello,
I'm not sure I understand the question. When you use set.seed, it will
have effect in all calls to the random number generator following it. So
the value for y is also fixed.
As for your code, you don't need the second set.seed. And though it is
not syntatically incorrect, the way you a
Dear all,
I’m trying to use the saveHTML() function with the svg graphics device (
ani.dev= “svg”) in order to obtain animation with vector graphics.
For this device the unit is in inch, while the units for ani.height and
ani.width options seems to be in px.
As a result, something like
ani.opti
Hello there,
I've got a netCDF file which I want to read, modify, and save, which works
well with 'ncdf4'-package.
The problem is, that the grid representing the data (temperature values) is
irregular, and I am not able to output an irregular grid. In the original
file I have two variables ('lat'
Thanks a lot, it works perfectly!
f.
--
Francesca Pancotto
Università degli Studi di Modena e Reggio Emilia
Palazzo Dossetti - Viale Allegri, 9 - 42121 Reggio Emilia
Office: +39 0522 523264
Web: https://sites.google.com/site/francescapancotto/
---
how do i use set.seed? for example i want to generate fix x with different
value of y each time i.e
genarate x<-rnorm(10)
generate y<-rnorm(10)
i want have x fix but y changes at each iteration. this what i try but is not
working
{
set.seed(100)
x<-10*runif(10)
}
x
set.seed(y<-rnorm(10))
y
On 20 Feb 2014, at 06:07 , IZHAK shabsogh wrote:
> please, how can i generate ten different estimated parameter with multiple y,
> response from the following response.
You did. You're just only printing the last one... (sapply(list, coef) to
extract the coefficients in matrix form, if that's
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