Also something like
## generate zero-inflated Poisson data
n - 100
x1 - rnorm(n)
x2 - rnorm(n)
ind - rbinom(n,1,1/(1+exp(-1-0.1*x1)))
y - ifelse(ind,rpois(n,lambda=exp(1+0.2*x2)),0)
There is at least zeroinfl in the CRAN package pscl and fmr in Lindsey's
non-CRAN package gnlm for fitting
?seq.Date
seq(as.Date(2000-01-01), len=5, by=1 mon)
[1] 2000-01-01 2000-02-01 2000-03-01 2000-04-01 2000-05-01
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Vasilchenko Aleksander
Sent: Friday, November 30, 2012 6:27 AM
Try R package 'brew'? From its package description
Description:brew implements a templating framework for mixing
text and R code for report generation. brew
template syntax is similar to PHP, Ruby's erb
module, Java Server
Here is one way using capture.output and regular expressions
cc - capture.output(ll) ## ll is your list object
kk - cc[cc!= ] ## remove blank lines
kk - gsub(\\$, , kk) ## remove '$' from names
writeLines(kk)
MU10
MU.16 MU.19 MU.21 meansd
0.8052791 0.4350489
to work just
fine.
I have no idea what is happening unless jagat.k.sheth is correct and
we
are hitting a buffer limit of some kind.
I'd suggest getting a decent editor and going with it. Working
directly
in an R terminal is enough to drive most people crazy.
Any of the editiors
Sounds like a finite mixture model. I haven't read your references but an
overall model for such an approach could be
f(Y=0; pi, kappa) = 1- pi + pi*f(Y=0|Z=1; kappa)
where pi=Pr(Z=1) is the probability of an event, z, and y is the value observed
when the event occurs and f is the probability
You probably hit a buffer limit in X11/xterm on your ubuntu machine with copy
and paste. I get that behavior with Putty using your vector or when pasting
(long) commands into Putty.
If you really prefer copy and paste for this vector then try something like
eval(parse(text=scan(clipboard,
You can also try ?rep and something like
dat - read.table(textConnection(ID record
120
. 30
. 25
2 26
. 15
3 21
),header=TRUE,na.strings=.)
ind - !is.na(dat$ID)
id - dat$ID[ind]
reps - diff(c(seq_len(nrow(dat))[ind],nrow(dat)+1))
dat$new.id -
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Fernando Bizuet
Sent: Friday, December 12, 2008 12:55 PM
To: r-help@r-project.org
Subject: [R] loop with dates
Hello,
I am trying to do a loop with dates, but when I try
This is almost a macro problem. It could be done in SAS language using
the WPS product (660 USD) I think. ...
OUCH! Why do it the complicated way??? Check out ?dir, ?list.files, and
then ?lapply for a simple start.
Don't give up so soon! When it comes to R there is no need to punt - you
can
You can get all the rownames by calling write.table once on something
like
# if all list elements have equal length
t(data.frame(MyList))
# otherwise, let NA extend where needed
t(do.call(cbind, lapply(MyList, ts)))
You can take care of NAs as needed on your end.
HTH
-Original
Note that length(m) = 16, but your m is only 4x4.
Try this m[is.na(m)] - 0
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Alex99
Sent: Monday, December 01, 2008 9:06 AM
To: r-help@r-project.org
Subject: [R] Error: subscript out of bounds
Hi All,
I
Maybe you can try one of Jim Lindsey's libraries available at
http://popgen.unimaas.nl/~jlindsey/rcode.html
If I recall, there is a function 'elliptic' in his 'growth' package that
may help you.
Otherwise, you can use one of R's nonlinear optimization functions, e.g.
'nlm', 'optim', etc.
Here's one way
x - sapply(seq(1,ncol(DF),by=2), function(i) DF[,i:(i+1)], simplify=F)
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Lauri Nikkinen
Sent: Tuesday, November 25, 2008 9:00 AM
To: [EMAIL PROTECTED]
Subject: [R] How to split DF into a list
On the help page for nlm (type ?nlm) check out the 'See Also' section.
It mentions other functions such as 'optim' and 'nlminb' which can do
constrained optimizations.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Jiang Peng
Sent: Monday, November 24,
You can use Mark's code by giving levels to the factor, e.g.
as.numeric(factor(unlist(strsplit(ABCDAXYZ, )), levels=LETTERS))
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of [EMAIL PROTECTED]
Sent: Monday, November 24, 2008 9:15 AM
To: [EMAIL
How about which(colSums(t-v) == 0) ?
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Salas, Andria Kay
Sent: Monday, November 24, 2008 10:04 AM
To: r-help@r-project.org
Subject: [R] matching matrix columns to a vector
I need help with (hopefully) just
Yes! my oversight ... thank you
-Original Message-
From: David Winsemius [mailto:[EMAIL PROTECTED]
Sent: Monday, November 24, 2008 11:48 AM
To: Sheth, Jagat K
Cc: [EMAIL PROTECTED]; r-help@r-project.org
Subject: Re: [R] matching matrix columns to a vector
Since negative values could
bEST is up to you to define. Here is one simple way
y.new - c(t(model.matrix(~factor(y)-1)))
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of zerfetzen
Sent: Saturday, November 22, 2008 12:00 PM
To: r-help@r-project.org
Subject: [R] What's the BEST way
Your code isn't changing the filename
Try this
for(i in seq_along(wull)) write(wull[i],
paste(C://Users//zuber//Documents//wull(,i,),.txt,sep=))
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of [EMAIL PROTECTED]
Sent: Friday, November 21, 2008 7:39 AM
No need to use a for loop. Try something like this
dim(EXAM1) - c(20,7420/20)
test.breakdown.list - as.list(data.frame(EXAM1))
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Rajasekaramya
Sent: Friday, November 21, 2008 9:48 AM
To:
?unlist
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Rajasekaramya
Sent: Friday, November 21, 2008 2:14 PM
To: r-help@r-project.org
Subject: [R] list of list objects
hi there,
I have a list of list objects i need to remove the top layer
[[1]]
?unlist
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Rajasekaramya
Sent: Friday, November 21, 2008 2:14 PM
To: r-help@r-project.org
Subject: [R] list of list objects
hi there,
I have a list of list objects i need to remove the top layer
[[1]]
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