Displaying 20 results from an estimated 1000 matches similar to: "error message - unexpected input"
2016 Nov 18
1
Letting you know about a broken link
Hi,
I appreciate you're busy so I just wanted to follow up on the email I sent you the other day, copy included below for reference.
On Sat, Nov 12, 2016 at 12:40 PM, Danielle James <danielle.james at whoishostingthismail.com> wrote:
Hi,
I just wanted to let you know about a link that seems to be broken on this page http://www.hdt-project.org/.
Here's the link
2011 Nov 11
1
Formula variable help
I have an R script with the following applicable lines:
xshort <- window(s, start=st, end=ed)
. . .
xshort <- ts(xshort, frequency=1, start=1)
. . .
m1 <- m2 <- m3 <- m4 <- m5 <- m6 <- NULL
m1 <- tslm(xshort ~ trend)
I get an error:
Error in get(dataname) : object 'xshort' not found
When I do traceback() I get:
3: get(dataname)
2: tslm(xshort ~
2011 Sep 20
1
Data
Hey everybody,
i am using the rugarch-package and its great!
I have a pretty easy problem, but i just dont get it, so thanks if you can
help me.
Normally i use:
/
data(DATANAME)
spec = ugarchspec()
fit = ugarchfit(data = x[,1], spec = spec)
fit
slotNames(fit)
names(fit at fit)
coef(fit)
infocriteria(fit)
likelihood(fit)
nyblom(fit)
signbias(fit)
head(as.data.frame(fit))
head(sigma(fit))
2000 May 02
2
Variable names in model formula
At 10:37 PM 5/1/00 -0400, E. S. Venkatraman wrote:
>I have the following problem. I have survival data (time, status) along
>with several covariates (X1, X2,..., Xn). I want to fit a Cox model for
>each of the covariate (univariately) and obtain the fitted probability of
>survival at a fixed time point t0 and covariate value Xi0. I tried to do
>this in a for loop where the index
2009 Nov 10
3
creating multiple plots using a splitting factor
Hello,
I am new to R. I often collect data at multiple sites and need to
create separate graphs (such as scatterplots or histograms) of specific
variables for each site. I have tried to do this by splitting the data
frame and then using lapply, but it seems that the graphing commands
cannot be called as functions. Here is a sample of my data, called
"seeddist2":
site
2010 Jul 27
6
Eval() or parse() do not work inside function
I am writing a function where the arguments are names of objects or variable
names in a data frame. To convert the strings to the objects I am using
eval(parse(text=name)):
f.graph.two.vbs<-function(dataname,v1){
val<-paste(dataname,v1,sep="$")
val<-eval(parse(text=val))
val
}
However running this returns an error:
2011 Oct 20
2
Expire plugin not executing for sieve plugin?
Running dovecot 1.2.12 on ubuntu 10.10. Here is the problem: I have
mysql backend for expire. I am using the dovecot-postfix package which
creates /etc/dovecot/conf.d/01-mail-stack-delivery.conf which looks like
this:
# Some general options
protocols = imap imaps managesieve
disable_plaintext_auth = no
ssl = yes
ssl_cert_file = /etc/ssl/certs/ssl-mail.pem
ssl_key_file =
2017 Dec 04
0
Dynamic reference, right-hand side of function
The generic rule is that R is not a macro language, so looping of names of things gets awkward. It is usually easier to use compound objects like lists and iterate over them. E.g.
datanames <- paste0("aa_", 2000:2007)
datalist <- lapply(datanames, get)
names(datalist) <- datanames
col1 <- lapply(datalist, "[[", 1)
colnum <- lapply(col1, as.numeric)
(The 2nd
2017 Dec 04
3
Dynamic reference, right-hand side of function
Hi R-users!
Being new to R, and a fairly advanced Stata-user, I guess part of my problem is that my mindset (and probably my language as well) is wrong. Anyway, I have what I guess is a rather simple problem, that I now without success spent days trying to solve.
I have a bunch of datasets imported from Stata that is labelled aa_2000 aa_2001 aa_2002, etc. Each dataset is imported as a matrix, and
2017 Dec 04
3
Dynamic reference, right-hand side of function
Hi!
Thanks for the replies!
I understand people more accustomed to R doesn't like looping much, and that thinking about loops is something I do since I worked with Stata a lot. The syntax from Peter Dalgaard was really clever, and I learned a lot from it, even though it didn't solve my problem (I guess it wasn't very well explained). My problem was basically that I have a data matrix
2011 May 05
1
Looping over graphs in igraph
Hi,
I'm trying to do some basic social network analysis with igraph in R, but
I'm new to R and haven't been able to find documentation on a couple basic
things:
I want to run igraph's community detection algorithms on a couple thousand
small graphs but don't know how to automate igraph looking at multiple
graphs described in a single csv file. My data look like something in
2017 Dec 04
0
Dynamic reference, right-hand side of function
Um, if you insist on doing it that way, at least use
assign(varname, as.vector(get(varname)))
-pd
> On 4 Dec 2017, at 22:46 , Love Bohman <love.bohman at sociology.su.se> wrote:
>
> Hi!
> Thanks for the replies!
> I understand people more accustomed to R doesn't like looping much, and that thinking about loops is something I do since I worked with Stata a lot. The
2011 Aug 17
1
multinomRob - error message
Hi,
I would like to use the multinomRob function to test election results.
However, depending on which independent variables I include and how
many categories I have in the dependent variable, the model cannot be
estimated.
My data look like this (there are 68 observations):
> head(database)
RESTE09 GAUCHE09 PDC09 PLR09 UDC09 MCG09 RESTE05 GAUCHE05 PDC05
D1 1455
2006 Jul 13
1
writeForeignSAS and potential extensions
Dear R-devel,
I've made some potential extensions to writeForeignSAS
in 'foreign' that I wanted to pass along if anyone is
interested. I've attached the diff -u output against
the version found in foreign_0.8-15 and an .R file
with my changes. (In this .R file, the function is named
writeForeignSAS7 to simplify testing/comparisons.)
I've tried to alter the current
2005 Feb 26
3
averaging within columns
I have a dataframe with names in the first column and wait times
between decisions in the second column. Since individuals make
multiple decisions, I want the average for each individual. For
example, the data might look like this
name wtime
jo 1
jo 2
jo 1
jo 3
tim 3
tim 2
tim 2
ro 1
ro 2
etc.
I'm hoping there is something like
mean(dataname$wtime[name])
which will just
2012 Oct 18
1
Getting a table of coefficients from R
Hi,?
I'm trying to obtain a table of coefficients and confidence intervals from a logistic regression analysis in R. My code is as follows:
# read in csv file
datafile<-read.csv("file.csv", row.names=1)
# read in the variable list
varlist<-names(datafile)[66:180]
models<-lapply(varlist, function(x) {glm(substitute(outcome ~ i, list(i=as.name(x))), data=datafile, family =
2003 Jan 09
7
Samba Authentication against NT domain
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2017 Dec 04
2
Dynamic reference, right-hand side of function
:-)
I don't insist on anything, I'm just struggling to learn a new language and partly a new way of thinking, and I really appreciate the corrections. I hope I someday will be able to handle lists in R as easy as I handle loops in Stata...
Thanks again!
Love
-----Ursprungligt meddelande-----
Fr?n: peter dalgaard [mailto:pdalgd at gmail.com]
Skickat: den 4 december 2017 23:09
Till:
2012 Apr 02
7
Calculating NOEL using R and logistic regression - Toxicology
Hello, I used the glm function in R to fit a dose-response relationship and
then have been using dose.p to calculate the LC50, however I would like to
calculate the NOEL (no observed effect level), ie the lowest dose above
which responses start occurring. Does anyone know how to do this?
[[alternative HTML version deleted]]
2006 Mar 28
0
Help with the code
library(survival) library(boot) data=NULL lambda=NULL result=NULL pat=rep(1:102,each=1) trt=rep(c(1,0),51) status=rep(1,102) site=rep(1:51, each=2) nr.datasets=100 seed=2006 beta=log(1/2) for (i in 1:51) { lambda[i]=1+((3-1)/50)*(i-1)} lambda1=rep(lambda, each=2) dummy=rep(c(exp(beta),1),51) elf=lambda1*dummy r=70 #the number of bootstrap replicates