Dear all,
the dat has missing values NA,
first.Name Name Department DCE DP date
5 Auction VideosYME 0.57 0.56 2013-09-30
18 Amish WivesTAS 0.59 0.56 2013-09-30
34 Ancient NationQLH 0.54 0.58 2013-09-30
53 Auction VideosYME NA
Hi,
I need help with fitting a non-linear mixed effects model (nested random
effect). I look at the relationship between 'wingbeat frequency' (beats per
sec) and 'wing length' in hummingbirds, and I will find a model with the
best fit.
I have posted the question at R-SIG-mixed-models list, but no
Yes, Peter's suggestion worked just right.
Here's an example that prints out the proposed cell content and will not
allow 'w' or '\n' to be entered into the table:
require(tcltk2)
tt<-tktoplevel(); tfr<-tkframe(tt); tkgrid(tfr)
tableData<-tclArray()
valChar<-function(S){
print(S); flush.consol
Perhaps more people write end-user-ready applications in R than I am aware of
and my bias is too strong. For working at the console I prefer not to have my
scripts installing packages on their own (one possible alternative execution
path), and it is too much trouble to implement multiple routes
Hi Dan
Were you able to find a way to access the % S values ??
Do you have any examples of how this works?
Thank you for your attention.
Cleber
Em 25/01/2016 08:21, peter dalgaard escreveu:
> It's been so long that I have forgotten how to get the package with the table
> widget installed on OS
On 11/17/2016 04:49 PM, Jeff Newmiller wrote:
require(tfplot)
tfplot(x.ts)
Would just like to point out that require() should not be treated as
interchangeable with library(). The former returns a logical status
indicating success or failure, while the latter throws an error if it
falls. You
> require(tfplot)
> tfplot(x.ts)
Would just like to point out that require() should not be treated as
interchangeable with library(). The former returns a logical status indicating
success or failure, while the latter throws an error if it falls. You should
reserve use of require() for cases w
On 11/17/2016 06:00 AM, r-help-requ...@r-project.org wrote:
Hi,
As I sit and learn how to work with time series, I've run into a problem
that is eluding a quick easy answer (and googling for answers seems to
really slow the process...)
Question #1--
In a simple example on R 3.3.1 (sorry my emp
> (1+ 0i)^ NA
[1] NA
> sessionInfo()
R version 3.3.1 (2016-06-21)
Platform: x86_64-apple-darwin13.4.0 (64-bit)
Running under: OS X 10.11.6 (El Capitan)
locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] stats graphics grDevices utils
[5] da
I tried on my computer.
> (1+ 0i)^ NA
[1] NaN+NaNi
On Thu, Nov 17, 2016 at 3:42 PM, Bert Gunter wrote:
> On Nov 17, 2016 11:55 AM, "Thierry Onkelinx"
> wrote:
> >
> > Dear Da,
> >
> > NA represents an unknown value x. 1 ^ x = 1 for all possible values of x.
> > Hence 1 ^ NA = 1.
> >
> That is
On Nov 17, 2016 11:55 AM, "Thierry Onkelinx"
wrote:
>
> Dear Da,
>
> NA represents an unknown value x. 1 ^ x = 1 for all possible values of x.
> Hence 1 ^ NA = 1.
>
That is false. For any n, n-1 of the nth roots of 1 differ from 1(they are
complex). I don't have my computer with me. What does (1+
On 17/11/2016 2:19 PM, Da Zheng wrote:
Hello,
I just realized that 1^NA outputs 1 while 1.1^NA outputs NA in R v3.3.1 and
R v3.2.3.
I tried other values such as 0^NA and 2^NA, and they all output NA.
I don't understand this inconsistency here. Shouldn't 1^NA output NA as
well? Why does R handle
Dear Da,
NA represents an unknown value x. 1 ^ x = 1 for all possible values of x.
Hence 1 ^ NA = 1.
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Klinieks
Hello,
I just realized that 1^NA outputs 1 while 1.1^NA outputs NA in R v3.3.1 and
R v3.2.3.
I tried other values such as 0^NA and 2^NA, and they all output NA.
I don't understand this inconsistency here. Shouldn't 1^NA output NA as
well? Why does R handle it differently? Or is this a bug in these
Hello,
I use R 3.0.2 on Win 7 through proxy server using ".Rprofile" in home
directory that includes "Sys.setenv(http_proxy=proxy_server:port)".
There has been no problem to access the internet for some years.
In this situation, I installed R 3.3.1 and then entered "update.packages
()", however, "
This should work. First create some reproducible data:
> set.seed(42)
> temp.matrix <- matrix(rnorm(255*13), 255, 13)
Then you need to send each row to t.test() using apply() extracting the
statistic with [[1]]:
> result <- apply(temp.matrix, 1, function(x) (t.test(x[3:7], x[8:12],
> paired=TR
On 17/11/2016 6:26 AM, Ulrik Stervbo wrote:
Hi Georg,
Your for loop iterates over just one value, to get it to work as you intend
use for(item in 1:length(kpis)){}
That usually works, but fails often enough that we recommend using
for (item in seq_along(kpis)) {}
(The failures happen if leng
Hi Ulrik,
oh no! What a mistake did I make. But I definitely did not see the
failure.
Many thanks for helping me.
Kind regards
Georg
Von:Ulrik Stervbo
An: g.maub...@weinwolf.de, r-help@r-project.org,
Datum: 17.11.2016 12:24
Betreff:Re: [R] for loop is looping only once
Hi Georg,
Your for loop iterates over just one value, to get it to work as you intend
use for(item in 1:length(kpis)){}
HTH
Ulrik
On Thu, 17 Nov 2016 at 12:18 wrote:
> Hi All,
>
> I need to execute a loop on variables to compute several KPIs.
> Unfortunately the for loop is executed only once
Hi All,
I need to execute a loop on variables to compute several KPIs.
Unfortunately the for loop is executed only once for the last KPI given.
The code below illustrates my current solution but is not completely
necessary to spot the problem. I just give an idea what I am doing
overall. Looks
Hello everybody,
I am a student at Unibas and I am trying to run a two sample t-test on a matrix
which is a
255*13 matrix (see attachment). I want to run the t-test, row-wise, with the
columns 3:7 being a part of the first group and columns
8-12 being a part of the second group.
I tried running
great minimal reproducible example, thanks. does something like this work?
#Log. Reg. model-all curric. concentrations including F1RTRCC as a predictor
allCC <-
svyglm(formula=F3ATTAINB~F1PARED+BYINCOME+F1RACE+F1SEX+F1RGPP2+F1HIMATH+F1RTRCC,family="binomial",design=elsq1ch_brr,subset=BYSCTRL==1
22 matches
Mail list logo
