o
the job, and the only thing I have been able to find is some proprietary
software HUDAP (Hebrew University Data Analysis Package) which may/ not
be compatible with R (or GNU/Linux for that matter).
Does anyone have information on how to do SSA using R?
Many thanks
Andrew
[[altern
,
Andrew
On 20/09/18 21:28, Eric Berger wrote:
> Hi Andrew,
> I don't have any experience in this area but I was intrigued by your
> question. Here is what I learned.
>
> 1, A bit of poking around turned up a thread on stats.stackexchange
> that mentions that "smalles
rop me a line to let me know?
Many thanks
Andrew
On 21/09/18 07:46, Martin Maechler wrote:
Eric Berger
on Thu, 20 Sep 2018 23:28:27 +0300 writes:
> Hi Andrew,
> I don't have any experience in this area but I was intrigued by your
> question. Here is what I l
visual
inspection of relative distance from different angles. I appreciate that
smacof will not do that, but as long as the analysis allows for the
graph to be plotted and analysed, that's what's important.
Thank you again, and to all of those who responded.
Best wishes
Andrew
On 21/0
Ha! Even better - thank you. Plenty here for me to play around with.
Many thanks
Andrew
On 23/09/18 15:22, Michael Friendly wrote:
On 9/22/2018 6:49 AM, Andrew wrote:
Hi Michael
This looks like it could be really helpful in moving my project
forwards thank you.
I remember many years ago
r is it 12"
[/ output]
How do I:
(a) reduce the gap between the reported number (i.e., 17, 9, 13) in each
of the lines? and
(b) ensure that in the case of the second run using 9 as the input, the
print is not over two lines?
I will try the switches function soon, which may yield a d
I'll dig into 'message' a little more, so that's a good one to know.
Rui, I liked the idea of the output vector and indexing - that will be
useful for longer/ more complicated branching
Thank you all again.
Best wishes
Andrew
On 18/12/2018 09:25, Ivan Krylov wrote:
On Tue,
**
I could use some help to create a robust and flexible workaround for RODBC's
asymmetric treatment of numeric data.
If there were some way to force RODBC sqlQuery to interpret all SQL Server
float datatypes as numeric my problem would be solved.
FWIW: RODBC does interpret the SQL Se
you could try stringsAsFactors in the read.table function
On Aug 17, 4:19 pm, rajclinasia wrote:
> Hi everyone,
>
> After reading the external file into R it will become a R dataset. Now my
> querry is i want to dispaly only character variables from that dataset.
> please help me in this aspect.
it is speedier to use sort than a combination of [] and order:
N<- 100
x <- runif(N)
> system.time(x[order(x)[c(N-1,N)]])
user system elapsed
1.030.001.03
> system.time(sort(x)[c(N-1,N)])
user system elapsed
0.280.000.28
On Sep 4, 11:17 am, Noah Silverman wrot
want something like
mapply(dim, ls())
but this just finds the dimension of the character array of the name
of the R object, not the actual size of the vector that has the same
name.
Any thoughts on how to do this please?
Regards,
Andrew.
__
R-help@r
Thanks Marc. I really appreciate your help. I'm going to try my function hack.
I forwarded your suggestion to Yohan at rmetrics.
Warm regards,
Andrew
--- On Fri, 7/10/09, Marc Schwartz wrote:
From: Marc Schwartz
Subject: Re: [R] strange strsplit gsub problem 0 is this a bug or a s
This is not a complete answer, but try the following in order
plot(c(1,1), ylim=c(0, 15), xlim=c(0, 13), axes = FALSE)
axis(1)
axis(2)
box()
If you want lines, use abline. Hacking something together can be done
with these commands.
On Sep 24, 9:36 am, Jason Rupert wrote:
> Using the following
both the following will probably do the trick.
?subset
?"["
Basically on the second one, you want to come down to something that
looks like
x[L]
where x is a matrix/vector, and L is a logical vector that has the
same dimension as x, but is TRUE on the values of x that you want to
select.
for i
?subset is probably what you want:
subset(x, x1 == 'A')
On Oct 2, 1:43 pm, Kavitha Venkatesan
wrote:
> Hi,
>
> I have a data frame that looks like this:
>
> >x
>
> x1 x2 x3
> A 1 1.5
> B 2 0.9
> B 3 2.7
> C 7 1.8
> D 7 1.3
>
> I want to "group" by the x1 column and in
The following is *significantly* easier to do than try and add in
dummy variables, although the dummy variable approach is going to give
you exactly the same answer as the factor method, but possibly with a
different baseline.
Basically, you might want to search the lm help and possibly consult a
Oh dear, that doesn't look right at all. I shall have a think about
what I did wrong and maybe follow my own advice and consult the doco
myself!
On Oct 21, 2:45 pm, andrew wrote:
> The following is *significantly* easier to do than try and add in
> dummy variables, although the dum
Sorry for this third posting - the second method is the same as the
first after all: the coefficients of the first linear model *is* a
linear transformation of the second. Just got confused with the
pasting, tis all.
On Oct 21, 2:51 pm, andrew wrote:
> Oh dear, that doesn't look righ
Change e2 to the following and it works
e2 <- function(a) a^2/2
The reason it doesn't is that e2 must be able to handle vector inputs
correctly. Your original function does not do this.
from ?integrate
f - an R function taking a numeric first argument and returning a
numeric vector of the sam
xs <- "this is string"
xsv <- paste(xs, 1:10)
sapply(xsv, function(x) strsplit(x, '\\sis\\s'))
This will split the vector of string "xsv" on the word 'is' that has a
space immediately before and after it.
On Oct 23, 1:34 pm, Jonathan Greenberg wrote:
> Quick question -- if I have a vector of s
the following works - double backslash to remove the "or"
functionality of | in a regex. (Bill Dunlap showed that you don't
need sapply for it to work)
xs <- "this is | string"
xsv <- paste(xs, 1:10)
strsplit(xsv, "\\|")
On Oct 23, 3:50 pm, Jonathan Greenberg wrote:
> William et al:
>
> Th
I don't seem to get a problem with this. Have you tried a Monte Carlo
approach to verify that you are getting incorrect answers?
For me, I get when the upper is 1 that
> integrate(e2, lower = 0, upper = 1)
-0.2820948 with absolute error < 5e-05
> sum(e2(runif(1)))/1
[1] -0.2825667
whic
.
Regards,
Andrew
On Oct 28, 5:42 am, Ken Ervin wrote:
> I have a data set of 6 or so ordered pairs, and I've been able to graph
> them and have decided to use a high-order polynomial regression. I've
> used the following piece of code:
>
> regression <-
hing like " myVar : 10"
I'd appreciate any suggestions.
Regards,
Andrew
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Henrique,
It works great. Perfect! Thank you.
Warm regards,
Andrew
--- On Wed, 11/18/09, Henrique Dallazuanna wrote:
Try this:
debugPrint <- function(x, ...){
print(sprintf("%s: %d", deparse(substitute(x)), x), ...)
}
On Wed, Nov 18, 2009 at 8:35 AM, Andrew wro
find the fitted value of the curve at any arbitrary month
within the series (i.e. no extraplolation is needed)
Warm regards,
Andrew
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
the plotmath help page should give the answer
http://www.stat.ucl.ac.be/ISdidactique/Rhelp/library/base/html/plotmath.html
On Mar 24, 2:53 pm, Daren Tan wrote:
> I need to have the maths symbol for >= in the legend, and to
> substitute threshold variable with its value. Somehow, various
> attemp
system elapsed
29.580.32 31.31
(yes, I have a pitifully slow computer)
HTH,
Andrew.
On Mar 29, 8:01 am, Tal Galili wrote:
> Yes Johannes - That helped, thank you.
>
> On Sat, Mar 28, 2009 at 11:50 PM, Johannes Huesing
> wrote:
>
>
>
> > Tal Galili [Sat, Mar 28, 2
< quantile(x)[2] - 1.5*IQR(x)) | (x > quantile(x)[4] + 1.5*IQR
(x))]
or if you prefer
is.outlier <- function(x) {(x < quantile(x)[2] - 1.5*IQR(x)) | (x >
quantile(x)[4] + 1.5*IQR(x))}
x[is.outlier(x)]
HTH.
Andrew.
On Apr 12, 8:51 am, Grześ wrote:
> Hello
> I have a problem l
or perhaps
agec <- 0*age
age[age<=46] <- 1
age[age>46 & age<=58 <- 2
age[age>58] <- 3
or perhaps a one liner
cut(x = age, breaks= c(0,46, 58,Inf), labels = c(1,2,3))
On Apr 24, 11:24 am, Jorge Ivan Velez
wrote:
> Dear Hollie,
> ifelse() is one alternative in this particular case:
>
> # Some
integrate offers some one-dimensional algorithms, but you need to
start with a smooth function to get it to converge properly. With a
cosine integral, there may be certain routines that offer better value
for money: the Clenshaw-Curtis integration, or perhaps the FFT. You
would have to recast you
05, 2009 10:18 AM
> To: r-h...@r-project.org
> Subject: Re: [R] FOURIER INTEGRALS IN R
>
> Ok thanks..
>
> No, my function is not smooth.
>
> Actually, I am dealing with an integral having the following form, for
> example:
>
> \int cos(tx) (1-t^2)^3 \exp(0.5*
Hello all
I am doing some text mining on a set of five plain text files and have
run into a snag when I run hclust in that there are just too many leaves
for anything to be read. It returns a solid black line.
The texts have been converted into a TDM which has a dim of 5,292 and 5
(as per 5 do
e able to produce a much simpler example.
e.g.
mycount.matrix <- matrix (rpois (25000, 20),, 5)
head (mycount.matrix, 3)
tail (mycount.matrix, 3)
plot (hclust (dist (mycount.matrix) ) )
On Tue, Sep 15, 2020 at 6:54 AM Andrew wrote:
Hello all
I am doing some text mining on a
Hello Jim
Thanks for that. I'll read up on it and will give it a go, either later
today or tomorrow. I am assuming this will work for both tf and tf-idf
weighted TDMs?
Much appreciated. :-)
Best wishes
Andy
On 18/09/2020 09:18, Jim Lemon wrote:
Hi Andrew,
>From your last email th
I am a Biologist and a beginner, please help me to solve this: please
anyone..its my homework and I dont have a clue abt R yet..
Write a function which does the following tasks:
(a) Calculates minimum and maximum value of a given argument x.
(b) If x is positive, some new vector gets the value
I think this should work
rgmm <- function(n, gmm) {
M <- sample(1:4, n, replace = TRUE, prob= gmm$weight)
mean <- gmm[M, ]$mean
sd <- gmm[M, ]$sd
return(gmm[M,]$sd*rnorm(n) + gmm[M,]$mean)
}
hist(rgmm(1, gmm), breaks = 500)
On Dec 19, 4:14 pm, "Bill McNeill
the SparseM package might be what you are looking for
http://www.econ.uiuc.edu/~roger/research/sparse/SparseM.pdf
On Jan 7, 11:36 am, Søren Højsgaard wrote:
> You can do
> mat[lower.tri(mat, diag=F)]
> Søren
>
>
>
> Fra: r-help-boun...@r-project.org på vegne af N
Hi Gundala,
The following should work.
x <- numeric(length(qr))
for(k in 1:3) x[k] <- which(repo %in% qr[k])
also, be careful about overwriting qr - it is an base package
function.
regards,
Andrew
On Jan 13, 1:41 pm, "Gundala Viswanath" wrote:
> Dear all,
>
> I tr
or this
x[,!(colSums(abs(x)) == 0)]
On Jan 15, 10:00 am, Marc Schwartz wrote:
> Careful:
>
> x <- matrix(c(1, 5, 3, 2, 1, 4, -1, 0, 1),
> ncol = 3, nrow = 3)
>
> > x
>
> [,1] [,2] [,3]
> [1,] 1 2 -1
> [2,] 5 1 0
> [3,] 3 4 1
>
> > x[, colSums(x) != 0]
>
Not sure if this helps, as I am attempting to do a mind read based on
ASCII (I get better results using unicode)
Regards,
Andrew
On Jan 15, 10:42 am, Duncan Murdoch wrote:
> rkevinbur...@charter.net wrote:
> > This is definitely a newbie question but from the documentation I have not
any idea why DDR2 667 MHz RAM isn't used instead of DDR? I thought
that DDR 400MHz was almost finished in production...
On Jan 27, 1:01 pm, zerfetzen wrote:
> What do you think of this:
>
> http://www.microway.com/whisperstation/whisperstation-r.html
>
> I'm considering ditching my Windows Vista
can you use the system function?
system("echo $PATH")
On Feb 2, 11:10 am, David Epstein wrote:
> I use a Mac (10.4.11 Mac Os X).
>
> In my .tcshrc I define an environmental variable MY.
> Is it possible to find out its value from inside R? When one loads
> R for Mac OS X Cocoa GUI written by:
I am not too sure what your question is, but try
?debug
or take a look at http://www.ats.ucla.edu/stat/r/library/R-debug-tools.pdf
On Feb 10, 9:07 am, Fuchs Ira wrote:
> I would like to have a function which gets data, does a calculation
> and prints a result and then waits some number of sec
Hi,
This looks like a Lyapanov equation. It is used in control theory I
think. There might be something if you search in those packages that
deal with this.
HTH,
On Feb 16, 9:25 am, �|珊�x wrote:
> Hi everyone,
>
> I have a question to solve linear equations with argument is a matrix.
> For ex
The PCA is just a singular value decomposition on a sample covariance/
correlation matrix. Do a search for ?svd and get the eigenvalues and
vectors from that function.
On Feb 14, 10:30 am, "glenn" wrote:
> Hi All, would appreciate an answer on this if you have a moment;
>
> Is there a function (
sqrt(svd(x)$d) maybe 2 more operations than princomp(covmat=x), but it
is hardly a chore.
On Feb 16, 9:15 pm, Mark Difford wrote:
> Hi Glen, Andrew,
>
> >> The PCA is just a singular value decomposition on a sample covariance/...
>
> I believe that Bjørn-Helge Mevik'
Hi all,
When running Rmdr using the demo data file using the following
commands:
data(mdrdata)
cvk<-10
nbr=2
res<-rmdr(mdrdata,10,2, randomize=TRUE)
I could not find the statistical results, but like this:
[1] 1
Cross Validation 1 Wed Feb 18 09:05:23 2009
The best set of loci is 13 17
Cross V
On Feb 24, 3:57 pm, jdeisenberg wrote:
> Λεωνίδας Μπαντής wrote:
>
> > 1. Suppose I have a<-c(1:10) (or a<-array(c(1:10),dim=c(1,10)))
>
> > and I want to end up with vector b=[0 0 0 0 0 1 1 1 1 1]. i.e. I want to
> > substitute alla elements that are <5 with 0 and >5 with 1.
>
> I think you
the following should work
library(lattice)
x <- seq(1,100)
y <- seq(1,100)
gr <- expand.grid(x,y)
gr$z <- x + y + rnorm(1,0,100)
cloud(z ~ x + y, data = gr)
also, look for the package rgl which does similar but with more
possiblities.
On Feb 27, 4:28 pm, Dipankar Basu wrote:
> Hi R Users,
>
actually, I just realised you also want a line in the plot. I am not
super-sure how to do this.
On Feb 27, 5:20 pm, andrew wrote:
> the following should work
>
> library(lattice)
> x <- seq(1,100)
> y <- seq(1,100)
> gr <- expand.grid(x,y)
> gr$z <- x + y + rno
perhaps this?
M <- dim(data_m)[2]
for(j in 1:M){
for (i in 4:T) {
data_q[i-3,j]=sum(data_m[(i-3):i,j])
}
}
of course, you can vectorize this and speed it up significantly, but
there is something evil about premature optimization.
On Mar 2, 1:29 pm, Mohammad Sabr wrote:
> I am t
in the seond link on the pdf you have cited is
http://www.vsthost.com/vstDocs/mgarchBEKK/
HTH.
On Mar 4, 1:52 pm, Mohammad Sabr wrote:
> Good day everyone,
>
> I tried to find a multivariate GARCH package and failed to find one. Although
> when I searched R I found the following link which d
use ?diff to calculate approximations for the slope.
HTH
Andrew
On Mar 7, 5:08 pm, Wei-han Liu wrote:
> Could somebody share some tips on implementing multivariate integration and
> partial differentiation in R?
> For example, for a trivariate joint distribution (cumulative density
&
Mean reverting model = autoregression? If so, then search for
?ar
or
?arima
to fit a time series.
On Mar 10, 4:36 am, Josuah Rechtsteiner wrote:
> dear useRs,
>
> i'm working with a mean reverting model of the following specification:
>
> y = mu + beta(x - mu) + errorterm, where mu is a cons
?ar to get full details.
On Mar 10, 9:18 am, Josuah Rechtsteiner wrote:
> hi andrew,
>
> the problem is that I don't know what kind of model this exactly is...
> I only know that I have to do it this way and how the model is
> structured.
>
>
>
> > Mean rever
ing to send it
> to you but the site where I wanted to go was busy. I think it's called
> planetmath.org or something like that.
>
> On Mon, Mar 9, 2009 at 7:54 PM, andrew wrote:
> > Autoregression is just X(n+1) = a X(n) + b + error. The mean
>
> reverting model is when |a| < 1
I saw this in
http://groups.google.com.au/group/r-help-archive/browse_thread/thread/7dca300a7fde5286/718dc5f1405618c9?lnk=gst&q=sort(+sapply(ls()%2Cfunction(x){object.size(get(x))}))#718dc5f1405618c9
The command is something like
sort( sapply(ls(),function(x){object.size(get(x))}))
very useful,
check out the interaction.plot. This *may* be what you are looking
for.
?interaction.plot
On Mar 15, 8:14 am, David Kaplan wrote:
> Greetings all,
>
> I have two questions. I have a data set that is arranged in the example
> below. I wish to obtain a plot of the performance of each ID over Ye
setwd
On Mar 18, 12:42 pm, Jonathan Greenberg wrote:
> I'm trying to redirect where temporary files go under R to
> D:\temp\somerandomname from its default (C:\Documents and
> Settings\username\Temp\somerandomname) -- how do I go about doing this?
>
> --j
>
> --
>
> Jonathan A. Greenberg, PhD
> P
.11 \\
5 & 0.12 & -0.11 & -0.14 & 1.50 & 0.12 \\
\hline
\end{tabular}
\end{center}
\end{table}
Or you could roll your own using for loops and the print function if
this format isn't what you want.
HTH,
Andrew.
On Mar 20, 12:48 pm, "Mary A. Marion" wrot
you might try the following.
Pochhammer_n <- function (a,b,c,n) {
if(n==0) return(1)
return(a*b*Pochhammer_n(a+1, b+1, c+1, n-1)/(c*n))
}
hypergeo_sum <- function (a,b,c,z,n) {
comb_sum <- 0
for (i in 0:n) {
comb_su
umeric(dupIDs$ID)-1)
> dupIDs
ID record ID2
1 1 20 1
2 . 30 1
3 . 25 1
4 2 26 2
5 . 15 2
6 3 21 3
HTH,
Andrew.
On Dec 15, 12:56 pm, "Zhixin Liu" wrote:
> Hi R helpers,
>
> If I have a dataset looks like:
> ID record
> 1
function onto the circle, you should be ok.
HTH,
Andrew.
##=
N <- 1000 #steps
x <- matrix(0, nrow=N+1, ncol =2)
R <- 1 #radius of circle
delta <- 0.5 #step size
inner <- function(x,y) {
if(length(x) != length(y)
Hello all,
Everything was working very well. Now when I try to load lmerTest using:
library("lmerTest"), I get this error:
Error: package or namespace load failed for ‘lmerTest’ in
loadNamespace(j <- i[[1L]], c(lib.loc, .libPaths()), versionCheck =
vI[[j]]):
there is no package called ‘purrr’
I'm trying to perform a mantel test that ignores specific pairs in my
distance matrices. The reasoning is that some geographic distances
below a certain threshold suffer from spatial autocorrelation, or
perhaps ecological relationships become less relevant that stochastic
processes above a certain
places where I am by no means certain that my
interpretation is correct. I have highlighted some of these below with a
“??”.
I look forward to learning from you.
Warmest regards,
J. Andrew Hoerner
** Standard function recognition:* recognizing some or all of a string code
as a function. (Part of
with this approach. They want to integrate the R models and graphics into
the real time portal.
I know SQL Server 2016 bundles R services, but how do you incorporate real
time R analytics and graphics in the existing MS/.Net stack?
Cheers
Andrew
__
R-h
Pro, however colleagues have run this on their machines and
not found the same result. gc() does not rectify the issue, although
re-starting R does.
Any help would be much appreciated.
AJL
--
Andrew Leach
Associate Professor of Natural Resources, Energy and Environment (NREE)
Academic Director, Ene
riate data=resids normal plot; id Barcode Grain_Yield__Mg_h_
pearsonresid; var resid;
proc print data=resids (obs=3);run;
Can someone please help me convert my code to R? Any help would be much
appreciated.
Thanks,
Andrew Harmon
[[alternative HTML version deleted]]
I have no problem setting up my mixed model, or performing anova or lsmeans
on my model’s outputs. However, performing lsd mean separation is giving me
fits.
So I do not have a problem when using two-way anova model. When using the
code:
fit.yield.add <- lm(data = ryzup, Yield ~ Rep + Nitrogen
--
-
Andrew J Perrin - Professor of Sociology, UNC-Chapel Hill
Director, Carolina Seminars http://carolinaseminars.unc.edu
Special Assistant to the Provost and Dean for Accreditation
and Curricular Innovation
andrew_per...@unc.edu http://perrin.socsci.unc.edu
, but anything that encourages me not to think about what
I'm doing in that realm seems like a bad idea.
Best wishes,
Andrew
On 27 July 2016 at 07:30, peter dalgaard wrote:
>
>> On 26 Jul 2016, at 22:26 , Hadley Wickham wrote:
>>
>> On Tue, Jul 26, 2016 at 3:2
You have selected the binomial family in the call to glm. You should
instead try something like
family=quasipoisson(link = "log")
I hope this helps
Andrew
On Tue, Jul 28, 2015 at 4:33 PM, Charlotte <
charlotte.hu...@griffithuni.edu.au> wrote:
> Hello
>
> I have c
d use min(Area) as a, figure out b from the maximum (I guess 2b+a is the
asymptote), and experiment with two values for year to retrieve c and d
uniroot might help?
Cheers
Andrew
--
Andrew Robinson
Deputy Director, CEBRA, School of Biosciences
Reader & Associate Professor in Applied
Hello All,
I have a question regarding building multiple models and then scoring new
data with these models. I have been tasked with converting existing SAS
modeling code into equivalent R code, but unfortunately I rarely use R so
I'm in unfamiliar territory. I've think I've found a good way to bu
an someone show me how to use the analysis of variance functions in R to get
the same result. Thanks.
Andrew McCulloch
Leeds Metropolitan University
>From 22 September 2014 Leeds Metropolitan University will become Leeds Beckett
>University.
Find out more at http://www.leedsbeckett.
thanks
Andrew
[[alternative HTML version deleted]]
__
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
ample with a much smaller data set (i.e., less than 2GB)
require(party)
data(iris)
cf <- cforest(Species ~ ., data=iris)
str(cf, max.level=2)
cf@data <- NULL # this fails
Andrew
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/list
I'm pretty sure you are using the package Picante. When I received the error
you described in my own project, it was due to the naming of species in the
community matrix and the phylogenetic tree being slightly different (one
separated latin binomials with a space, the other with an underscore). Wh
method = yaiMethod, bootstrap = bootstrap, :
object 'xcvRefs' not found
If anybody could tell me what I am doing wrong.
Cheers
Andrew
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and m
= ~x, psi = NA, control =
seg.control(K=1))
I get the following error.
Error in if (psi == Inf) psi <- median(XREGseg) :
missing value where TRUE/FALSE needed
Any advice would be greatly appreciated.
Kind regards
Andrew
[[alternative HTML version dele
error?
Kind regards
Andrew
On Thu, Oct 8, 2015 at 7:11 AM, David Winsemius
wrote:
>
> On Oct 7, 2015, at 6:50 AM, andrew haywood wrote:
>
> > Dear List,
> >
> > I am trying to run a simple pieewise regression using the segmented
> package.
> >
> > When
seg.Z = ~x, psi = NA, control =
seg.control(K = 1))
I am unsure how to interpret this.
In addition when I run the command
options(error="browser")
I get the following error
Error in options(error = "browser") : invalid value for 'error'
Any help would be great
, xa, y, wts, yaiMethod, bootstrap,
...))
1: varSelection(x = x, y = y, nboot = 5, yaiMethod = "randomForest",
useParallel = FALSE)
Any help/guidance would be greatly appreciated.
Kind regards
Andrew
-- Forwarded message --
From: andrew haywood
Date: Thu
en as
nuisance variables, and test the interaction using the model that includes
them.
BTW, your question might better be located with the mixed-effects models
special interest group.
https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
Best wishes
Andrew
On Mon, Nov 23, 2015 at 9:19 PM, angel
Hi,
I’m trying to solve what looks like the same issue as stack overflow article,
but within an lapply:
http://stackoverflow.com/questions/18939254/cant-use-a-variable-as-an-argument-but-can-use-its-value
I’ve replicated the issue with partialPlot below in ‘testFunc’. The lines up to
the final
se the parent
environment, the subtitue should remain.
See the resolution here (jcheng beat r-help to it this time!)
https://groups.google.com/forum/?utm_medium=email&utm_source=footer#!topic/shiny-discuss/cIZJzQmw8tQ
On 11 May 2016 at 08:48, David Winsemius wrote:
>
> > On May 9
one of
the black arts of statistical fitting. Good luck! And don't forget to
check for sensitivity.
Andrew
On 9 October 2016 at 22:21, Pinglei Gao wrote:
> Hi,
>
> I have some data that i'm trying to fit a double exponential model: data.
> Frame (Area=c (521.5, 689.78,
Without access to the data, or commented, minimal, self-contained,
reproducible code, it's pretty hard to speculate. I suggest that you
reframe your question so that we can see what you can see.
Andrew
On 25 October 2016 at 03:03, Santiago Bueno wrote:
> Dear people:
>
>
>
e
I am so new, I am not sure what the necessary steps are to complete this
conversion without a lot of NA.
--
Andrew D. Lachance
Chief of Service, Bates Emergency Medical Service
Residence Coordinator, Hopkins House
Bates College Class of 2017
alach...@bates.edu
(207) 620-4854
[[alter
ompute PCNMs = 0.82 sec Error in if (temp2.test[1, 5] <=
alpha) { : argument is of length zeroTiming stopped at: 1.06 0.05 1.19
I do not understand the error message coming up and would appreciate
some advice.
Andy
--
Andrew Halford Ph.D
Research Scientist (Kimberley Marine Parks)
Dep
Hello all,
Thank you for the extremely helpful information. As a follow up, some of
the nested elements are of the form below:
-
I've been having trouble extracting this information and was wondering if
anyone had any suggestions.
Thank you,
Andrew
On Thu, Jan 5, 2017 at 7:
out the names of the
thing that is being lapplied?
Thanks,
Andrew
[[alternative HTML version deleted]]
__
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
Thanks everyone for your suggestions (and sorry for the delay in the
acknowledgement). Jorge and Jim, thanks for pointing out your approach.
Andrew
On Fri, Jul 18, 2008 at 7:02 PM, Jorge Ivan Velez
<[EMAIL PROTECTED]>wrote:
>
> Dear Andrew,
>
> Following Jim Holtman
break;
}
and the output from running in batch mode that I get is this:
Da:
0.18034.988e-017
PING
and here the program crashes. I've tried this in multiple places and
sometimes the error is thrown sometimes not. Does anyone have an idea of
what is going on. I could not find any di
Switching all of my printfs to Rprintf fixed the problem the errors now
proceed correctly. I was unable to reproduce the error, but now that is
irrelevant, at least to me. Thanks for the help.
-Andrew
[[alternative HTML version deleted]]
__
R
Hello--
I am a relatively new user to R and I cannot find the information I
need. Please help.
I have a very large data set with values including letters, numbers,
and symbols (sometimes within the same vector value [ie X9-].
I've imported the data using read.fwp and it arrives in list fo
ample)
lme(Y ~ A + AC, random = ~ 1 | Block, data = example)
##########
Are we doing anything obviously wrong? Is there another approach to
the goal that we are trying to achieve?
Many thanks,
Andrew
--
Andrew Robinson
On Mon, Aug 04, 2008 at 10:17:38AM +0200, Peter Dalgaard wrote:
> Andrew Robinson wrote:
> >Dear R colleagues,
> >
> >a friend and I are trying to develop a modest workflow for the problem
> >of decomposing tests of higher-order terms into interpretable sets of
> &g
1 - 100 of 793 matches
Mail list logo
