On Mon, Sep 17, 2012 at 2:34 PM, jpm miao wrote:
> To dear Dr Sarkar and anyone that knows about Lattice package,
>
>I make 4 graphs by Lattice package. Each of the graphs has two time
> series. All the series are plotted in plain lines by default, and I would
> like one series to be in plain
Hi
Does anyone know whether the rms package provides a confidence interval for
the bootstrap-corrected Dxy or c-index?
I have fitted a logistic model, and would like to obtain the 95% confidence
interval of the bootstrap-corrected area under the ROC curve estimate.
Thanks.
[[alter
hi list,
I googled this invalid lables issue. It seems different people doing different
analysis encountered this problem. So I guess this is not about the MICE
package.
However, in general, they have categorical variables, in my case, I double
checked, the bulg_1, bulg_2, and bulg_3 are conti
Dear R users,
I'm using optim to optimize a pretty complicated function. This function takes
the parameter vector "theta" and within its body I use instructions like
sigma<-theta[a:b]; computations with sigma...
out<-c()
for (i in 1:d){
a<-theta[(3*d+i):c]
out[i]<-evaluation of an expressi
On 20/09/2012 09:24, Gildas Mazo wrote:
Dear R users,
I'm using optim to optimize a pretty complicated function. This function takes the
parameter vector "theta" and within its body I use instructions like
sigma<-theta[a:b]; computations with sigma...
out<-c()
for (i in 1:d){
a<-theta[(3*d+i):
Dear all
I have written a function in C++ , equil_distC, that I am calling from an R
script.
In the last few days, R has repeatedly crashed when calling this function, or
delivered obviously wrong outputs. However, when I restarted R after the crash,
the results turned out to be OK most of t
First, the posting guide asks you to send questions about compiled code
to the R-devel list.
These are all classic symptoms of the use of uninitialized variables or
writing out of bounds, but of which you can catch by using valgrind: see
'Writing R Extensions'. They are much harder to find on
Hi,
Finally I could resolve. I understood how you can use dummy variables in lm().
Thanks!
Eva
--- El jue, 20/9/12, Eva Prieto Castro escribió:
De: Eva Prieto Castro
Asunto: Re: Dummy Variable : Doubt
Para: R-help@r-project.org
Fecha: jueves, 20 de septiembre, 2012 11:27
Sorry, I could wri
Hello all,
I am working with high-frequency hydrological time series, from
automatic loggers. These have a known instrumental error, determined
from laboratory tests. Could anybody please advise on methods (an indeed
R packages/functions) that I could use to remove part of this
instrumental no
Sorry, I could write Dummy and not Gummy.
Regards
--- El jue, 20/9/12, Eva Prieto Castro escribió:
De: Eva Prieto Castro
Asunto: Gummy Variable : Doubt
Para: R-help@r-project.org
Fecha: jueves, 20 de septiembre, 2012 11:13
Hi,
Â
I have a system in which I analyze 2 subjects and 1 vari
Hi,
Â
I have a system in which I analyze 2 subjects and 1 variable, so I have
2 models as follow:
Â
y ~ x_1[, 1] + x_2[, 1] + x_1[, 2] + x_2[, 2]
Â
Where
Â
x_1[, i] = cos(2 * pi * t / T_i)
x_2[, i] = sin(2 * pi * t / T_i)
Â
i = 1, 2
Â
Data have two columns: t and y.
Â
As
Hello,
I have being trying to estimate the parameters of the generalized exponential
distribution. The random number generation for the GE distribution
is x<-(-log(1-U^(1/p1))/b), where U stands for uniform dist. The data i have
generated to estimate the parameters is right censored and the code
Hmmm, I'm not sure what the problem is, but I suspect it's related to
the fact the xmin and xmax have different factors levels and there are
some bugs in ggplot2 related to combining factors in some situations
(it's basically impossible to always do it right).
Explicitly ensuring the levels were t
Depending on an R computation I would like to include an Sweave documents
in the main Sweave document.
How can I do it?
So I was thinking to use Latex features :
\newif\ifpaper
\ifpaper
\SweaveInput{"my1.Rnw"}
\else
\SweaveInput{"my2.Rnw"}
\fi
But how do I set paper to true or false give
On 20-09-2012, at 13:46, Christopher Kelvin wrote:
> Hello,
> I have being trying to estimate the parameters of the generalized exponential
> distribution. The random number generation for the GE distribution is
> x<-(-log(1-U^(1/p1))/b), where U stands for uniform dist. The data i have
> gene
Dear all,
I would like to add mixed effects in a multinomial model and I am trying
to use MCMCglmm for that.
The main problem I face: my data set consits of a trapping data set,
where the observation at eah trap (1 or 0 for each species) have been
aggregated per traplines. Therefore we have
Mohamed Radhouane Aniba gmail.com> writes:
>
>
> Thank you Thomas,
>
> So you think a t-test is more adequate to use in this case ?
>
> Rad
No, because a t-test makes even stronger parametric assumptions.
You were given more specific advice on stackoverflow
http://stackoverflow.com/questi
On 20/09/2012 8:47 AM, Witold E Wolski wrote:
Depending on an R computation I would like to include an Sweave documents
in the main Sweave document.
How can I do it?
So I was thinking to use Latex features :
\newif\ifpaper
\ifpaper
\SweaveInput{"my1.Rnw"}
\else
\SweaveInput{"my2.Rnw"}
Dear R users,
I am trying to estimate a median regression with fixed effects. I have an
unbalanced panel data set with 5,000 individuals and 10 years, resulting in a
total of 20,000 observations.
When I try to add individual (firmid) fixed effects to the quantile regression
using the followin
Hi everyone,
Running the vif() function from the car package like
> reg2 <- lm(CARsPur~Delay_max10+LawChange+MarketTrend_20d+MultiTrade,
data=data.frame(VarVecPur))
> vif(reg2)
Delay_max10 LawChange MarketTrend_20d MultiTrade
It does not offer that feature. That is an area of active research. If
anyone knows of a paper that has solved this problem please let me know.
Frank
Whee Sze Ong wrote
> Hi
>
>
>
> Does anyone know whether the rms package provides a confidence interval
> for
> the bootstrap-corrected Dxy or
Hi there,
I hope you don't mind me asking a general stats rather than a R-specific
question.
I'm estimating a median regression model and would like to interpret my (mainly
categorical) regression coefficients in terms of percentage changes in the
dependent variable. There are about 10-15%(nat
On 20-09-2012, at 15:17, Christopher Kelvin wrote:
> By your suggestion if i understand you, i have changed (p[1]) and (p[2]) and
> have also corrected the error sum(t), but if i run it, the parameters
> estimates are very large.
> Can you please run it and help me out? The code is given below.
On 20-09-2012, at 16:06, Berend Hasselman wrote:
>
> On 20-09-2012, at 15:17, Christopher Kelvin wrote:
>
> .
> A final remark on function z:
>
> - do not calculate things like n*sum(s) repeatedly: doing something like
> A<-n*sum(s) and reusing A is more efficient.
> - same thing for log(
On Mon, Sep 17, 2012 at 3:49 PM, Christian Hoffmann
wrote:
> Hi,
>
> The browser opens, when the command
>
> * checking whether package 'cwhmisc' can be installed ... OK
>
> is executed.
>
>> sessionInfo()
> R version 2.15.1 (2012-06-22)
> Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)
>
> loc
By your suggestion if i understand you, i have changed (p[1]) and (p[2]) and
have also corrected the error sum(t), but if i run it, the parameters estimates
are very large.
Can you please run it and help me out? The code is given below.
p1<-0.6;b<-2
n=20;rr=5000
U<-runif(n,0,1)
for (i in 1:rr){
Hello,
I am working with a dataset with three variables and one binomial parameter.
The glm function provides coefficients for these three variables, e.g.
-1.5 | 27.2 | -2.9
If I'm not mistaken, $fitted.values gives me an estimate of how likely my
parameter is to be true/1 . I would like to apply
Very much a rookie at R, and have only recently started using it again so
pardon the simple question. I am trying to produce a box plot from one data
set and then overlay a line plot from another data set. The box plot data
set is made up of 20 sets of 30 data points, or 600 total data points. The
my reproducible example
test<-structure(list(site = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L,
3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L), .Label = c("A",
"B",
Thank you very much for everything. Your suggestions were very helpful.
Chris
- Original Message -
From: Berend Hasselman
To: Christopher Kelvin
Cc: "r-help@r-project.org"
Sent: Thursday, September 20, 2012 10:06 PM
Subject: Re: [R] Problem with Newton_Raphson
On 20-09-2012, at 15:
Hi Heramb,
No "Help(anova)" does not work either. Thanks for the suggestions. I put
the question out there in case anyone else was having similar problems. I
think I will throw in the towel and install the latest version of R to see
if that resolves the issue.
Thanks
On Thu, Sep 20, 2012 at 12:4
Windows XP (SP3) , R 2.15.1 32bit
Hi ...
I have a script which fails and closes my R session.
Unfortunately, it bombs out at a different point each time I run it.
I'm guessing that it may be something to do with memory management, or
perhaps it's to do with the various .C dll's the script
Dear UseRs,
i have a matrix of 365 rows and 444 columns. i drew each column of this matrix
against the number of days in a year, which are obviously 365. now i have 444
curves and i want to Use Fourier analysis for the approximation of the average
values.
does anyone know how to do it?
any help
Thank you very much, especially Milan and Bert!
I will do some speedtests and fit the function to my needs.
I think the best way would be a modified function in C...
But i am not familiar enough with C. Perhaps this would be a simple
but useful extension.
If someone has a solution, i would appreci
Dear Martin,
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Martin H. Schmidt
> Sent: Thursday, September 20, 2012 8:52 AM
> To: r-help@r-project.org
> Subject: [R] Variance Inflation Factor VIC() with a matrix
>
> Hi everyon
If you want to program Sweave documents, you can try the knitr
package. This case will be something like:
<<>>=
paper <- TRUE # or change it to FALSE
@
<>=
@
i.e. you use the logical variable 'paper' to control which child
document to include in the parent document. See
http://yihui.name/knitr/
On Sep 20, 2012, at 6:55 AM, SirRon wrote:
> Hello,
> I am working with a dataset with three variables and one binomial parameter.
> The glm function provides coefficients for these three variables, e.g.
> -1.5 | 27.2 | -2.9
>
> If I'm not mistaken, $fitted.values gives me an estimate of how lik
On Thu, Sep 20, 2012 at 9:01 AM, David Winsemius wrote:
(In response to an OP's aplogy for an "awkwardly worded question"):
> Awkwardly worded questions will get much better answers if they are
> accompanied by some test data.
Fortune nomination!
Cheers,
Bert
> --
> David Winsemius, MD
> Ala
Maybe it is longer, but it's also more general, it issues an error if
the tables are not 1-dim. That's where most of the function's extra
lines are. Otherwise it's the same as your first solution. The second
one has the problem you've mentioned.
Rui Barradas
Em 20-09-2012 16:46, Stefan Th. Gri
On Thu, Sep 20, 2012 at 10:57 AM, Stefan Th. Gries wrote:
> >From my book on corpus linguistics with R:
>
> # (10) Imagine you have two vectors a and b such that
> a<-c("d", "d", "j", "f", "e", "g", "f", "f", "i", "g")
> b<-c("a", "g", "d", "f", "g", "a", "f", "a", "b", "g")
>
> # Of these vecto
On Thu, Sep 20, 2012 at 4:20 PM, Basil Iannone wrote:
> Hi Heramb,
>
> No "Help(anova)" does not work either. Thanks for the suggestions. I put
> the question out there in case anyone else was having similar problems. I
> think I will throw in the towel and install the latest version of R to see
>
Hi,
In Wilcoxon test , we look for medians rather than the means. Ratio of
medians should be more coherent with P value.
On Thu, Sep 20, 2012 at 6:30 PM, Ben Bolker wrote:
> Mohamed Radhouane Aniba gmail.com> writes:
>
> >
> >
> > Thank you Thomas,
> >
> > So you think a t-test is more adequat
On Thu, Sep 20, 2012 at 4:34 PM, Robert Douglas Kinley
wrote:
> Windows XP (SP3) , R 2.15.1 32bit
>
> Hi ...
>
> I have a script which fails and closes my R session.
>
> Unfortunately, it bombs out at a different point each time I run it.
>
> I'm guessing that it may be something to do with m
If you want help understanding the theory of what you want to do, that is of
topic here.
If you understand what you want to do, or just want to see what resources you
can leverage in R, may I recommend the RSiteSearch function. I do think you
should be wary of applying a tool you don't understan
I am creating graphs for a publication and would like them to have the same
font size... but when I create a figure with multiple plots, the font size
decreases even though I haven't changed the tiff() resolution or
pointsize specifications, I have increased the figure size according to how
many pl
I want to count attributes of IDs:
--8<---cut here---start->8---
z <- data.frame(id=c(10,20,10,30,10,20),
a1=c("a","b","a","c","b","b"),
a2=c("x","y","x","z","z","y"),
stringsAsFactors=FALSE)
> z
id a1 a2
1 10 a
I don't entirely understand what you want as an alternative. What is
wrong with relation ="same", the default?
In any case, it sounds like you'll need to write your own panel
function. If you look at panel.dotplot(), you'll see it's fairly
straightforward, so modification should not be difficult.
HI Stefan,
Thanks for the solutions.
Just to add 1 more:
f.a<-table(a); f.b<-table(b)
c(f.a[!names(f.a)%in%names(f.b)],f.b[!names(f.b)%in%names(f.a)],xtabs(f.a[names(f.a)%in%names(f.b)]+f.b[names(f.b)%in%names(f.a)]~
names(f.a[names(f.a)%in%names(f.b)])))
#e i j a b d f g
#1 1 1 3 1 3 5 5
A.K.
Hello,
Works with me after correcting your lines() instruction. Your code
doesn't say what columns to use as coordinates, just where to look for them.
Also, (1) allways explicitly close the device using dev.off(). (2) The
grid lines were over the boxes. A way to avoid this is to plot the
boxes
I expect that the coordinate system being set up and used by boxplot
is different from what you are expecting. See the ?boxplot and ?bxp
help pages for details. You may be able to have the boxplots drawn
where you expect by using the "at" argument (you may want to specify
"xlim" as well).
On Thu
Thanks for the reply! Using predict() on new data works just fine. What I'm
interested in is, if I can use the coefficients or other data, to develop my
own formula which does the same as predict().
--
View this message in context:
http://r.789695.n4.nabble.com/Applying-glm-coefficients-Beginne
I apologize. I meant to type "help(anova)" and not "Help(anova)". I am
installing the newest version later today after I complete some analyses. I
will let everyone know if doing so solves my problem.
Thanks,
On Thu, Sep 20, 2012 at 11:46 AM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrot
Hi all,
I have the following sql query that I am executing on a machine with single
core. I want to know how can I execute the same sqery on a maching that is
running with 4 cores. Please provide me the code.
NEW_TABLE <- rhive.query("SELECT A, B, COUNT(C) FROM TABLE_A WHERE
A>='01-01-2012'")
On Sep 20, 2012, at 02:43 , Thomas Lumley wrote:
> On Thu, Sep 20, 2012 at 5:46 AM, Mohamed Radhouane Aniba
> wrote:
>> Hello All,
>>
>> I am writing to ask your opinion on how to interpret this case. I have two
>> vectors "a" and "b" that I am trying to compare.
>>
>> The wilcoxon test is gi
On Sep 20, 2012, at 10:58 AM, SirRon wrote:
> Thanks for the reply! Using predict() on new data works just fine. What I'm
> interested in is, if I can use the coefficients or other data, to develop my
> own formula which does the same as predict().
Of course you can. Extract the coefficients, ap
Hi,
http://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test
We can clearly see that null hypothesis is median different or not.
One way of proving non difference is P(X>Y) = P(X wrote:
>
> On Sep 20, 2012, at 02:43 , Thomas Lumley wrote:
>
>> On Thu, Sep 20, 2012 at 5:46 AM, Mohamed Radhouane Anib
Hiii
I have tried to measure ES with cornish fisher expansion using
PerformanceAnalytics package, but i still confuse because to measure volatility
i use GARCH model and i don't know how to consolidate it with ES in
PerformanceAnalytics package..
i have measured ES under normality using
Okay. I installed the newest version of R and I still cannot link to help
files. Is anyone else having this problem? Again, I am using Window's XP.
Thanks for the help.
On Thu, Sep 20, 2012 at 12:23 PM, Basil Iannone wrote:
> I apologize. I meant to type "help(anova)" and not "Help(anova)". I a
On 20/09/2012 9:05 AM, Duncan Murdoch wrote:
On 20/09/2012 8:47 AM, Witold E Wolski wrote:
> Depending on an R computation I would like to include an Sweave documents
> in the main Sweave document.
> How can I do it?
>
> So I was thinking to use Latex features :
>
> \newif\ifpaper
>
> \ifpap
On Fri, Sep 21, 2012 at 6:43 AM, avinash barnwal
wrote:
> Hi,
>
> http://en.wikipedia.org/wiki/Wilcoxon_signed-rank_test
>
> We can clearly see that null hypothesis is median different or not.
> One way of proving non difference is P(X>Y) = P(X ordered.
Avinash. No.
Firstly, the Wikipedia link
Is result3 what you are looking for?
test<-structure(list(site = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L,
3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L), .Label = c
On 20/09/2012 2:14 PM, Basil Iannone wrote:
Okay. I installed the newest version of R and I still cannot link to help
files. Is anyone else having this problem? Again, I am using Window's XP.
I don't think you've told us what output you get when you ask for help.
Here's what I see from ?mean:
Several years ago there were R implementations of a socio-linguistics
analysis method called Variable Rule Analysis namely rbrul and r-varb. Both
neither of the sites listed (in the method's WikiPedia page
http://en.wikipedia.org/wiki/Variable_rules_analysis ) appear to be online
any more (one was
You've stumbled across the answer to your question --
while lm() supports y~X formulas without a data=argument
and y~ X1+X2+X3 formulas with one, you can't depend on
all contributed functions to do the same.
As John pointed out, the advantage of car::vif over other
implementations is that it cor
On Thu, 20 Sep 2012, Bert Gunter wrote:
On Thu, Sep 20, 2012 at 9:01 AM, David Winsemius wrote:
(In response to an OP's aplogy for an "awkwardly worded question"):
Awkwardly worded questions will get much better answers if they are accompanied
by some test data.
Fortune nomination!
Yes,
On Thu, Sep 20, 2012 at 4:39 PM, Mike Spam wrote:
> Thank you very much, especially Milan and Bert!
> I will do some speedtests and fit the function to my needs.
>
> I think the best way would be a modified function in C...
> But i am not familiar enough with C. Perhaps this would be a simple
> bu
Thanks for your help! Unfortunately, I am now getting this:
> pdf(file="boxplot_tmax_2012091912.pdf", height=10, width=12)
>
> soton.df = read.table ( tmax.final.text, header=TRUE )
Error in read.table(tmax.final.text, header = TRUE) :
object 'tmax.final.text' not found
Execution halted
The
Hi,
Try this:
z1<-aggregate(z,list(id=z$id),FUN=paste,sep=",")
dat1<-data.frame(id=z1[,1],a1.total=unlist(lapply(z1[,3],length)),a1.val1=unique(z$a1),a1.num=unlist(lapply(lapply(z1[,3],table),`[`,1)),a1.val2=unlist(lapply(z1[,3],`[`,3)),a1.num2=unlist(lapply(lapply(z1[,3],table),`[`,2)),a2.total=u
Hi,
I need some help with making a function a bit more elegant. How would you
all suggest avoiding the problem I've made myself below - I've written a
function that creates a temporary matrix by subseting a larger one I assign
it. I then call vectors from that matrix, add each item in the vector t
It was a bit hard for me to follow the thread and figure out exactly
what the problem is that you're having, but I think it has something
to do with the ticks on the x axis not appearing in the correct order?
It's probably related to this issue:
https://github.com/hadley/ggplot2/issues/577
I belie
Enclose the file name tmax.final.text in quotes. Otherwise R is
looking for a variable named tmax.final.text, not the file name named
"tmax.final.text".
HTH
Peter
On Thu, Sep 20, 2012 at 1:02 PM, gfishel wrote:
> Thanks for your help! Unfortunately, I am now getting this:
>
>> pdf(file="boxplot
Hello,
Inline.
Em 20-09-2012 22:48, Benjamin Caldwell escreveu:
Hi,
I need some help with making a function a bit more elegant.
Yes you do!
Below, your first function, for instance, becomes a one liner.
Trick: R is vectorized. Use functions that act on whole vectors,
avoiding loops.
How
On 12-09-20 6:58 PM, Basil Iannone wrote:
Hi Duncan,
The output I see after typing ?mean is
starting httpd help server ... done
Then the browser opens but goes nowhere. I just see a black IE page with no
address in the address bar.
OK. I tried what you suggested. I typed the following
tools:
Hi Duncan,
The output I see after typing ?mean is
starting httpd help server ... done
Then the browser opens but goes nowhere. I just see a black IE page with no
address in the address bar.
OK. I tried what you suggested. I typed the following
tools:::httpdPort
[1] 17081
I then pasted the fol
I'm working with some data from which a client would like to make a decision
tree predicting brand preference based on inputs such as price, speed, etc.
After running the decision tree analysis using rpart, it appears that this data
is not capable of predicting brand preference.
Here's the d
Not very sure what the problem is as I was not able to take your data for run.
You might want to use dput() command to present the data.
Now on the programming side. As we can see that we have more than 2 levels for
the brands and hence method = class is not able to able to understand what you
Hi there,
I am having difficulties with what seems like a very simple thing.
My objective is to plot a distribution map for a species.
I have produced a plot with spplot which uses a raster, a few shapefiles and
xy points which are the species coordinates.
It all works fine until I want to add co
Thanks! Here's the dput output:
> dput(test.df)
structure(list(BRND = structure(c(1L, 12L, 16L, 17L, 18L, 19L,
20L, 21L, 22L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 13L,
14L, 15L), .Label = c("Brand 1", "Brand 10", "Brand 11", "Brand 12",
"Brand 13", "Brand 14", "Brand 15", "Brand 16", "Bran
I am trying to do parallel programming and I tried this
library(doSNOW)
library(foreach)
testfunc<-function(x){
x<-x+1
x
}
noc<-2
cl <- makeCluster(do.call(rbind,rep(list("localhost"),noc)), type = "SOCK")
registerDoSNOW(cl)
clusterExport(cl=cl,c("testfunc.r"))
testl<-foreach(pp=1:2) %dopar%
Then don't do that.
Use your script file to define functions. Source that file before the loop to
load them into memory. Call those functions from within your loop.
---
Jeff NewmillerThe .
Bhupendrashinh, thanks very much! I ran J48 on a respondent-level data set and
got a 61.75% correct classification rate!
Correctly Classified Instances 988 61.75 %
Incorrectly Classified Instances 612 38.25 %
Kappa statistic
One possible way to think of it is using " variable reduction" before going for
J48. You may want to use several methods available for that. Again prediction
for brands is more of a business question to me.
Two solution which I can think of.
1. Variable reduction before decision tree.
2. Let th
Very good. Could you point me in a couple of potential directions for variable
reduction? E.g. correlation analysis?
On Sep 20, 2012, at 10:36 PM, Bhupendrasinh Thakre wrote:
> One possible way to think of it is using " variable reduction" before going
> for J48. You may want to use several m
Hi,
just to add a few points to the discussion:
- rpart() is able to deal with responses with more than two classes.
Setting method="class" explicitly is not necessary if the response is a
factor (as in this case).
- If your tree on this data is so huge that it can't even be plotted, I
woul
On Thu, Sep 20, 2012 at 7:48 PM, maxbre wrote:
> my reproducible example
>
> test<-structure(list(site = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
> 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L,
> 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
> 4L, 4L, 4L, 4L, 4L,
85 matches
Mail list logo
