Thank you very much, Benilton and Prof. Ripley, for the speedy replies!
Looking forward to the fix!
Tao
From: Prof Brian Ripley [EMAIL PROTECTED]
To: Benilton Carvalho [EMAIL PROTECTED]
CC: Tao Shi [EMAIL PROTECTED], [EMAIL PROTECTED],
r-help@r-project.org
Subject: Re: [R]
I need an object created in a user defined function to be accessible to
another user defined function. I am fairly certain the object is correctly
created as it prints when I use the return() function, however the 2nd
function seems unable to use it.
I am guessing objects created in functions are
dear list
I am student M.S. statistics in department statistics . I am working in the
function nls in the [R 2.3.1] with 246 data and want to fit the exp model
to vectors( v and u ) but I have
a problem to use it
u
5.00e-13 2.179057e+03 6.537171e+03 1.089529e+04 1.525340e+04
Hello
If you want the multivariate t-distribution,
use rmvt() of the mvtnorm package.
If you want the Wishart distribution, one
can write a little nonce function:
library(mvtnorm)
rwis - function(n,sigma){
crossprod(rmvnorm(n,sigma=sigma))
}
I'm not sure what you mean by the multivariate
Dear all,
As a response to the comment of Mr. Romain Francois,
I wrote a small set of REST web APIs for R Graphical Manuals.
It has become possible to get image URLs for a given package/function
name via the web API.
http://cged.genes.nig.ac.jp/RGM2/webAPI.php
Any comment would be gratefully
Hi, in order to define multivariate distribution functions with known
marginals, you can also use the copula package, e.g.,
library(copula)
x - mvdc(normalCopula(0.75), c(gamma, gamma),
list(list(shape = 2, rate = 3), list(shape = 3, rate = 2)))
x.samp - rmvdc(x, 100)
dmvdc(x, x.samp)
You need better starting values:
f - function(x) { c0 - x[1]; ce - x[2]; ae - x[3]
sum((v - c0+(ce*(1-exp((-u)/ae^2)
}
g - 4^(0:8)
g - c(-g, g)
g - expand.grid(c0 = g, ce = g, ae = g)
start - g[which.min(apply(g, 1, f)), ]
nls(v ~ c0+(ce*(1-exp((-u)/ae))), data = data1, start = start)
On
Hi
System: Linux Open SuSE 10.2
I upgraded to R 2.6.0 and wanted to upgrade my packages as well. I tried
to use the checkBuild=TRUE option in update.packages, it asked me if I
wanted to update the package and I answered with yes, and finally ig
tave me the error message unknown parameter
Hi,
May I ask how I can save the coefficients and the p values into a table?
thanks.
jiong
The email message (and any attachments) is for the sole...{{dropped:11}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
Dear Olivier,
I had the same problem the other day.
It is important that you set the PATH in the right order.
I use the Rtools set from http://www.murdoch-sutherland.com/Rtools/ and created
a little batch file to set the PATH in the right order, please see below:
azadeh sadeghian a.sadeghian1386 at yahoo.com writes:
I am student M.S. statistics in department statistics . I am working in the
function nls in the [R 2.3.1]
You should update to a more recent version of R (2.6.0 is current)
with 246 data and want to fit the exp model to vectors( v and
Great idea, thanks!
小笠原理 wrote:
Dear all,
As a response to the comment of Mr. Romain Francois,
I wrote a small set of REST web APIs for R Graphical Manuals.
It has become possible to get image URLs for a given package/function
name via the web API.
Van Campenhout Bjorn wrote:
I upgraded to R 2.6.0 and wanted to upgrade my packages as
well. I tried to use the checkBuild=TRUE option in
update.packages, it asked me if I wanted to update the
package and I answered with yes, and finally ig tave me the
error message unknown parameter
This question teeters on the brink of general statistics rather than an
R-specific issue, but I'm hoping it boils down to how I coded VR's EM
algorithm for finite mixture modeling and not something worse (i.e. that
my plan is impossible).
Briefly, I'm trying to simulate an experiment where
Dear all,
I would like to be able to subset a data.frame in a special way. I will put
here an example:
Score Name
88 19_0070
88 19_0070
87 19_0070
79 002127_0658
79 002127_0658
77 002127_0658
So, for the
Thanks, Steve. I have read that document a bit, seems I need read it more.
I have 6 variables, all factors. 3^2 and 4^4
the following is the code
require(AlgDesign)
set.seed(1)
levels = c(v1=2,v2=2, v3=4,v4=4,v5=4,v6=4)
Hi,
I am trying to get a title on a plot that contains both some formatting
and prints the value of an object. What I have been using to get the
italics is:
title(sub=expression(paste(Log-rank test ,italic(p),-value = ,p.val)))
But this prints p.val rather than the object value. I have tried
Hi all,
Suppose I have the following data.frame, with an id column and two
variables columns :
idX Y
0001 NA 21
0002 NA 13
0003 000145
0004 NA 71
0005 000320
What I would like to do is to create a new variable Z whose values are
the
Alberto Monteiro wrote:
Prof Brian Ripley wrote:
Read ?par and the descriptiuon in 'An Introduction to R'. din,
fin, mai, omi, pin and usr are relevant.
Is there any hope that, instead of fin, din, pin, etc someday
we will have fmm, dmm, pmm?
Why worry about that, Alberto, when you can
?model.matrix
On Tue, 9 Oct 2007, Ajay Shah wrote:
I am using the clever formula notation of R to first do an OLS. E.g. I
say
m - lm(y ~ x + f)
where f is a factor, and R automatically constructs the dummy
variables. Very nice.
I need to then go on to do some other ML estimation using
On Wed, 10 Oct 2007, Rainer M Krug wrote:
Hi
System: Linux Open SuSE 10.2
I upgraded to R 2.6.0 and wanted to upgrade my packages as well. I tried
to use the checkBuild=TRUE option in update.packages, it asked me if I
wanted to update the package and I answered with yes, and finally ig
Rob,
It might be worth upgrading at least mgcv and MASS to the current versions
(latest mgcv is 1.3-28, just gone to CRAN). Way back I vaguely remember that
there was an issue with glmmPQL (called by gamm) not picking up correlation
structure variables from the dataframe, but I can't now find
On Tue, 2007-10-09 at 20:04 +0530, Ajay Shah wrote:
I am using the clever formula notation of R to first do an OLS. E.g. I
say
m - lm(y ~ x + f)
where f is a factor, and R automatically constructs the dummy
variables. Very nice.
I need to then go on to do some other ML estimation
I am using the clever formula notation of R to first do an OLS. E.g. I
say
m - lm(y ~ x + f)
where f is a factor, and R automatically constructs the dummy
variables. Very nice.
I need to then go on to do some other ML estimation using the same
design matrix that's used for the OLS. I could,
Hi,
I was looking for a way to make Sweave output not only EPS and PDF,
but also PNG. Searching through the R mailing list archives, I found
the following post from Thibaut Jombard:
https://stat.ethz.ch/pipermail/r-help/2006-March/102122.html
and
I have 400 cycles of a noisy waveform. I would like to find the average
(over the 400 cycles) of one cycle. Please tell me to do it in R. Any
suggestions welcome!
Thanks very much for any help.
Bill
__
R-help@r-project.org mailing list
Hello,
I want to use the quantile function so I read the doc but I don't understand
with this
qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1))
[1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39 63160.27
63205.65 63250.33 63295.04 63340.48 63387.48 63437.03 63490.53 63550.14
We do recommend
Rcmd build mpkg
to build a source package, followed by
Rcmd INSTALL --build mypkg_ver.tar.gz
to install it and zip up a version for distribution.
It is 'check' and 'build', and I believe R should refuse to use CHECK or
BUILD (as in does on all other platforms and at one time
On 10/9/07, Christos Hatzis [EMAIL PROTECTED] wrote:
There are at least a couple of versions of such plots.
Search for forest plots:
help.search(forest plot)
Hi Christos
Could you be more specific? On my system I found nothing:
help.search(forest plot)
No help files found matching 'forest
I'm trying to plot two distance matrices against each other
(74x74,phylogenetic distance and phenotypic distances). However R
gives an error message:Error in plot.new() : figure margins too
large.
Is it because I have too many points to plot?
No. It *sounds* like you don't have distance
elyakhlifi mustapha wrote:
Hello,
I want to use the quantile function so I read the doc but I don't
understand with this
qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1))
[1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39
63160.27 63205.65 63250.33 63295.04 63340.48
elyakhlifi mustapha wrote:
Hello,
I want to use the quantile function so I read the doc but I don't understand
with this
qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1))
[1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39 63160.27
63205.65 63250.33 63295.04
Bill Simpson wrote:
I have 400 cycles of a noisy waveform. I would like to find the average
(over the 400 cycles) of one cycle. Please tell me to do it in R. Any
suggestions welcome!
Hi Bill,
I recently had to do something like this for an approach to target
curve. The steps were:
1) Create
Elyakhlifi Mustapha wrote:
I want to use the quantile function so I read the doc but I don't
understand with this
qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1))
[1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39
63160.27 63205.65 63250.33 63295.04 63340.48 63387.48
On 10/10/2007 6:41 AM, Prof Brian Ripley wrote:
We do recommend
Rcmd build mpkg
to build a source package, followed by
Rcmd INSTALL --build mypkg_ver.tar.gz
to install it and zip up a version for distribution.
It is 'check' and 'build', and I believe R should refuse to use CHECK or
Try this:
transform(d, z = y[match(x, id)])
On 10/10/07, Julien Barnier [EMAIL PROTECTED] wrote:
Hi all,
Suppose I have the following data.frame, with an id column and two
variables columns :
idX Y
0001 NA 21
0002 NA 13
0003 000145
0004
Thanks Jim for the help.
I forgot to say that one of my main snags is figuring how how to chop up
the one long (44100 samples) vector into pieces (400 of them).
This is my C-inspired way of thinking but there must be a better way in
R.
cyclelength=44100/400
for i= 1 to 400
for j= 1 to
It really depends on what you want to do with the values. If you are
computing these in a loop, then I would suggest that you store the
output of 'summary' in a list and you can then extract the values
later. Can you provide an idea of what you want to do with them and
how you are computing
On 10-Oct-07 10:59:43, elyakhlifi mustapha wrote:
Hello,
I want to use the quantile function so I read the doc but I don't
understand with this
qchisq(seq(0.05,0.95,by=0.05),df=(length(don)-1))
[1] 62667.11 62795.62 62882.42 62951.47 63010.74 63064.00 63113.39
63160.27 63205.65 63250.33
Hi,
mods - lapply(lapply(df[,which(sapply(df, is.factor))],
function(reg)lm(df$value~reg)), summary)
res - lapply(mods, [, c(4,10))
And for each model adjusted:
1-pf(res[[2]][[2]][1], res[[2]][[2]][2], res[[2]][[2]][3])
1-pf(res[[1]][[2]][1], res[[1]][[2]][2], res[[1]][[2]][3])
On
Henrique Dallazuanna wrote:
Hi,
mods - lapply(lapply(df[,which(sapply(df, is.factor))],
function(reg)lm(df$value~reg)), summary)
res - lapply(mods, [, c(4,10))
And for each model adjusted:
1-pf(res[[2]][[2]][1], res[[2]][[2]][2], res[[2]][[2]][3])
1-pf(res[[1]][[2]][1],
Hi Petr,
d$z-NA
d$z[d$x %in% d$id] - d$y[d$id %in% d$x]
works in this particular case but it means you do not have multiple same
ids and X
Thanks for the idea. But the problem is that I can have multiple
ids...
In fact in the meantime I found a solution by using row names :
R d
id
try:
y - matrix(x, ncol=400, byrow=TRUE)
On 10/10/07, Bill Simpson [EMAIL PROTECTED] wrote:
Thanks Jim for the help.
I forgot to say that one of my main snags is figuring how how to chop up
the one long (44100 samples) vector into pieces (400 of them).
This is my C-inspired way of thinking
Dear R users,
I'm using the reldist add-on package to calculate relative distribution in R
as part of my research project. The subject is a general mental health score
ranging from 0 to 12 (integer values only) with 0 indicating no mental health
problem and positive values meaning some or
Is this what you want?
x - read.table(textConnection(Score Name
+ 88 19_0070
+ 88 19_0070
+ 87 19_0070
+ 79 002127_0658
+ 79 002127_0658
+ 77 002127_0658), header=TRUE)
# return best scores
best - by(x, x$Name,
On 10/10/07, Jim Lemon [EMAIL PROTECTED] wrote:
Alberto Monteiro wrote:
Prof Brian Ripley wrote:
Read ?par and the descriptiuon in 'An Introduction to R'. din,
fin, mai, omi, pin and usr are relevant.
Is there any hope that, instead of fin, din, pin, etc someday
we will have fmm,
Try bquote as in:
http://tolstoy.newcastle.edu.au/R/e2/help/07/09/26353.html
On 10/10/07, Daniel Brewer [EMAIL PROTECTED] wrote:
Hi,
I am trying to get a title on a plot that contains both some formatting
and prints the value of an object. What I have been using to get the
italics is:
Hadley Wickham wrote:
Is there any hope that, instead of fin, din, pin, etc someday
we will have fmm, dmm, pmm?
Why worry about that, Alberto, when you can use my ammazing function:
mm2in(x) return(x/25.4)
thus:
par(pin=mm2in(126))
But be sure to use it consistently!
Thanks. That works great if I do this:
title(sub=bquote(italic(p)-value == .(p.val)))
But if I add text to the beginning e.g.
title(sub=bquote(Log rank test italic(p)-value == .(p.val)))
I get an error message saying,
Error: syntax error, unexpected SYMBOL, expecting ',' in
title(sub=bquote(Log
Petr PIKAL/CTCAP napsal dne 10.10.2007 15:03:28:
[EMAIL PROTECTED] napsal dne 10.10.2007 13:55:32:
Hi Petr,
d$z-NA
d$z[d$x %in% d$id] - d$y[d$id %in% d$x]
works in this particular case but it means you do not have multiple
same
ids and X
Thanks for the idea. But the
Hello,
is there a possibility to set default options to qplot?
I need to draw a lot of graphs and would like to have all of them as point
plot but with a greater size and a fixed color for all dots.
Thanks for help.
Christoph
__
R-help@r-project.org
Hi All,
I entered a R statement, e.g. 1:20 = x or log(a) on an HTML
form and passed it to a R-CGI script. Obviously, neither of both
is a correct R statement or expression. However, my R-CGI script
could not return and report the error message to the Web site even
though it received the
Hi,
I have some problems when using AlgDesign-optFederov() generating
designs.
I have 6 variables, all factors. 3^2 and 4^4, I want to have a design that
can take care of main effects and two interactions within 2 pair of
variables v3-v4 and v5-v6, the following is the code
Armin,
RSiteSearch(forest plot meta analysis)
will point you to the metaplot function in the rmeta package.
There is also a forestplot function in the same package that offers more
flexibility.
These plots are useful in meta analysis in general (e.g. showing odds ratios
from different studies).
The expression in plotmath must be valid R syntax.
plot(1)
p.val - .1
title(sub=bquote(Log ~ rank ~ test ~ (italic(p)-value == .(p.val
On 10/10/07, Daniel Brewer [EMAIL PROTECTED] wrote:
Thanks. That works great if I do this:
title(sub=bquote(italic(p)-value == .(p.val)))
But if I add
On 10/10/07, Christoph Krammer [EMAIL PROTECTED] wrote:
Hello,
is there a possibility to set default options to qplot?
I need to draw a lot of graphs and would like to have all of them as point
plot but with a greater size and a fixed color for all dots.
Sure. It's a bit of hack, but:
You need to follow the posting guide and provide commented, minimal,
self-contained, reproducible code.
I can only guess at what your code looks like, but the following
catches the error:
z - try(eval(parse(text=1:20 - x)))
Error in eval(expr, envir, enclos) : object x not found
str(z)
Class
Hello all,
I read the following variable
x
x
1 1_A1_ML1_a.DLL
2 11_B1_ML2_a.DLL
3 4_A1_ML3_a.DLL
4 55_C1_ML4_a.DLL
5 14_C1_ML5_a.DLL
I would like to disperse it in three variable such as
x1
[1] 1 11 4 55 14
x2
[1] A1 B1 A1 C1 C1
x3
[1] ML1 ML2 ML3 ML4 ML5
Howard,
though it received the statement. I tried to use some R built-in
functions of try, tryCatch, eval, expression, as.expression,
parse, deparse, etc. None of them worked.
How did they not work? What did you attempt? Did you use silent = TRUE
in try?
Please be more specific about the
Hello all:
I'm sure this is a trivial request, but I'm still a beginner at this,
and haven't been able to find it. I need to create simulated data based
on some empirical distributions of a single variable. I've found R
functions to help me simulate data based on analytical distributions, or
Dear R users,
I am trying to ´detect´ the trend in an artificial time series created
by the simple function
x=seq(pi,10*pi,0.1)
my.ts=0.1*x+sin(x)
my.ts=ts(my.ts,start=1800)
plot(my.ts)
I have tried stl(my.ts), but because I don´t have ´replications´ at
every time point, this somehow doesn´t
Hi all,
I'm trying to download a file from the net, and the download.file
command leaves it corrupted, i.e excel cannot open it. It seems as if
it's going ok, however.
Does anyone have an idea of what's going on?
regards,
Gustaf Rydevik
Dear UseRs,
I wrote following function in order to solve Data Envelopment Analysis.
Reason for posting is that the function is slow when nrow(dat) is large.
I wonder if other functions could substitute the for() loop in the
code, such as mapply().
Can anybody help to rewrite the dea() function
On Wed, 2007-10-10 at 16:42 +0200, Birgit Lemcke wrote:
Hello and sorry that I still haven´t found a solution for my problem.
I need to extract the lower and upper triangle from a square matrix
including the diagonal. This diagonal is not zero in that special case.
How can a site have a
jim holtman wrote:
Is this what you want?
x - read.table(textConnection(Score Name
+ 88 19_0070
+ 88 19_0070
+ 87 19_0070
+ 79 002127_0658
+ 79 002127_0658
+ 77 002127_0658), header=TRUE)
# return
If all you're doing is fitting different responses to the same X data, then
you don't need model.matrix. See ?update, ?update.formula.
Bert Gunter
Genentech Nonclinical Statistic
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Ajay Shah
Sent: Tuesday,
On 10/10/07, Birgit Lemcke [EMAIL PROTECTED] wrote:
Hello and sorry that I still haven´t found a solution for my problem.
I need to extract the lower and upper triangle from a square matrix
including the diagonal. This diagonal is not zero in that special case.
I tried with as.dist
On Wed, 10 Oct 2007, Gustaf Rydevik wrote:
Hi all,
I'm trying to download a file from the net, and the download.file
command leaves it corrupted, i.e excel cannot open it. It seems as if
it's going ok, however.
Does anyone have an idea of what's going on?
That is a binary file. You needed
This are distances but between male and female plants of the same
species.
species A B C D E -- male
A
B
C
D
|
female
I would like to do a Manteltest.
Birgit
Am 10.10.2007 um 16:59 schrieb Gustaf Rydevik:
Tom Sgouros tomfool at as220.org writes:
What I have is twelve bins of data, and the population
in each bin. The top bin is open-ended, and the whole distribution is
more or less poisson-ish.
I can think of a couple of ways to fake this ok, but is there a real R
way to do it?
Hi Jim,
What you told me works well.
I have tried using 'eval(parse(text=1:20 = x))' as well as
'try(parse(text=1:20 = x))' before but not using eval anfd try
functions together. I just added try function following your
suggestion, e.g. 'try(eval(parse(text=1:20 = x)))'and then it
worked
Hello,
I problem is in the format of the date, my time series is like this:
2006070100 1244 6162
2006070101 1221 6060
2006070102 1214 6060
2006070103 1194 5959
2006070104 1182 5858
2006070105 1178
See ?rgb and its alpha argument, for example.
Uwe Ligges
Markus Loecher wrote:
Dear R graphics experts,
Do you know of any way of plotting semi-transparent points in R ?
I face this problem any time I want to e.g. plot thousands of points
that belong to two or three classes which I color
You may be better off using a complete distance matrix and a model
matrix describing plant sex, as explained in Legendre and Fortin 1989
and elsewhere.
Otherwise you will have to write your own Mantel code - all the examples
I know of (vegan and ecodist, primarily) assume that diagonals are
zero
See ?rgb and the article by Paul Murrell in R-news 2004-2. (The details
of where it is supported have changed in the last three years, though.)
On Wed, 10 Oct 2007, Markus Loecher wrote:
Dear R graphics experts,
Do you know of any way of plotting semi-transparent points in R ?
I face this
Rob,
Thanks for sending the example...
m1=gamm(present~s(week,bs=cc)+s(week,bs=cc,by=y1),random=list(garden=~1
) ,
correlation=corAR1(form=~1|garden),family=binomial,data=count.data2)
(ie removing the week term). This model proceeds - to a point...
Maximum number of PQL iterations: 20
It can be as simple as:
plot( rnorm(1000), rnorm(1000), col=#ff22, pch=16,cex=3)
Where the color is #RRGGBBAA and the AA portion is the
opacity/trasparency. Of course this only works if you are using a
graphics device that supports transparency. The windows device under R
2.6.0 does (but
Hi,
I tried to install the caTools package manually
from the main R Project website and I get the
following error message when typing library('caTools')
Error in library(caTools) :
'caTools' is not a valid package -- installed
2.0.0?
What am I doing wrong? Also, why is this package
Chris Jones-35 wrote:
Hi all,
I seem to be having a difficult time using the 'ape' package in R
when it comes to rooting trees. Here's a short screenshot:
[snip]
Can you post your dataset, or a subset of it that replicates
the problem, somewhere? (On the list if short,
One of the things that you should do is to use Rprof to see where time
is being spent. I would guess that is the not the 'for' loop, but
instead what is being done inside it. My guess it that most of the
time is being spent in the number of times that 'lp' is being called.
So the real problem
Hello,
I have a question about splitting label in regression tree.
for example, by using the following command
plot(m.regression.tree.pr,uniform=T,margin=0.05,branch=0.7,compress=F)
text(m.regression.tree.pr,pretty=0)
I get a tree with one label in some splitting node like this
bcu0.004969
Now
Here you go
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=114
Also have a look @
http://www.bmj.com/cgi/reprint/322/7300/1479.pdf
HTH
Francois
- Original Message -
From: Armin Goralczyk [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Cc: R-help@r-project.org
Sent: Wednesday,
Joao,
I can't reproduce your results. Try:
str(A)
When I read in the data frame A, I see
str(A)
'data.frame':10 obs. of 4 variables:
$ date: int 2006070100 2006070101 2006070102 2006070103 2006070104
2006070105 2006070106 2006070107 2006070108 2006070109
$ my1 : int 1244 1221 1214
Dear R,
I am writing a simple function to extract the sign of values and apply it
to a data frame (see below). Whatever I do I get error-messages... What is
wrong?
Thanking you in advance,
Cheers, Georg.
***
Georg Ehret
Institute of Genetic Medicine
Johns Hopkins University
Maura:
I looked at the scatter plots you sent.
A few thoughts:
1- Patient 3 data has a lot of missing data. This will make
doing a good grouping against your cases an issue.
Missing data is so common and much work has been done in this area.
One can do the trivial approach, forward fill
Hi Georg,
Georg Ehret wrote:
Dear R,
I am writing a simple function to extract the sign of values and apply it
to a data frame (see below). Whatever I do I get error-messages... What is
wrong?
The main problem here is you are ignoring existing functions that will
do the job much better.
On 10/10/2007 2:03 PM, Alberto Monteiro wrote:
Jim Holtman wrote:
One of the things that you should do is to use Rprof
to see where time is being spent.
Rprof is great! I didn't know such functionality existed.
But I am a very destructive user; I think I found a way
to break Rprof: if I
As usual, explicit code would help (see the posting guide)-- what kinds of
graphs (R has at least three separate systems)? What graphs/maps in
particular?.
But presumably a careful reading of ?par with particular attention to the
mar parameter (possibly also xlim/ylim) might solve your problem .
The squishplot function in the TeachingDemos package will move the axes
and other annotation closer to the plot so there is less whitespace
within the plot region (there will still be whitespace in the margin
areas beyond the axes and labels). Is this what you want?
To completely eliminate all
Hi all!
I am trying to fit a 2-level hierarchical lme().
I can extract ranef() and coef() without problems for levels=1:2.
However, when it comes to resid() at level=2, the resulting list has some
unnamed entries (label=NA). I checked with my data (NAs have been omitted)
and I found out that
in case you are not aware: the function psi is available without a port;
see help(digamma).
On Wed, 10 Oct 2007, Bernardo Lagos Alvarez wrote:
Hi All,
Anybody know as run the function psi on
http://www.alglib.net/translator/dl/specialfunctions.psi.csharp.zip
or
Hi Bill,
You get cyclelemgth = 110.
If this is true, i.e if every 110 points of data
represent an entire cycle then everything is all
right, but are you sure this is true? Think what
happens if the true cycle length was 110.5 data
points.
Regards,
Moshe.
--- Bill Simpson [EMAIL PROTECTED]
Hello Wayne,
Welch Test (as suggested by M.S.) can be a good
choice, but since your sample is very small you must
ask yourself what can you assume about the two samples
(i.e. do they come from more or less normal
distributions, etc.).
Regards,
Moshe.
--- Wayne Aldo Gavioli [EMAIL PROTECTED]
Please supply enough information to allow us to reproduce the error.
Also, nirK.tree is a different object to nirk.tree, which might be
confusing things.
Simon.
On Wed, 2007-10-10 at 15:07 +0200, Chris Jones wrote:
Hi all,
I seem to be having a difficult time using the 'ape' package in R
Hi all, I have a datasei like this :
ID Date Price
aa 01/01/07 12
aa01/02/07 13
bb 01/01/0723
bb01/02/0712
Now I want to write them in following format :
ID Date Price ID Date
I want to create a data-frame. any idea?
Moshe Olshansky [EMAIL PROTECTED] wrote: What do you mean by write - write
to file, create a
data.frame?
--- Megh Dal wrote:
Hi all, I have a datasei like this :
ID Date Price
aa 01/01/07 12
aa 01/02/07 13
bb 01/01/07 23
bb 01/02/07 12
Now
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