Hello,
We are searching for information resources for wholesale and retail data
analysis
tasks: the tasks that are relevant and demanded from statists of
wholesale and retail companies. The reason for this search for us, a
sfotware development company is simple: one wholesale
company showed
I am a new user and I have been struggling for hours. So finally I
decide to ask you:
If I have a matrix P, and P.2 = P%*%P, and P.3=P.2%*%P
is there a function to calculate the power of a matrix?? if not how can
i do:
for (i in 1:10) {P.i=P^i}
after this I need to sum them up and my
Hi everyone,
I've got quite a few labels along the x axis and ggplot2 basically just
crams them on top of each other.
Is it possible to reduce the font size and/or text direction?
Stretching the windows device window manually also helps, but I found that
setting the parameters for the pdf
Hi Mikhail,
You can reduce the text size using the grid.gedit approach described
at the end of the ggplot book, available on Hadley Wickham's website:
http://had.co.nz/ggplot2/book.pdf
You could use something like:
grid.gedit(gPath(xaxis, labels), gp=gpar(fontsize=6))
I'm not aware of a good
Hello.
Has anyone any idea how a function would look like of a model based bootstrap,
when the underlying time series follows an ARIMA(1,1,1)-process?
A pure AR-process is no problem, but what is, if the time series need to be
differentiated of order one or above and the additional MA-part?
Bonjour Christophe,
You can pass ... to a list and then extract its names, so something like:
if( ! uuu %in% names(list(...)) )
Good luck with your keyboard,
Romain
[EMAIL PROTECTED] wrote:
Hi the list
Is it possible to add one argument to the arguments contain in ... ?
I would like to do
As the answers you've received suggest, you
can use a list. Or you could have two
vectors: one with the data, the other with the
group identity. The latter format is likely more
convenient for a lot of analyses.
Since your data are not inherently rectangular,
it is probably best to get the
Hi,
I have no problem performing the regression using R, and I successfully
obtain the parameter estimates using the function nls(). However, how do I
obtain the ANOVA output, r, r^2 and adj. r^2?
Thanks Regards,
Guru
[[alternative HTML version deleted]]
Hi,
My problem is simple: since having updated the lattice package, I cannot
load lattice anymore. If I type in the command 'library(lattice)' the
loading fails with the following message:
--- cut here ---
Error in library.dynam(lib, package, package.lib) :
shared library 'lattice' not
CA == chockri adnen [EMAIL PROTECTED]
on Wed, 30 Apr 2008 13:34:42 +0200
CA My problem in a few words is as folow:
CA I used the fCopulae packages because i have 2 series which
CA are already
CA transformed in the uniform domain (the space of the copulas
CA functions) and i
my post may have slipped through the bank holiday and be forgotten by
now, I'm still hoping for some pointers. Please let me know if I need
to clarify the problem.
baptiste
On 4 May 2008, at 16:39, baptiste Auguié wrote:
DeaR list,
I'm running an external program that computes some
Try installing again by
install.packages(lattice, .Library)
(from an account with suitable privileges).
If that still fails, we need to see the output produced during
installation.
On Tue, 6 May 2008, K. Elo wrote:
Hi,
My problem is simple: since having updated the lattice package, I
Hi,
thanks for the quick reply :)
Prof Brian Ripley kirjoitti viestissään (06.05.2008):
Try installing again by
install.packages(lattice, .Library)
(from an account with suitable privileges).
Tried (as root) - not working :(
If that still fails, we need to see the output produced during
Prof Brian Ripley kirjoitti viestissään (06.05.2008):
Does starting R --vanilla help?
I am wondering if you have another corrupt copy of lattice somewhere.
The latter was the problem, many thanks for this! I use Rkward as GUI
and obviously some packages have been installed into the user
Guru S guru.rcom at rediffmail.com writes:
I have no problem performing the regression using R, and I successfully
obtain the parameter estimates using the function nls(). However, how do I
obtain the ANOVA output, r, r^2 and adj. r^2?
This is a feature, not a bug. See Douglas Bates's
I'm trying to set up a lattice plot with two y-axes for each panel. (Yes,
I know that multiple y-axes are generally a bad idea; the graph is for
someone else and they want it that way.) I've used a custom
yscale.component in xyplot to achieve this:
myyscale.component - function(...)
{
ans
Is it possible to resize the labels in a dendrogram without applying
circles and triangles to edges?
I tried cex.labels:
plot(scoreDendogramObj, horiz=TRUE, axes=FALSE, cex.labels=0.8)
but that didn't have any effect.
Thanks,
Alex
__
Using R 2.6.0 patched I was able to calculate the variance of each row
of a matrix without error, even if some rows had only NAs. I would just
get NAs back as the variance for those rows. Now with R 2.7.0 patched I
get an error no complete element pairs if any one row has all NAs.
Can anyone
pascal vrolijk [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
Hello best helpers,
I am a new user and I have been struggling for hours. So finally I
decide to ask you: If I have a matrix P, and P.2 = P%*%P, and
P.3=P.2%*%P is there a function to calculate the power of a matrix??
if
When I plot a survival fit using rpart for the classification tree, for each
node, there is a decimal based number above the event/total. I tried to see if
it's the exponential ratio or logrithmics, neither seem to be the case. I'm
wondering if anyone knows what they are.
-
It is
Melanie Murphy mamurphy at turbonet.com writes:
I was wondering if anyone has developed (or is developing) an
implementation for gravity models (spatial interaction) in R. I
conducted several searches on the CRAN website with no luck. Currently
I am estimating parameters via
Hi R,
par(mfrow=c(2,2))
x1=(1:5)^1; x2=(1:5)^2; x3=(1:5)^3; x4=(1:5)^4
I need to write a single plot statement, which creates 4 plots (for x1,
x2, x3 and x4) in the graphics window, without using 'for' loop. Is this
possible? Does 'do.call' help in this context? Or do I have any option
in
Dear list,
I am trying to perform a significance analysis of a microarray experiment
with survival data using the {samr} package. I have a matrix containing my
data which has 17816 rows corresponding to genes, and 286 columns
corresponding to samples. The name of this matrix is data.matrix2. Some
On 5/6/08, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
I'm trying to set up a lattice plot with two y-axes for each panel. (Yes,
I know that multiple y-axes are generally a bad idea; the graph is for
someone else and they want it that way.) I've used a custom
yscale.component in xyplot to
Try plot.zoo in which case you don't need the par:
library(zoo)
plot(zoo(cbind(x1, x2, x3, x4)), nc = 2)
or
plot(zoo(outer(1:5, 1:4, ^)), nc = 2)
See ?plot.zoo, ?xyplot.zoo and the three vignettes in
the zoo package.
On Tue, May 6, 2008 at 9:47 AM, Shubha Vishwanath Karanth
[EMAIL PROTECTED]
Thank you very much Gabor...Zoo is very powerful...
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
-Original Message-
From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
Hi R,
par(mfrow=c(2,2))
x1=(1:5)^1; x2=(1:5)^2; x3=(1:5)^3; x4=(1:5)^4
I need to write a single plot statement, which creates 4 plots (for
x1, x2, x3 and x4) in the graphics window, without using 'for' loop.
Hi all,
I have following problem :
a = b = seq(1, 5, by=500)
v = matrix(0, nrow=length(a), ncol=length(a))
for (i in 1:length(a))
{
for (j in 1:length(a))
{
d = c(17989*a[i], -18109*b[j])
v[i,j] = t(d) %*% matrix(c(0.0001741, 0.0001280, 0.0001280,
Wonderful...This works...
lapply(list(x1,x2,x3,x4),plot,type=l)
Thanks a lot!
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of David Winsemius
Sent: Tuesday, May 06, 2008 7:55 PM
To: [EMAIL PROTECTED]
Subject: Re: [R] single plot statement, multiple plots
On 5/6/2008 10:31 AM, Megh Dal wrote:
Hi all,
I have following problem :
a = b = seq(1, 5, by=500)
v = matrix(0, nrow=length(a), ncol=length(a))
for (i in 1:length(a))
{
for (j in 1:length(a))
{
d = c(17989*a[i], -18109*b[j])
v[i,j] = t(d) %*%
Hi,
Suppose
a=matrix(1:9,3,3)
a
[,1] [,2] [,3]
[1,]147
[2,]258
[3,]369
Now,
class(a[1:2,])
[1] matrix
class(a[1:3,])
[1] matrix
class(a[,1:2])
[1] matrix
class(a[,1:3])
[1] matrix
But,
class(a[1,])
[1] integer
Thank you very much Mark! That worked Just a question, ?[ does give an
error to me...how do I find it?
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
-Original Message-
Dear list:
I ask for your help in a simple problem in which I'm not figuring out
the solution
My data looks like:
dat- data.frame(date=c(12/12/1980, 03/11/1994, 15/11/1999,
31/10/2000, 20/03/2007, 05/01/2001),
var1=c(A, A, B, D, C, A), var2=runif(6))
I was wondering if I could split the column
Hi Eleni --
Although samr is not a Bioconductor package, you might have more luck
asking on the Bioconductor mailing list, http://bioconductor.org. The
obvious place to start, and probably you have already done this, is to
ensure that the class of the objects passed to the function agree with
the
Hi,
I have 3 vectors,
x=rnorm(10); y=rnorm(20); z=rnorm(30).
I would like to plot 3 vectors side by side (like a
bar plot) with scatter plot something similar to the
following:
.. *** ;;;
.. ** ;;;
.. *** ;;;
.. *** ;;;
... *** ;;
x y z
How can I do this with Plot()?
Dear useRs,
This is to announce that the maintainers of the various distributions
have decided to provide experimental up-to-date versions of the
following R related packages on Debian stable and Ubuntu (i386 and
amd64 architectures):
littler
rkward
python-rpy
Try this:
subset(dat, format(date, %Y-%m) == 1999-11)
On Tue, May 6, 2008 at 12:25 PM, [EMAIL PROTECTED] wrote:
Dear list:
I ask for your help in a simple problem in which I'm not figuring out
the solution
My data looks like:
dat- data.frame(date=c(12/12/1980, 03/11/1994, 15/11/1999,
On 5/6/2008 11:46 AM, A Ezhil wrote:
Hi,
I have 3 vectors,
x=rnorm(10); y=rnorm(20); z=rnorm(30).
I would like to plot 3 vectors side by side (like a
bar plot) with scatter plot something similar to the
following:
.. *** ;;;
.. ** ;;;
.. *** ;;;
.. *** ;;;
... *** ;;
x
Dear R-users,
I have the following problem
In a lab experiment I have to mix three solutions to get different
concentrations of various molecules in a cuvette
I've used R to calculate the necessary µliters for each of the level of
the experiment and I must confess that it is more useful and
Indeed both options are salable, though I agree the latter may be more
convenient.
Merci!
Patrick Burns wrote:
As the answers you've received suggest, you
can use a list. Or you could have two
vectors: one with the data, the other with the
group identity. The latter format is likely
Hi all,
I have issues using some basic functions in R such as these ones :
pp.test(R) (where is a vector of returns)
Error in .C(R_approx, as.double(x), as.double(y), as.integer(nx), xout =
as.double(xout), :
C symbol name R_approx not in DLL for package base
boxcox(reg,plotit=T)
My summary of Bates' comments cited below is as follows:
1. ANOVA is an excellent tool but requires nested models.
You can do this fairly easily, but it is not so easily automated.
2. The standard definition of R^2 loses its meaning with
nonlinear models.
hie all
i am trying to carry out a categorical data analysis but my problem is that
when in i use the chi squared test some of my expected values are less than
5. is there a test that can handle this situation. the data is not a 2*2
table. its more from the social sciences where you have
On 5/6/2008 12:07 PM, Dr. Ottorino-Luca Pantani wrote:
Dear R-users,
I have the following problem
In a lab experiment I have to mix three solutions to get different
concentrations of various molecules in a cuvette
I've used R to calculate the necessary µliters for each of the level of
the
f - (structure(list(X = structure(96:97, .Label = c(119DAmm, 119DN,
119DNN, 119DO, 119DOC, 119Flow, 119Nit, 119ON, 119OPhos,
119OrgP, 119Phos, 119TKN, 119TOC, 148DAmm, 148DN,
148DNN, 148DO, 148DOC, 148Flow, 148Nit, 148ON, 148OPhos,
148OrgP, 148Phos, 148TKN, 148TOC, 179DAmm, 179DN,
179DNN, 179DO,
Try:
x - unlist(f[2, 4:26])
y - unlist(f[1, 4:26])
plot(x, y)
On Tue, May 6, 2008 at 12:43 PM, stephen sefick [EMAIL PROTECTED] wrote:
f - (structure(list(X = structure(96:97, .Label = c(119DAmm, 119DN,
119DNN, 119DO, 119DOC, 119Flow, 119Nit, 119ON, 119OPhos,
119OrgP, 119Phos, 119TKN, 119TOC,
Hi Ottorino,
You could just use the modulus operator %% as follows:
x-c(1803.02, 193.51, 3.47);
x-x%%c(50,5,1) #just using the modulus operator
[1] 1800 1903
thanks
Dr. Ottorino-Luca Pantani wrote:
Dear R-users,
I have the following problem
In a lab experiment I have to mix three
Dear all,
With R.2.7.0 (on windows XP) I have encountered what seems to be a negative
memory size. If I use gc() afterwards the R goes down. I use R_alloc for
*most* memory allocations in my C-routines:
memory.size()
[1] 11.08132
dyn.load(rconipm.dll)
dyn.unload(rconipm.dll)
Subject pretty much says it all. I am running 64-bit Ubuntu 8.04, i.e. Hardy
Heron, have openmpi installed, and get the following error message with
attempted install of Rmpi. sessionInfo() follows.
Mark
checking for ANSI C header files... yes
checking for sys/types.h... yes
checking for
cast the two vectors as.matrix-- see here:
plot(as.matrix(f[2,4:26]), as.matrix(f[1,4:26]))
y
stephen sefick wrote:
f - (structure(list(X = structure(96:97, .Label = c(119DAmm, 119DN,
119DNN, 119DO, 119DOC, 119Flow, 119Nit, 119ON, 119OPhos,
119OrgP, 119Phos, 119TKN, 119TOC, 148DAmm,
I am running 64-bit Ubuntu 8.04 and when I invoke rggobi the interactive
graph displays but R crashes. See my sessionInfo() and a short example
below. Ggobi and rggobi installed without complaints. Mark
sessionInfo()
R version 2.7.0 Patched (2008-05-04 r45620)
x86_64-unknown-linux-gnu
locale:
Hi,
I'm having trouble plotting populations as separate colors and points in the
3d scatterplot package. I have a column with 4 different population names
and 3 columns with my data. I want to plot each population with a different
color and pch. In addition, I want to use the type=h in my
#Install library rgl
#here is the function which you need to run first:
rgl.plot3d-function(z, x, y, cols=red,axes=T,new=T)
{xr-range(x)
x01-(x-xr[1])/(xr[2]-xr[1])
yr-range(y)
y01-(y-yr[1])/(yr[2]-yr[1])
zr-range(z)
z01-(z-zr[1])/(zr[2]-zr[1])
if(new) rgl.clear()
if(axes)
Thanks for the help- It worked just fine:
This will cast my ignorance across the table, but here it goes. Why do I
need to make them a matrix? because they are in a row? if
d - as.matrix(f)
f.t - t(d)
f.m - as.data.frame(f.t)
now I could use just the following
plot(f.m[c(rownumber), column],
?gsub()
Match the start (_) followed by anything (.*) and replace by
gsub(_.*,,Name)
Weidong Gu,
Department of Medicine
University of Alabama, Birmingham
1900 University Blvd., Birmingham, Alabama 35294
PH: (205)-975-9053
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL
the data frames work as follows: columns= variables and rows= records. when
you plot with a data frame it's logical to plot one variable against another
not one record against another. with casting into a matrix and rotating then
casting back into a dataframe, you made your records variables..
Hello, I have come across a result that I cannot explain, and am hoping that
someone else can provide an answer. A student fitted a mixed model using
the lme function: out- lme(fixed=Y~A+B+A:B, random=~1|Site). Y is a
continuous variable while A and B are factors. The data set is balanced
with
I'm trying to use combine c('a','b','c','a','c') into 'a, b, c', the order
does not matter.
paste(c('a','b','c','a','c'), collapse=', ') yields 'a, b, c, a, c'.
Any idea?
--
Regards,
Anh Tran
[[alternative HTML version deleted]]
__
Hello,
I have a set of one-liners (many thanks to previous responses from this
list) that I use to look at newly imported data sets with functions like
dim(), names(), str(), etc. within lapply(). Generally, these commands
work for me but, I am apparently still missing some aspect of list
3) Bill Venables offered this about a week ago in this list:
--
This is probably as good a way as any way for this kind of problem.
First define a binary operator:
%^% - function(x, n)
with(eigen(x), vectors %*% (values^n * t(vectors)))
This example only works for
Hi Anh,
Try this,
x=c('a','b','c','a','c')
paste(unique(x),collapse=, )
[1] a, b, c
HTH,
Jorge
On Tue, May 6, 2008 at 2:22 PM, Anh Tran [EMAIL PROTECTED] wrote:
I'm trying to use combine c('a','b','c','a','c') into 'a, b, c', the order
does not matter.
paste(c('a','b','c','a','c'),
Hi Pascal,
I think the function could be better but try this:
# Function: M is your matrix and n MUST be an integer0
mat.pow-function(M,n) {
result-M
if(n1){
for ( iter in 2:n) result-M%*%result
result
}
else {result}
result
}
# The matrix
m -
Thanks Mark,
Your suggestion led me to this:
!is.null(lapply(ls(pattern='bn'), function(y) cat(y, dim(get(y)),
\t, names(get(y)), \n)))
bn1993 2885 11 oplt rplt rsiz tree bd ht oaz odst raz rdst spr
bn1994 3158 7oplt tree bd ht spr stat dam
bn1995 734 7 oplt tree bd ht spr stat dam
probably you want:
stg - c('a', 'b', 'c', 'a', 'c')
paste(unique(stg), collapse = , )
I hope it helps.
Best,
Dimitris
--
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel:
Hello,
Still a newbie with R, though I have learned a lot from reading
this list. I'm hoping someone can help with this question:
I have two vectors, one for variables, and one for bits.
I want to build a string (really a formula) based on the values in my
vector of 1s and 0s in bits. If I
Hi Esmail,
Try this:
vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5')
bits=c(1, 0, 1, 1, 0)
paste(vars[which(bits==1)],collapse=+)
HTH,
Jorge
On Tue, May 6, 2008 at 3:06 PM, Esmail Bonakdarian [EMAIL PROTECTED]
wrote:
Hello,
Still a newbie with R, though I have learned a lot from reading
this
Esmail Bonakdarian wrote:
Hello,
Still a newbie with R, though I have learned a lot from reading
this list. I'm hoping someone can help with this question:
I have two vectors, one for variables, and one for bits.
I want to build a string (really a formula) based on the values in my
vector of
I would actually go with this:
bits=c(1, 0, 1, 1, 0)
paste(X, which(bits==1), sep=.,collapse=+)
No need for the vars variable. Though admittedly it breaks down if
bits is identically 0.
Haris Skiadas
Department of Mathematics and Computer Science
Hanover College
On May 6, 2008, at 3:11 PM,
Hi there,
I've tried to install the A2R package using the files from
http://addictedtor.free.fr/packages/A2R/lastVersion/
This is the error I get when trying to load the library:
library(A2R)
Error in library(A2R) :
'A2R' is not a valid package -- installed 2.0.0?
Can anyone please help?
Marc Schwartz wrote:
Esmail Bonakdarian wrote:
Hello,
Still a newbie with R, though I have learned a lot from reading
this list. I'm hoping someone can help with this question:
I have two vectors, one for variables, and one for bits.
I want to build a string (really a formula) based on the
Fisher's exact test works with small cells. See ?fisher.test
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of raymond chiruka
Sent:
Le mar. 06 mai à 14:23, Alberto Monteiro a écrit :
3) Bill Venables offered this about a week ago in this list:
--
This is probably as good a way as any way for this kind of problem.
First define a binary operator:
%^% - function(x, n)
with(eigen(x), vectors %*% (values^n
Thanks to all of you that helped me with the issues of bootstrapping
and downloading packages to a local disk.
As an starter I'm in the lower side of the learning curve, but this R
software is awesome. What I like most is this kind of forums when
people share their problems and we can find
Jorge Ivan Velez wrote:
Hi Esmail,
Try this:
vars=c('X.1', 'X.2', 'X.3', 'X.4', 'X.5')
bits=c(1, 0, 1, 1, 0)
paste(vars[which(bits==1)],collapse=+)
HTH,
Jorge
Wow .. that is beautiful :-) .. and exactly what I was looking
for (and suspected existed).
I ended up doing this:
Hi,
Is there away to print a short table out along side with a plot?
I'm thinking about doing a
par(mfrow = c(1,2))
Then, the plot is on one side, summary result on the other.
Is there any quick way to print out a data.frame in table format? Thanks
--
Regards,
Anh Tran
[[alternative
Hi Anh,
Take a look at http://finzi.psych.upenn.edu/R/Rhelp02a/archive/128041.html
HTH,
Jorge
On Tue, May 6, 2008 at 3:46 PM, Anh Tran [EMAIL PROTECTED] wrote:
Hi,
Is there away to print a short table out along side with a plot?
I'm thinking about doing a
par(mfrow = c(1,2))
Then, the
Look at the addtable2plot function in the plotrix package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Anh Tran
Sent: Tuesday, May
Hi Birgit,
I'm not sure that I understand your question. I'll try to answer
anyways. Regression trees and therefore also RandomForests are invariant
to monotonic transformations in the independent variables. There are no
distributional assumptions for the independent variables. The dependent
Hi Xavier,
Changing the grid settings has worked!
Thanks a lot!
Mikhail
PS Perhaps it'd still be really useful to be able to change text direction
in labels.
Hadley, what do you think?
--
View this message in context:
On Tue, May 6, 2008 at 10:32 AM, Mark Kimpel [EMAIL PROTECTED] wrote:
I am running 64-bit Ubuntu 8.04 and when I invoke rggobi the interactive
graph displays but R crashes. See my sessionInfo() and a short example
below. Ggobi and rggobi installed without complaints. Mark
sessionInfo()
R
Hi All.
I've run into a problem with the plinear algorithm in nls that is confusing
me.
Assume the following reaction time data over 15 trials for a single unit.
Trials are coded from 0-14 so that the intercept represents reaction time in
the first trial.
trl RT
01132.0
1 630.5
How did you install it?
You need to get A2R_0.0-4.tar.gz and do R CMD INSTALL A2R_0.0-4.tar.gz,
after reading the 'R Installation and Administration manual, especially
the sections for your OS.
The package in A2R/lastVersion/ was built for Unix on R 2.2.1. An expert
might be able to get it
I suspect you have more than one version of R installed and are mixing
them up. Those symbols have been in package stats for quite a while.
Try starting R with --vanilla, and if that works, clean out your startup
files (see ?Startup). If not, remove all your R installations and
reinstall R
recall that 0 ^{-.2} = 1/0^{.2}, and that dividing by 0 gives Inf.
so when 0 is in trl, part of your model for RT is Inf:
trl - 0:14
p - -.2
cbind(1,trl, trl^p)
trl
[1,] 1 0 Inf
[2,] 1 1 1.000
[3,] 1 2 0.8705506
[4,] 1 3 0.8027416
[5,] 1 4 0.7578583
[6,] 1 5
Alex
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Alex Reynolds
Sent: Wednesday, 7 May 2008 12:26 a.m.
To: r-help@r-project.org
Subject: [R] Dendrogram label size
Is it possible to resize the labels in a dendrogram without
applying circles
0^(-0.2) = Inf, so you started with an infinite prediction for your first
point and hence an infinite sum of squares.
On Tue, 6 May 2008, Rick DeShon wrote:
Hi All.
I've run into a problem with the plinear algorithm in nls that is confusing
me.
Assume the following reaction time data over
Thanks to all of you that helped me with the issues of bootstrapping
and downloading packages to a local disk.
As an starter I'm in the lower side of the learning curve, but this R
software is awesome. What I like most is this kind of forums when
people share their problems and we can find
I am looking for a way to simulate genotypes of cases and control at a
disease locus in R. I am supposed to set the allele frequency as
control/cases. for each of the column below simulate 200 snp dataset.
I am looking at treesim function from popgen to stimulate the genotypes in
R.
Here is
Dear Bill,
I expect that the problem is in the contrasts that your student used for A
and B, though I haven't thought specifically about the context of a mixed
model. If he or she used the default contr.treatment(), then the contrasts
for different factors (and the interaction) are not orthogonal
Hi R users
I am trying to create a spatial join between two datasets.
The first data set is large and contains descriptive data including x
and y co-ordinates.
The second dataset is small and has been selected spatially. The only
data contained within the second dataset is the x and y
I believe this function matches the description in OOO:
mround - function(number, multiple) multiple * round(number/multiple)
I've implemented a slightly more general form in the reshape package:
round_any - function (x, accuracy, f = round) {
f(x/accuracy) * accuracy
}
Hadley
--
Thanks. I used the command you mention and it works fine in Linux. But I
need to get it to work for Windows XP as well (currently running R-2.7.0).
Any idea if it's possible?
On Tue, May 6, 2008 at 5:18 PM, Prof Brian Ripley [EMAIL PROTECTED]
wrote:
How did you install it?
You need to get
Andrew
?merge
HTH ..
Peter Alspach
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Andrew McFadden
Sent: Wednesday, 7 May 2008 9:23 a.m.
To: r-help@r-project.org
Subject: [R] Spatial join between two datasets using x and y
co-ordinates
In Dr. Wood's book on GAM, he suggests in section 4.1.6 that it might be
useful to shrink a single smooth by adding S=S+epsilon*I to the penalty
matrix S. The context was the need to be able to shrink the term to zero if
appropriate. I'd like to do this in order to shrink the coefficients towards
vector dat1.select is the selected records from dat1 by dat2.
dat1.select- dat1$x1 %in% dat2$x2 dat1$y1 %in% dat2$y2
dat1[dat1.select,]
x1 y1 descript
1 1824615 5980732 cat
2 1823650 5983220 dog
Andrew McFadden wrote:
Hi R users
I am trying to create a spatial
Hi Andrew,
Try also:
x1-c(1824615,1823650,1821910)
y1-c(5980732,5983220,5990931)
descript-c(cat, dog, horse)
dat1-data.frame(x1,y1,descript)
x2-c(1824615,1823650)
y2-c(5980732,5983220)
dat2-data.frame(x2,y2)
colnames(dat2)=c('x1','y1')
merge(dat1,dat2,by=c('x1','y1'))
HTH,
Jorge
On Tue,
Hello R-list. I am a long time listener - first time caller who has
been using R in research and graduate teaching for over 5 years. I
hope that my question is simple but not too foolish. I've looked
through the FAQ and searched the R site mail list with some close hits
but no direct
Hi all,
I'm looking for a way to use loess model (?loess) with a different
percentile (instead of median).
I hope I made myself clear enough. Here's and example:
a-data.frame(x=seq(1:200),y=200*rnorm(200))
lss-loess(y~x,a)
predict(lss, 120) will give a local median of that dataset.
How about a
Shubha Vishwanath Karanth [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
Thank you very much Mark! That worked Just a question, ?[ does
give an error to me...how do I find it?
Try:
?[
__
R-help@r-project.org mailing list
-Original Message-
From: [EMAIL PROTECTED] on behalf of Yasir Kaheil
Sent: Tue 5/6/2008 4:24 PM
To: r-help@r-project.org
Subject: Re: [R] Spatial join between two datasets using x and y co-ordinates
vector dat1.select is the selected records from dat1 by dat2.
dat1.select-
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