Tim Churches wrote:
Finny Kuruvilla [EMAIL PROTECTED] wrote:
I've been a R-user for quite some time. The graphic on the home page
looks a bit in need of polish so I applied some antialiased
transformations that Peter Dalgaard has previously posted to R-help
for improving graphic quality.
I have following syntax for putting a legend :
legend(bottom, fill=c(red,blue), legend=expression(p==0.30, p==0.50),
bty=n)
However what I want is that : the value 0.30 should be a value of a variable
instead of a constant, so that I can put the name of this variable and in
legend it's value
Oh yes.
I did searched for help, but Ijust didn't read carefully, I read
MAAS, instead of MASS..
Carlos
On 9/25/07, Peter Dalgaard [EMAIL PROTECTED] wrote:
Carlos Guâno Grohmann wrote:
Hello,
After a long time, I needed the truehist function, but my system
couldn't found it. I tried to
(Ted Harding) wrote:
Pat Altham (now retired) developed extensive teaching (and
other) materials in R at the Cambridge University Statistical
Laboratory. From her personal web page:
Some of the computer languages I have had to try to
learn since graduating in 1964: Cambridge autocode,
Dear R ussers,
I noticed a small problem when ussing for example the function
plotMeamns under Rcmdr.
In this case a graph will be ploted, giving the means on the Y-axis, by a
factor on the X-axis.
All is correct when doing this, ussing 'sd' for error bars.
The problem occures when ussing
Hello,
I've created an application (IDE for ecological modelling) on the basis of
Eclipse which uses Rserve
http://www.rforge.net/Rserve/ http://www.rforge.net/Rserve/
to archieve this.
I also implemented a small api to send or access data from or to R and Java.
The transfer is very fast.
How about
a - .33
b - .55
legend(bottom, fill=c(red,blue),
legend=c(bquote(p == .(a)), bquote(p == .(b))), bty=n)
or look at ?substitute
- Peter Ehlers
stat stat wrote:
I have following syntax for putting a legend :
legend(bottom, fill=c(red,blue), legend=expression(p==0.30, p==0.50),
Hello,
I wrote
setwd(D:/)
dd - read.table(file=test.txt,header=TRUE)
attach(dd)
boxplot(x)
outlier - function(y){
+ out - boxplot(y, range = 1)$out
+ outliers - which(y == out)
+ dev.off()
+ return(out,outliers)
+ }
outlier(x)
$out
[1] 1.950208 2.082025 4.768637
Not sure which of the questions yo want answered in your email.
However, if its the one regarding the boxplot try:
dd - read.table(test.txt,header=T)
attach(dd)
boxplot(x)
outlier - function(y){
out - boxplot(y, range = 1)$out
outliers - which(y %in% out)
http://www.broad.mit.edu/~finnyk/Rhome.jpg
If you run Eric Lecoutre's code to produce the graphic, available at
http://www.r-project.org/misc/acpclust.R, unchanged except for the
addition of these lines:
library(Cairo)
Cairo(600,400,file=Rlogo_swiss.png,type=png,bg=white)
then you
So I applied my corrected margins to Tim's Cairo trick and voila:
http://www.broad.mit.edu/~finnyk/Rlogo_swiss.png
This is hands-down the best version, in my opinion!
Best,
Finny
On Wed, Sep 26, 2007 at 06:35:40AM -0500, [EMAIL PROTECTED] wrote:
http://www.broad.mit.edu/~finnyk/Rhome.jpg
On Tue, 25 Sep 2007 20:16:05 -0400,
Wiebke Timm (WT) wrote:
You might want to check if there is a neural gas algorithm in R.
kmeans generally has a high variance since it is very dependent on
the initialization. Neural gas overcomes this problem by using a
ranked list of
On Wed, 26 Sep 2007, Friedrich Leisch wrote:
On Tue, 25 Sep 2007 20:16:05 -0400,
Wiebke Timm (WT) wrote:
You might want to check if there is a neural gas algorithm in R.
kmeans generally has a high variance since it is very dependent on
the initialization. Neural gas overcomes this
?wordReport gives a lot of information, but I think it makes us wish for
more :-)
Where can I find all the ways to write Word documents using R?
Namely:
(1) is there any way to open a new document and save it automatically?
The sequence:
WordOpen(new_file.doc); WordInsertText(R rulez!\n);
On 26.09.2007, at 09:54, Prof Brian Ripley wrote:
On Wed, 26 Sep 2007, Friedrich Leisch wrote:
On Tue, 25 Sep 2007 20:16:05 -0400,
Wiebke Timm (WT) wrote:
You might want to check if there is a neural gas algorithm in R.
kmeans generally has a high variance since it is very dependent
We have been asked to write a paper on the snow package for
parallel computing in R for a parallel computing journal and
would like to include some references to examples of the use of
snow in practice beyond our own use. If you think you have a
good example we would like to hear from you.
In accordance with Venables and Ripley, SAS documentation and other
sources AIC with sigma^2 unknown is calculated as:
AIC = -2LL + 2* #parameters = n log(RSS/n) + 2p
For the fitness data:
(http://support.sas.com/ctx/samples/index.jsp?sid=927), SAS gets an AIC
of 64.534 with model oxygen =
Dear List:
Below is how I specify an axis:
axis(2, at=c(0.5, 0.0005))
R displays the numbers in scientific notation. What
argument/parameter should I use to tell R to display the numbers as
specified rather than in scientific notation?
version
_
platform
On 9/26/2007 11:24 AM, Jacques Wagnor wrote:
Dear List:
Below is how I specify an axis:
axis(2, at=c(0.5, 0.0005))
R displays the numbers in scientific notation. What
argument/parameter should I use to tell R to display the numbers as
specified rather than in scientific notation?
joe,
some procs in SAs calculates log likelihood differently than what it
is supposed to be. try using proc nlmixed and specifying the LL
explicitly.
in your case, I has stronger faith in R result instead of SAS result.
On 9/26/07, Joe Yarmus [EMAIL PROTECTED] wrote:
In accordance with Venables
I'm using rpart to fit a tree using a large dataset: 7000 observations, 4651
variables. All but one of the variables (age) are binary. When I run the
code:
fit1 - rpart(lowergi ~ ., data=dset,method=class)
I get the error:
Error in dimnames(X) -list(dn[1L]], unlist(collabs, use.names=FALSE)) :
On 9/26/07, Wensui Liu [EMAIL PROTECTED] wrote:
RDCOMClient and rcom are great. but I can't find much manual or
tutorial about them.
They both have home pages that will lead you to whatever
information exists on them:
http://sunsite.univie.ac.at/rcom/
http://www.omegahat.org/RDCOMClient/
So I applied my corrected margins to Tim's Cairo trick and voila:
http://www.broad.mit.edu/~finnyk/Rlogo_swiss.png
This is hands-down the best version, in my opinion!
Yes, it is definitely much nicer than the version on www.r-project.org
now. :-)
--e
Transition matrices are Markov transition matrices among different
life stages of organisms -- in the simplest case (Leslie matrices,
Ben,
Thanks for your clear explanation and plot examples. I like the dotplots alot
and added a few modifications below. Since I often compare rows and
if your regression under gaussian assumption, then you can't
constraint your predicted to be positive.
I don't know much about your dep in the model. but given more
appropriate distribution assumption, the constraint is doable. One
possibility that I can think of is poisson.
On 9/25/07, Westley
You seem to be assuming that 'regression' has to do with 'gaussian
assumption'. However, I presume WLS stands for 'weighted least squares',
and 'regression' is historically associated with fitting linear models by
least squares.
I don't see why even in the model-based framework you assert
On Wed, 26 Sep 2007, Juned Siddique wrote:
I'm using rpart to fit a tree using a large dataset: 7000 observations, 4651
variables. All but one of the variables (age) are binary. When I run the
code:
fit1 - rpart(lowergi ~ ., data=dset,method=class)
I get the error:
Error in dimnames(X)
I believe the argument to format is scientific i.e.
axis(2, at=at, labels=format(at, scientific=FALSE))
Best regards,
Francisco
Duncan Murdoch wrote:
On 9/26/2007 11:24 AM, Jacques Wagnor wrote:
Dear List:
Below is how I specify an axis:
axis(2, at=c(0.5, 0.0005))
R displays the
On 26.09.2007, at 13:46, Olena Morozova wrote:
I am very new to R and statistical analysis in general. I am trying
to plot
a matrix of several hundred rows against a vector of 4 values. This
all has
to be on the same graph with different rows represented by
different color.
This
Thank you Wensui and Professor Ripley for your responses.
Prof. Ripley, your assumptions regarding the context in which I'm using 'WLS'
and 'regression' are correct. The function solve.QP in the quadprog package
sounds like a great way to go. Thank you, and I will try this method.
Westley A.
R-listers,
Given a polygon and a circle defined by its center coordinates and a
radius, I would like to calculate the area of overlap. I know that I
can create a polygon from the circle and then use available packages to
get the area of the intersection. However, because the polygon is of a
I would appreciate confirmation that the function vcov(model.nlme)
gives the var-cov matrix of the fixed effects in an nlme model.
Presumably the random-effects var-cov matrix is given by cov(ranef
(model.nlme)?
Rob Forsyth
__
R-help@r-project.org
Prof Ripley,
Thank you for your post. Below is the output of traceback(). I'm still not
sure what the problem is. Thank you.
4: as.matrix.data.frame(frame)
3: as.matrix(frame)
2: rpart.matrix(m)
1: rpart(lowergi ~ ., data=dset, method=class)
On 9/26/07, Prof Brian Ripley [EMAIL PROTECTED]
At 7:25 AM -0700 9/26/07, Samuel Okoye wrote:
Hello,
I have got the following problem:
setwd(C:/temp)
library(xlsReadWrite)
MyData - read.xls(file=Mappe1.xls, colNames = TRUE,dateTimeAs
= isodatetime)
attach(MyData)
MyData
name value times
1 A1 2
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Alberto Monteiro wrote:
?wordReport gives a lot of information, but I think it makes us wish for
more :-)
Where can I find all the ways to write Word documents using R?
Using either of the R/DCOM client packages (rcom and RDCOMClient)
you
On Wed, 26 Sep 2007, Francisco J. Zagmutt wrote:
I believe the argument to format is scientific i.e.
axis(2, at=at, labels=format(at, scientific=FALSE))
Yes, but abbreviations of argument names are allowed.
Here setting a small positive value of options(scipen) (e.g. 2) gives
floating-point
I am trying to generate a fourth vector,z, given three known and fixed
vectors, x1,x2,x3 with corresponding known and fixed correlations with
themeselves and with z. That is, all correlations are known and
prespecified. How can I do this?
Thank you,
ben
Hi,
I want to be able to create a vector of z-scores from a vector of
continuous data, conditional on a group membership vector.
Say you have 20 numbers distributed normally with a mean of 50 and an sd
of 10:
x - rnorm(20, 50, 10)
Then you have a vector that delineates 2 groups within x:
On Wed, 2007-09-26 at 15:18 -0500, Jacques Wagnor wrote:
Dear List,
Is there any way to password-protect script files (either within R or
otherwise)?
You might want to review this thread:
http://thread.gmane.org/gmane.comp.lang.r.general/94290/
which covers perhaps a similar concern.
Here is one approach (someone with better linear algebra skills may be
able to shorten this, some steps could be combined, but I wanted to show
each step):
# create x1-x3
x1 - rnorm(100, 50, 3)
x2 - rnorm(100) + x1/5
x3 - rnorm(100) + x2/5
# find current correlations
cor1 - cor(
I came across a post by Karl Knoblick regarding the modeling of longitudinal
data (see https://stat.ethz.ch/pipermail/r-help/2007-May/132137.html). I am
often asked by physicians to perform what Karl refers to in his post as option
1: to perform paired t-tests against baseline at each follow
Hello -
First, I doubt you really want to cbind() those two vectors within the
data.frame() function call.
test.data - data.frame(x, group) is probably what you want. That may
be the source of your trouble.
If you really want a vector returned, the following should work given
your
Dear All,
I hope this is not a FAQ, but my online research was not fruitful.
Consider a standard 2D plot generated with the plot command.
I would like to introduce inside the graph some text with an arrow
pointing to a specific position of the plotted function.
Is this doable in R? Can anyone
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Destination vacances
[2]Merveilles de l'Egypte
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[rect-jaune.gif] [4]Athènes et ses joyaux
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I feel like I must be missing something rather plain, but I don't get
it. how is one supposed to use R as a PgSQL client on Windows? Assume
my windows desktop is on the same network as a PgSQL server, and I just
need to use R to connect and pull down some data.
The thing that is
Hello all:
A question from a new user. I have data on a geo-referenced curvilinear
grid. This is a grid with 75 rows and 51 columns, is not aligned
north-south, and the rows and columns are not straight. (And the
coordinates are in meters.) I want to make image plots of this data,
but where
You might be able to do this with the ggplot2 package - see for
example http://had.co.nz/ggplot2/coord_map.html, which shows plots on
map coordinate systems. Because the design of ggplot2 means the
coordinate system and geom (eg. points vs tiles) operate
independently, you can draw image plots
hadley wickham [EMAIL PROTECTED] wrote:
You might be able to do this with the ggplot2 package - see for
example http://had.co.nz/ggplot2/coord_map.html, which shows plots on
map coordinate systems. Because the design of ggplot2 means the
coordinate system and geom (eg. points vs tiles)
On Wed, 26 Sep 2007, mfrumin wrote:
hey all,
I feel like I must be missing something rather plain, but I don't get it.
how is one supposed to use R as a PgSQL client on Windows? Assume my
windows desktop is on the same network as a PgSQL server, and I just need to
use R to connect and
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