Hi,
I believe this question has been asked before, but I cant find and don't
remember the answer.
The problem is simple, calling 'plot.data.frame(x)' gives a nice 'matrix of
scatterplots' for each pair of columns in x. for example;
x -
data.frame(a=jitter(01:20),
b=jitter(20:01),
modify the content of the title
boxes without changing the names of the variables, e.g. myParameter
instead of a?
Yup, that is easy :-D
Try the following (on the code already given);
plot(x, labels=paste(hello\n, names(x)))
See ?pairs for the really cool things that you can do.
Dan Bolser
Hi,
I am generating a beautiful plot with the 'levelplot' function over my
square matrix of data. In order to help visualise the data I would
like to draw a diagonal line on the matrix. Because the plot is
actually a trellis object, I am having difficulty working out how to
do this. I have been
My question was thus;
Given
library(lattice)
my.m - matrix(seq(1,100,1),nrow=10)
levelplot(my.m)
How can I add a diagonal line onto the resulting 'color square'?
The answer I found was to hack the 'panel.levelplot' function. Here is the
diff between the old (panel.levplot) and the new
On 04/04/07, Sundar Dorai-Raj [EMAIL PROTECTED] wrote:
Dan Bolser said the following on 4/4/2007 7:52 AM:
My question was thus;
Given
library(lattice)
my.m - matrix(seq(1,100,1),nrow=10)
levelplot(my.m)
How can I add a diagonal line onto the resulting 'color square
Hi,
I am trying to create a triangular plot to show the 'composition' of a
set of items with three variables (historically the percent sand, silt
and clay in soil).
So far I have tried the 'soil texture triangle plot' in the package
plotrix and the 'ternary or triangular plots' in the
hadley wickham wrote:
Thing is, for one reason or another, the number of marbles per bag may
systematically vary with age too. However, I am not interested in the
number of marbles per bag, so I would like to group the students into 8
groups such that each group has the same total number of
Gabor Grothendieck wrote:
On 3/19/06, Dan Bolser [EMAIL PROTECTED] wrote:
Gabor Grothendieck wrote:
On 3/18/06, Dan Bolser [EMAIL PROTECTED] wrote:
Gabor Grothendieck wrote:
If you are just looking for something simple that may be good enough
then assign the largest one to group 1
for the beautifully clear examples!
Dan.
Regards, Adai
On Sun, 2006-03-19 at 18:58 +, Dan Bolser wrote:
Adaikalavan Ramasamy wrote:
Do you by any chance want to sample from each group equally to get an
equal representation matrix ?
No.
I want to make groups of equal sizes, where size isn't
Gabor Grothendieck wrote:
On 3/18/06, Dan Bolser [EMAIL PROTECTED] wrote:
Gabor Grothendieck wrote:
If you are just looking for something simple that may be good enough
then assign the largest one to group 1, the second largest to group 2,
..., the 8th largest to group 8 and then start over
with weights via:
out3 - sample.by.group(mydf, mydf$grp, k=c(20, 20, 30, 30), replace=T)
table( out3$grp )
Regards, Adai
On Fri, 2006-03-17 at 16:01 +, Dan Bolser wrote:
Hi,
I have tuples of data in rows of a data.frame, each column is a variable
for the 'items' (one per row
/17/06, Dan Bolser [EMAIL PROTECTED] wrote:
Dan Bolser wrote:
Hi,
I have tuples of data in rows of a data.frame, each column is a variable
for the 'items' (one per row).
One of the variables is the 'size' of the item (row).
I would like to cut my data.frame into groups such that each group has
Hi,
I have tuples of data in rows of a data.frame, each column is a variable
for the 'items' (one per row).
One of the variables is the 'size' of the item (row).
I would like to cut my data.frame into groups such that each group has
the same *total size*. So, assuming that we order by size,
Dan Bolser wrote:
Hi,
I have tuples of data in rows of a data.frame, each column is a variable
for the 'items' (one per row).
One of the variables is the 'size' of the item (row).
I would like to cut my data.frame into groups such that each group has
the same *total size*. So
Marc Schwartz (via MN) wrote:
On Mon, 2006-03-06 at 15:40 +0100, Roland Kaiser wrote:
How can i set a rotation for the names.arg in barplot?
See R FAQ 7.27 How can I create rotated axis labels?:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-create-rotated-axis-labels_003f
Hi,
Since I started to make some 'final' plots of my data I found that I
have tons of questions related to 'the little things'. Rather than
bother the list with all the questions (ahem), or search the archives
for similar questions and translate the context, I would like to find a
FAQ for
Romain Francois wrote:
Le 03.03.2006 13:13, Dan Bolser a écrit :
Hi,
Since I started to make some 'final' plots of my data I found that I
have tons of questions related to 'the little things'. Rather than
bother the list with all the questions (ahem), or search the archives
for similar
Marc Schwartz (via MN) wrote:
On Tue, 2005-12-13 at 10:53 +, Dan Bolser wrote:
Hi, I am plotting a distribution of (ordered) values as a barplot. I
would like to label groups of bars together to highlight aspects of the
distribution. The label for the group should be the range of values
Hi, I am plotting a distribution of (ordered) values as a barplot. I
would like to label groups of bars together to highlight aspects of the
distribution. The label for the group should be the range of values in
those bars.
As this is hard to describe, here is an example;
x - rlnorm(50)*2
Duncan Murdoch wrote:
Over the weekend I wrote a small package to evaluate poker hands and to
do some small simulations with them. If anyone is interested in looking
at it, I'd appreciate comments and/or contributions.
How do I install this package?
A README or a hint on th webpage below
Meta:
This question is somewhat long and has two parts, I would be very happy
for someone just to nudge me in the right direction with the manual /
tutorial, as I am somewhat lost...
1) How do I fit a curve of the form y = k1 * x^k2 ?
I want to estimate values of k1 and k2 given the x/y data I
On Wed, 10 Aug 2005, S.O. Nyangoma wrote:
I see that
log(y)=log(k1)+k2*log(x)
use lm?
Thats a nice solution in this instance, but in general how do I get R to
fit a particular function (formula) and return the parameters?
Cheers,
Dan.
- Original Message -
From: Dan Bolser [EMAIL
On Sun, 24 Jul 2005, John Wilkinson wrote:
Dan,
Another tweak !
If you want the 'legend' to look pretty you can resize it by adding,say,
'cex=0.6' into the legend code; try---
legend(topleft, #inset=-1,
legend = do.call(expression, L),
bg='white',
ncol = 2,
On Fri, 22 Jul 2005, Marc Schwartz (via MN) wrote:
Ok guys,
So I played around with this a bit, going back to Dan's original
requirements and using Thomas' do.call() approach with legend(). Gabor's
approach using sapply() will also work here. I have the following:
# Note the leading spaces here
On Sat, 23 Jul 2005, Dan Bolser wrote:
On Fri, 22 Jul 2005, Marc Schwartz (via MN) wrote:
Ok guys,
So I played around with this a bit, going back to Dan's original
requirements and using Thomas' do.call() approach with legend(). Gabor's
approach using sapply() will also work here. I have
On Thu, 21 Jul 2005, Marc Schwartz (via MN) wrote:
[Note: the initial posts have been re-arranged to attempt to maintain
the flow from top to bottom]
Dan Bolser writes:
I would like to annotate my plot with a little box containing the slope,
intercept and R^2 of a lm on the data
numbers, 2) remove excessive white space.
I like the above because it dosn't require me to calculate exactly where
to put each piece of text.
I just want to annotate a plot :(
On 7/22/05, Dan Bolser [EMAIL PROTECTED] wrote:
On Thu, 21 Jul 2005, Marc Schwartz (via MN) wrote:
[Note
I would like to annotate my plot with a little box containing the slope,
intercept and R^2 of a lm on the data.
I would like it to look like...
++
| Slope : 3.45 +- 0.34 |
| Intercept : -10.43 +- 1.42 |
| R^2 : 0.78 |
--
Christoph Buser [EMAIL PROTECTED]
Seminar fuer Statistik, LEO C13
ETH (Federal Inst. Technology) 8092 Zurich SWITZERLAND
phone: x-41-44-632-4673fax: 632-1228
http://stat.ethz.ch/~buser/
--
Dan Bolser writes
On Tue, 7 Jun 2005, Sander Oom wrote:
Hi Romain,
You have stuck your neck out which is great. You are however not
responsible for all the work! Let others join in!
Thus a wiki could provide opportunities for other people to contribute!
I was suggesting that the wiki be used as a platform for
On Tue, 7 Jun 2005, Martin Maechler wrote:
Dan == Dan Bolser [EMAIL PROTECTED]
on Mon, 6 Jun 2005 14:48:50 +0100 (BST) writes:
Dan On Mon, 6 Jun 2005, Douglas Bates wrote:
On 6/6/05, Dan Bolser [EMAIL PROTECTED] wrote:
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1
On Tue, 7 Jun 2005, David Forrest wrote:
On Tue, 7 Jun 2005, Dan Bolser wrote:
...
I agree, but its hard to get people contributing without something
centralized (some central autohrity). The problem with the existing wiki
http://fawn.unibw-hamburg.de/cgi-bin/Rwiki.pl?RwikiHome
This 'strange behaviour' manifest itself within some quite complex
code. When I created a *very* simple example the behaviour dissapeared.
Here is the simplest version I have found which still causes the strange
behaviour (it could be quite unrelated to the boot library, however).
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
Signif. codes: 0 8098***8099 0.001 8098**8099 0.01
8098*8099 0.05 8098.8099 0.1 8098 8099 1
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
hmm... they go away when I paste them in...
On Mon, 6 Jun 2005, Romain Francois wrote:
Hello all,
It seems that the next improvement to the R Graph Gallery is
categorization of the graphics, that way each graph will be easier to
find. That step should be done *carefully* if we want to avoid the
opposite side-effect : graph not
On Mon, 6 Jun 2005, Roger D. Peng wrote:
What is your operating system?
Very sorry for lack of details...
I am RH 9, unix pc i386.
-roger
Dan Bolser wrote:
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
Signif. codes: 0 8098***8099 0.001 8098**8099 0.01
8098*8099 0.05
On Mon, 6 Jun 2005, Douglas Bates wrote:
On 6/6/05, Dan Bolser [EMAIL PROTECTED] wrote:
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
Signif. codes: 0 8098***8099 0.001 8098**8099 0.01
8098*8099 0.05 8098.8099 0.1 8098 8099 1
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05
On Mon, 6 Jun 2005, Sander Oom wrote:
Barry Rowlingson wrote:
Romain Francois wrote:
Graphics will be classified in :
- categories
- sub-categories within those categories
So far so good.
Maybe, maybe not! Would a system of keywords work better than strict
hierarchical
On Mon, 6 Jun 2005, David Forrest wrote:
On Mon, 6 Jun 2005, Dan Bolser wrote:
On Mon, 6 Jun 2005, Romain Francois wrote:
...
It seems that the next improvement to the R Graph Gallery is
categorization of the graphics, that way each graph will be easier to
find. That step should be done
I want to include missing values in my barplot to get the correct x-axis,
for example,
x - c(1,2,3,4, 9)
y - c(2,4,6,8,18)
barplot(y)
The above looks wrong because the last height in y should be a long way
over.
So I want to do something like...
x - c(1,2,3,4,5,6,7,8, 9)
y -
On Sat, 4 Jun 2005, Uwe Ligges wrote:
Dan Bolser wrote:
I want to include missing values in my barplot to get the correct x-axis,
for example,
x - c(1,2,3,4, 9)
y - c(2,4,6,8,18)
barplot(y)
The above looks wrong because the last height in y should be a long way
over.
So I want
On Sat, 4 Jun 2005, Uwe Ligges wrote:
Dan Bolser wrote:
On Sat, 4 Jun 2005, Uwe Ligges wrote:
Dan Bolser wrote:
I want to include missing values in my barplot to get the correct x-axis,
for example,
x - c(1,2,3,4, 9)
y - c(2,4,6,8,18)
barplot(y)
The above looks wrong because the last
On Sat, 4 Jun 2005, Marc Schwartz wrote:
On Sat, 2005-06-04 at 14:50 +0100, Dan Bolser wrote:
snip
This must be because of the log='y' option that I am using here.
y - c(2,4,6,8,NA,NA,NA,NA,18)
barplot2(y,log='y')
Above fails.
I appreciate that what I am trying to do is somewhat
/server irc.freenode.net
/join #R
:)
On Fri, 3 Jun 2005, Robert Citek wrote:
Is there an IRC channel for R?
I have a bunch of relatively simple questions that I can't find
answers to. So it'd be nice to send them to an IRC channel rather
than clutter up the mailing list. I'm sure
Any simple way to take a (2D) table and 'rotate' it so all the rows become
columns and the columns rows?
I'll wager there is a simple way ;) - but I can't find it :(
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
The same problem I am having has been reported here
http://tolstoy.newcastle.edu.au/R/devel/04a/0344.html
Namely that using par(mfg=...) with a postscript (eps) for inclusion with
latex makes the figure appear upside down and back to front (flipped)!
Converting the dvi to ps makes matters
Should I post this to 'bugs'?
On Tue, 26 Apr 2005, Dan Bolser wrote:
The same problem I am having has been reported here
http://tolstoy.newcastle.edu.au/R/devel/04a/0344.html
Namely that using par(mfg=...) with a postscript (eps) for inclusion with
latex makes the figure appear upside down
If I integrate over the result of the density() funcion, is the result 1?
For example
x - rnorm(1000)
plot(density(x))
Does the area under the curve I see sum to 1?
What I really want to know is if I can directly compare two particular
curves, generated like this
x - rnorm(1000)
On Fri, 18 Feb 2005, Romain Francois wrote:
Hello Sander,
That's a good idea and i am up to it.
Right now i am in an exam period, so it's not really the better time,
give me a couple of weeks and i will come up with a specific format of R
files to submit to me that i could post-process to
Try one of the sci.* news groups (google groups) you should be able to
find a specific group and ask this question there.
On Wed, 19 Jan 2005, Erin Hodgess wrote:
Dear R People:
Here is another off topic question, please:
Does anyone know where to find some archaelogical data (carbon
dating),
This may sound crazy but...
I have data like this...
results.matrix
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 949 932 11 26 2076 10 5 0 3
[2,] 1233 124 24 35 58 57 17 21 31191121
Which is the result
On Tue, 11 Jan 2005, Marc Schwartz wrote:
On Tue, 2005-01-11 at 14:59 +, Dan Bolser wrote:
On 10 Jan 2005, Peter Dalgaard wrote:
Dan Bolser [EMAIL PROTECTED] writes:
Cheers. This is really me just being lazy (as usual). The latex
function
in Hmisc allows me to make a .ps file
Hello,
I am making some use of ROC curve analysis.
I find much help on the mailing list, and I have used the Area Under the
Curve (AUC) functions from the ROC function in the bioconductor project...
http://www.bioconductor.org/repository/release1.5/package/Source/
ROC_1.0.13.tar.gz
On Wed, 12 Jan 2005, Berton Gunter wrote:
R-Listers.
The following is a rant originally sent privately to Frank Harrell in
response to remarks he made on this list. The ideas are not new or original,
but he suggested I share it with the list, as he felt that it might be of
wider interest,
On 10 Jan 2005, Peter Dalgaard wrote:
Dan Bolser [EMAIL PROTECTED] writes:
Cheers. This is really me just being lazy (as usual). The latex function
in Hmisc allows me to make a .ps file then grab a screen shot of that ps
and make a .png file.
I would just like to use plot so I can wrap
I would like R to produce some tabulated data in a graphical output. When
I say tabulated data, what I mean is a table with rows and columns. This
would be useful when reading in a big file, performing some analysis on
it, and then wanting to display the results as a table.
Something like
On Mon, 10 Jan 2005, Marc Schwartz wrote:
On Mon, 2005-01-10 at 11:05 +, Dan Bolser wrote:
I would like R to produce some tabulated data in a graphical output. When
I say tabulated data, what I mean is a table with rows and columns. This
would be useful when reading in a big file
Cheers. This is really me just being lazy (as usual). The latex function
in Hmisc allows me to make a .ps file then grab a screen shot of that ps
and make a .png file.
I would just like to use plot so I can wrap it in a png command and not
have to use the 'screen shot' in between.
I can use
Hi, I would like to bound the lower limit of my y scale to zero, and let R
chose an upper limit.
Something like
plot(x,ylim=c(0,))
or
plot(x,ylim=c(0,na))
but nither of these do the job. I searched the docs, but I can't see a way
to do this.
Naturally its nothing I can't do 'by hand', I
below, how would I go about it?
Cheers,
plot(x,y,ylim=c(0,max(y)))
Does it work for you ?
Cordialement. Romain.
Dan Bolser a écrit :
Hi, I would like to bound the lower limit of my y scale to zero, and let R
chose an upper limit.
Something like
plot(x,ylim=c(0,))
or
plot(x,ylim=c(0,na
On Sun, 9 Jan 2005, Thomas Lumley wrote:
On Sun, 9 Jan 2005, Dan Bolser wrote:
On Sun, 9 Jan 2005, [ISO-8859-1] Romain François wrote:
Hello Dan,
Look at the code of the plot.default function, you'll see that's not
possible to specify one limit, nevertheless, you can do :
Suppose I
Hi,
I want to use hist with non-equi-spaced breaks, picked such that the
fraction of the data points falling in the cells (defined by 'breaks') is
roughly equal accross all cells.
Is there such a function that will automatically try to determine the
breaks to fullfill this requirement?
--
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dan Bolser
Sent: Friday, January 07, 2005 11:21 AM
To: R mailing list
Subject: [R] hist function to give each cell equal area
Hi,
I want to use hist with non-equi-spaced breaks, picked
? formatC
Else (if you are linux) the following shell command can be useful...
rename file_name_ file_name_0 file_name_[0-9]_bleh_*.png
Where file_name_[0-9]_bleh_*.png is supposed to match all those files
with a single digit, and the rename command adds a zero before that digit.
rename can
I was doing something very similar.
I found it tricky to work out how to find a confidence interval for the
'percentage' of the outcome (I call it proportion).
Some of my bins had all zeroes or all ones, so I couldn't work out how to
make a variance that was sensible. Also some bins had few
On Tue, 28 Dec 2004, Dan Bolser wrote:
On Thu, 23 Dec 2004, Dan Bolser wrote:
I have data that looks (very roughly) like this...
Declarative:
Several 'groups', each group with a very variable number of
data points associated.
Procedural:
v.1 - c(rep(50,1), rep(5,5), rep(2,10)) # Set up
v
I want to add values onto the end of a vector, for example...
x - vector
for (i in 1:5){
add_to_end_of_vector(i,x)
}
I just cant find the answer to this question!
Sorry for such a basic question, I tried...
x - c()
for (i in 1:5) x[length(x)] - i
but it didn't work.
I want to use boot.ci to generate confidence intervals over the
bootstrapped mean(s) of a group of observations (i.e. I have 10
observations and I want to know how confident I can be on the value for
the mean).
I don't know (or want to know) the details of bootstrapping - I just have
the
I found this great search
source(http://www.statslab.cam.ac.uk/~djw1005/Stats/Interests/search.R;)
helpHTML()
Has (or will) this become the standard search method?
Is R 'Free Software'? The dependence on Java seems a bit of a pain for
'freeness'.
Did the above make it into CRAN?
Cheers,
Hello,
I have several ordered groups (20), each with several observations. Each
group has fewer observations roughly linearly from 30 to 0. Each
observation is a proportion. As I know the max and min values for a
proporion are 1 and 0, I am adding these values to each group to allow
bootstrap
I am trying the following code...
for(i in 1:bins){
print(cat(c(i, length(var1[var1==i]
}
Where var1 is a vector with zero or more values = {1, 2, ..., bins} (or
something like that).
The result I get is...
1 33NULL
2 28NULL
3 39NULL
4 27NULL
5 32NULL
6 30NULL
7 23NULL
8 16NULL
9 10NULL
On 4 Jan 2005, Peter Dalgaard wrote:
Dan Bolser [EMAIL PROTECTED] writes:
I am trying the following code...
for(i in 1:bins){
print(cat(c(i, length(var1[var1==i]
}
Where var1 is a vector with zero or more values = {1, 2, ..., bins} (or
something like that).
The result I get
On Tue, 4 Jan 2005, Dan Bolser wrote:
On 4 Jan 2005, Peter Dalgaard wrote:
Dan Bolser [EMAIL PROTECTED] writes:
I am trying the following code...
for(i in 1:bins){
print(cat(c(i, length(var1[var1==i]
}
Where var1 is a vector with zero or more values = {1, 2, ..., bins
On Thu, 23 Dec 2004, Dan Bolser wrote:
I have data that looks (very roughly) like this...
Declarative:
Several 'groups', each group with a very variable number of
data points associated.
Procedural:
v.1 - c(rep(50,1), rep(5,5), rep(2,10)) # Set up
v.2 - c('a','b','c','d','e','f','g','h
I have data that looks (very roughly) like this...
Declarative:
Several 'groups', each group with a very variable number of
data points associated.
Procedural:
v.1 - c(rep(50,1), rep(5,5), rep(2,10)) # Set up
v.2 - c('a','b','c','d','e','f','g','h', # the
On Wed, 15 Dec 2004, Wiener, Matthew wrote:
It sounds like clara in package cluster might help.
Cheers, this looks just the ticket. How should I choose k though?
Dan.
Regards,
Matt Wiener
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dan Bolser
Hi,
I have ~40,000 rows in a database, each of which contains an id column and
20 additional columns of count data.
I want to cluster the rows based on these count vectors.
Their are ~1.6 billion possible 'distances' between pairs of vectors
(cells in my distance matrix), so I need to do
On Fri, 26 Nov 2004, Martin Maechler wrote:
Dan == Dan Bolser [EMAIL PROTECTED]
on Thu, 25 Nov 2004 22:35:22 + (GMT) writes:
Dan On Thu, 25 Nov 2004, Uwe Ligges wrote:
Dan Bolser wrote:
Is this an impossible task?
How about just problem 2 below, having one pch
Hello,
I am using code like the following to create as simple plot...
plot(x,y,type='b')
lines(lowess(x,y),lwd=3,lty=3,col=2)
I want to add a legend which shows lines looking exactly like those used
in my plot, i.e. a thin black line with gaps taken up by circles (the
default for type='b',
Is this an impossible task?
How about just problem 2 below, having one pch in one legend entry, but no
pch in the second?
On Thu, 25 Nov 2004, Dan Bolser wrote:
Hello,
I am using code like the following to create as simple plot...
plot(x,y,type='b')
lines(lowess(x,y),lwd=3,lty=3,col=2
On Thu, 25 Nov 2004, Uwe Ligges wrote:
Dan Bolser wrote:
Is this an impossible task?
How about just problem 2 below, having one pch in one legend entry, but no
pch in the second?
Please be at least a little bit patient! This is not a hotline! People
are not working 24 hours a day just
Sorry for the dumb question, but I cant work out how to do this.
Quick version,
How can I re-bin a given frequency distribution using new breaks without
reference to the original data? Given distribution has integer valued
bins.
Long version,
I am loading a frequency table into R from a
On Sun, 21 Nov 2004 [EMAIL PROTECTED] wrote:
On 21-Nov-04 Dan Bolser wrote:
Sorry for the dumb question, but I cant work out how to do this.
Quick version,
How can I re-bin a given frequency distribution using new breaks
without reference to the original data? Given distribution has
On Sun, 7 Nov 2004, [ISO-8859-15] Thomas Schönhoff wrote:
Hello,
right now I'm thinking about running R 2.0.0 on box A (Debian SID) but
at the same time having access to the ressources of box B (Ubuntu
Linux) regarding disk capacity, RAM, idle CPU cycles . Is there anyone
of you that has
On Fri, 5 Nov 2004, Pierre BADY wrote:
hi,
you can see these links:
http://pbil.univ-lyon1.fr/R/enseignement.html
http://zoonek2.free.fr/UNIX/48_R/all.html
http://www.ceremade.dauphine.fr/~xian/Noise.html
http://statwww.epfl.ch/davison/teaching/ProbStat/20032004/PDF
On Wed, 3 Nov 2004, Marc Schwartz wrote:
On Wed, 2004-11-03 at 09:55, Dan Bolser wrote:
This has been asked before, but all the answers are hidiously complex.
The
legend.text=TRUE
option of barplot is almost exactly what I need, except I need a
legend.placement='tl'
(top left
On Wed, 3 Nov 2004, Gabor Grothendieck wrote:
Dan Bolser dmb at mrc-dunn.cam.ac.uk writes:
:
: This has been asked before, but all the answers are hidiously complex.
:
: The
:
: legend.text=TRUE
:
: option of barplot is almost exactly what I need, except I need a
:
: legend.placement='tl
Hello, supposing that I have two or three clear categories for my data,
lets say pet preferece across fish, cat, dog. Lets say most people rate
their preference as being mostly one of the categories.
I want to do pca on the data to see three 'groups' of people, one group
for fish, one for cat
.
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
The business of the statistician is to catalyze the scientific learning
process. - George E. P. Box
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dan Bolser
Sent
their is a way to code this, but I don't want to write code - at
least not code that I have to look at when what I want to see is my data.
From: Gabor Grothendieck [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: Re: [R] Legend placement in barplot?
Date: Wed, 3 Nov 2004 18:48:48 + (UTC)
Dan Bolser dmb
This has been asked before, but all the answers are hidiously complex.
The
legend.text=TRUE
option of barplot is almost exactly what I need, except I need a
legend.placement='tl'
(top left) option. This option would be in contrast to the default
placement which we could call 'tr' (top
On 13 Oct 2004, Jari Oksanen wrote:
On Wed, 2004-10-13 at 09:51, Prof Brian Ripley wrote:
On Wed, 13 Oct 2004, Dan Bolser wrote:
I have a complex distance matrix, and I am thinking about how to cluster
it and how to visualize the quality of the resulting clusters.
Using PCA and plotting
On Tue, 12 Oct 2004, Martin Maechler wrote:
Dan == Dan Bolser [EMAIL PROTECTED]
on Mon, 11 Oct 2004 16:21:53 +0100 (BST) writes:
Dan Gives strange results.
Dan I get 'weird' dendrograms with canberra / binary distance metric and
Dan median / centroid cluster methods
If I perform PCA on the 'eurodist' data, should I get an accurate
geographic layout of the cities with biplot?
(barring inversions, i.e. their is no way to define north.. but you get
the idea...)
I have a complex distance matrix, and I am thinking about how to cluster
it and how to visualize
I use the following code to scan a (limited) parameter space of clustering
strategies ...
data - read.table(...
dataTranspose - t(data)
distMeth - c(euclidean,
maximum,
manhattan,
canberra,
binary
)
clustMeth - c(ward,
Gives strange results.
I get 'weird' dendrograms with canberra / binary distance metric and
median / centroid cluster methods.
Is this just my data?
Dan
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PLEASE do read
I have the following contingency table
dat - matrix(c(1,506,13714,878702),nr=2)
And I want to test if their is an association between events
A:{a,not(a)} and B:{b,not(b)}
| b | not(b) |
+-++
a | 1 | 13714 |
+-++
not(a) | 506 |
Why can't I just use Log odds? Does the standard error of the logs score
depend on a similar chisq assumption?
On Sat, 9 Oct 2004, Dan Bolser wrote:
I have the following contingency table
dat - matrix(c(1,506,13714,878702),nr=2)
And I want to test if their is an association between events
On Wed, 6 Oct 2004, Gabor Grothendieck wrote:
Dan Bolser dmb at mrc-dunn.cam.ac.uk writes:
:
: Is there an R wiki?
:
: Looking at the huge amount of traffic on this list, I think wiki could be
: an exelet outlet for all the constructive enthusiasm here.
:
: I don't think it would be too hard
On Thu, 7 Oct 2004, Gabor Grothendieck wrote:
Tony Plate tplate at blackmesacapital.com writes:
:
: At Thursday 11:29 AM 10/7/2004, Dan Bolser wrote:
: [snip]
: I just added some pages... I think it would be great if people could get
: motivated to contribute to something like this. Its one
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