I think the fact that the grid package does not support cross-hatching is a
feature not a bug (or deficiency), and I hope that this is not "fixed".
Tufte's book (The Visual Display of Quantitative Information) has a section on
why cross-hatching should be avoided (unless of course your goal is
t: Monday, January 19, 2009 5:37 PM
> To: Greg Snow
> Cc: hadley wickham; R-help
> Subject: Re: [R] Bar Plot ggplot2 Filling bars with cross hatching
>
> what is your suggestion for distinguishing between many bars without
> color? I have grown up in the time of standarized te
I don't know how efficient you would consider this, but here is one solution:
library(TeachingDemos)
ms.2circ <- function(r=1, adj=0, col1='blue', col2='red', npts=180) {
tmp1 <- seq( 0, pi, length.out=npts+1) + adj
tmp2 <- seq(pi, 2*pi, length.out=npts+1) + adj
polygon(
Use either the 'clip' function from the graphics package, or the 'clipplot'
function from the TeachingDemos package.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-b
Here is one way:
tmpmat <- cbind( c(1,1), c(1,1), c(2,3) )
layout(tmpmat)
with(iris, plot(Sepal.Width, Sepal.Length,
col=c('red','green','blue')[Species]))
with(iris, plot(Sepal.Width, Sepal.Length, col=c('red','green','blue')[Species],
xlim=c(2.5,3), ylim=c(6,6.5)))
with
Read the help page for the quantile function (the whole page, there is a lot of
good detail in there), the 2nd reference on the page should also be a helpful
read.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Origin
The loess function in R uses the tricube weights that are described in my copy
of Cleveland, so that may do what you want. If you really want to do the same
general idea but with a different weight function, then it is not that hard to
write your own function to do the estimating (but I doubt t
I believe the original thread was about whether the function returns NA or
stops with an error when given an invalid date (such as Feb 29 in a non-leap
year). Your question was about how as.Date returned something different from
what you expected. Related, but different enough that it probably
I don't see right off why the one works and the other doesn't, but this looks
like one of those cases that would be better done using a list rather than
global variables.
Instead of assigning the variables in the global workspace, create a list and
assign them there. Then you can use lapply in
Comments interspersed below
From: Marie Sivertsen [mailto:mariesiv...@gmail.com]
Sent: Thursday, January 22, 2009 1:17 PM
To: Greg Snow
Cc: r-h...@stat.math.ethz.ch
Subject: Re: [R] Unexpected behaviour of the as.Date (was: Error as.Date on
Invalid Dates)
[snip]
For your question, the help
ps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r_lo...@web.de [mailto:r_lo...@web.de]
> Sent: Thursday, January 22, 2009 1:27 PM
> To: Greg Snow; r-help@R-project.org
> Subject
I like:
Applied Linear Statistical Models by Neter, Kutner, Nachtsheim, and Wasserman
(McGraw Hill)
It is not specific to any stats package, but it gives a good mix of theory
behind the routines and how to apply them and covers a good breadth of material.
A must have for statistics and R is:
Z ~ (A+B+C+D)^3 means give all main effects and interactions up to the 3 way
interactions, but not above (change ^3 to ^2 to limit to 2 way interactions).
You can do a semi manual stepwise procedure using the add1 and drop1 commands
or the addterm and dropterm commands in the MASS package. But
You may want to consider a dotchart instead of a barplot. Then you can
distinguish between groups by using symbols, grouping, and labels rather than
depending on colors/shades of grey.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.
This is a known issue, the documentation of clip talks about some plotting
functions resetting the clipping region and some don't. abline is apparently
one of those that plots first, then resets the clipping region (so the first
time it doesn't acknowledge it, but does afterwards). The functio
If you do ?Startup at the command line, you will get a help file that will give
you several options for doing this.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.or
Use cat instead of print, or better yet:
winProgressBar
tkProgressBar (tcltk package)
txtProgressBar
Also for simple date stamps you can just use the date() function.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Origi
If the time until change is exponentially distributed with a mean of 3, then
the probability of changing in the first day is:
> pexp(1,1/3)
[1] 0.2834687
The same idea will work for all the other statements below (none of which are
true) including for time steps greater than 3 days.
Hope this
We don't have your data, so we cannot reproduce what you are doing and the plot
was stripped off before we saw it (only certain types of attachments are
allowed, and some e-mail programs don't give the correct information about
attachments so even those types can be stripped if it is not clear w
Sigbert,
The plot2script function in the TeachingDemos package does essentially what
Duncan talks about for you. Create your plot then run the function giving it a
filename to save the info into (or run without arguments and then past into a
script window or text editor (only tested on windows
What you are describing is actually a permutation test rather than a bootstrap
(related concepts but with a subtle but important difference).
The way to do a permutation test with multiple x's is to fit the reduced model
(use all x's other than x1 if you want to test x1) on the original data and
; From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Thomas Mang
> Sent: Thursday, January 29, 2009 3:52 PM
> To: r-h...@stat.math.ethz.ch
> Subject: Re: [R] bootstrapping in regression
>
> Greg Snow wrote:
> > What you are describing
There has already been good discussion on this topic, but here are a couple of
other things to think about:
1. is it your job to convince your IT department, or is it your job to convince
your boss, and your boss's job to convince/dictate to the IT department
(getting your boss on your side cou
If you want to keep the functions, why not move them to a different environment
so that they don't get deleted when you delete everything else (this will also
work better if you want to use these same functions in other R sessions).
The most comprehensive way to do this is to create a package wi
If you are interested in rolling dice with R (as opposed to using this as a
simple test case to start writing your own programs), then you may want to look
at the "dice" function from the TeachingDemos package. The core line in this
function is basically the same as Gabor's suggestion, but with
?lattice::shingle
Hope that helps, if not, give more detail/example.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On B
It is not transposing (it just looks that way). The first result is a vector
which is 1 dimensional, but is neither a row or a column. The printed version
of it looks like a row, because that is a more compact representation. If you
sample enough points you will see it wrap around and be repr
Without knowing how your data is formatted or how you intend to link the
information together, all we can do is reconsctruct the plot from scratch.
Here is one way to do that (change whatever values you want to tweek the look):
plot( c(1,10), c(0,95), type='n', axes=F, xlab='', ylab='', xlim=c(
Let me guess, you are using MSwindows and using notpad as the editor.
Notepad has this annoying "feature" that if you save a file as a text file then
it automatically adds '.txt' to the end of the file name. So when you saved as
a text file it actually named the file "Rbatch.bat.txt" and the co
You can run a function from a package by doing something like:
> Rcmdr::reliability(cov(DavisThin))
This will load the package in the background, but not run the gui and other
things. So you can use the function(s) that you want without running
everything like when you do library(Rcmdr).
Hope
Look at the nws package, it has tools for passing data among multiple instances
of R (and waiting for data to be ready).
There are other packages that provide some of the same, but from what I
remember, nws was fairly simple to set up on a single computer.
Hope this helps,
--
Gregory (Greg) L
Try:
> b_tmp %in% a_tmp
You may also want to use the all or any function with the above.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailt
Googling for Denmark Shapefile leads to this website:
http://www.vdstech.com/map_data.htm
Where a few clicks lead to downloading a set of shapefiles (different levels of
subdividing the country, hopefully you know more about what they correspond to
than I do, about all I know about Denmark is fr
The runif function meets the stated criteria, if that is not good enough, then
give us more detail.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help
Tom,
Bill Venables gave you references to important tools for dealing with
for loops in R and they may be all the solution that you need. But here
is a little more detail on what is going on in case you want/need more
control in the future.
Note that the R for loop is what some programers call
Yes, here is one way:
> plot(1:10, pch=1:2)
> par(xpd=NA)
> tmp.u <- par('usr')
> legend(tmp.u[1], tmp.u[4], xjust=0, yjust=0, c('a','b'), pch=1:2)
>
You will probably want to increase the margins for a real case. See
?par and the 'xpd' entry for details on the clipping and ?legend for
more det
Type "?par" at the command line to bring up the help page, then scroll down and
read the sections on xpd and usr.
From: Edwin Sendjaja [mailto:[EMAIL PROTECTED]
Sent: Fri 4/11/2008 9:56 PM
To: Greg Snow; r-help@r-project.org
Subject: Re: [R] Legen
od enough? or do you need the bars to look much more like the barplot?
From: Anne-Katrin Link [mailto:[EMAIL PROTECTED]
Sent: Sat 4/12/2008 7:11 AM
To: Greg Snow
Cc: r-help@r-project.org
Subject: RE: [R] two graphs in one figure?
Dear Greg, dear all,
thank yo
I think that the problem is with the dev.off commands.
The first time running b there are not a device 2 and device 3 so
dev.off does nothing (or if they did exist then they are turned off).
Then you set the graphics event callback which attaches to the current
graph window (number 3 in general).
same capability, or Prof. Ripley
writes a better version to include in a core package).
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
From: Anne-Katrin
The "type" argument to plotting functions determines the type of plot
(plot lines or points or both or ...). To specify the type of line you
need to use the "lty" argument. See the help page for "par" (?par) for
details on the types of line you can use (also "lwd" is the width of the
lines).
B
Read the help page on '[', then read it again a couple more times (it
can be subtle). I think you want to use fp[[1]] rather than fp[1], the
first will return a single element from the list (a data frame in the
case below), the second returns a list of length 1 (with that element
being a data fram
When I first started using an early port of S, the main way to create
graphs was with the "printer" graphics device which created plots like
this (back then that was impressive). S-PLUS does still have the
printer graphics device to do things like this, but I don't think that
anyone has ever imple
Look at the hexbin package (bioconductor I think).
-Original Message-
From: "Georg Ehret" <[EMAIL PROTECTED]>
To: "r-help" <[EMAIL PROTECTED]>
Sent: 4/15/08 3:23 PM
Subject: [R] heavy graphs
Dear R community,I am creating large graphs with hundreds of
thousands of datapoints. My u
This is a case of you can't please everyone. A while back there was
some complaint that "Introduction to R" spent to much time on talking
about the different types of variables, just the opposite complaint of
yours.
There are several other sources of documentation (look under the books
link on th
You seem to be confusing several issues.
First, being object oriented does not mean that it matches C++ syntax or
that every object is guarenteed to have specific methods. What lead you
to think that a recordedplot would have a metafile method? Nothing else
has that method, and the only function
Look at the logspline package. It has a different way of estimating
densities that allows for limits to be specified (i.e. probability is 0
beyond the point(s) you specify).
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801)
It is not clear exactly what you want to do by subtracting a vector from a
matrix, but look at the "sweep" function, it may do what you want.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From
Can you show us the code you used for the 5 to 10 points? (either
generate some random data, or use a sample dataset).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PROTECTED]
> [
Read FAQ 7.22
--
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 qian z
> Sent: Wednesday, April 23, 2008 10:49 AM
> To: r-help@r-pr
First off there are multiple definitions of "best", you need to decide
which "best" is best for you.
Second, for reasonable definitions of "best", deciding between, linear,
polynomial, and ... Requires backgroud knowledge and real thought.
R can fit many different models and give you numerical an
You may want to read the help page for '[' and '[[' a few more times
(the differencese can be subtle).
The '[[' only returns a single element from a data structure (but in the
case of a list, that single element can be quite complex). So something
like list[[1:2]] does not make sense because you
You might want to look at the distr package (and its relatives). It
provides methods for calculating the distribution function of
combinations (the sum is one) of other distributions. I'm not sure how
you would convert your percentiles to a distribution function, but there
may be a way in the doc
So you want mylist[[1:2]] to return something (other than an error) when mylist
is a list.
What if mylist <- list( 1:10, 101:110 , some.other.things) so the first 2
elements are vectors of length 10. then mylist[1:2] makes sense as still being
a list with the 2 vectors. What should mylist[[1:
, but I think it will work better for you at this point.
From: Beck, Kenneth (STP) [mailto:[EMAIL PROTECTED]
Sent: Mon 4/28/2008 8:53 AM
To: Greg Snow; r-help@r-project.org
Subject: RE: [R] Use of recordPlot
Thanks for the clarification, this helps a lot
This is for the same reason that:
3
+4
Does not give 7. R is optimized for interactive use, so if a statement
can be considered complete, it evaluates it instead of waiting for more.
In your case:
if(mxx>mxy)
mxy=mxx
Is a complete statement and is evaluated without waiting to see if there
is
Does the following do what you want (or at least move in that
direction)?
u <- array(NA,c(5,8))
t <- seq(from=0.5, to=0.11,length=15)
t2 <- seq(from=(-0.7),to=(-0.1),length=25)
u[1:15] <- t
u[16:40] <- t2
v <- array(NA,c(5,8))
y <- seq(from=(-0.9), to=(-0.01),length=40)
v[1:40] <- y
library(Teac
In addition to what others have pointed to regaurding buffering and cat,
there are also a couple of new functions in R 2.7.0 that help with
showing the progress of computations. They are winProgressBar (windows
only) and TkProgressBar (in the tcltk package). You can use these to
create and update
Is this what you are trying to do?
> tt <- matrix( 5:10, ncol=2 )
> df <- as.data.frame.table(tt)
> df2 <- df[ rep( 1:nrow(df), df$Freq ), ]
And is that elegant enough?
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -
?symbols
--
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 Irene Mantzouni
> Sent: Friday, May 02, 2008 7:52 AM
> To: [EMAIL PROTEC
Try:
> tmp <- rep( list( 0:1 ), 5 )
> out <- do.call(expand.grid, tmp)
Then change the 5 to whatever you want.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PRO
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:
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,
The last example in ?fisher.test is not a 2x2 table, in fact it uses levels
with a natural ordering similar to the original question. Why would this not
be applicable to the situation?
From: [EMAIL PROTECTED] [EMAIL PROTECTED] On Behalf Of David Winsemiu
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Esmail Bonakdarian
> Sent: Sunday, May 11, 2008 7:25 AM
> To: Prof Brian Ripley
> Cc: [EMAIL PROTECTED]
> Subject: Re: [R] Random number generation
[snip]
> What I read doesn't seem to be incorrect ho
I would have thought that:
> lm( C1 ~ M^2, data=DF )
Would give the main effects and 2 way interaction(s) (but a quick test did not
match my expectation). Possibly a feature request is in order if people plan
to use this a lot.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermou
You probably want to look at "Sweave" in the "utils" package or the "odfWeave"
package.
Both let you set up a planned set of commands interspersed with text (notes,
explanations, full report, etc.) and then you process the file and get the
output (and commands) in either a LaTeX file or an Open
?sweep
-Original Message-
From: "Yukihiro Ishii" <[EMAIL PROTECTED]>
To: "r-help@r-project.org"
Sent: 5/13/08 8:30 AM
Subject: [R] A Very Simple Question
Hi Rusers!
I am ashed of asking such a simple question.
X<-matrix(rnorm(24), 4)
X0<-apply(X,2,mean)
What I want is a matrix which
> -Original Message-
> From: Esmail Bonakdarian [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, May 13, 2008 8:13 AM
> To: Greg Snow
> Cc: Prof Brian Ripley; [EMAIL PROTECTED]
> Subject: Re: [R] Random number generation
>
> Greg Snow wrote:
> >> -Original M
Using just base graphics you can use the floating.pie function from the plotrix
package, or the my.symbols function from the TeachingDemos package with your
own symbol/plotting function. Pie charts are usually not very useful, there
may be another type of symbol (see the symbols function) or pl
SDF3 <- SDF1/SDF2[ rep(1:12,12), ]
-Original Message-
From: "Bert Jacobs" <[EMAIL PROTECTED]>
To: "'Henrique Dallazuanna'" <[EMAIL PROTECTED]>
Cc: "r-help@r-project.org"
Sent: 5/14/08 7:41 AM
Subject: Re: [R] Dividing Two Dataframes
Hi Henrique,
I think I understand your formula, bu
Use logical subscripts rather than which:
> k <- c(1,1,1,2,2,1,1,1)
> k[ k != 1 ]
[1] 2 2
> k[ k != 2 ]
[1] 1 1 1 1 1 1
> k[ k != 3 ]
[1] 1 1 1 2 2 1 1 1
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -O
If the main goal is to include graphs created by R in LaTeX documents and have
them look nice, and using xfig was just one way of attempting this, then here
are a couple other options that may or may not work better.
Use the postscript graphics device and use psfrag in LaTeX to replace text in
The R2HTML package has tools for creating an HTML log of your session (see
?HTMLStart). Or the TeachingDemos package has a text based set of tools (see
?txtStart) for creating a log of your session. Neither is perfect (and
imperfect in different ways), but could be what you are looking for.
H
It is not clear what exactly you are trying to do, but you may want to look at
?matplot for an alternative to your loop, then look at ?legend for adding
explanations.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-81
21,
+ col=c('red','green','blue'))
>
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: Bruno C. [mailto:[EMAIL PROTECTED]
> Sent: Monday, May 19, 2008
Mathematicians like to have axes cross at 0, the general rule for statistics is
to have the axes positioned so that they help you understand the data, but
don't interfere with the actual points (or force too much whitespace by being
put to far away from the data), so the default positioning foll
t: Tuesday, May 20, 2008 10:37 AM
> To: Greg Snow; r-help@r-project.org
> Subject: Re: Alignment of axes intersection
>
>
> Agreed. The main reason I wanted the change in alignment was
> that I had three curves that were converging to a asymptote,
> and when I drew the horizontal
Here is one example:
> y <- rnorm(12)
> x <- 1:12
>
> par(mar=c(10, 4, 4, 1)+0.1)
> plot(x,y, xlab='', xaxt='n')
> axis(1, at=x, labels=month.name, las=2)
>
See help on par and axis for details.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(8
Here is an alternative approach to saving the itterations:
testfunc <- function(data) {
ms.mat <- matrix(nrow=0, ncol=2)
f1 <- function(ms, data){
ms.mat <<- rbind(ms.mat, ms)
-sum( dnorm(data, mean=ms[1], sd=ms[2], log=TRUE) )
}
ou
An alternative way to draw the symbols (or some approximation of them) is to
use the my.symbols function from the TeachingDemos package along with ms.male
and ms.female (or your improvement of these, also from the TeachingDemos
package). See the 4th example from ?ms.male (or ?ms.female).
Hope
That is actually from August 2007 (typo in the fortunes package).
If you do
> RSiteSearch('Thou Shalt Not Even Think Graph Spreadsheet')
Then it will bring up the actual post and you can read it in context (along
with the posts that lead to it and those that followed).
Basically, Ted was tryin
Mark,
Others have given answers to the question that you asked, in the spirit of
fortune(108) I am going to answer some of the questions that you should have
asked:
Q1: Should I ever do this?
Short answer: No
Less short answer: probably not
Longer answer: You should only do this once you fully
An easy and good way to make a bunch of variables is to store them in a list.
For example:
> mylist <- replicate(100, sample(1:100, 10), simplify=FALSE)
> id <- paste('id.',1:100, sep='')
> names(mylist) <- id
> sapply(mylist, median)
id.1 id.2 id.3 id.4 id.5 id.6 id.7 id.8 id.
For the first question, what do you want to do with all the subsets? There are
tools like split, by, lapply, lmList, etc. that make working with all the
subsets easy. If you tell us what your final goal is, we may be able to help
with a simple solution that does not need the intermediate "tabl
Does this do what you want?
Bill <- data.frame(A= c(1,1,1,2,3,4,5,6,6,7),
B=c(5,5,6,7,7,7,8,9,10,11),
group=rep(1:2, each=5) )
with(Bill, plot(A,B, pch=group) )
mns <- with(Bill, tapply(B, group, mean))
tmp <- par('usr') # or range(Bill$A)
med <- with(Bill, mean( c(max(A[group==
One simple way is to do something like:
> fit <- lm(y ~ I(x1-x2) + x3, data=mydata)
The first coeficient (after the intercept) will be the slope for x1, the slope
for x2 will be the negative of that. This model is nested in the fuller model
with x1 and x2 fit seperately and you can therefore t
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Patrick Burns
> Sent: Friday, June 06, 2008 12:04 PM
> To: Daniel Folkinshteyn
> Cc: r-help@r-project.org
> Subject: Re: [R] Improving data processing efficiency
>
> That is going to be situation depende
> -Original Message-
> From: Gabor Grothendieck [mailto:[EMAIL PROTECTED]
> Sent: Friday, June 06, 2008 12:33 PM
> To: Greg Snow
> Cc: Patrick Burns; Daniel Folkinshteyn; r-help@r-project.org
> Subject: Re: [R] Improving data processing efficiency
>
> On Fri, Jun
, June 06, 2008 12:58 PM
> To: Gabor Grothendieck
> Cc: Greg Snow; r-help@r-project.org
> Subject: Re: [R] Improving data processing efficiency
>
> My guess is that number 2 is closest to the mark.
> Typing too fast is unfortunately not one of my habitual attributes.
>
> Gabor Gr
Try this (you need tcl 8.5 and the TeachingDemos package):
library(teachingDemos)
tmpplot <- function(col1='red', col2='yellow', col3='green'){
plot(1:10,1:10, type='n')
rect(1,1,4,4, col=col1)
rect(1,4,4,7, col=col2)
rect(1,7,4,10, col=col3)
rect(6,1,9,4, col=col2grey(col1))
rect(6,4,9,7, c
You may also want to look at the "show.colors" function in the "DAAG" package
to get candidate colors.
From: [EMAIL PROTECTED] [EMAIL PROTECTED] On Behalf Of Michael Friendly [EMAIL
PROTECTED]
Sent: Friday, June 06, 2008 2:37 PM
To: R-Help
Subject: [R] co
Here is an example that may get you started:
point1 <- c(1.1, 1.7, 255)
point2 <- c(2.2, 1.5, 180)
point3 <- c(1.8, 2.2, 60)
mydf <- as.data.frame( rbind(point1, point2, point3) )
names(mydf) <- c('x1','x2','red')
fit <- lm(red~x1+x2, data=mydf)
df2 <- expand.grid( x1=seq(min(mydf$x1), max(mydf
here is one approach:
res <- cbind( c(10, 5, 1, 12, 3, 8, 7, 2, 10, 1),
c(90,15,79,38,7,92,13,78,40,9) )
line <- gl(5,1,length=10, labels=LETTERS[1:5])
qa <- gl(2,5)
fit <- glm( res ~ line*qa, family=binomial )
summary(fit)
anova(fit, test='Chisq')
The interaction terms measure the differe
9:16 PM
To: Greg Snow
Cc: R-help forum
Subject: Re: [R] Comparing two groups of proportions
Your approach tacitly assumes --- as did the poster's question --- that
the probability of passing an item by one method is *independent* of
whether it is passed by the other method. Which makes t
Look at the "image" function and the "levelplot" function in the lattice
package.
-Original Message-
From: "sumit gupta" <[EMAIL PROTECTED]>
To: "r-help@r-project.org"
Sent: 6/10/08 6:55 AM
Subject: [R] Hello
Hello,
I am facing a problem in drawing heat map using R.
I have a 70X3 mat
s are (error component).
>
> Thanks,
> Ivan
>
> > cheers,
> >
> > Rolf Turner
> >
> > On 10/06/2008, at 2:57 PM, Greg Snow wrote:
> > > here is one approach:
> > >
> > > res <- cbind( c(10, 5, 1, 12, 3,
?tapply
?aggregate
--
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 Daren Tan
> Sent: Tuesday, June 10, 2008 9:37 AM
> To: r-help@r-pr
Look at ?print.trellis in the lattice package (specifically the split and more
argumnets).
--
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 Beh
In your plot and lines function call you specify type="b" which says to use
"b"oth lines and points with gaps in the lines around the points. If you
change that to type="o" then it will "o"verplot the points and lines (no gaps).
The different types are documented in ?plot.default.
Hope this he
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