Try this:
plot(1:100, xaxt = "n")
axis(1, c(1, 20, 40, 60, 80, 100))
# and optionally add this third line
axis(1, 1:100, FALSE, tcl = -0.3)
See ?par (for the xaxt argument to plot) and ?axis . ?plot and
?plot.default have info on the plot command. A good source of
sample code for graphics is:
Try this. It captures the output of x and sets idx to the line numbers
of the coefficients, rearranging their order in the next line and printing
them out in the line after that.
my.print.summary.lm <- function(x, ...) {
out <- capture.output(x)
idx <- seq(grep("Intercept", out),
know of a format
> convention that gets it back to numeric (although conventions are
> documented for forcing to character). Comment?
>
> Regards,
>
> Tim.
>
>
> Gabor Grothendieck wrote:
> > If you use
> >
> > Sys.putenv(TZ = "GMT")
> >
> >
If you use
Sys.putenv(TZ = "GMT")
at the beginning of your session then local time zone and GMT time
zone will be the same so you should not have a problem. This was
not possible, at least on Windows, at the time the R News article
was written.
On 4/5/07, Tim Bergsma <[EMAIL PROTECTED]> wro
Try this:
x <- 1:100
xyplot(dnorm(x, 50, 10) + dnorm(x, 55, 12) + dnorm(x, 60, 15) ~ x, type = "l")
By the way, you can save classic graphics, at least on windows, like this:
### your code
parms <- data.frame(ID=c(1,2,3),mu=c(50,55,60),sigma=c(10,12,15))
curve(dnorm(x,mean=parms$mu[1],sd=parms$s
Check out:
http://tolstoy.newcastle.edu.au/R/help/06/02/21634.html
On 4/5/07, Robert McFadden <[EMAIL PROTECTED]> wrote:
> Hi R Users,
>
> In Windows I can clear console using CTRL-L, but can I do this by certain
> command in my programs? e.g
>
> for (i in 1:10){
>
> something
>
> clear conso
What version of R are you using? When I copy and paste that into
R 2.4.1 on Windows XP I get 2001, 2002, ..., 2007
On 4/3/07, Henrik Andersson <[EMAIL PROTECTED]> wrote:
> Hello fellow R people,
>
> I don't understand the default behavior of the axis labeling when
> plotting dates.
>
> I would ex
e: Run Length Endcoding
>
> dat$IDorder1 <- unlist(sapply(y$lengths, FUN=function(x) seq(1,x)))
>
> dat
>
>
>
> # 2st (by Chaliraos Skiadas)
>
> dat$IDorder2 <- unlist(tapply(dat$ID,factor(dat$ID), function(x)
>
> 1:length(x)),use.names=FALSE)
>
>
Assuming the ID's are contiguous, as in your example:
transform(dat, IDorder = seq(ID) - match(ID, ID) + 1)
On 4/1/07, Nguyen Dinh Nguyen <[EMAIL PROTECTED]> wrote:
> Dear R helpers
> I have a data set sth like this:
> set.seed(123);dat <- data.frame(ID= c(rep(1,2),rep(2,3), rep(3,3), rep(4,4
See gpar in grid:
library(grid)
?gpar
library(lattice)
xyplot(rivers ~ rivers, type = "l", lwd = 10, lineend = 1)
On 4/1/07, John Bullock <[EMAIL PROTECTED]> wrote:
>
> I would like to use llines() to draw lines with
> square endings in lattice plots. But the default
> behavior seems to be to d
On 4/1/07, Laurent Valdes <[EMAIL PROTECTED]> wrote:
> Hi everybody,
>
> really sorry !!!
> this is RODBC indeed. But I'm not sure if the package or R version makes a
> real difference.
> In fact, I used fast=FALSE, and I'm now pretty sure the error is due to that
> parameter.
> It may be working f
Try this:
aggregate(a[3], a[1:2], max)
On 3/31/07, Deepak Manohar <[EMAIL PROTECTED]> wrote:
> Hi team,
> I have the data of the form:
>
> > a<- data.frame(x=c(1,2,1,4,3), y=c(1,2,1,4,3), z=c(1,2,3,4,5))
>
> I need the output of the form
>
> > b<- data.frame(x=c(1,2,3,4), y=c(1,2,3,4), z=(3,2,
= Plot 2 ===
> # system TZ is in "Pacific Daylight Time"
> # POSIXct object is in "GMT"
> dp <- seq(Sys.time(),len=10,by="day")
> plot(as.POSIXct(format(dp),"GMT"),1:length(dp))
> # Shifted by 7 hours
>
> #=== Plot 3 ===
> #
On 3/31/07, Martin Maechler <[EMAIL PROTECTED]> wrote:
> >>>>> "SteT" == Stephen Tucker <[EMAIL PROTECTED]>
> >>>>> on Fri, 30 Mar 2007 18:41:39 -0700 (PDT) writes:
>
> [..]
>
>SteT> For dates, I usually store th
Try this:
'my dog named "Spot" and my cat named "Kitty" fight like cats and dogs'
On 3/31/07, Laurent Valdes <[EMAIL PROTECTED]> wrote:
> Hi everybody,
>
> I'm doing a sqlSave() in R, to insert a big data frame of 1 rows.
> However, there is problems, since several rows contains quotations ma
ot;, NA, "d", "e")),
stringsAsFactors = FALSE)
DF$a <- factor(DF$a)
DF$c <- factor(DF$c)
DF <- na.roughfix(DF)
DF$c <- as.character(DF$c)
DF
On 3/30/07, Sergio Della Franca <[EMAIL PROTECTED]> wrote:
> This is that i obtained.
>
> There isn't
I assume you are referring to na.roughfix in randomForest. I don't think it
works for logical vectors or for factors outside of data frames:
> library(randomForest)
> DF <- data.frame(a = c(T, F, T, NA, T), b = c(1:3, NA, 5))
> na.roughfix(DF)
Error in na.roughfix.data.frame(DF) : na.roughfix onl
The unit of time for a "ts" class object is deltat(ldeaths).
See the
?deltat
help page.
On 3/29/07, tom soyer <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am using ccf but I could not figure out how to calculate the actual lag in
> number of periods from the returned results. The documentation for cc
Please read the last line on every post to r-help.
On 3/28/07, Alfonso Sammassimo <[EMAIL PROTECTED]> wrote:
> Is there a way of aggregating 'zoo' daily data according to day of week? eg
> all Thursdays
>
> I came across the 'nextfri' function in the documentation but am unsure how
> to change thi
Try this:
library(gsubfn)
pat <- "([[:upper:]][[:lower:]]*) "
s <- "Aaaa 3 x 0 Bbbb"
strapply(s, pat, backref =-1)[[1]]
or (not quite as general but works in this case):
pat <- "([[:upper:]][[:lower:]]*) "
s <- "Aaaa 3 x 0 Bbbb"
s.idx <- gregexpr(pat, s)[[1]]
substring(s, s.idx, s.idx + attr(s
I am not 100% sure of what you want but maybe this will help. I
have modified your example data to omit one point so that there
is a holiday among the rows:
library(zoo)
# read data
Lines <- "DateOpenHigh Low CloseNifty
2004-01-01 1880.35 1917.05 1880.351912.25
2004-01-02 1912.25
The solution in my post has the advantage of not using
eval or character conversions (except in setting up
L where you want such character conversions to
build up the names, as your more general sitution
shows). The following is the same as in that post except
the line setting L in that post is re
As indicated in ?boxplot it passes certain arguments to bxp so check out
?bxp where you will find various out... arguments:
boxplot(c(1:10, 20), outlty = 2, outcex = 0 )
On 3/27/07, AA <[EMAIL PROTECTED]> wrote:
> Dear Users
>
> Is there any way to generate lines instead of points for outliers i
Try substitute:
e <- expression(u1 + u2 + u3)
L <- list(u2 = as.name("x"), u3 = 1)
as.expression(do.call(substitute, list(as.call(e), L))[[1]])
On 3/27/07, Daniel Berg <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> Suppose I have a very long expression e. Lets assume, for simplicity, that it
> is
>
On 3/27/07, Charles Dupont <[EMAIL PROTECTED]> wrote:
> What are the preferred date, and data/time classes for R?
See the help desk article in R News 4/1.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do re
You could use RDCOMClient or rcom packages to update an Excel
spreadsheet in place and you would not need any VBA at all. Search
through the archives for the keyword Excel.Application .
On 3/27/07, Moshe Olshansky <[EMAIL PROTECTED]> wrote:
> OK.
>
> By the way, I only thought that I could do wh
Try these:
search()
sessionInfo()
loadedNamespaces()
The first two show the attached packages (and some other info) and the
last one shows the namespaces that are loaded. Note that detaching
a package does not unload its namespace and unloading a namespace
does not de-register its methods. This
Try adding this argument to your xyplot call:
par.settings = list(axis.line = list(col = 0))
The subparameters oif axis.line are:
trellis.par.get()$axis.line
in case you want to temporarily set others.
On 3/25/07, John Bullock <[EMAIL PROTECTED]> wrote:
>
> I am trying to eliminate panel
Try this:
aggregate(atest[3:4], atest[1:2], sum)
Use a data base and SQL is you don't otherwise have enough
computer resources.
On 3/24/07, Delcour Libertus <[EMAIL PROTECTED]> wrote:
> Hello!
>
> Given is an Excel-Sheet with actually 11,000 rows and 9 columns. I want
> to work with the data in
Read the help desk article in R-News 4/1 and see
?as.Date
?strptime (for setting the as.Date format= argument)
Also, you might be interested in the zoo package
library(zoo)
?read.zoo
vignette("zoo")
vignette("zoo-quickref")
On 3/23/07, Andreas Tille <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm on
path.expand("~") also seems to work on Windows XP.
On 3/23/07, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
> But the request was for a *generic* solution. On Windows there
> might not be anything corresponding to a home directory (and the rw-FAQ
> discusses the concept and how R resolves this).
On 3/23/07, Liaw, Andy <[EMAIL PROTECTED]> wrote:
> From: Gabor Grothendieck
> >
> > See:
> >
> > ?R.home
>
> That's not what Alberto wanted: It gives the location of the R
> installation, not where user's home directory is. AFAIK Windows doe
See:
?R.home
?dput
On 3/23/07, Alberto Monteiro <[EMAIL PROTECTED]> wrote:
> Is there any generic function that gets the "home" directory? This
> should return /home/ in Linux and
> x:/Documents and Settings/ (or whatever) in Windows XP.
>
> Another (unrelated) question: what is the _simplest_ wa
Looks like there is code in the appendix.
On 3/23/07, Jan Wijffels <[EMAIL PROTECTED]> wrote:
> Dear useRs,
> I very much like the effect display of the proportional odds model on
> page 29 (Figure 8) of the following paper by John Fox:
> http://socserv.mcmaster.ca/jfox/Papers/logit-effect-display
Try pt.lwd= instead of lwd= in your legend call.
On 3/22/07, Robin Hankin <[EMAIL PROTECTED]> wrote:
> Hi
>
> I have a scatterplot of points with pch=1 and a single point with
> pch=3, lwd=3.
> It has a high line width to attract attention to it.
>
> The following script
>
>
>
> plot(rnorm(10),rno
Try
sprintf("%.2f", m)
On 3/22/07, Baoqiang Cao <[EMAIL PROTECTED]> wrote:
> Dear All,
>
> I was trying to format a numeric vector (100*1) by using
> outd <- format(x=m, sci=F, digits=2)
>
> > outd[1:10]
> [1] " 0.01787758" "-0.14760306" "-0.45806041" "-0.67858525" "-0.64591748"
> [6] "-0.0591
Package dynlm (and dyn) are used to align the time series in the dependent
and independent portions of the equations so that one can perform regressions
on lagged and differenced versions of the dependent and independent variables.
They compensate for the fact that lm (and in the case of dyn lm, gl
Sometimes but its also easy to forget about simple solutions.
On 3/21/07, Alberto Monteiro <[EMAIL PROTECTED]> wrote:
> Gabor Grothendieck wrote:
> >
> > Could you call yours Edges?
> >
> Is it a good idea to have two different functions, whose name
> differs by
Could you call yours Edges?
On 3/20/07, Søren Højsgaard <[EMAIL PROTECTED]> wrote:
> I am writing a package which uses the Rgraphviz package which in turn uses
> the graph package, but my question does not (I believe) pertain specifically
> to the these packages so therefore I dare to post the q
On 3/20/07, enrico.foscolo <[EMAIL PROTECTED]> wrote:
> Dear participants to the list,
>
> this is my problem: I want to obtain an expression that represents the second
> derivative of one function.
> With "deriv3" (package "stats") it is possible to evaluate the second
> derivative, but I do not k
Along the lines you mention, check out:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/32297.html
On 3/19/07, Dalphin, Mark <[EMAIL PROTECTED]> wrote:
> On Mon, 19 Mar 2007, Thomas Lumley wrote:
> >> On 3/19/07, Thomas Lumley <[EMAIL PROTECTED]> wrote:
> >>> On Thu, 15 Mar 2007, Andrew Perrin wr
I think there are a lot of misconceptions regarding what is possible on Windows.
On Windows you just right click the graphic in R and choose Copy as Metafile
which puts a vector graphic representation on the clipboard and then
ctrl-V in Word to copy it in. (Alternately save all your graphics as f
On 3/19/07, Thomas Lumley <[EMAIL PROTECTED]> wrote:
> On Mon, 19 Mar 2007, Gabor Grothendieck wrote:
>
> > On 3/19/07, Thomas Lumley <[EMAIL PROTECTED]> wrote:
> >> On Thu, 15 Mar 2007, Andrew Perrin wrote: (in part)
> >>>
> >>> 2.) Yes,
On 3/19/07, Marc Schwartz <[EMAIL PROTECTED]> wrote:
> On Mon, 2007-03-19 at 11:43 -0400, Gabor Grothendieck wrote:
> > On 3/19/07, Thomas Lumley <[EMAIL PROTECTED]> wrote:
> > > On Thu, 15 Mar 2007, Andrew Perrin wrote: (in part)
> > > >
> > > &
On 3/19/07, Thomas Lumley <[EMAIL PROTECTED]> wrote:
> On Thu, 15 Mar 2007, Andrew Perrin wrote: (in part)
> >
> > 2.) Yes, by all means you should use linux instead of windows. The
> > graphics output is completely compatible with whatever applications you
> > want to paste them into on Windows.
>
Here is one way. This matches strings which contain those characters
found in a number, converting each such string to numeric.
library(gsubfn)
strapply(x, "[-0-9+.E]+", as.numeric)
On 3/19/07, Robin Hankin <[EMAIL PROTECTED]> wrote:
> Hi.
>
> Is there a straightforward way to convert a charact
You could do this:
f <- function(x) {
"This is some
text that I
would like to
have here."
x + 1
}
f(2) # 3
On 3/18/07, Wensui Liu <[EMAIL PROTECTED]> wrote:
> Dear Lister,
> I understand I can put '#' to put comment line by line. But is there a
> way to put comment with multiple lines without ha
ng needed to
> be within quotes. What is happening here, exactly? Why the use of
> "~"? I tried without and it no longer works.
>
> Thanks in advance,
>
> Denis
> Le 07-03-18 à 08:59, Gabor Grothendieck a écrit :
>
> > Sorry, legend= was omitted:
> >
Sorry, legend= was omitted:
plot(1:10)
legend("topleft", legend = This ~ study ~ italic(n) == 3293)
On 3/18/07, Chabot Denis <[EMAIL PROTECTED]> wrote:
> Thank you Marc, Jim and Gabor,
>
> I like the solution with "expression", nice and simple. Gabor, your
> solution did not work, probably just a
Try this:
plot(1:10)
legend("topleft", This ~ study ~ italic(n) == 3293)
On 3/17/07, Chabot Denis <[EMAIL PROTECTED]> wrote:
> Hi,
>
> As part of the legend to a plot, I need to have the "n" in italics
> because it is a requirement of the journal I aim to publish in:
> "This study, n = 3293"
>
>
Try this (or use xyplot.zoo and write a panel function for that):
library(zoo)
set.seed(1)
tt <- as.Date(paste(2004, rep(1:2, 5), sample(28, 10), sep = "-"))
foo <- zoo(matrix(rnorm(100), 10), tt)
pnl <- function(x, y, ...) {
lines(x, y, ...)
abline(h = mean(y))
}
plot(foo, panel
On 3/16/07, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> 1. Here is your example redone using proto:
>
> library(proto)
>
> parent <- proto()
> child <- proto(a = 1)
> parent$child1 <- child
> child$parent.env <- parent
This last line should have been:
ode object type with a
> default behavior for the "<-" operator of its member variables being
> referencing rather than copying? Any good reference material/ similar
> code examples?
>
>Thanks.
>
> Gabor Grothendieck wrote:
> > Lists are not good for this. Th
it via references as requested then lists
are not appropriate.
On 3/16/07, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Lists are not good for this. There is an example in section 3.3 of
> the proto vignette of using proto objects for this. That section
> also references an S4 e
Lists are not good for this. There is an example in section 3.3 of
the proto vignette of using proto objects for this. That section
also references an S4 example although its pretty messy with S4.
You might want to look at the graph, RBGL and graphviz packages
in Bioconductor and the dynamicgrap
Regarding time series manipulation in R you could read the
two vignettes that come with the zoo package:
library(zoo)
vignette("zoo")
vignette("zoo-quickref")
Note that zoo is an entirely different system than rmetrics and timeDate so
if your question is specifically aimed at timeDate this does n
Such a calculation would be dominated by the time spent inside a call
to an offf-the-shelf C matrix inversion library used by R and is not really
any test of R itself.
On 3/9/07, Richard Morey <[EMAIL PROTECTED]> wrote:
> My adviser has a Mac notebook that he bought 6 months ago, and I have a
> PC
Read the FAQ 7.21:
http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f
Note that it points out that you don't really want to do this anyways.
On 3/9/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> Hi All
>
> I am pretty new to R but saw stata and sas's
On 3/9/07, Maciej Radziejewski <[EMAIL PROTECTED]> wrote:
> Hello,
>
> I do some computations on datasets that come from climate models. These data
> are huge arrays, significantly larger than typically available RAM, so they
> have to be accessed row-by-row, or rather slice-by slice, depending on
Try this:
library(gsubfn)
x <- "CB01_0171_03-27-2002-(Sample 26609)-(126)"
unlist(strapply(x, "..-..-"))
The gsubfn home page is at:
http://code.google.com/p/gsubfn/
On 3/9/07, Shawn Way <[EMAIL PROTECTED]> wrote:
> I have a set of character strings like below:
>
> > data3[1]
> [1] "CB01_01
Try this:
C1 <- 1:3
C2 <- 4:6
sapply(ls(pattern = "C"), get)
On 3/8/07, BOISSON, Pascal <[EMAIL PROTECTED]> wrote:
> Dear all,
>
>
>
> It seems to be a recurrent problem to me and I am asking your help to
> get over it once for all ...
>
>
>
> My idea is :
>
>
>
> I have variables C_1, C_2, C_3 .
Read R News 4/1 help desk article about dates.
Also see ?as.yearmon in the zoo package.
> library(zoo)
> dd <- Sys.Date() + seq(1, 1000, 100)
> dd
[1] "2007-03-09" "2007-06-17" "2007-09-25" "2008-01-03" "2008-04-12"
[6] "2008-07-21" "2008-10-29" "2009-02-06" "2009-05-17" "2009-08-25"
> ym <- as
Sorry. An extra line got in there. It should have been:
library(gsubfn)
gsubfn("(?\\d+)/(?\\d+)/",
month + day ~ sprintf("%s/%s/", day, month),
backref = -2, British.dates, perl = TRUE)
On 3/8/07, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> This does n
This does not really answer the specific question you posted but
gsubfn can do it without using \\1 and \\2 like this (you probably
realized that already but I thought I would point it out just in case):
library(gsubfn)
gsubfn("(?\\d+)/(?\\d+)/",
month + day ~ sprintf("%s/%s/", day, month),
Decompose your code into small understandable functions.
On 3/7/07, Aimin Yan <[EMAIL PROTECTED]> wrote:
> Dear R list,
> I have a question in R, it could be very simple, but I don't know how to do
> it?
>
> for example:
> I assign 6 to x in beginning of of my R script code
> > x<-6
> ..
> A
Try na.locf from the zoo package and then use merge with specified suffixes:
library(zoo)
f <- function(x) {
rownames(x) <- NULL
merge(x, na.locf(x[-1], na.rm = FALSE), by = 0, suffixes = c("", ".by"))[-1]
}
do.call("rbind", by(x, x$id, f))
On 3/7/07, Jon Olav Vik <[EMAIL PROTECTED]> wrote
If your problem is small enough just use a grid of starting values
and run your optimization on each one and then take the best.
On 3/6/07, Dae-Jin Lee <[EMAIL PROTECTED]> wrote:
> Hi all !
>
> I've been trying to maximize a likelihood using optim( ) function, but it
> seems that the function has
Hi, Your response to my post was EXTREMELY useful.
For georegistering I used the site:
http://www.runningmap.com
which lets one click on a a Yahoo! map and get lat/long in lower right.
As my data is in UTM I converted the lat/long to UTM using:
http://home.hiwaay.net/~taylorc/toolbox/geog
Is there any R software that create an image from Yahoo maps together
with points of known UTM coordinates (or lat/long marked? Note that
my region of interest is not covered in sufficient detail by Google maps.
It actually does not have to be Yahoo maps as long as it has sufficient
coverage of my
On 3/2/07, Ted Harding <[EMAIL PROTECTED]> wrote:
> I think the first developments which could be recognised as
> "statistical cmputing" (as opposed to using computers to do
> statistics) were the pioneering GENSTAT and GLIM (1973-4,
> though developed over some years previously). Possibly
> what
Change the name of n to something else. Also please using spacing
in your code for readability.
> x <- c(10.2, 7.7, 5.1, 3.8, 2.6)
> y <- c(9, 8, 3, 2, 1)
> n. <- c(10, 9, 6, 8, 10)
>
>
> derfs4 <- function(b, x, y, n.)
+ {
+ b0 <- b[1]
+ b1 <- b[2]
+ c=b0 +b1 * x
+
See the aspect argument, asp, in ?plot.default . Also eqscplot in MASS.
On 3/2/07, Thomas Steiner <[EMAIL PROTECTED]> wrote:
> I want to plot something (eg a circle) with a fixed ratio of the x and
> y axis, or (even better) with a fixed size when I print it. Output
> should then be a circle (act
Check out:
https://stat.ethz.ch/pipermail/r-help/2006-October/114431.html
On 3/2/07, Ido M. Tamir <[EMAIL PROTECTED]> wrote:
> Charilaos Skiadas wrote:
>
> > On Mar 2, 2007, at 9:42 AM, Ido M. Tamir wrote:
> >>
> >> But how do I get from a function to its name?
> >
> > Can you do this with any ob
Please read the last line of every message to r-help and follow it when
posting. You ought to have provided small examples for A and B and
the result.
The following lists every row in z that is at a date in zs or is up to
4 rows later:
library(zoo)
z <- zoo(matrix(101:148, 24), 2001:2024)
zs <-
Suggest you review the help desk article on dates in
R News 4/1. It mentions that POSIXlt objects are
lists with 9 components and other facts about date
objects in R.
On 3/2/07, Sérgio Nunes <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm having a weird result with the length() function:
>
> >a
> [...
There is an example in the example section of plotting two time series
on the same plot with different left hand and right hand scales here:
library(zoo)
?plot.zoo
On 3/2/07, Berta <[EMAIL PROTECTED]> wrote:
>
> Hi R-Users,
>
> I am trying to plot two time series in the same plot, but they measur
On Windows XP it worked for me on both 2.4.1 and 2.5.0. I did notice
that on 2.4.1 it says "using Synchronous WinInet calls" but does not
say this on 2.5.0. See below for the two transcripts.
> ftp <- "ftp://anonymous:[EMAIL PROTECTED]/edgar/full-index/company.idx"
> download.file(url=ftp, destf
On 3/1/07, Bart Joosen <[EMAIL PROTECTED]> wrote:
> Dear All,
>
> thanks for the replies, Jim Holtman has given a solution which fits my
> needs, but Gabor Grothendieck did the same thing,
> but it looks like the coding will allow faster processing (should check this
&g
Read in the data using readLines, extract out
all desired lines (namely those containing only
numbers, dots and spaces or those with the
word Time) and remove Retention from all
lines so that all remaining lines have two
fields. Now that we have desired lines
and all lines have two fields read the
You can evaluate it, differentiate it, pick apart its components,
use it as a title or legend in a plot, use it as a function body
and probably other things too:
e <- expression(x+y)
eval(e, list(x = 1, y = 2)) # 3
D(e, "x")
e[[1]][[1]] # +
e[[1]][[2]] # x
e[[1]][[3]] # y
plot(1:10, main = e)
Try rq in quantreg using the default value for tau.
On 2/28/07, lalitha viswanath <[EMAIL PROTECTED]> wrote:
> Hi
> I am looking for suitable packages in R that do
> regression analyses using least median squares method
> (or better). Additionally, I am also looking for
> packages that implement a
The FAQ does mention your point already.
On 2/27/07, Greg Snow <[EMAIL PROTECTED]> wrote:
> Others have pointed you to the answer to your question, but both FAQ
> 7.21 and the assign help page should really have a big banner at the top
> saying "Here Be Dragons".
>
> Using a loop or other automate
Try this:
plot(cbind(observed = z$random +
z$trend * z$seasonal, trend = z$trend, seasonal = z$seasonal,
random = z$random), main = "My title")
Change the * to + if your setup is additive.
On 2/27/07, Alberto Monteiro <[EMAIL PROTECTED]> wrote:
> Is there any way to give a "decent" t
You can use par("usr") to get the min/max coords of the plot and then
use that. For example, this will plot a red dot in the middle of the
plot area regardless of the coordinates:
plot(1:10) # sample plot
usr <- par("usr")
points(mean(usr[1:2]), mean(usr[3:4]), pch = 20, col = "red") # red dot
On 2/26/07, Thaden, John J <[EMAIL PROTECTED]> wrote:
> I had written
>
> >> ...someattributes() does not seem to exist.
>
> And Haris Skiadas replied
>
> > My help shows it as "moreattributes", not
> > "someattributes". (MacOSX), though doesn't
> > sound like it should be platform-specific).
>
> T
Its a FAQ.
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f
On 2/25/07, Monika Kerekes <[EMAIL PROTECTED]> wrote:
> Dear members,
>
>
>
> I have started to work with R recently and there is one thing which I could
> not solve so far. I don't know how to de
There is a fourth possibility too:
4. `scale colour`
I guess my preference is #1, #2, #3 and #4 in that order with #1 best. Even
though #1 can conflict with S3 it usually does not and its historically what R
used so I usually just stay consistent with historical precedent. #2 is
otherwise best
Try this:
plot(1)
mtext(quote(A ~ bold(bold) ~ word), cex = 1.3)
On 2/24/07, Joseph Retzer <[EMAIL PROTECTED]> wrote:
> Hi everyone,
> I suspect this is an easy task however I've not been able to accomplish it.
> I'd like to create an mtext title which has certain words bold, the rest not
> bo
parser. So I would double-check
> to make sure this doesn't happen.
>
> Do you have any control on where those XML files are generated
> though? It sounds to me it might be easier to fix the utility
> generating those XML files, since it clearly is doing something wrong.
>
&g
I assume is known.
This removes any occurrence .* where .* does not
contain or .
The regular expression, re, matches , then does a greedy
match (?U) for anything followed by but uses a zero
width lookahead subexpression (?=...) for the second
so that it it can be rematched again. gsubfn in p
Try this:
tab <- crossprod(as.matrix(ratings[,-1]))
tab <- tab - diag(diag(tab))
tab
tab / nrow(ratings)
On 2/22/07, Michael Wexler <[EMAIL PROTECTED]> wrote:
> Using R version 2.4.1 (2006-12-18) on Windows, I have a dataset which
> resembles this:
>
> idatt1att2att3
> 11
Its not clear from your post what the framework is that you are working with
but assuming that you have sourced a file and want its name place this
fn <- parent.frame(2)$ofile
# other code
in a file a.R, say, and from within R source it:
source("a.R")
This is not very safe and could eas
Try this:
DF[c(FALSE, tail(DF$Open, -1) > head(DF$High, -1)), ]
or using zoo objects just compare the Open to the reverse lag of the High.
Lines <- "Date Open HighLowClose
1/15/2000 10 11 8 10
1/16/2000 12 12 10 11
1/17/2000 12 12 10
If you call splom with do.call then your solution should work:
do.call("splom", list(~data[cols], groups = as.symbol(groups), data = data,
panel = panel.superpose, col = colors))
or use as.name where as.symbol is which also works.
On 2/14/07, Roberto Perdisci <[EMAIL PROTECTED]> wrote:
>
vcov can simplify it slightly:
se <- sqrt( t(a) %*% vcov(m) %*% a )
( a %*% beta ) / se
On 2/14/07, Abhijit Dasgupta <[EMAIL PROTECTED]> wrote:
> look at the function "estimable" in the package gmodels. This does the
> t.stat that you mention below.
>
> Otherwise, if you wanted to do this by han
ace(DFs[[1]], TRUE, medians)
On 2/13/07, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Suppose our data frames are called DF1, DF2 and DF3. Then
> find the least number of rows, n, among them. Create a
> list, DFs, of the last n rows of the data frames and another
> list, mats,
Suppose our data frames are called DF1, DF2 and DF3. Then
find the least number of rows, n, among them. Create a
list, DFs, of the last n rows of the data frames and another
list, mats, which is the same but in which each component is a
matrix. Create a parallel median function, pmedian, analogo
Also try gplot in the sna package to see if it does what you want.
Here are some examples from the r-help archives:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/87003.html
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/84442.html
On 2/13/07, Jarrett Byrnes <[EMAIL PROTECTED]> wrote:
> Hey,
Actually this simpler replacement for the b <- ... line would work just as well:
fo <- as.formula(sprintf("y ~ s(x0) + s(x1) + ns(%s, 3)", names(Mydata)[i]))
b <- gam(fo, data = Mydata)
On 2/13/07, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> The call to library(
The call to library(splines) is missing and also try replacing the
line b <- ... with
fo <- as.formula(sprintf("y ~ s(x0) + s(x1) + ns(%s, 3)", names(Mydata)[i]))
b <- do.call("gam", list(fo, data = Mydata))
to dynamically recreate the formula on each iteration of the loop
with the correct name
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