Re: [R] Trying to write a linear regression using MLE and optim()
On 5/30/05, Ajay Shah <[EMAIL PROTECTED]> wrote:
>
> I wrote this:
>
> # Setup problem
> x <- runif(100)
> y <- 2 + 3*x + rnorm(100)
> X <- cbind(1, x)
>
> # True OLS --
> lm(y ~ x)
>
> # OLS likelihood function --
> ols.lf <- function(theta, K, y, X) {
> beta <- theta[1:K]
> sigma <- exp(theta[K+1])
> e <- (y - X%*%beta)/sigma
> logl <- sum(log(dnorm(e)))
> return(logl)
> }
>
> optim(c(2,3,0), ols.lf, gr=NULL,
> method="BFGS", lower=-Inf, upper=Inf,
> control=list(trace=2, fnscale=-1),
> # Now for the "..." stuff
> K, y, X)
The last line should be:
K=K, y=y, X=X)
and also you have to set K.
>
>
> I get:
>
> Error in fn(par, ...) : argument "X" is missing, with no default
> In addition: Warning message:
> numerical expression has 100 elements: only the first used in: 1:K
> Execution halted
>
> If someone can show me the way, it'll be most appreciated. :-)
>
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[R] Installing RGL on SuSE 9.3
I am running R 2.10 on SuSE 9.3. When I tried to install the package "rgl", I got error message: ... In file included from pixmap.cpp:13: pngpixmap.h: In static member function `static void PNGPixmapFormat::Load::info_callback(png_struct*, png_info*)': pngpixmap.h:149: error: invalid conversion from `long unsigned int*' to ` png_uint_32*' pngpixmap.h:149: error: invalid conversion from `long unsigned int*' to ` png_uint_32*' make: *** [pixmap.o] Error 1 ERROR: compilation failed for package 'rgl' ** Removing '/usr/local/lib/R/library/rgl' ** Restoring previous '/usr/local/lib/R/library/rgl' ... Any ideas why? Shige __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Trying to write a linear regression using MLE and optim()
I wrote this:
# Setup problem
x <- runif(100)
y <- 2 + 3*x + rnorm(100)
X <- cbind(1, x)
# True OLS --
lm(y ~ x)
# OLS likelihood function --
ols.lf <- function(theta, K, y, X) {
beta <- theta[1:K]
sigma <- exp(theta[K+1])
e <- (y - X%*%beta)/sigma
logl <- sum(log(dnorm(e)))
return(logl)
}
optim(c(2,3,0), ols.lf, gr=NULL,
method="BFGS", lower=-Inf, upper=Inf,
control=list(trace=2, fnscale=-1),
# Now for the "..." stuff
K, y, X)
I get:
Error in fn(par, ...) : argument "X" is missing, with no default
In addition: Warning message:
numerical expression has 100 elements: only the first used in: 1:K
Execution halted
If someone can show me the way, it'll be most appreciated. :-)
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Re: [R] Formatting numbers with a limited amount of digits consistently
On 5/30/05, Duncan Murdoch <[EMAIL PROTECTED]> wrote: > Gabor Grothendieck wrote: > > On 5/30/05, Duncan Murdoch <[EMAIL PROTECTED]> wrote: > > > >>Henrik Andersson wrote: > >> > >>>I have tried to get signif, round and format to display numbers like > >>>these consistently in a table, using e.g. signif(x,digits=3) > >>> > >>>17.01 > >>>18.15 > >>> > >>>I want > >>> > >>>17.0 > >>>18.2 > >>> > >>>Not > >>> > >>>17 > >>>18.2 > >>> > >>> > >>>Why is the last digit stripped off in the case when it is zero! > >> > >>signif() changes the value; you don't want that, you want to affect how > >>a number is displayed. Use format() or formatC() instead, for example > >> > >> > x <- c(17.01, 18.15) > >> > format(x, digits=3) > >>[1] "17.0" "18.1" > >> > noquote(format(x, digits=3)) > >>[1] 17.0 18.1 > >> > > > > > > That works in the above context but I don't think it works generally: > > > > R> f <- head(faithful) > > R> f > > eruptions waiting > > 1 3.600 79 > > 2 1.800 54 > > 3 3.333 74 > > 4 2.283 62 > > 5 4.533 85 > > 6 2.883 55 > > > > R> format(f, digits = 3) > > eruptions waiting > > 1 3.60 79 > > 2 1.80 54 > > 3 3.33 74 > > 4 2.28 62 > > 5 4.53 85 > > 6 2.88 55 > > > > R> # this works in this case > > R> noquote(prettyNum(round(f,1), nsmall = 1)) > > eruptions waiting > > [1,] 3.6 79.0 > > [2,] 1.8 54.0 > > [3,] 3.3 74.0 > > [4,] 2.3 62.0 > > [5,] 4.5 85.0 > > [6,] 2.9 55.0 > > > > and even that does not work in the desired way (which presumably > > is not to use exponent format) if you have some > > large enough numbers like 1e6 which it will display using > > the e notation rather than using ordinary notation. > > formatC with format="f" seems to work for me, though it assumes you're > specifying decimal places rather than significant digits. It also wants > a vector of numbers as input, not a dataframe. So the following gives > pretty flexible control over what a table will look like: > > > data.frame(eruptions = formatC(f$eruptions, digits=2, format='f'), > +waiting = formatC(f$waiting, digits=1, format='f')) >eruptions waiting > 1 100.1179.0 > 2 1.8054.0 > 3 3.3374.0 > 4 2.2862.0 > 5 4.5385.0 > 6 2.8855.0 > > > > > I have struggled with this myself and have generally been able > > to come up with something for specific instances but I have generally > > found it a pain to do a simple thing like format a table exactly as I want > > without undue effort. Maybe someone else has figured this out. > > I think that formatting tables properly requires some thought, and R is > no good at thinking. You can easily recognize a badly formatted table, > but it's very hard to write down rules that work in general > circumstances. It's also a matter of taste, so if I managed to write a > function that matched my taste, you would find you wanted to make changes. > > It's sort of like expecting plot(x, y) to always come up with the best > possible plot of y versus x. It's just not a reasonable expectation. > It's better to provide tools (like abline() for plots or formatC() for > tables) that allow you to tailor a plot or table to your particular needs. > Thanks. That seems to be the idiom I was missing. One thing that would be nice would be if formatC could handle data frames. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Problem going back to a viewport with gridBase
I am setting up base plots -- one in viewport A and and one in B. This part
works fine. But if I go back to A after having done B and add
horizontal lines it seems
to not use the correct coordinates. How do I tell it to resume using A's
coordinates? I am already using par(fig = gridFIG()) but it seems that that's
not enough to reestablish them. What happens is that when I go back to
A it draws the horizontal lines as if its relative to B's coordinates
rather than
restablishing A's coordinates. As a result the horizontal lines are
drawn near the
bottom of the graph instead of at the correct heights. Try running the code
below to see what I mean.
I have also tried to use baseViewports with this but did not have any
success.
How do I modify this example so that the horizontal red lines come out
at the appropriate levels?Note that this is just an example and in
the future I will want to have multiple viewports each with a base plot and
add arbitrary additional line or point plots to them so the solution needs
to be sufficiently general that I can so generalize it.
Thanks.
library(gridBase)
opar <- par(no.readonly = TRUE)
grid.newpage()
# two columns, one row
unit. <- unit(c(1,1), c("null","null"))
pushViewport(viewport(layout = grid.layout(1, 2, widths = unit.)))
# draw green graph in first column (viewport A)
pushViewport(viewport(layout.pos.col = 1, name = "A"))
par(fig = gridFIG()); par(new = TRUE)
plot(1:10, col = "green", pch = 20)
upViewport(1)
# draw purple graph in second column (viewport B)
pushViewport(viewport(layout.pos.col = 2, name = "B"))
par(fig = gridFIG()); par(new = TRUE)
plot(1:100, col = "purple", pch = 18)
upViewport()
# go back to A and add horizontal grid lines
seekViewport("A")
par(fig = gridFIG())
abline(h=1:10, col = "red") THESE DO NOT GET DRAWN AS EXPECTED
popViewport()
# go back to B and add vertical grid lines
seekViewport("B")
par(fig = gridFIG())
abline(v=1:10, col = "red")
popViewport()
par(opar)
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RE: [R] R GUI for Linux?
John: Thank you for your interest. After more investigation I see that my problem is with linking tcltk to R. tcl 8.4.7-2 and tk 8.4.7-2 are tuned to fc3 (Fedora Core 3) and part of the standard installation. However, required file locations are different than what is expected by R. I set the locations for tclConfig.sh and tkConfig.sh using ./configure but R tells me something to the effect that Tcl and Tk major or minor versions disagree. Your help or the help of others on the R-List would be appreciated. FYI, I am also including the locations of important files and directories. tclConfig.sh: /usr/lib tkConfig.sh: /usr/lib tcl library: /usr/lib/tcl8.4 (symbolically linked from /usr/share/tcl8.4) tk library: /usr/lib/tk8.4 (separate copy in /usr/share/tk8.4) tcl.h:I can't find it with full drive search tk.h: I can't find it with full drive search Thanks for your time. Chuck -Original Message- From: John Fox [mailto:[EMAIL PROTECTED] Sent: Mon 5/30/2005 4:20 PM To: White, Charles E WRAIR-Wash DC Cc: [EMAIL PROTECTED]; r-help@stat.math.ethz.ch Subject: RE: [R] R GUI for Linux? Dear Charles, > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of White, > Charles E WRAIR-Wash DC > Sent: Monday, May 30, 2005 10:52 AM > To: r-help@stat.math.ethz.ch > Cc: [EMAIL PROTECTED] > Subject: [R] R GUI for Linux? > . . . I haven't tried the Rcmdr package with FC3, but I ran it under an older version of Red Hat Linux some time ago, and the current version (Rcmdr 1.0-2) under Quantian. Can you be more specific about the problems that you encountered? Are these general to the tcltk package or specific to the Rcmdr? I'm sorry that you're experiencing problems. John __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: Re: [R] labels on map
thanks Duncan, thats what I was looking for :)
p.s. xlab in map() gives the same result as in axis()
j. silva
jose silva wrote:
> dear all:
>
> Im trying to obtain maps on R, under mapdata library, but I cannot define the
> labels.
> Here is an example:
>
> library(mapdata)
> map("worldHires", c("portugal","spain"),ylim=c(34,46),xlim=c(-14,3.5))
> axis(1,at=seq(-12,3,3))
> axis(2)
>
> when I try the parameter xlab or ylab in axis, i get:
> parameter "ylab" couldn't be set in high-level plot() function
>
> any suggestion? thanks in advance for your always useful advices
Despite their names, x and y axis labels aren't really part of the axis,
they're titles in the margins of the plot. Use the title() function to
set them.
You could probably also set them in map(), but I don't have it installed
to check. mtext() is another general purpose way to put text in the
margins.
Duncan Murdoch
[[alternative HTML version deleted]]
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RE: [R] labels on map
What makes you think that there is a ylab parameter?
> args(map)
function (database = "world", regions = ".", exact = FALSE, boundary = TRUE,
interior = TRUE, projection = "", parameters = NULL, orientation = NULL,
fill = FALSE, col = 1, plot = TRUE, add = FALSE, namesonly = FALSE,
xlim = NULL, ylim = NULL, wrap = FALSE, resolution = if (plot) 1 else 0,
type = "l", bg = par("bg"), mar = c(0, 0, par("mar")[3],
0.1), border = 0.01, ...)
NULL
>
Just because there are ellipses does not mean that ylab is an appropriate
parameter. If you type map you'll see the function. Go looking for where the
ellipses are used and you find
polygon(coord, col = col, ...)
lines(coord, col = col, type = type, ...)
So how why would these functions take a ylab parameter, they are for use within
a plot. You have to remember that each function will only do what the author
has programmed it to do.
So if you need to put a label somewhere you could always use mtext(side =
2,line = 1,"A label on side 2")
Tom
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] Behalf Of jose silva
> Sent: Tuesday, 31 May 2005 8:12 AM
> To: r-help@stat.math.ethz.ch
> Subject: [R] labels on map
>
>
> dear all:
>
> Im trying to obtain maps on R, under mapdata library, but I
> cannot define the labels.
> Here is an example:
>
> library(mapdata)
> map("worldHires", c("portugal","spain"),ylim=c(34,46),xlim=c(-14,3.5))
> axis(1,at=seq(-12,3,3))
> axis(2)
>
> when I try the parameter xlab or ylab in axis, i get:
> parameter "ylab" couldn't be set in high-level plot() function
>
> any suggestion? thanks in advance for your always useful advices
>
> j. silva
> [[alternative HTML version deleted]]
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
>
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Re: [R] labels on map
jose silva wrote:
dear all:
Im trying to obtain maps on R, under mapdata library, but I cannot define the
labels.
Here is an example:
library(mapdata)
map("worldHires", c("portugal","spain"),ylim=c(34,46),xlim=c(-14,3.5))
axis(1,at=seq(-12,3,3))
axis(2)
when I try the parameter xlab or ylab in axis, i get:
parameter "ylab" couldn't be set in high-level plot() function
any suggestion? thanks in advance for your always useful advices
Despite their names, x and y axis labels aren't really part of the axis,
they're titles in the margins of the plot. Use the title() function to
set them.
You could probably also set them in map(), but I don't have it installed
to check. mtext() is another general purpose way to put text in the
margins.
Duncan Murdoch
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Re: [R] Piecewise Linear Regression
It is conventional to fit piecewise linear models by assuming Gaussian error and using least squares methods, but one can argue that median regression provides a more robust approach to this problem. You might consider the following fit: x = c (6.25,6.25,12.50,12.50,18.75,25.00,25.00,25.00,31.25,31.25,37.50,37.50,5 0.00,50.00,62.50,62.50,75.00,75.00,75.00,100.00,100.00) y = c (0.328,0.395,0.321,0.239,0.282,0.230,0.273,0.347,0.211,0.210,0.259,0.186 ,0.301,0.270,0.252,0.247,0.277,0.229,0.225,0.168,0.202) library(quantreg) plot(x,y) fit <- rqss(y ~ qss(x)) plot(fit) it gives 5 segments not 3, but this can be controlled by the choice of lambda in the qss function, for example, try: fit <- rqss(y ~ qss(x,lambda=3) plot(fit,col="red") which gives a fit like you suggest might be reasonable with only three segments. url:www.econ.uiuc.edu/~rogerRoger Koenker email [EMAIL PROTECTED] Department of Economics vox:217-333-4558University of Illinois fax:217-244-6678Champaign, IL 61820 On May 30, 2005, at 6:38 PM, Abhyuday Mandal wrote: Hi, I need to fit a piecewise linear regression. x = c (6.25,6.25,12.50,12.50,18.75,25.00,25.00,25.00,31.25,31.25,37.50,37.50 ,50.00,50.00,62.50,62.50,75.00,75.00,75.00,100.00,100.00) y = c (0.328,0.395,0.321,0.239,0.282,0.230,0.273,0.347,0.211,0.210,0.259,0.1 86,0.301,0.270,0.252,0.247,0.277,0.229,0.225,0.168,0.202) there are two change points. so the fitted curve should look like \ \ /\ \/ \ \ \ How do I do this in R ? Thank you, Abhyuday __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting- guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] persp, add lines/highlights
Hi
[EMAIL PROTECTED] wrote:
Hello R-sters,
I'm trying to add several lines to a response surface that I've plotted
using persp(). I've tried lines() using the "trans3d" function but I've
been unsuccessful in getting it to work (R v2.0.1). Essentially, I'm
trying to highlight one or more of the surface wireframe lines in a
bolder (or different) color. Any tips from those of you who have some
experience with this would be greatly appreciated. [Would it be easier
using wireframe() in library(lattice) instead?]
Here's an example that just overlays two persp() plots ...
x <- seq(-10, 10, length= 30)
y <- x
f <- function(x,y) { r <- sqrt(x^2+y^2); 10 * sin(r)/r }
z <- outer(x, y, f)
z[is.na(z)] <- 1
op <- par(bg = "white")
persp(x, y, z, theta = 30, phi = 30, expand = 0.5, col = "lightblue",
zlim=range(z))
# overlay plot with just "highlighted" surface area
par(new=TRUE)
z2 <- matrix(NA, ncol=30, nrow=30)
xi <- 15:18
yi <- 13:14
z2[xi, yi] <- z[xi, yi]
persp(x, y, z2, theta = 30, phi = 30, expand = 0.5,
zlim=range(z), border="red", col="pink", box=FALSE, axes=FALSE)
And, any suggestions on how to add text outside of the persp() plot next
to the highlighted line would be much appreciated.
Do you mean like a legend?
Paul
--
Dr Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
[EMAIL PROTECTED]
http://www.stat.auckland.ac.nz/~paul/
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[R] labels on map
dear all:
Im trying to obtain maps on R, under mapdata library, but I cannot define the
labels.
Here is an example:
library(mapdata)
map("worldHires", c("portugal","spain"),ylim=c(34,46),xlim=c(-14,3.5))
axis(1,at=seq(-12,3,3))
axis(2)
when I try the parameter xlab or ylab in axis, i get:
parameter "ylab" couldn't be set in high-level plot() function
any suggestion? thanks in advance for your always useful advices
j. silva
[[alternative HTML version deleted]]
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Re: [R] Piecewise Linear Regression
Abhyuday, There are a number of answers in the archives: http://www.google.com/u/newcastlemaths?q=piecewise+linear+regression&sa=Google+Search Do any of those meet your needs? Sean - Original Message - From: "Abhyuday Mandal" <[EMAIL PROTECTED]> To: Sent: Monday, May 30, 2005 7:38 PM Subject: [R] Piecewise Linear Regression Hi, I need to fit a piecewise linear regression. x = c(6.25,6.25,12.50,12.50,18.75,25.00,25.00,25.00,31.25,31.25,37.50,37.50,50.00,50.00,62.50,62.50,75.00,75.00,75.00,100.00,100.00) y = c(0.328,0.395,0.321,0.239,0.282,0.230,0.273,0.347,0.211,0.210,0.259,0.186,0.301,0.270,0.252,0.247,0.277,0.229,0.225,0.168,0.202) there are two change points. so the fitted curve should look like \ \ /\ \/ \ \ \ How do I do this in R ? Thank you, Abhyuday __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Piecewise Linear Regression
Hi, I need to fit a piecewise linear regression. x = c(6.25,6.25,12.50,12.50,18.75,25.00,25.00,25.00,31.25,31.25,37.50,37.50,50.00,50.00,62.50,62.50,75.00,75.00,75.00,100.00,100.00) y = c(0.328,0.395,0.321,0.239,0.282,0.230,0.273,0.347,0.211,0.210,0.259,0.186,0.301,0.270,0.252,0.247,0.277,0.229,0.225,0.168,0.202) there are two change points. so the fitted curve should look like \ \ /\ \/ \ \ \ How do I do this in R ? Thank you, Abhyuday __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R GUI for Linux?
On Mon, 30-May-2005 at 08:29PM +0200, Sander Oom wrote: |> .. I have |> given up on any instructions that tell me to run 'make'. Linux |> distributions are just to[o] idiosyncratic for it to be worth the effort. That might be true of Linux distributions in general, but installing R is *very* easy. There are a couple of lines you can paste from the README file to get the configure and make thing working. Depending on your machine, it might take some time to do all its stuff which is extensive and thorough. Installing ESS is only slightly different. Simply extract the tar.gz file, find the README file and adjust your .emacs file to let it know where you've installed ESS. Granted, it does take a bit to get used to the way Emacs does things, but you need only about 1% of its capability and then it is extremely easy to use. Many questions asked on this list just don't arise. best -- Patrick Connolly HortResearch Mt Albert Auckland New Zealand Ph: +64-9 815 4200 x 7188 ~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~ I have the world`s largest collection of seashells. I keep it on all the beaches of the world ... Perhaps you`ve seen it. ---Steven Wright ~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] persp, add lines/highlights
Hello R-sters, I'm trying to add several lines to a response surface that I've plotted using persp(). I've tried lines() using the "trans3d" function but I've been unsuccessful in getting it to work (R v2.0.1). Essentially, I'm trying to highlight one or more of the surface wireframe lines in a bolder (or different) color. Any tips from those of you who have some experience with this would be greatly appreciated. [Would it be easier using wireframe() in library(lattice) instead?] And, any suggestions on how to add text outside of the persp() plot next to the highlighted line would be much appreciated. Thanks, Jeff Jorgensen __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] R GUI for Linux?
Dear Charles, > -Original Message- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of White, > Charles E WRAIR-Wash DC > Sent: Monday, May 30, 2005 10:52 AM > To: r-help@stat.math.ethz.ch > Cc: [EMAIL PROTECTED] > Subject: [R] R GUI for Linux? > . . . > > Rcmdr: There are all sorts of things in FC3 that seem to be > tcl/tk related but Rcmdr doesn't seem to work with them. > Since some are part of the base FC3 installation, I'm > nervious about replacing them or installing competing > software. Potentially conflicting software in FC3 are listed below: > > tcl.i386 8.4.7-2 > installed > tclx.i3868.3.5-4 > installed > db4-tcl.i386 4.2.52-6 base > postgresql-tcl.i386 7.4.8-1.FC3.1 > updates-released > ruby-tcltk.i386 1.8.2-1.FC3.1 > updates-released > tcl-devel.i386 8.4.7-2base > tcl-html.i3868.4.7-2base > tclx-devel.i386 8.3.5-4base > tclx-doc.i3868.3.5-4base > . . . I haven't tried the Rcmdr package with FC3, but I ran it under an older version of Red Hat Linux some time ago, and the current version (Rcmdr 1.0-2) under Quantian. Can you be more specific about the problems that you encountered? Are these general to the tcltk package or specific to the Rcmdr? I'm sorry that you're experiencing problems. John __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] "FANNY" function in R package "cluster"
Dear All, I am attempting to use the FANNY fuzzy clustering function in R (Kaufman & Rousseeuw, 1990), found in the "cluster" package. I have run into a variety of difficulties; the two most crucial difficulties are enumerated below. 1. Where is the 'm' parameter in FANNY? In _Finding Groups in Data: An Introduction to Cluster Analysis_ (1990) by Kaufman & Rousseeuw, the authors discuss the FANNY algorithm. On pages 189-190, they attempt to demonstrate an equivalence between Fuzzy c-Means (FCM) and FANNY. In doing so they, appear to be assuming that the value of the 'm' parameter in FCM (a measure of the fuzziness) is fixed at m=2. Although this is how FCM was originally formulated, it eventually became apparent that m should be a user-specified parameter and not a fixed value of m=2. My question, then, is twofold. Firstly, am I correct in saying that Kaufman & Rousseeuw have assumed m=2 in their formulation of Fuzzy c-Means and FANNY? Secondly, is it possible to modify the FANNY algorithm to allow user-specification of the m (fuzziness) parameter? 2. What do I do with training data? Is there any way to use FANNY for assigning clustering membership values to new, test data? In Fuzzy c-Means, new data is compared to the cluster centers in order to assign clustering membership values to the test data. However, in FANNY these centers do not exist. Is there, then, any way to compute the FANNY clustering membership values of a test data point without affecting the clustering membership values of the training data? Perhaps there are enough conditions to use the objective function as a way of computing the membership values of the test data? Aamir M University of Toronto __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Unique arrangements of a vector
Dear List, Running on a PC (Windows 2000) with 256 MB RAM, Version R1.9.1 I have a relatively simple problem, which I can solve for relatively small datasets, but run into difficulties with larger ones. I believe that my approach is a hack rather than something elegant and I was hoping that somebody on this list might help me improve my code. Basically, given a vector of values (e.g., 0,0,1,1), I want to generate all of the unique arrangements of these values, of which there are 4!/(2!2!) = 6. 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 0 Using unique() in conjunction with expand.grid(), and later filtering impossible results, I can obtain the answer. However, this is slow, and does not work for large initial vectors and is difficult to filter when using values beyond 0,1. Is there some mathematically elegant method for doing this? I'd hope to have initial vectors significantly longer than the demonstrated 4 values (e.g., thousands). Any help is appreciated and I will gladly SUM afterwards. Thank you, Tarmo __ Tarmo Remmel Ph.D. GUESS Lab, Department of Geography University of Toronto at Mississauga Mississauga, Ontario, L5L 1C6 Tel: 905-828-3868 Fax: 905-828-5273 Skype: tarmoremmel http://eratos.erin.utoronto.ca/remmelt __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Formatting numbers with a limited amount of digits consistently
Gabor Grothendieck wrote: On 5/30/05, Duncan Murdoch <[EMAIL PROTECTED]> wrote: Henrik Andersson wrote: I have tried to get signif, round and format to display numbers like these consistently in a table, using e.g. signif(x,digits=3) 17.01 18.15 I want 17.0 18.2 Not 17 18.2 Why is the last digit stripped off in the case when it is zero! signif() changes the value; you don't want that, you want to affect how a number is displayed. Use format() or formatC() instead, for example > x <- c(17.01, 18.15) > format(x, digits=3) [1] "17.0" "18.1" > noquote(format(x, digits=3)) [1] 17.0 18.1 That works in the above context but I don't think it works generally: R> f <- head(faithful) R> f eruptions waiting 1 3.600 79 2 1.800 54 3 3.333 74 4 2.283 62 5 4.533 85 6 2.883 55 R> format(f, digits = 3) eruptions waiting 1 3.60 79 2 1.80 54 3 3.33 74 4 2.28 62 5 4.53 85 6 2.88 55 R> # this works in this case R> noquote(prettyNum(round(f,1), nsmall = 1)) eruptions waiting [1,] 3.6 79.0 [2,] 1.8 54.0 [3,] 3.3 74.0 [4,] 2.3 62.0 [5,] 4.5 85.0 [6,] 2.9 55.0 and even that does not work in the desired way (which presumably is not to use exponent format) if you have some large enough numbers like 1e6 which it will display using the e notation rather than using ordinary notation. formatC with format="f" seems to work for me, though it assumes you're specifying decimal places rather than significant digits. It also wants a vector of numbers as input, not a dataframe. So the following gives pretty flexible control over what a table will look like: > data.frame(eruptions = formatC(f$eruptions, digits=2, format='f'), +waiting = formatC(f$waiting, digits=1, format='f')) eruptions waiting 1 100.1179.0 2 1.8054.0 3 3.3374.0 4 2.2862.0 5 4.5385.0 6 2.8855.0 I have struggled with this myself and have generally been able to come up with something for specific instances but I have generally found it a pain to do a simple thing like format a table exactly as I want without undue effort. Maybe someone else has figured this out. I think that formatting tables properly requires some thought, and R is no good at thinking. You can easily recognize a badly formatted table, but it's very hard to write down rules that work in general circumstances. It's also a matter of taste, so if I managed to write a function that matched my taste, you would find you wanted to make changes. It's sort of like expecting plot(x, y) to always come up with the best possible plot of y versus x. It's just not a reasonable expectation. It's better to provide tools (like abline() for plots or formatC() for tables) that allow you to tailor a plot or table to your particular needs. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] R GUI for Linux?
Hi Charles, Warm felt sympathies for your struggles. I consider myself a happy GUI user and have also struggled with the 'command line' history and lack of out-of-the-box functionality associated with Linux. However, Linux does have many, many advantages over other OS's, so I will stick to it. That said, I am reluctantly suggesting it to others, as there are lots of other things one can do on a Saturday afternoon. ;-) After many long afternoons, I have come to the conclusion that accepting what comes as standard is the best approach to using Linux if you do have better things to do. I tried ESS, but I found it impenetrable at my first try, so gave up. I run SuSE linux and R without any problem. The RPM was downloaded from CRAN and installed without any errors. I run update.packages to download, install and refresh all contributed and other packages. I also looked for an R GUI, but must admit I stopped way before you did. I ended up using my favorite editor (Kate, comes standard with KDE) and cut and paste code into an R console (available as standard in the Kate window). Gives a nice clean window for writing code and running R. Of course Kate has code highlighting facilities (as standard). I soon realized that the console has a limited buffer for commands, such that long code sequences are abruptly ended part way through. Thus I have a second console open from which I source whole R files. The setup gives a relatively comfortable code debugging environment! I was amazed when I read the instruction for JGR on Linux. I thought the whole point of Java was to create platform independent software. I have given up on any instructions that tell me to run 'make'. Linux distributions are just to idiosyncratic for it to be worth the effort. I'll just wait until Novell packs JGR with the new version of SuSE Linux. Even if I have to pay something for the added bonus! Will keep following the GUI developments with interest, Sander. White, Charles E WRAIR-Wash DC wrote: I feel your pain. I am a new Linux user who has spent most of the weekend trying to get a functional R setup. When I installed Fedora Core 3 (FC3) on my home computer, I thought using R in a terminal would be a snap. I installed R using the rpm packages and tried to use it with the FC3 default terminal (GNOME Terminal 2.7.3). Before long, I found out the terminal was rudely discarding output beyond a set number of lines. I could increase the number of lines kept by the terminal but that didn't strike me as an acceptable solution. Cutting to my stress relieving intermediate solution, I am currently using xemacs with ESS as my R programming environment under FC3. Eventually, I will want to run JGR as my programming environment and Rcmdr as both a teaching tool and means to distribute code to some of my clients. On my way to xemacs, I also tried to install emacs and gnomeGUI. I will briefly document my experience with trying to install each of these packages below: XEMACS with ESS: XEMACS is within the 'walled garden' of packages tuned specifically to run under FC3 and XEMACS provides a tuned installation for ESS. Since I had already compiled R from source with shared libraries enabled (the rmp does not enable shared libraries), I don't know if XEMACS will work with the rpm version of R. Note also that I installed this package using yum; 'Add or Remove Applications' lists xemacs but wouldn't allow me to install. JGR: I have installed jdk1.5.0_03 and MOST of the the output from make looks like JGR is compiling correctly. JavaGD and rJava are not finding jini.h. I don't see an explicit statement of how to start JGR but I assume that is done by typing JGR (or maybe jgr) in a terminal window. Nothing happens. Two potential problems are: (a) I never should have downloaded JavaGD and rJava from CRAN (they won't uninstall, deleting the directories doesn't stop the problem, and I can't use yum to remove R to start over because yum doesn't recognise that R is installed.) or (b) I need to uninstall some of the stray versions of java littering my hard drive. I haven't removed the symbolic link between jre1.5.0_02 and firefox. Rcmdr: There are all sorts of things in FC3 that seem to be tcl/tk related but Rcmdr doesn't seem to work with them. Since some are part of the base FC3 installation, I'm nervious about replacing them or installing competing software. Potentially conflicting software in FC3 are listed below: tcl.i386 8.4.7-2installed tclx.i3868.3.5-4installed db4-tcl.i386 4.2.52-6 base postgresql-tcl.i386 7.4.8-1.FC3.1 updates-released ruby-tcltk.i386 1.8.2-1.FC3.1 updates-released tcl-devel.i386 8.4.7-2base tcl-html.i3868.4.
Re: [R] Formatting numbers with a limited amount of digits consistently
On 5/30/05, Duncan Murdoch <[EMAIL PROTECTED]> wrote: > Henrik Andersson wrote: > > I have tried to get signif, round and format to display numbers like > > these consistently in a table, using e.g. signif(x,digits=3) > > > > 17.01 > > 18.15 > > > > I want > > > > 17.0 > > 18.2 > > > > Not > > > > 17 > > 18.2 > > > > > > Why is the last digit stripped off in the case when it is zero! > > signif() changes the value; you don't want that, you want to affect how > a number is displayed. Use format() or formatC() instead, for example > > > x <- c(17.01, 18.15) > > format(x, digits=3) > [1] "17.0" "18.1" > > noquote(format(x, digits=3)) > [1] 17.0 18.1 > That works in the above context but I don't think it works generally: R> f <- head(faithful) R> f eruptions waiting 1 3.600 79 2 1.800 54 3 3.333 74 4 2.283 62 5 4.533 85 6 2.883 55 R> format(f, digits = 3) eruptions waiting 1 3.60 79 2 1.80 54 3 3.33 74 4 2.28 62 5 4.53 85 6 2.88 55 R> # this works in this case R> noquote(prettyNum(round(f,1), nsmall = 1)) eruptions waiting [1,] 3.6 79.0 [2,] 1.8 54.0 [3,] 3.3 74.0 [4,] 2.3 62.0 [5,] 4.5 85.0 [6,] 2.9 55.0 and even that does not work in the desired way (which presumably is not to use exponent format) if you have some large enough numbers like 1e6 which it will display using the e notation rather than using ordinary notation. R> f[1,1] <- 1e6 + 0.11 R> noquote(prettyNum(round(f,1), nsmall = 1)) eruptions waiting [1,] 1.0e+06 79.0 [2,] 1.8e+00 54.0 [3,] 3.3e+00 74.0 [4,] 2.3e+00 62.0 [5,] 4.5e+00 85.0 [6,] 2.9e+00 55.0 I have struggled with this myself and have generally been able to come up with something for specific instances but I have generally found it a pain to do a simple thing like format a table exactly as I want without undue effort. Maybe someone else has figured this out. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] problem in R2.2-dev
ronggui wrote: > when i install the R2.2-dev(customed install,the with the chinese translation > msg ,etc.),i found the tanslated GUI can not display correctly.totally unlike > Chinese characters. I don't think you've given nearly enough information for anyone to act on this. Please describe what OS you're on, what customizations you did during installation, and give an example of something that should work but doesn't. Hopefully someone who uses a Chinese locale will be able to help you. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Formatting numbers with a limited amount of digits consistently
Henrik Andersson wrote: I have tried to get signif, round and format to display numbers like these consistently in a table, using e.g. signif(x,digits=3) 17.01 18.15 I want 17.0 18.2 Not 17 18.2 Why is the last digit stripped off in the case when it is zero! signif() changes the value; you don't want that, you want to affect how a number is displayed. Use format() or formatC() instead, for example > x <- c(17.01, 18.15) > format(x, digits=3) [1] "17.0" "18.1" > noquote(format(x, digits=3)) [1] 17.0 18.1 Is this a "feature" of R or did I miss something? I'd say both. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] problem in R2.2-dev
when i install the R2.2-dev(customed install,the with the chinese translation msg ,etc.),i found the tanslated GUI can not display correctly.totally unlike Chinese characters. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to invert the matrix with quite small eigenvalues
On Mon, 30 May 2005, huang min wrote:>
My intention is to invert the covariance matrix to perform some
algorithm which is common in the estimating equations like GEE.
In that case there is no benefit in being able to invert very extreme
covariance matrices. The asymptotic approximations to the distribution of
regression coefficients will break down really badly with such extreme
working covariance matrices.
I think in a case like this you should either
1) report an error and stop
2) shrink the covariance matrix towards a diagonal one, eg increase the
diagonal entries until the condition number becomes reasonable.
3) Use a one-step estimator from the independence working model (which is
asymptotically equivalent to the full solution and better behaved).
Remember that in GEE the matrix V^{-1} is just a set of weights, chosen to
get good efficiency. Your matrix solve(a) is not a good set of weights.
I think in an earlier thread on this topic Brian Ripley recommended using
the singular value decomposition if you really have to compute something
like D^TV^{-1}D. In your example this still isn't good enough.
Another strange point is that my friend use the LU decomposition in
Fortran to solve the inverse matrix of aa for me. He used double
precision of course, otherwise no inverse matrix in Fortran too. He
show that the product of the two matrix is almost identity(The biggest
off-digonal element is about 1e-8). I copy his inverse matrix(with 31
digits!) to R and read in aa I sent to him(16 digits). The product is
also not an identity matrix. It is fairly strange! Any suggestion?
It isn't that strange. The system is computationally singular, meaning
that you should expect to get different answers with apparently similar
computations on different systems.
Also remember that what you care about for GEE is the result of
solve(a,y-mu), rather than solve(a,a). Getting one of these right is no
guarantee of getting the other one right.
If you really had to work with this matrix in double precision you would
need to track very carefully the error bounds on all your computations,
which would be very difficult. Fortunately this is almost never necessary
in statistics, and I don't think it's necessary in your case.
A good habit to get into when you aren't tracking error bounds carefully
is to think of the last couple of digits of the result of any calculation
as random.
-thomas
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Re: [R] How to access to sum of dissimilarities in CLARA
Hello, #Example: data(xclara) p <- clara(xclara,3) names(p) p$diss Best, Matthias > Dear All , > Since dissimilarity is one of quality measures in clustering > , I'm trying to access to the sum of dissimilarity as a whole > measure. But after running my data using CLARA I obtain : > 1128 dissimilarities, summarized : > Min. 1st Qu. Median Mean 3rd Qu. Max. > 0.033155 0.934630 2.257000 2.941600 4.876600 8.943700 > But I can not find the sum of dissimilarity.How can i access to it? > > Thanks a lot > Safari > > __ > > > > [[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read > the posting guide! http://www.R-project.org/posting-guide.html > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R GUI for Linux?
I feel your pain. I am a new Linux user who has spent most of the weekend trying to get a functional R setup. When I installed Fedora Core 3 (FC3) on my home computer, I thought using R in a terminal would be a snap. I installed R using the rpm packages and tried to use it with the FC3 default terminal (GNOME Terminal 2.7.3). Before long, I found out the terminal was rudely discarding output beyond a set number of lines. I could increase the number of lines kept by the terminal but that didn't strike me as an acceptable solution. Cutting to my stress relieving intermediate solution, I am currently using xemacs with ESS as my R programming environment under FC3. Eventually, I will want to run JGR as my programming environment and Rcmdr as both a teaching tool and means to distribute code to some of my clients. On my way to xemacs, I also tried to install emacs and gnomeGUI. I will briefly document my experience with trying to install each of these packages below: XEMACS with ESS: XEMACS is within the 'walled garden' of packages tuned specifically to run under FC3 and XEMACS provides a tuned installation for ESS. Since I had already compiled R from source with shared libraries enabled (the rmp does not enable shared libraries), I don't know if XEMACS will work with the rpm version of R. Note also that I installed this package using yum; 'Add or Remove Applications' lists xemacs but wouldn't allow me to install. JGR: I have installed jdk1.5.0_03 and MOST of the the output from make looks like JGR is compiling correctly. JavaGD and rJava are not finding jini.h. I don't see an explicit statement of how to start JGR but I assume that is done by typing JGR (or maybe jgr) in a terminal window. Nothing happens. Two potential problems are: (a) I never should have downloaded JavaGD and rJava from CRAN (they won't uninstall, deleting the directories doesn't stop the problem, and I can't use yum to remove R to start over because yum doesn't recognise that R is installed.) or (b) I need to uninstall some of the stray versions of java littering my hard drive. I haven't removed the symbolic link between jre1.5.0_02 and firefox. Rcmdr: There are all sorts of things in FC3 that seem to be tcl/tk related but Rcmdr doesn't seem to work with them. Since some are part of the base FC3 installation, I'm nervious about replacing them or installing competing software. Potentially conflicting software in FC3 are listed below: tcl.i386 8.4.7-2installed tclx.i3868.3.5-4installed db4-tcl.i386 4.2.52-6 base postgresql-tcl.i386 7.4.8-1.FC3.1 updates-released ruby-tcltk.i386 1.8.2-1.FC3.1 updates-released tcl-devel.i386 8.4.7-2base tcl-html.i3868.4.7-2base tclx-devel.i386 8.3.5-4base tclx-doc.i3868.3.5-4base EMACS with ESS: A version of EMACS is tuned to FC3 but ESS has to be obtained elsewhere. Installing ESS requires editing the hidden .emacs file. I know how to see hidden files but this one does not appear to be where the ESS directions say to look. gnomeGUI: Error message says that I don't have Gnome installed. It would be nice and GNU to have a working copy of gnome GUI. I hope you have found my message to be entertaining. If anybody can stop me before I do something else stupid, I'd appreciate it. Thanks. Chuck Charles E. White, Senior Biostatistician, MS Walter Reed Army Institute of Research 503 Robert Grant Ave., Room 1w102 Silver Spring, MD 20910-1557 Personal/Professional Site: http://users.starpower.net/cwhite571/professional/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to access to sum of dissimilarities in CLARA
Dear All , Since dissimilarity is one of quality measures in clustering , I'm trying to access to the sum of dissimilarity as a whole measure. But after running my data using CLARA I obtain : 1128 dissimilarities, summarized : Min. 1st Qu. Median Mean 3rd Qu. Max. 0.033155 0.934630 2.257000 2.941600 4.876600 8.943700 But I can not find the sum of dissimilarity.How can i access to it? Thanks a lot Safari __ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Incompatibility with VGAM
On Mon, 30 May 2005, Thomas Yee wrote: Hello, It seems that if glm used a namespace then the conflict would be avoided? No. glm does use a namespace, so this can't be true. Remember that R passes arguments by value, and consider glm(y~x, family=poisson()) The namespace ensures that functions called from glm are looked up in the stats namespace, but family() not called from glm(). It is called by the user and the result is passed as an argument to glm(). The exception is when no family argument is supplied. In that case the default argument is created by a call to family() from inside glm(), which should always find stats::family -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] a question about read.marrayRaw
Hi, everyone, I ran the test case in "Introduction to the Bioconductor marrayInput package first. When I ran the session in (read.marrayRaw) I got the following error messages. Other sessions work well. The functions and errors are labled with blue and red respectively. >mraw <- read.Spot(path = datadir, layout = galinfo$layout, gnames = >galinfo$gnames, target = swirlTargets) Error in file(con, "r") : unable to open connection In addition: Warning message: cannot open file `C:/PROGRA~1/R/rw1091/library/marray/swirldata/81' > fnames <- as.vector([EMAIL PROTECTED],1]) > swirl.raw <- read.marrayRaw(fnames, path = datadir, + name.Gf = "Gmean", name.Gb = "morphG", + name.Rf = "Rmean", name.Rb = "morphR", + layout = swirl.layout, + gnames = swirl.gnames, + targets = swirlTargets) Error in file(con, "r") : unable to open connection In addition: Warning message: cannot open file `C:/PROGRA~1/R/rw1091/library/marray/swirldata/81' > array1.raw <- read.Spot(fnames, path=datadir) Error in file(con, "r") : unable to open connection In addition: Warning messawell cannot open file `C:/PROGRA~1/R/rw1091/library/marray/swirldata/81' I copy and paste the whole set of examples. Is it possible something is not right with the spot files? I really need your help to solve this problem. Thank you very much!! Best regards Zang Shizhu - ×¢²áÊÀ½çÒ»Á÷Æ·ÖʵÄÑÅ»¢Ãâ·ÑµçÓÊ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] rbind wastes memory
Rather than 'rbind' in a loop, try putting your dataframes in a list and
then doing something like 'do.call("rbind", list.of.data.frames")'.
-roger
[EMAIL PROTECTED] wrote:
Hello everybody,
if I try to (r)bind a number of large dataframes I run out of memory because R
wastes memory and seems to "forget" to release memory.
For example I have 10 files. Each file contains a large dataframe "ds" (3500 cols
by 800 rows) which needs ~20 MB RAM if it is loaded as the only object.
Now I try to bind all data frames to a large one and need more than 1165MB (!)
RAM (To simplify the R code, I use the same file ten times):
start example 1 __
load(myFile)
ds.tmp <- ds
for (Cycle in 1:10) {
ds.tmp <- rbind(ds.tmp, ds)
}
end example 1 __
Stepping into details I found the following (comment shows RAM usage after this line
was executed):
load(myFile)# 40MB (19MB for R itself)
ds.tmp <- ds# 40MB; => only a pointer seems to be
copied
x<-rbind(ds.tmp, ds) # 198MB
x<-rbind(ds.tmp, ds) # 233MB; the same instruction a second time leads to
# 35MB more RAM usage - why?
Now I played around, but I couldn't find a solution. For example I bound each dataframe
step by step and removed the variables and cleared memory, but I still need 1140MB(!)
RAM:
start example 2 __
tmpFile<- paste(myFile,'.tmp',sep="")
load(myFile)
ds.tmp <- ds
save(ds.tmp, file=tmpFile, compress=T)
for (Cycle in 1:10) {
ds <- NULL
ds.tmp <- NULL
rm(ds, ds.tmp)
gc()
load(tmpFile)
load(myFile)
ds.tmp <- rbind(ds.tmp, ds)
save(ds.tmp,file=tmpFile, compress=T)
cat(Cycle,': ',object.size(ds),object.size(ds.tmp),'\n')
}
end example 1 __
platform i386-pc-solaris2.8
arch i386
os solaris2.8
system i386, solaris2.8
status
major1
minor9.1
year 2004
month06
day 21
language R
How can I avoid to run in that memory problem? Any ideas are very appreciated.
Thank you in advance & kind regards,
Lutz Thieme
AMD Saxony/ Product Engineering AMD Saxony Limited Liability Company & Co. KG
phone: + 49-351-277-4269 M/S E22-PE, Wilschdorfer Landstr. 101
fax:+ 49-351-277-9-4269 D-01109 Dresden, Germany
__
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
--
Roger D. Peng
http://www.biostat.jhsph.edu/~rpeng/
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Re: [R] how to "singlify" entries
On Mon, May 30, 2005 at 09:09:27AM -0400, Gabor Grothendieck wrote :
> Try using reshape, e.g. if dd is your data frame:
>
> reshape(dd, dir = "wide", idvar = "F1", timevar = "F2",
> varying = list(c("VX","VY")))
Thank you very much, and to Petr Pikal too. Reshape is exactly what I had
forgotten.
Now the bad news is that I have simplified my example ; I am in a
slightly more complex situation :
I have three factors, and one value
> count_per_tc[1:10,]
rna lib tc x
1 CAB 114BA T01F00380F47 1
2 CAE 114BB T01F00381273 1
3 CAJ 114BA T01F0048F6D1 1
4 CAB 114BC T01F0048F6D1 1
5 CAB 114BA T01F00498689 2
6 CAC 114BA T01F00498689 1
7 CAE 114BA T01F00498689 2
8 CAG 114BA T01F00498689 2
9 CAH 114BA T01F00498689 1
10 CAI 114BA T01F00498689 2
I would like a data frame where I have the value of x for each combination of
"rna" and "lib", for each "tc"
> reshape(count_per_tc[1:10,], direction="wide", timevar="tc",
> idvar=c("rna","lib"))
rna lib x.T01F00380F47 x.T01F00381273 x.T01F0048F6D1 x.T01F00498689
1 CAB 114BA 1 NA NA 2
2 CAE 114BB NA 1 NA NA
3 CAJ 114BA NA NA 1 NA
4 CAB 114BC NA NA 1 NA
6 CAC 114BA NA NA NA 1
7 CAE 114BA NA NA NA 2
8 CAG 114BA NA NA NA 2
9 CAH 114BA NA NA NA 1
10 CAI 114BA NA NA NA 2
oops, the other way round :
> t(reshape(count_per_tc[1:10,], direction="wide", timevar="tc",
> idvar=c("rna","lib")))
1 2 3 4 6 7 8 9
10
rna"CAB" "CAE" "CAJ" "CAB" "CAC" "CAE" "CAG" "CAH"
"CAI"
lib"114BA" "114BB" "114BA" "114BC" "114BA" "114BA" "114BA" "114BA"
"114BA"
x.T01F00380F47 " 1"NA NA NA NA NA NA NA
NA
x.T01F00381273 NA " 1"NA NA NA NA NA NA
NA
x.T01F0048F6D1 NA NA " 1"" 1"NA NA NA NA
NA
x.T01F00498689 " 2"NA NA NA " 1"" 2"" 2"" 1"
" 2"
The ultimate goal is (after proper renaming of the columns) to do things like
plot(CAA-114BA[CAA-114BA >0 & CAA-114BB > 0], CAA-114BB[CAA-114BA >0 &
CAA-114BB > 0])
(this combination will appear if I reshape the whole data frame, which has
200,000 rows.)
and then proper statistical tests (which I still have to learn / remember from
12 years ago).
once again, thank you, and please warn me if I am doing something stupid with
this transposition of the reshaped table.
Best regards,
--
Charles
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[R] Formatting numbers with a limited amount of digits consistently
I have tried to get signif, round and format to display numbers like these consistently in a table, using e.g. signif(x,digits=3) 17.01 18.15 I want 17.0 18.2 Not 17 18.2 Why is the last digit stripped off in the case when it is zero! Is this a "feature" of R or did I miss something? - Henrik Andersson Netherlands Institute of Ecology - Centre for Estuarine and Marine Ecology P.O. Box 140 4400 AC Yerseke Phone: +31 113 577473 [EMAIL PROTECTED] http://www.nioo.knaw.nl/ppages/handersson __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Reference Card?
> Hello! > > For LaTeX I found a reference Card at > http://www.cs.ualberta.ca/~c603/LaTeX_docs/Symbol_Source/latex > _symbols.pdf > > Is there something available for R? Hello Martin, See the reference cards on http://cran.r-project.org/other-docs.html Best, Matthias > PLEASE do read > the posting guide! http://www.R-project.org/posting-guide.html > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Reference Card?
Hi, Have you tried looking under Documentation -> Contributed, under CRAN? Kev Martin Klaffenboeck wrote: Hello! For LaTeX I found a reference Card at http://www.cs.ualberta.ca/~c603/LaTeX_docs/Symbol_Source/latex_symbols.pdf Is there something available for R? thanks, Martin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Ko-Kang Kevin Wang PhD Student Centre for Mathematics and its Applications Building 27, Room 1004 Mathematical Sciences Institute (MSI) Australian National University Canberra, ACT 0200 Australia Homepage: http://wwwmaths.anu.edu.au/~wangk/ Ph (W): +61-2-6125-2431 Ph (H): +61-2-6125-7407 Ph (M): +61-40-451-8301 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Reference Card?
Hello! For LaTeX I found a reference Card at http://www.cs.ualberta.ca/~c603/LaTeX_docs/Symbol_Source/latex_symbols.pdf Is there something available for R? thanks, Martin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Egarch
Hallo, I got to implement parameter estimation for Egarch(p,q) processes. My question: has anyone done that so far or are there implementations for other asymmetric garch models? Andreas [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] rbind wastes memory
[EMAIL PROTECTED] wrote:
Hello everybody,
if I try to (r)bind a number of large dataframes I run out of memory because R
wastes memory and seems to "forget" to release memory.
For example I have 10 files. Each file contains a large dataframe "ds" (3500 cols
by 800 rows) which needs ~20 MB RAM if it is loaded as the only object.
Now I try to bind all data frames to a large one and need more than 1165MB (!)
RAM (To simplify the R code, I use the same file ten times):
start example 1 __
load(myFile)
ds.tmp <- ds
for (Cycle in 1:10) {
ds.tmp <- rbind(ds.tmp, ds)
}
end example 1 __
Stepping into details I found the following (comment shows RAM usage after this line
was executed):
load(myFile)# 40MB (19MB for R itself)
ds.tmp <- ds# 40MB; => only a pointer seems to be
copied
x<-rbind(ds.tmp, ds) # 198MB
x<-rbind(ds.tmp, ds) # 233MB; the same instruction a second time leads to
# 35MB more RAM usage - why?
I'm guessing your problem is fragmented memory. You are creating big
objects, then making them bigger. This means R needs to go looking for
large allocations for the replacements, but they won't fit in the spots
left by the things you've deleted, so those are being left empty.
A solution to this is to use two passes: first figure out how much
space you need, then allocate it and fill it. E.g.
for (Cycle in 1:10) {
rows[Cycle] <- some calculation based on the data ...
}
ds.tmp <- data.frame(x=double(sum(rows)), y=double(sum(rows)), ...
for (Cycle in 1:10) {
ds.tmp[ appropriate rows, ] <- new data
}
Duncan Murdoch
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Re: [R] rbind wastes memory
[EMAIL PROTECTED] wrote:
> Hello everybody,
>
> if I try to (r)bind a number of large dataframes I run out of memory because R
> wastes memory and seems to "forget" to release memory.
>
> For example I have 10 files. Each file contains a large dataframe "ds" (3500
> cols
> by 800 rows) which needs ~20 MB RAM if it is loaded as the only object.
> Now I try to bind all data frames to a large one and need more than 1165MB (!)
> RAM (To simplify the R code, I use the same file ten times):
>
> start example 1 __
> load(myFile)
> ds.tmp<- ds
> for (Cycle in 1:10) {
> ds.tmp <- rbind(ds.tmp, ds)
> }
> end example 1 __
>
>
>
> Stepping into details I found the following (comment shows RAM usage after
> this line
> was executed):
> load(myFile) # 40MB (19MB for R itself)
> ds.tmp<- ds # 40MB; => only a pointer seems to be
> copied
> x<-rbind(ds.tmp, ds) # 198MB
> x<-rbind(ds.tmp, ds) # 233MB; the same instruction a second time
> leads to
> # 35MB more RAM usage - why?
>
>
> Now I played around, but I couldn't find a solution. For example I bound each
> dataframe
> step by step and removed the variables and cleared memory, but I still need
> 1140MB(!)
> RAM:
>
> start example 2 __
> tmpFile<- paste(myFile,'.tmp',sep="")
> load(myFile)
> ds.tmp <- ds
> save(ds.tmp, file=tmpFile, compress=T)
>
> for (Cycle in 1:10) {
> ds <- NULL
> ds.tmp <- NULL
> rm(ds, ds.tmp)
> gc()
> load(tmpFile)
> load(myFile)
> ds.tmp <- rbind(ds.tmp, ds)
> save(ds.tmp,file=tmpFile, compress=T)
> cat(Cycle,': ',object.size(ds),object.size(ds.tmp),'\n')
> }
> end example 1 __
>
>
> platform i386-pc-solaris2.8
> arch i386
> os solaris2.8
> system i386, solaris2.8
> status
> major1
> minor9.1
> year 2004
> month06
> day 21
> language R
>
>
>
>
> How can I avoid to run in that memory problem? Any ideas are very
> appreciated.
> Thank you in advance & kind regards,
If you are going to look at the memory usage you should use gc(), and
perhaps repeated calls to gc(), before checking the memory footprint.
This will force a garbage collection.
Also, you will probably save memory by treating your data frames as
lists and concatenating them, then converting the result to a data frame.
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[R] rbind wastes memory
Hello everybody,
if I try to (r)bind a number of large dataframes I run out of memory because R
wastes memory and seems to "forget" to release memory.
For example I have 10 files. Each file contains a large dataframe "ds" (3500
cols
by 800 rows) which needs ~20 MB RAM if it is loaded as the only object.
Now I try to bind all data frames to a large one and need more than 1165MB (!)
RAM (To simplify the R code, I use the same file ten times):
start example 1 __
load(myFile)
ds.tmp <- ds
for (Cycle in 1:10) {
ds.tmp <- rbind(ds.tmp, ds)
}
end example 1 __
Stepping into details I found the following (comment shows RAM usage after this
line
was executed):
load(myFile)# 40MB (19MB for R itself)
ds.tmp <- ds # 40MB; => only a pointer seems to be copied
x<-rbind(ds.tmp, ds)# 198MB
x<-rbind(ds.tmp, ds)# 233MB; the same instruction a second time
leads to
# 35MB more RAM usage - why?
Now I played around, but I couldn't find a solution. For example I bound each
dataframe
step by step and removed the variables and cleared memory, but I still need
1140MB(!)
RAM:
start example 2 __
tmpFile<- paste(myFile,'.tmp',sep="")
load(myFile)
ds.tmp <- ds
save(ds.tmp, file=tmpFile, compress=T)
for (Cycle in 1:10) {
ds <- NULL
ds.tmp <- NULL
rm(ds, ds.tmp)
gc()
load(tmpFile)
load(myFile)
ds.tmp <- rbind(ds.tmp, ds)
save(ds.tmp,file=tmpFile, compress=T)
cat(Cycle,': ',object.size(ds),object.size(ds.tmp),'\n')
}
end example 1 __
platform i386-pc-solaris2.8
arch i386
os solaris2.8
system i386, solaris2.8
status
major1
minor9.1
year 2004
month06
day 21
language R
How can I avoid to run in that memory problem? Any ideas are very appreciated.
Thank you in advance & kind regards,
Lutz Thieme
AMD Saxony/ Product Engineering AMD Saxony Limited Liability Company & Co. KG
phone: + 49-351-277-4269 M/S E22-PE, Wilschdorfer Landstr. 101
fax:+ 49-351-277-9-4269 D-01109 Dresden, Germany
__
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] values of bars in barplot
On 30 May 2005 at 15:19, luc tardieu wrote: > Hi, > > I couldn't find how to have the values written on the > top of each bar in a barplot. When using hist(), it is > possible to use labels=T, but this option does not > seem to exist for barplot(). Hallo >From help page Value: A numeric vector (or matrix, when 'beside = TRUE'), say 'mp', giving the coordinates of _all_ the bar midpoints drawn, useful for adding to the graph. e.g. > dat<-trunc(runif(5)*10) > dat [1] 5 7 1 9 5 > bpl<-barplot(dat, ylim=c(0,10)) > text(bpl,dat+.2, dat) HTH Petr > > Is there a trick I could use to do that ? > > Thanks to all > > Luc > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] values of bars in barplot
luc tardieu wrote: Hi, I couldn't find how to have the values written on the top of each bar in a barplot. When using hist(), it is possible to use labels=T, but this option does not seem to exist for barplot(). Is there a trick I could use to do that ? Thanks to all Luc __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html something like: x <- 1:3 #ensure some free space above the bar and remember midpoints: bmp <- barplot(x, ylim = c(0, 1.1*max(x))) text(bmp, x + .03*max(x), x) should do. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] values of bars in barplot
On 5/30/05, luc tardieu <[EMAIL PROTECTED]> wrote: > Hi, > > I couldn't find how to have the values written on the > top of each bar in a barplot. When using hist(), it is > possible to use labels=T, but this option does not > seem to exist for barplot(). > > Is there a trick I could use to do that ? > > Thanks to all > Here is an example you can modify: x <- c(44,56,34,35,44,51,55) nams <- LETTERS[1:7] bp <- barplot(x,horiz=T,col="light blue",xlim=c(0,60)) text(30,-1.25,xpd=NA,"Networks",cex=1.5) # x title text(-6,4,xpd=NA,"Research Areas",cex=1.5,srt=90) # y title text(x,bp,x,pos=4) # place numbers to right of bars text(0,bp,nams,cex=1.2,pos=4) # label bars right on the bars themselves __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to "singlify" entries
On 30 May 2005 at 21:56, Charles Plessy wrote:
> On Mon, May 30, 2005 at 09:15:32AM +, zhihua li wrote :
> > hi netters
> >
> > I have a rather simple question. I have a data frame with two
> > variables X and Y, both of which are factors. X has 100 levels while
> > Y has 10 levels only. The data frame has 100 rows in all, so for X
> > the values are unique, and Y has many replicate values. Now I wanna
> > reduce the data frame into 10 rows only, according to the 10 levels
> > of Y. I don't care which value of X is in the same row with Y in
> > the final data frame, as long as it is in agreement with the
> > original data frame.
>
> Dear list,
>
> I am a new subscriber, using R to analyse genomics data. I have a
> similar question, maybe even identical, but I am not sure...
Hallo
Me neither.
?reshape
> dat<-read.table("clipboard", header=T)
> dat
F1 F2 V
1 A X 3
2 A Y 6
3 B X 5
4 C X 9
5 C Y 3
> reshape(dat, idvar="F1", timevar="F2",direction="wide")
F1 V.X V.Y
1 A 3 6
3 B 5 NA
4 C 9 3
Homework: change NA to zero.
HTH
Petr
>
> >From a data frame with two factors and one value, I would like to
> >obtain a data
> frame with one factor and one value per level in the removed factor.
>
> For instance:
>
> F1 F2 V
> -
> A X 3
> A Y 6
> B X 5
> C X 9
> C Y 3
>
> Would become:
>
> F1 VX VY
> --
> A 3 6
> B 5 0
> C 9 3
>
> I am sure I have seen a tool to do this some time ago, but I do not
> remember its name.
>
> Can somebody help me ?
>
> Best regards,
>
> --
> Charles Plessy, Ph.D. - Genome Science Laboratory
> The Institute for Physical and Chemical Research (RIKEN)
> 2-1 Hirosawa, Wako, Saitama 351-0198, Japan
> [EMAIL PROTECTED] -- Fax: 048-462-4686 -- Tel: 048-467-9515
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
Petr Pikal
[EMAIL PROTECTED]
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[R] values of bars in barplot
Hi, I couldn't find how to have the values written on the top of each bar in a barplot. When using hist(), it is possible to use labels=T, but this option does not seem to exist for barplot(). Is there a trick I could use to do that ? Thanks to all Luc __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to "singlify" entries
On 5/30/05, Charles Plessy <[EMAIL PROTECTED]> wrote:
> On Mon, May 30, 2005 at 09:15:32AM +, zhihua li wrote :
> > hi netters
> >
> > I have a rather simple question. I have a data frame with two variables X
> > and Y, both of which are factors. X has 100 levels while Y has 10 levels
> > only. The data frame has 100 rows in all, so for X the values are unique,
> > and Y has many replicate values. Now I wanna reduce the data frame into 10
> > rows only, according to the 10 levels of Y. I don't care which value of X
> > is in the same row with Y in the final data frame, as long as it is in
> > agreement with the original data frame.
>
> Dear list,
>
> I am a new subscriber, using R to analyse genomics data. I have a
> similar question, maybe even identical, but I am not sure...
>
> >From a data frame with two factors and one value, I would like to obtain a
> >data
> frame with one factor and one value per level in the removed factor.
>
> For instance:
>
> F1 F2 V
> -
> A X 3
> A Y 6
> B X 5
> C X 9
> C Y 3
>
> Would become:
>
> F1 VX VY
> --
> A 3 6
> B 5 0
> C 9 3
>
> I am sure I have seen a tool to do this some time ago, but I do not remember
> its name.
>
Try using reshape, e.g. if dd is your data frame:
reshape(dd, dir = "wide", idvar = "F1", timevar = "F2",
varying = list(c("VX","VY")))
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Re: [R] how to "singlify" entries
On Mon, May 30, 2005 at 09:15:32AM +, zhihua li wrote : > hi netters > > I have a rather simple question. I have a data frame with two variables X > and Y, both of which are factors. X has 100 levels while Y has 10 levels > only. The data frame has 100 rows in all, so for X the values are unique, > and Y has many replicate values. Now I wanna reduce the data frame into 10 > rows only, according to the 10 levels of Y. I don't care which value of X > is in the same row with Y in the final data frame, as long as it is in > agreement with the original data frame. Dear list, I am a new subscriber, using R to analyse genomics data. I have a similar question, maybe even identical, but I am not sure... >From a data frame with two factors and one value, I would like to obtain a data frame with one factor and one value per level in the removed factor. For instance: F1 F2 V - A X 3 A Y 6 B X 5 C X 9 C Y 3 Would become: F1 VX VY -- A 3 6 B 5 0 C 9 3 I am sure I have seen a tool to do this some time ago, but I do not remember its name. Can somebody help me ? Best regards, -- Charles Plessy, Ph.D. - Genome Science Laboratory The Institute for Physical and Chemical Research (RIKEN) 2-1 Hirosawa, Wako, Saitama 351-0198, Japan [EMAIL PROTECTED] -- Fax: 048-462-4686 -- Tel: 048-467-9515 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Re: Vector Manipulation
x[cumsum(x!=0)!=0] or x[!!cumsum(!!x)] will also do it. On 5/30/05, ManojW <[EMAIL PROTECTED]> wrote: > OK...x[min(which(x!=0)):length(x)] does the trick! > > I guess the coffee is slowly but surely working! . > > Manoj > - Original Message - > From: ManojW > To: R-help > Sent: Monday, May 30, 2005 2:12 PM > Subject: Re: Vector Manipulation > > > I should clarify that I tried x[cumsum(x)!=0] but the problem is that I > might have negative numbers in the vector that can potentially make cumsum(x) > equal to zero somewhere down the line in the vector. > > Manoj >- Original Message - >From: ManojW >To: R-help >Sent: Monday, May 30, 2005 2:00 PM >Subject: Vector Manipulation > > >Dear All, >For any given vector, I want to extract a sub-vector such that the new > vector skips all zeros, if any , at the start of vector. Is it possible to > achieve this w/o looping? > >E.G : >x= c(0,0,1,2,3,4,5,0,0,8,9) > >y= somefunc(x); >> y >[1]1 2 3 4 5 0 0 8 9 > >In the example above, I want to skip the two leading zeroes till I hit > the non-zero (1 in the above case). I also want to retain all zero after the > first non-zero digit (again 1 in this case). > >Hope I am not confusing you guys. Thanks in advance for your help. > >Regards > >Manoj > > > > >[[alternative HTML version deleted]] > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] sapply following using by with a list of factors
McClatchie, Sam (PIRSA-SARDI) wrote:
Background:
OS: Linux Mandrake 10.1
release: R 2.0.0
editor: GNU Emacs 21.3.2
front-end: ESS 5.2.3
-
Colleagues
I am having some trouble extracting results from the function by, used to
average variables in a data.frame first by one factor (depth) and then by a
second factor (station). The real data.frame is quite large
dim(data.2001)
[1] 32049 11
Here is a snippet of code:
## bin density data for each station into 1 m depth bins, containing means
data.2001.test$integer.Depth <- as.factor(round(data.2001.test$Depth,
digits=0))
attach(data.2001.test)
binned.data.2001 <- by(data.2001.test[,5:11], list(depth=integer.Depth,
station=Station), mean)
and here is a snippet of the data.frame
dim(data.2001.test)
[1] 150 11
dump("data.2001.test", file=stdout())
data.2001.test <-
structure(list(Cruise = structure(as.integer(c(1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)), .Label =
"Ngerin01", class = "factor"),
Station = structure(as.integer(c(2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 5, 5)), .Label = c("a1", "a2", "a3",
"a4", "a5", "a6", "a7", "a8", "a9", "b1", "b2", "b3", "b4",
"b5", "b6", "b7", "c1", "c2", "c3", "c4", "c5", "c6", "c7",
"d1", "d2", "d3", "d4", "d5", "d6", "e1", "e2", "e3", "e4",
"e5", "e6", "e7", "f1", "f2", "f3", "f4", "f5", "f6", "f7",
"f8", "f9", "g1", "g10", "g11", "g2", "g3", "g4", "g5", "g6",
"g7", "g8", "g9", "gsvc1", "gsvc2", "gsvc3", "gsvc4", "gsvc5",
"gsvc6", "gsvc7", "gsvc8", "gsvc9", "gsvd1", "gsvd2", "gsvd3",
"gsvd4", "gsvd5", "gsvd6", "gsvd7", "gsvd8", "h1", "h11",
"h2", "h3", "h4", "h5", "h6", "h7", "h8", "h9", "i1", "i10",
"i2", "i3", "i4", "i5", "i6", "i7", "i8", "i9", "j1", "j10",
"j2", "j3", "j4", "j5", "j6", "j7", "j8", "j9", "k1", "k10",
"k2", "k3", "k4", "k5", "k6", "k7", "k8", "k9", "l1", "l10",
"l11", "l12", "l2", "l3", "l4", "l5", "l6", "l7", "l8", "l9",
"m1", "m10", "m11", "m2", "m3", "m4", "m5", "m6", "m7", "m8",
"m9", "mx", "n1", "n10", "n11", "n2", "n3", "n4", "n5", "n6",
"n7", "n8", "n9", "nx", "o1", "o10", "o11", "o13", "o2",
"o3", "o4", "o5", "o6", "o7", "o8", "o9", "ox", "p10", "p11",
"p2", "p3", "p4", "p5", "p6", "p7", "p8", "p9", "px", "q1",
"q10", "q11", "q12", "q13", "q2", "q3", "q4", "q5", "q6",
"q7", "q8", "q9", "qx", "r1", "r10", "r11", "r12", "r13",
"r14", "r15", "r2", "r3", "r4", "r5", "r6", "r7", "r8", "r9",
"rx", "s1", "s10", "s11", "s12", "s13", "s14", "s15", "s16",
"s2", "s3", "s4", "s5", "s6", "s7", "s8", "s9", "sgc1", "sgc2",
"sgc3", "sgc4", "sgc5", "sgc6", "sgc7", "sx", "t1", "t10",
"t11", "t12", "t13", "t14", "t15", "t16", "t2", "t3", "t4",
"t5", "t6", "t7", "t8", "t9", "tx", "u1", "u10", "u11", "u12",
"u13", "u14", "u15", "u2", "u3", "u4", "u5", "u6", "u7",
"u8", "u9", "ux", "v1", "v10", "v11", "v2", "v3", "v4", "v5",
"v6", "v7", "v8", "v9", "vx", "w2", "w3", "w4", "w5", "w6",
"w7", "w8", "w9", "wx", "x2", "x3", "x4", "x5", "x6", "x7",
"x8"), class = "factor"), Lon = c(138.421, 138.421, 138.421,
138.421, 138.421, 138.421, 138.421, 138.421, 138.421, 138.421,
138.421, 138.421, 138.421, 138.421, 138.421, 138.421, 138.421,
138.421, 138.421, 138.421, 138.421, 138.421, 138.421, 138.421,
138.421, 138.421, 138.421, 138.421, 138.421, 138.421, 138.421,
138.421, 138.421, 138.421, 138.421, 138.421, 138.421, 138.421,
138.421, 138.421, 138.421, 138.421, 138.421, 138.421, 138.421,
138.421, 138.421, 138.421, 138.421, 138.421, 138.421, 138.421,
138.421, 138.421, 138.352, 138.352, 138.352, 138.352, 138.352,
138.352, 138.352, 138.352, 138.352, 138.352, 138.352, 138.352,
138.352, 138.352, 138.352, 138.352, 138.352, 138.352, 138.352,
138.352, 138.352, 138.352, 138.352, 138.352, 138.352, 138.352,
138.352, 138.352, 138.352, 138.352, 138.352, 138.352, 138.352,
138.352, 138.352, 138.352, 138.352, 138.352, 138.352, 138.352,
138.352, 138.352, 138.352, 138.352, 138.352, 138.352, 138.352,
138.352, 138.352, 13
Re: [R] how to invert the matrix with quite small eigenvalues
On 30-May-05 huang min wrote: > Maybe I should state more clear that I define b to get the > orthogonal matrix bb$vectors. OK. Certainly bbv<-bb$vectors is close to orthogonal: bbv%*%bbv differs from the unit matrix only in that the off-diagonal terms are O(10^(-16)). > We also can define diag(b)<-diag(b)+100, which will make the > eigenvalues of b much bigger to make sure the orthogonal matrix is > reliable. > > My intention is to invert the covariance matrix to perform some > algorithm which is common in the estimating equations like GEE. Which comes back to my general question: why do you need to ensure that you get correct results for matrices like these? This matrix is very nearly singular, and in many contexts you would interpret it as exactly singular (e.g. if I got such a matrix as the empirical covariance matrix from a sample, or by computation from the structure of a model such as a design matrix, I would not normally want to preserve this very small margin of non-singularity: I would replace it with the lower-dimensional singular version, e.g. by decomposing it with eigen() or svd() and reconstructing the singular version after setting very small eigenvalues to 0). But then of course there would be no inverse, which might be required by the further computational expressions you want to evaluate. However, in that case (depending on your application), you may be able to proceed satisfactorily by using ginv() (see package MASS) instead of solve(). But if you take that route, then you have to bear in mind that a generalised inverse is not a true inverse (if only because the latter does not exist): the only property required for G=ginv(M) is that M%*%G%*%M = M If such a matrix G will work for the rest of your calculations, then you are OK, in particular if what you need is a possible solution to an under-determined system of linear equations. If not, then you should seriously consider whether your computational strategy is suitable for the problem you have in hand. > [...] > Another strange point is that my friend use the LU decomposition > in Fortran to solve the inverse matrix of aa for me. He used > double precision of course, otherwise no inverse matrix in > Fortran too. He show that the product of the two matrix is > almost identity (The biggest off-digonal element is about 1e-8). > I copy his inverse matrix(with 31 digits!) to R and read in aa > I sent to him(16 digits). The product is also not an identity > matrix. It is fairly strange! Any suggestion? These phenomena are yet another instance of the fact that at the margins of computability the results will be anomalouus. Your friend is simply using a narrower margin, but it is still not exact! Not strange at all. Best wishes, Ted. E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 094 0861 Date: 30-May-05 Time: 11:41:28 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] joining files after canonical correlation
Brett Stansfield wrote: Dear R, I recently did a canonical correlation analysis on two subsets of data (location and weather). So I now have canonical scores for location and weather. but I'd now like to do a scatterplot matrix using the pairs statement. Is there a way to somehow join location.U and weather.V to become a new data set from which I could undertake a scatterplot matrix of the canonical variates? cbind() can create a matrix by binding together the columns of two (or more) input matrices. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Chars as numbers
Package 'sfsmisc'
has a function AsciiToInt() and a few useful related "R-code"
only functions such as chars8bit(); see the help pages once
you've installed and attached the package.
But do note that these things do depend on the encoding, as Uwe
Ligges has already told you.
Things work fine for pure ASCII {independently of encoding}
but already differ for Umlauts / Accents
between iso-latin-1 or Unicode.
Martin Maechler, ETH Zurich
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to "singlify" entries
Hallo On 30 May 2005 at 9:15, zhihua li wrote: > hi netters > > I have a rather simple question. I have a data frame with two Well, I do not understand you simple question fully. You have something like that dat<-data.frame(X=1:100, Y=sample(1:10,10)) dat$X<-factor(dat$X) dat$Y<-factor(dat$Y) dat$Y[5]<-10 > variables X and Y, both of which are factors. X has 100 levels while Y > has 10 levels only. The data frame has 100 rows in all, so for X the > values are unique, and Y has many replicate values. Now I wanna > reduce the data frame into 10 rows only, according to the 10 levels of > Y. I don't care which value of X is in the same row with Y in the > final data frame, as long as it is in agreement with the original data > frame. Do you want to choose only some rows from your data frame to get unique Y and any corresponding X? dat[!duplicated(dat$Y),] Or do you want something different? HTH Petr > > I think this task can be carried out with some function like > aggregate. but I failed in figuring it out. Could anybody give me a > hint? > > Thanks a lot! > > __ > R-help@stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html Petr Pikal [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] how to invert the matrix with quite small eigenvalues
Maybe I should state more clear that I define b to get the orthogonal matrix bb$vectors. We also can define diag(b)<-diag(b)+100, which will make the eigenvalues of b much bigger to make sure the orthogonal matrix is reliable. My intention is to invert the covariance matrix to perform some algorithm which is common in the estimating equations like GEE. I meet difficulty to invert the covariance matrix. Two possibilities here: 1. The rounding error in defining the covariance matrix make the eigenvalue to small. 2. The solve function in R can not cope with the matrix with so small an eigenvalue. For the first possibility, I think it can not be improved unless we can define more precise number than the double precision. So I ask for the possiblity of coping with the second. I can not find the default way to invert the matrix with solve(). For the symmetric matrix, I wonder if there are some algorithm which can naturally make the inverse matrix symmetric and make sure it is the inverse in the sense that the product is an identity matrix. I know there are many decompositions which can be used to find the inverse of a matrix. QR, SVD, Chol, and maybe some iterative method. I wonder anyone can suggest me which algorithm might be good. Another strange point is that my friend use the LU decomposition in Fortran to solve the inverse matrix of aa for me. He used double precision of course, otherwise no inverse matrix in Fortran too. He show that the product of the two matrix is almost identity(The biggest off-digonal element is about 1e-8). I copy his inverse matrix(with 31 digits!) to R and read in aa I sent to him(16 digits). The product is also not an identity matrix. It is fairly strange! Any suggestion? Regards, Huang On 5/30/05, Ted Harding <[EMAIL PROTECTED]> wrote: > On 30-May-05 huang min wrote: > > Dear all, > > > > I encounter some covariance matrix with quite small eigenvalues > > (around 1e-18), which are smaller than the machine precision. The > > dimension of my matrix is 17. Here I just fake some small matrix for > > illustration. > > > > a<-diag(c(rep(3,4),1e-18)) # a matrix with small eigenvalues > > b<-matrix(1:25,ncol=5) # define b to get an orthogonal matrix > > b<-b+t(b) > > NB: b is not *orthogonal*! Each row of b is equal to the preceding > row plus (row2 - row1) (and similar for the columns), and b has > rank 2 (though eigen(b) taken literally says that it has 5 non-zero > eugenvalues; however, 3 of these are snaller that 10^(-14)). > > Perhaps you meant "define b to get a symmetric matrix". > > > bb<-eigen(b,symmetric=T) > > aah<-bb$vectors%*%diag(1/sqrt(diag(a))) > > aa<-aah%*%t(aah) # aa should have the same eigenvalues as a and > > should be #invertable,however, > > solve(aa) # can not be solved > > Well, I did get a (non-symmetric) result for solve(aa) ... > > > solve(aa,tol=1e-19) # can be inverted, however, it is not symmetric > > and furthermore, > > and an idenitical (to solve(aa)) result for this. > > > solve(aa,tol=1e-19)%*%aa # deviate much from the identity matrix > > But here I agree with you! > > > I have already define aa to make sure it is symmetric. So the inverse > > should be symmetric. > > > > Does the problem come from the rounding error since the eigenvalue is > > smaller than the machine precision? In fact, the eigenvalue of aa is > > negative now, but at least, it is still invertable. How can I get the > > inverse? Thanks. > > It does indeed, like the eigenvalue result for b above, come from > the rounding error. > > You should clarify in your mind why you want to ensure that you get > correct results for matrices like these. > > You are (and in your example deliberately so) treading on the very > edge of what is capable of being computed, and results are very likely > to lead to unexpected gross anomalies (such as being unable to > invert a mathematically invertible matrix, or getting a non-symmetric > inverse to a symmetric matrix [depending on the algorithm], or > getting non-zero values for eigenvalues which should be zero, or > the gross difference from the identity matrix which you expected). > > It is like using a mechanical ditch-digger to peel an apple. > > Exactly what will happen in any particular case can only be > determined by a very fine-grained analysis of the operation > of the numerical algorithm, which is beyond your reach in the > normal usage of R. > > Best wishes, > Ted. > > > > E-Mail: (Ted Harding) <[EMAIL PROTECTED]> > Fax-to-email: +44 (0)870 094 0861 > Date: 30-May-05 Time: 09:43:56 > -- XFMail -- > __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Menu Language
Hafner, Reinhold (Risklab) wrote: Hello, In my windows xp system the regional settings are set to German. However, I still would like to use the English version of R with English menu items and texts. RWinEdt tells me this is necessary in order to use the mdi framework. How can this be achieved with the 2.1 version. In 2.0.1 this wasn't a problem. Yes, because R-2.0.1 has not had any translations. You can set an envrionment variable LANGUAGE=en for exmaple... See the new documentation and the most recent R Newsletter which has some nice articles on languages and internationalization by Brian Ripley. Uwe Ligges Reinhold [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] how to "singlify" entries
hi netters I have a rather simple question. I have a data frame with two variables X and Y, both of which are factors. X has 100 levels while Y has 10 levels only. The data frame has 100 rows in all, so for X the values are unique, and Y has many replicate values. Now I wanna reduce the data frame into 10 rows only, according to the 10 levels of Y. I don't care which value of X is in the same row with Y in the final data frame, as long as it is in agreement with the original data frame. I think this task can be carried out with some function like aggregate. but I failed in figuring it out. Could anybody give me a hint? Thanks a lot! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
RE: [R] how to invert the matrix with quite small eigenvalues
On 30-May-05 huang min wrote: > Dear all, > > I encounter some covariance matrix with quite small eigenvalues > (around 1e-18), which are smaller than the machine precision. The > dimension of my matrix is 17. Here I just fake some small matrix for > illustration. > > a<-diag(c(rep(3,4),1e-18)) # a matrix with small eigenvalues > b<-matrix(1:25,ncol=5) # define b to get an orthogonal matrix > b<-b+t(b) NB: b is not *orthogonal*! Each row of b is equal to the preceding row plus (row2 - row1) (and similar for the columns), and b has rank 2 (though eigen(b) taken literally says that it has 5 non-zero eugenvalues; however, 3 of these are snaller that 10^(-14)). Perhaps you meant "define b to get a symmetric matrix". > bb<-eigen(b,symmetric=T) > aah<-bb$vectors%*%diag(1/sqrt(diag(a))) > aa<-aah%*%t(aah) # aa should have the same eigenvalues as a and > should be #invertable,however, > solve(aa) # can not be solved Well, I did get a (non-symmetric) result for solve(aa) ... > solve(aa,tol=1e-19) # can be inverted, however, it is not symmetric > and furthermore, and an idenitical (to solve(aa)) result for this. > solve(aa,tol=1e-19)%*%aa # deviate much from the identity matrix But here I agree with you! > I have already define aa to make sure it is symmetric. So the inverse > should be symmetric. > > Does the problem come from the rounding error since the eigenvalue is > smaller than the machine precision? In fact, the eigenvalue of aa is > negative now, but at least, it is still invertable. How can I get the > inverse? Thanks. It does indeed, like the eigenvalue result for b above, come from the rounding error. You should clarify in your mind why you want to ensure that you get correct results for matrices like these. You are (and in your example deliberately so) treading on the very edge of what is capable of being computed, and results are very likely to lead to unexpected gross anomalies (such as being unable to invert a mathematically invertible matrix, or getting a non-symmetric inverse to a symmetric matrix [depending on the algorithm], or getting non-zero values for eigenvalues which should be zero, or the gross difference from the identity matrix which you expected). It is like using a mechanical ditch-digger to peel an apple. Exactly what will happen in any particular case can only be determined by a very fine-grained analysis of the operation of the numerical algorithm, which is beyond your reach in the normal usage of R. Best wishes, Ted. E-Mail: (Ted Harding) <[EMAIL PROTECTED]> Fax-to-email: +44 (0)870 094 0861 Date: 30-May-05 Time: 09:43:56 -- XFMail -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Menu Language
Hello, In my windows xp system the regional settings are set to German. However, I still would like to use the English version of R with English menu items and texts. RWinEdt tells me this is necessary in order to use the mdi framework. How can this be achieved with the 2.1 version. In 2.0.1 this wasn't a problem. Reinhold [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] how to invert the matrix with quite small eigenvalues
Dear all, I encounter some covariance matrix with quite small eigenvalues (around 1e-18), which are smaller than the machine precision. The dimension of my matrix is 17. Here I just fake some small matrix for illustration. a<-diag(c(rep(3,4),1e-18)) # a matrix with small eigenvalues b<-matrix(1:25,ncol=5) # define b to get an orthogonal matrix b<-b+t(b) bb<-eigen(b,symmetric=T) aah<-bb$vectors%*%diag(1/sqrt(diag(a))) aa<-aah%*%t(aah) # aa should have the same eigenvalues as a and should be #invertable,however, solve(aa) # can not be solved solve(aa,tol=1e-19) # can be inverted, however, it is not symmetric and furthermore, solve(aa,tol=1e-19)%*%aa # deviate much from the identity matrix I have already define aa to make sure it is symmetric. So the inverse should be symmetric. Does the problem come from the rounding error since the eigenvalue is smaller than the machine precision? In fact, the eigenvalue of aa is negative now, but at least, it is still invertable. How can I get the inverse? Thanks. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
