[R] Re: pls regression - optimal number of LVs
On Thursday 24 July 2003 05:02, Dowkiw, Arnaud wrote: Dear R-helpers, I have performed a PLS regression with the mvr function from the pls.pcr package an I have 2 questions : 1- do you know if mvr automatically centers the data ? It seems to me that it does so... Yup, it does... common practice. 2- why in the situation below does the output say that the optimal number of latent variables is 4 ? In my humble opinion, it is 2 because the RMS increases and the R2 decreases when 3 LVs are considered : Many criteria exist and for some data sets they agree, for most they do not. The criterion applied here checks whether the decrease in cross-validated error is significant; Hastie et al. use it in their book The elements of statistical learning. It is described in the man page, and like all criteria, it is not guaranteed to satisfy all users. If you feel better using 2LVs, you can do that. Ron -- Ron Wehrens Dept. of Chemometrics University of Nijmegen Email: [EMAIL PROTECTED] Toernooiveld 1 http://www-cac.sci.kun.nl/cac/ 6525 ED NijmegenTel: +31 24 365 2053 The Netherlands Fax: +31 24 365 2653 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] nls.control in gnls
Hi, I've made a selfStart function for use with gnls and the following piece of code works nicely: check1 - gnls(y ~ spot.shape.fct(xcord, ycord, background, spotintensity, rho, sigma, delta, mux, muy), start=getInitial(y ~ spot.shape.fct(xcord, ycord, background, spotintensity, rho, sigma, delta, mux, muy), data=claus), data=claus ) spot.shape.fct is the selfStart function and I later want to include a correlation term which is why I'm using gnls instead of nls. I want to get rid of the start= option, but I keep getting the following error message: Error in nls(formula = y ~ spot.shape.fct(xcord, ycord, background, spotintensity, : number of iterations exceeded maximum of 50 From the help page and the glns code I can see that the initial start values are improved by estimating the parameters in a model without the correlation structures that are possible with gnls. This is quite reasonable and will quite likely improve the convergence succes and time once a correlation is introduced. Obviously the call to nls does not converge before the standard 50 iterations specified in nls.control so I was wondering if it is possible to pass nls.control information on to nls when calling gnls? Several work-around exists: * Write a better initial function to get more precise starting values, i.e., fewer iterations needed for nls. * Make a new function my.own.gnls where nls.control(maxiter=100) is set in the call to nls. Maybe it would even be a good idea to pass nlsMaxIter from gnlsControl on to the initial call to nls? As far as I can see from the code it is only used in maximization after the call to get the starting value. Happy R'ing Claus -- * Claus Thorn Ekstrøm [EMAIL PROTECTED] Dept of Mathematics and Physics, KVL Thorvaldsensvej 40 DK-1871 Frederiksberg C Denmark Phone:[+45] 3528 2341 Fax: [+45] 3528 2350 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R]: performing marginal tests to glm objects
Dear all, I wonder if it is possible to obtain marginal tests for effects in generalized linear models. Indeed, the anova function produces sequential tests and it doesn't have any type argument to specify that we would like marginal tests instead, as in the similar anova function for lme objects. Thanks a lot for your help! Eve CORDA Office national de la chasse et de la faune sauvage (http://www.oncfs.gouv.fr) BP 20 78612 Le Perray en Yvelines Cedex FRANCE Phone: +33 (0)1 30 46 60 64 Fax: +33 (0)1 30 46 60 99 [EMAIL PROTECTED] [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R]: performing marginal tests to glm objects
?drop1 for the valid marginal tests. If you want `type III' tests (which aren't marginal in the usual sense) see Anova in package car, which also does `type II' tests (as performed by drop1). On Thu, 24 Jul 2003, Eve Corda wrote: I wonder if it is possible to obtain marginal tests for effects in generalized linear models. Indeed, the anova function produces sequential tests and it doesn't have any type argument to specify that we would like marginal tests instead, as in the similar anova function for lme objects. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Confidence Band for empirical distribution function
Dear Kjetil,
As I already mentioned, it appears that there isn't a function
available calculating the quantiles directly (at least, it doesn't appear
in the C source of ctest). So as I already suggested, uniroot (or a similar
C routine which calls the corresponding C code directly) is probably the
best you can do (apart from writing it completely yourself).
I didn't program this using uniroot, but I'd certainly try the following
for speed-up:
- For symmetry reasons, you only need to compute half of the quantiles.
- The quantiles depend smoothly on the probabilities (of your reference
distribution). Therefore, calculating only a few for probabilities
between 0 and 0.5, and using (e.g. linear) interpolation should be
satisfying.
I am sorry not be of more help.
HTH anyway
Thomas
-Original Message-
From: kjetil brinchmann halvorsen [mailto:[EMAIL PROTECTED]
Sent: 23 July 2003 15:46
To: Hotz, T.
Subject: RE: [R] Confidence Band for empirical distribution function
On 22 Jul 2003 at 11:37, Hotz, T. wrote:
Dear Leif,
If you look at the definition of ks.test, you'll find the lines
pkstwo - function(x, tol = 1e-06) {
if (is.numeric(x))
x - as.vector(x)
else stop(Argument x must be numeric)
p - rep(0, length(x))
p[is.na(x)] - NA
IND - which(!is.na(x) (x 0))
if (length(IND) 0) {
p[IND] - .C(pkstwo, as.integer(length(x)), p =
as.double(x[IND]),
as.double(tol), PACKAGE = ctest)$p
}
return(p)
}
which calls C code to calculate the p-values given the test
statistic.
You'll find explanations on what this function does in the
original C file
src/library/ctest/src/ks.c
I haven't tried that but I assume that you could use it to
calculate p-values
given the test-statistics yourself.
That could certainly be done, but what was asked for is the inverse,
which can be calculated, using for instance uniroot(). I tried that,
but it is to slow, .C() will be called repeatedly in a loop. For me
it took several minutes.
Kjetil Halvorsen
Please also note that ks.test() returns the p-value as well.
If you need quantiles, I assume you need to invert the cdf yourself,
e.g. using uniroot().
HTH
Thomas
---
Thomas Hotz
Research Associate in Medical Statistics
University of Leicester
United Kingdom
Department of Epidemiology and Public Health
22-28 Princess Road West
Leicester
LE1 6TP
Tel +44 116 252-5410
Fax +44 116 252-5423
Division of Medicine for the Elderly
Department of Medicine
The Glenfield Hospital
Leicester
LE3 9QP
Tel +44 116 256-3643
Fax +44 116 232-2976
-Original Message-
From: Leif.Boysen [mailto:[EMAIL PROTECTED]
Sent: 21 July 2003 14:42
To: [EMAIL PROTECTED]
Subject: [R] Confidence Band for empirical distribution function
Hi,
I was trying to draw an empirical distribution function
with uniform
confidence bands. So I tried to find a way to calculate
values of the
Kolmogorov-Smirnov Distribution but failed.
I guess it must be hidden somewhere (since the ks-test is
implemented),
but I was unable to find it.
Is there any way to do this?
Thanks
Leif Boysen
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[R] How can I represent a shape using 1-D discrete waveletdescriptor s?
I only concern about the boundary of the shape and I dont concern gray level. U[m] = x[m] + jy[m] m= 0,1,,N-1 is the N points on the boundary. It is easy to express U as a discrete Fourier series. But it seems like that wd in the wavethresh package cannot deal with complex values. And it is not interesting to transfer U to a matrix. Thanks a lot Qin __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] R won't connect to the internet on Linux!
OK, I really am struggling with this one! Forgive me if I am being stupid I am running R 1.7.1 on Suse Linux 8.1. I connect to the internet through a proxy so I have: IAHC-LINUX03:~ # echo $http_proxy wwwcache.bbsrc.ac.uk:8080 IAHC-LINUX03:~ # echo $HTTP_PROXY wwwcache.bbsrc.ac.uk:8080 just in case ;-) SO, i go into R and I get: source(http://www.bioconductor.org/getBioC.R;) unable to connect to 'www.bioconductor.org' on port 80. Error in file(file, r) : cannot open URL `http://www.bioconductor.org/getBioC.R' OK so is R just not picking up my proxy setting? It seems to be trying port 80 on something, and I have specifically set it to port 8080 in my environment variables. As far as I can see I have followed the reference manual suggestion, so does anyone else have one? Thanks Mick __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Passing references to data objects into R functions
One way is to use an object-oriented design and wrap up the reference
functionality in a common superclass. At
http://www.maths.lth.se/help/R/ImplementingReferences/ I have got some
discussions which are in line what you are trying to achieve and that
you might be able to adopt.
Also, note that passing huge objects as arguments to functions is NOT
expensive (considering memory or time) in R if they are used for
read-only purposes. It only becomes expensive if you assign a new value
to the argument. In such cases R *has to* copy the whole object to make
sure you only modify a local instance of the object. Thus, objects can
be though of being passed by reference to functions as long as they are
not modified, if modified they are passed by value. This is intentional
as R is a (one-threaded) functional language.
Best wishes
Henrik Bengtsson
Lund University
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of David
Khabie-Zeitoune
Sent: den 23 juli 2003 18:18
To: [EMAIL PROTECTED]
Subject: [R] Passing references to data objects into R functions
Hi.
I have the following question about reading from large data
objects from within R functions; I have tried to simplify my
problem as much as possible in what follows.
Imagine I have various large data objects sitting in my
global environment (call them data1, data2, ...). I want
to write a function extract that extracts some of the rows
of a particular data object, does some further manipulations
on the extract and then returns the result. The function
takes the data object's name and an index vector -- for
example the following call would return the first 3 rows of
object data1.
ans = extract(data1, 1:3)
I could write a simple function like this:
extract1 = function(object.name, index) {
temp = get(object.name, envir = .GlobalEnv)
temp = temp[index, , drop=FALSE]
# do some further manipulations here
return(temp)
}
The problem is that the function makes a copy temp of the
object in the function frame, which (in my application) is
very memory inefficient as the data objects are very large.
It is especially inefficient when the length of the index
vector is much smaller than the number of rows in the data
object. What I really would like to do is to be able to read
from the underlying data object directly (in other
programming languages this would be achieved by passing a
pointer to the object instead), without making a copy.
Given the rules of variable name scoping in R, I could avoid
making a copy with the following call:
extract2 = function(object.name, index) {
eval(parse(text = temp = , object.name, [index, , drop=FALSE],
sep=))
# do some further manipulations here
return(temp)
}
But this seems very messy. Is there a better way?
Thanks for your help
David Khabie-Zeitoune
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[R] median and joint distribution
Dear R-helpers! May I kindly ask the pure statistics-experts to help me for a purpose which first part is not directly concerned with R. Consider two distribution functions, say f and g. For both, the median is smaller than a half. Now, the multiplicative or additive linkage of both distribution leads to a new distribution function, say h, whereas the median of h is greater than a half. Does anybody know under which circumstances such a construction of h is possible (my intuition is that it depends on the correlation of f and g) or can anybody advice a helpful literature. Furthermore, does anybody know whether or how such a construction can be done with R. Thanks in advance. s. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] S3 and S4 classes
I've made some changes (in bold) following your comments :
# class definition :
setClass(MyClass, representation(mynumber=numeric));
# the initilization method :
setMethod(initialize,MyClass,
function(.Object)
{
[EMAIL PROTECTED] - 10;
return(.Object);
}
);
# defining the generic function generates a warning message (already
generic function)- but everything works fine without it - so it seems
better not to use it
*setGeneric(perform, function(object) standardGeneric(perform));
*
# the perform method definition - this alone seems sufficient for me,
and S4-looking
*setMethod(perform,MyClass,**
function(object) {
[EMAIL PROTECTED] - [EMAIL PROTECTED] + 10;
return(object);
} **);*
I replaced my unlucky /.Object/ by /object/ but for method initialize.
The execution of my program used to begin with an output like :
Creating a new generic function for perform
With the setMethod only, there is not this output any longer. Does this
mean that the execution becomes faster ?
Thank you for your fast and useful answers. I've found an interesting R
help page I had never seen before at
http://www.maths.lth.se/help/R/S3toS4 which gives additionnal
information about S3 and S4 classes.
Regards,
Laurent
http://www.maths.lth.se/help/R/S3toS4/
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Re: [R] R won't connect to the internet on Linux!
On Thu, 24 Jul 2003, michael watson (IAH-C) wrote: OK, I really am struggling with this one! Forgive me if I am being stupid I am running R 1.7.1 on Suse Linux 8.1. I connect to the internet through a proxy so I have: IAHC-LINUX03:~ # echo $http_proxy wwwcache.bbsrc.ac.uk:8080 IAHC-LINUX03:~ # echo $HTTP_PROXY wwwcache.bbsrc.ac.uk:8080 just in case ;-) SO, i go into R and I get: source(http://www.bioconductor.org/getBioC.R;) unable to connect to 'www.bioconductor.org' on port 80. Error in file(file, r) : cannot open URL `http://www.bioconductor.org/getBioC.R' OK so is R just not picking up my proxy setting? Your setting is wrong, so it is being ignored. The help page says quite explicitly The form of `http_proxy' should be `http://proxy.dom.com/;' or `http://proxy.dom.com:8080/;' where the port defaults to `80' and the trailing slash may be omitted. It seems to be trying port 80 on something, and I have specifically set it to port 8080 in my environment variables. As far as I can see I have followed the reference manual suggestion, so does anyone else have one? The problem is in your seeing, it seems. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] geoR size limit problem
Hi all, I tried to produce some kriged surfaces with geoR (latest version). The size of the grid should be around 900 x 650 cells (what I find is not a very big grid), and the number of points is around 2500. The command krige.conv stopped after arround 5 min saying it can not allocate a vector with around 1.5 billion units. Sounds reasonable. Is there a workaround? How would I partition the map into several smaller pieces to krige them separately and retain the smooth transitions from one part to another? Is there a possibility to apply a mask to the grid in a way that it only interpolates the cells of interest? My machine has 512MB of memory and additional 2000MB of swap. Thanks in advance, Miha Staut __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Condition indexes and variance inflation factors
Thanks for all the help. Juergen Gross supplied a program which does just what Belsley suggested. Chuck Cleland, John Fox and Andy Liaw all made useful programming suggestions. John Fox asked (1) I've never liked this approach for a model with a constant, where it makes more sense to me to centre the data. I realize that opinions differ here, but it seems to me that failing to centre the data conflates collinearity with numerical instability. Opinions do differ. A few years ago, I could have given more details (my dissertation was on this topic, but a lot of the details have disappeared from memory); I think, though, that Belsley is looking for a measure that deals not only with collinearity, but with several other problems, including numerical instability (the subtitle of his later book is Collinearity and Weak Data in Regression). I remember being convinced that centering was generally not a good idea, but there are lots of people who disagree and who know a lot more statistics than I do. (2) I also disagree with the comment that condition indices are easier to interpret than variance-inflation factors. In either case, since collinearity is a continuous phenomenon, cutoffs for large values are necessarily arbitrary. While any cutoff is arbitrary (and Belsley advises against using a cutoff rigidly) he does provide some evidence of how regression models with different condition indices are affected by them. (3) If you're interested in figuring out which variables are involved in each collinear relationship, then (for centred and scaled data) you can equivalently (and to me, more intuitively) work with the principal-components analysis of the predictors. This would also work. (4) I have doubts about the whole enterprise. Collinearity is one source of imprecision -- others are small sample size, homogeneous predictors, and large error variance. Aren't the coefficient standard errors the bottom line? If these are sufficiently small, why worry? I think (correct me if I am wrong) that the s.e.s and the condition indices serve very different purposes. The condition indices are supposed to determine if small changes in the input data could make big differences in the results. Belsley provides some examples where a tiny change in the data results in completely different results (e.g., different standard errors, different coefficients (even reversing sign) and so on). Peter Peter L. Flom, PhD Assistant Director, Statistics and Data Analysis Core Center for Drug Use and HIV Research National Development and Research Institutes 71 W. 23rd St www.peterflom.com New York, NY 10010 (212) 845-4485 (voice) (917) 438-0894 (fax) __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] median and joint distribution
For distribution functions F and G we have the (Frechet) bounds:
max{0, F(x)+G(y) -1} = H(x,y) = min{F(x),F(y)}
where H is the joint df of (X,Y) having marginals F and G. If
X and Y are comonotonic (Schmeidler (Econometrica, 1989)), that is
if there is a random variable Z, such that X = f(Z), and Y=g(Z)
for monotone f and g, then the upper Frechet bound holds and the
quantile function of X+Y is the sum of the quantile functions of
X and Y. For multiplicative linkage take logs, presuming, of
course, that I'm not misinterpreting this phrase. One can think
of comonotone X and Y as perfectly concordant in the language
of rank correlation.
url:www.econ.uiuc.edu/~roger/my.htmlRoger Koenker
email [EMAIL PROTECTED] Department of Economics
vox:217-333-4558University of Illinois
fax:217-244-6678Champaign, IL 61820
On Thu, 24 Jul 2003, Salvatore Barbaro wrote:
Dear R-helpers!
May I kindly ask the pure statistics-experts to help me for a
purpose which first part is not directly concerned with R.
Consider two distribution functions, say f and g. For both, the
median is smaller than a half. Now, the multiplicative or additive
linkage of both distribution leads to a new distribution function,
say h, whereas the median of h is greater than a half. Does
anybody know under which circumstances such a construction of h is
possible (my intuition is that it depends on the correlation of f
and g) or can anybody advice a helpful literature. Furthermore,
does anybody know whether or how such a construction can be done
with R. Thanks in advance.
s.
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RE: [R] median and joint distribution
Dear Salvatore, Assuming that you mean convolution when you write additive linkage, the answer is that there is no general answer. It will depend heavily on the joint distribution of the two random variables. Just to give a simple example, let X~f, Y~g, and P(X=0.4)=P(Y=0.4)=1. Then, X and Y are independent, their medians are 0.5, but there sum has a median 0.5. Its different for multiplicative, since in general from X~f, Y~f, P(X=0.5)=0.5, and P(Y=0.5)=0.5 it follows that P(XY=0.5) = P(X=p or Y=q) if p*q=0.5. Thus, if there are numbers p and q with that property, such that P(X=p) + P(Y=q) = 0.5, then the median of XY will be =0.5. You might argue that there is a relationship between additive and multiplicative scale through a log-transformation (note that the median is stable under monotone transformations). However, I assume there is no obvious formulation of the above statement on the additive scale. There is no way of carrying convolutions out in R directly; you'd need to do numerical integration to do that, e.g. using integrate(). HTH Thomas --- Thomas Hotz Research Associate in Medical Statistics University of Leicester United Kingdom Department of Epidemiology and Public Health 22-28 Princess Road West Leicester LE1 6TP Tel +44 116 252-5410 Fax +44 116 252-5423 Division of Medicine for the Elderly Department of Medicine The Glenfield Hospital Leicester LE3 9QP Tel +44 116 256-3643 Fax +44 116 232-2976 -Original Message- From: Salvatore Barbaro [mailto:[EMAIL PROTECTED] Sent: 24 July 2003 12:56 To: [EMAIL PROTECTED] Subject: [R] median and joint distribution Dear R-helpers! May I kindly ask the pure statistics-experts to help me for a purpose which first part is not directly concerned with R. Consider two distribution functions, say f and g. For both, the median is smaller than a half. Now, the multiplicative or additive linkage of both distribution leads to a new distribution function, say h, whereas the median of h is greater than a half. Does anybody know under which circumstances such a construction of h is possible (my intuition is that it depends on the correlation of f and g) or can anybody advice a helpful literature. Furthermore, does anybody know whether or how such a construction can be done with R. Thanks in advance. s. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] R won't connect to the internet on Linux!
Hello Professor If you are suggesting that I am simply missing the http://; part of my cache URL, or that I am missing a trailing /, then I pre-empted this response and it still doesn't work. I have tried setting both http_proxy and HTTP_PROXY to all of: wwwcache.bbsrc.ac.uk:8080 http://wwwcache.bbsrc.ac.uk:8080 wwwcache.bbsrc.ac.uk:8080/ http://wwwcache.bbsrc.ac.uk:8080/ and I still get the same response - R cannot open the URL. And yes, that is thw right proxy address, I copied it straight from Netscape on the same computer, and Netscape connects to the internet fine. Thanks Mick -Original Message- From: Prof Brian Ripley [mailto:[EMAIL PROTECTED] Sent: 24 July 2003 13:17 To: michael watson (IAH-C) Cc: '[EMAIL PROTECTED] ' Subject: Re: [R] R won't connect to the internet on Linux! On Thu, 24 Jul 2003, michael watson (IAH-C) wrote: OK, I really am struggling with this one! Forgive me if I am being stupid I am running R 1.7.1 on Suse Linux 8.1. I connect to the internet through a proxy so I have: IAHC-LINUX03:~ # echo $http_proxy wwwcache.bbsrc.ac.uk:8080 IAHC-LINUX03:~ # echo $HTTP_PROXY wwwcache.bbsrc.ac.uk:8080 just in case ;-) SO, i go into R and I get: source(http://www.bioconductor.org/getBioC.R;) unable to connect to 'www.bioconductor.org' on port 80. Error in file(file, r) : cannot open URL `http://www.bioconductor.org/getBioC.R' OK so is R just not picking up my proxy setting? Your setting is wrong, so it is being ignored. The help page says quite explicitly The form of `http_proxy' should be `http://proxy.dom.com/;' or `http://proxy.dom.com:8080/;' where the port defaults to `80' and the trailing slash may be omitted. It seems to be trying port 80 on something, and I have specifically set it to port 8080 in my environment variables. As far as I can see I have followed the reference manual suggestion, so does anyone else have one? The problem is in your seeing, it seems. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] S3 and S4 classes
-Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Laurent Faisnel Sent: den 24 juli 2003 14:04 To: John Chambers Cc: [EMAIL PROTECTED]; Duncan Murdoch Subject: Re: [R] S3 and S4 classes [snip] Thank you for your fast and useful answers. I've found an interesting R help page I had never seen before at http://www.maths.lth.se/help/R/S3toS4 which gives additionnal information about S3 and S4 classes. Beware that I haven't updated this page since early 2002, it is based on early versions of the 'methods' package and my early understandings of it and is likely to be out of date (I'll put a note about this on the page as soon as I get access to the server). Regards, Laurent Have a nice day! Henrik Bengtsson Lund University __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: Re[2]: [R] lattice: how to format axis labels?
On 23 Jul 2003 09:12:07 -0500 Douglas Bates [EMAIL PROTECTED] wrote: High-level control of axes in xyplot is implemented by the scales argument to xyplot. You can include components 'at' and 'labels' in a list given as the scales argument. See ?xyplot. Wladimir Eremeev [EMAIL PROTECTED] writes: jhcc check out 'sprintf' for formating in a specific way. This will not solve the problem. I will have to specify the argument like labels=(...). I would like to avoid it. I wonder if there a key or option to make automatically appearing labels be formatted in the mentioned way. I haven't found it in the documentation. = jhcc I draw graphics with xyplot() function. jhcc Labels on the y axis are appearing as follows: 1.5, 1, 0.5, 0 jhcc I'd like to have them to be 1.5, 1.0, 0.5, 0.0, i.e. with fixed jhcc number of digits after the dot (one in this case). jhcc Is there any way to do this without implicit specifying labels? I don't think so. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- [EMAIL PROTECTED] 1-670-322-6580 Alan E. Davis, PMB 30, Box 10006, Saipan, MP 96950-8906, CNMI I have steadily endeavored to keep my mind free, so as to give up any hypothesis, however much beloved -- and I cannot resist forming one on every subject -- as soon as facts are shown to be opposed to it. -- Charles Darwin (1809-1882) The right to search for truth implies also a duty; one must not conceal any part of what one has recognized to be true. -- Albert Einstein As we enjoy great advantages from the inventions of others we should be glad of an opportunity to serve others by any invention of ours, and this we should do freely and generously. -- Benjamin Franklin __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Condition indexes and variance inflation factors
Dear Peter, At 08:24 AM 7/24/2003 -0400, Peter Flom wrote: (1) I've never liked this approach for a model with a constant, where it makes more sense to me to centre the data. I realize that opinions differ here, but it seems to me that failing to centre the data conflates collinearity with numerical instability. Opinions do differ. A few years ago, I could have given more details (my dissertation was on this topic, but a lot of the details have disappeared from memory); I think, though, that Belsley is looking for a measure that deals not only with collinearity, but with several other problems, including numerical instability (the subtitle of his later book is Collinearity and Weak Data in Regression). I remember being convinced that centering was generally not a good idea, but there are lots of people who disagree and who know a lot more statistics than I do. To elaborate my remark slightly, in most problems the intercept is not of much interest. When the data are far from the origin, it's natural that the intercept isn't well estimated. When data are very far from the origin, computations with the uncentred data may be numerically unstable (depending upon how the computations are done) because of collinearity with the intercept. If the real interest is in the coefficients other than the intercept, this seems to me purely a numerical artefact. The possibly more generally interesting sense of collinearity is imprecision in estimation due to strong relationships among the predictors. . . . (4) I have doubts about the whole enterprise. Collinearity is one source of imprecision -- others are small sample size, homogeneous predictors, and large error variance. Aren't the coefficient standard errors the bottom line? If these are sufficiently small, why worry? I think (correct me if I am wrong) that the s.e.s and the condition indices serve very different purposes. The condition indices are supposed to determine if small changes in the input data could make big differences in the results. Belsley provides some examples where a tiny change in the data results in completely different results (e.g., different standard errors, different coefficients (even reversing sign) and so on). Indeed, ill-conditioned data produce unstable numerical solutions (even affected by how the data are rounded), but condition indices aren't a particularly effective way of looking for instability in a more general sense. Consider, for example, Anscombe's famous simple-regression examples, which are in the data frame Quartet in the car package. The fourth example has a highly influential data point (number 8): Quartet[, c(x4, y4)] x4y4 1 8 6.58 2 8 5.76 3 8 7.71 4 8 8.84 5 8 8.47 6 8 7.04 7 8 5.25 8 19 12.50 9 8 5.56 10 8 7.91 11 8 6.89 The regression of y4 on x4 isn't especially ill-conditioned (using the function I posted yesterday): mod - lm(y4 ~ x4) belsley(mod) Singular values: 1.394079 0.2377891 Condition indices: 1 5.86267 Variance-decomposition proportions (Intercept)x4 1 0.028 0.028 2 0.972 0.972 but the 8th observation has an infinite Cook's D: round(cooks.distance(mod), 2) 123456789 10 11 0.01 0.06 0.02 0.14 0.09 0.00 0.12 Inf 0.08 0.03 0.00 Regards, John - John Fox Department of Sociology McMaster University Hamilton, Ontario, Canada L8S 4M4 email: [EMAIL PROTECTED] phone: 905-525-9140x23604 web: www.socsci.mcmaster.ca/jfox __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Dismal R performance of Athlon moble CPU?
On Wed, 23 Jul 2003, Jason Liao wrote: Thanks for Prof. Ripley and Andy for your technical explantion. It seems that that the real CPU speed has not advanced as fast as these Ghz or other performance indicator suggest. Yes, my program is totally CPU intensive. It may not be even if you think it is. Note that much of the speed increase that ATLAS achieves comes from optimal use of various levels of cache. It's quite possible that your code is actually limited by memory bandwidth. In any case, I think R is often limited by integer rather than floating point peformance. Back in the days when I timed things in R, its relative performance across PCs and Sparcs suggested that floating point wasn't a big deal. -thomas Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] R won't connect to the internet on Linux!
You're right, I have tried keeping the double quotes in there too and that doesn't work
What is clear is that R is completely ignoring my http_proxy environment variable, as
it's error says unable to connect on port 80, when every single one of my http_proxy
attempts has stipulates port 8080 I know for a fact that my proxy isn't going to
work on port 80.
So what I need to know is why R is choosing to ignore http_proxy
-Original Message-
From: Bashir Saghir (Aztek Global) [mailto:[EMAIL PROTECTED]
Sent: 24 July 2003 15:06
To: 'michael watson (IAH-C)'
Subject: RE: [R] R won't connect to the internet on Linux!
I'm not 100% sure but I think the double quotes are needed too. You've probably done
this too.
Saghir
-Original Message-
From: michael watson (IAH-C) [SMTP:[EMAIL PROTECTED]
Sent: Thursday, 24 July, 2003 3:16 PM
To: 'Prof Brian Ripley'
Cc: '[EMAIL PROTECTED] '
Subject:RE: [R] R won't connect to the internet on Linux!
Hello Professor
If you are suggesting that I am simply missing the http:// http:// part
of my cache URL, or that I am missing a trailing /, then I pre-empted this response
and it still doesn't work.
I have tried setting both http_proxy and HTTP_PROXY to all of:
wwwcache.bbsrc.ac.uk:8080
http://wwwcache.bbsrc.ac.uk:8080 http://wwwcache.bbsrc.ac.uk:8080
wwwcache.bbsrc.ac.uk:8080/
http://wwwcache.bbsrc.ac.uk:8080/ http://wwwcache.bbsrc.ac.uk:8080/
and I still get the same response - R cannot open the URL.
And yes, that is thw right proxy address, I copied it straight from Netscape
on the same computer, and Netscape connects to the internet fine.
Thanks
Mick
-Original Message-
From: Prof Brian Ripley [ mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] ]
Sent: 24 July 2003 13:17
To: michael watson (IAH-C)
Cc: '[EMAIL PROTECTED] '
Subject: Re: [R] R won't connect to the internet on Linux!
On Thu, 24 Jul 2003, michael watson (IAH-C) wrote:
OK, I really am struggling with this one! Forgive me if I am being
stupid
I am running R 1.7.1 on Suse Linux 8.1. I connect to the internet
through a proxy so I have:
IAHC-LINUX03:~ # echo $http_proxy
wwwcache.bbsrc.ac.uk:8080
IAHC-LINUX03:~ # echo $HTTP_PROXY
wwwcache.bbsrc.ac.uk:8080
just in case ;-)
SO, i go into R and I get:
source( http://www.bioconductor.org/getBioC.R
http://www.bioconductor.org/getBioC.R )
unable to connect to 'www.bioconductor.org' on port 80.
Error in file(file, r) : cannot open URL ` http://www.bioconductor.org/getBioC.R'
http://www.bioconductor.org/getBioC.R'
OK so is R just not picking up my proxy setting?
Your setting is wrong, so it is being ignored. The help page says
quite explicitly
The form of `http_proxy' should be ` http://proxy.dom.com/
http://proxy.dom.com/ ' or
` http://proxy.dom.com:8080/ http://proxy.dom.com:8080/ ' where the port
defaults to `80' and
the trailing slash may be omitted.
It seems to be trying
port 80 on something, and I have specifically set it to port 8080 in my
environment variables. As far as I can see I have followed the
reference manual suggestion, so does anyone else have one?
The problem is in your seeing, it seems.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] data.frame subsetting
Hi, is there advice how to subset a data.frame, where until R version 1.7.0 it was possible to write data[[subset]] which meant the same as data$subset (data is a data.frame). From 1.7.1 on this does not seem to work longer. Could there be a bug in downwards compatibility? sincerely yours Dirk __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] trellis plot question
Greetings, Does anyone know how to get an id number in the little header above each individual plot within a trellis plot? The default seems to be to print the word id and add a line indicating on a linear scale where the current id sits. Thanks in advance for any help you can send, Suzette = Suzette Blanchard, Ph.D. Research Scientist Frontier Science Foundation 1244 Boylston St. Suite 303 Chestnut Hill, MA 02467 Email: [EMAIL PROTECTED] Phone: (617) 632-2007 Fax:(617) 632-2001 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Problem w/ source
I'm trying to use the source command to run commands from a file. For instance: source(do.R), where do.R is a file in the same directory in which I am running R. The contents of do.R are: ls() print(hello) sum(y1) mean(y1) After source(do.R), all I see is: source(do.R) [1] hello I'm using the X11 version of R for Mac OS X (downloadable binary). Does anyone know how to get source to work? Thanks! Peter __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] R benchmark, moble Pentium III, 1.13 GHs
I got the following using the benchmark program at http://www.sciviews.org/other/benchmark.htm under Windows 2000 prof., 256 MB of RAM R Benchmark 2 = Number of times each test is run__: 3 I. Matrix calculation - Creation, transp., deformation of a 1500x1500 matrix (sec): 1.97 800x800 normal distributed random matrix ^1000__ (sec): 2.80 Sorting of 2,000,000 random values__ (sec): 1.20 700x700 cross-product matrix (b = a' * a)___ (sec): 1.84 Linear regression over a 600x600 matrix (c = a \ b') (sec): 2.43 Trimmed geom. mean (2 extremes eliminated): 2.06825841067831 II. Matrix functions FFT over 800,000 random values__ (sec): 1.61 Eigenvalues of a 320x320 random matrix__ (sec): 1.20 Error in eval(expr, envir, enclos) : couldn't find function det.Matrix Timing stopped at: 0 0 0 NA NA = Jason G. Liao, Ph.D. Division of Biometrics University of Medicine and Dentistry of New Jersey 335 George Street, Suite 2200 New Brunswick, NJ 08903-2688 phone (732) 235-8611, fax (732) 235-9777 http://www.geocities.com/jg_liao __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Plotting math functions
Hi all, I was wondering whether it is possible to plot math functions, for example sin, cos or a Gaussian type function, in R, and if so, how to do it. I have been searching through the archives and the R manual but had no luck in finding any hints on how to go about this. Any help is much appreciated! Thanks, Jonck __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Plotting math functions
Jonck van der Kogel wrote: Hi all, I was wondering whether it is possible to plot math functions, for example sin, cos or a Gaussian type function, in R, and if so, how to do it. I have been searching through the archives and the R manual but had no luck in finding any hints on how to go about this. Any help is much appreciated! Thanks, Jonck __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help See ?curve Uwe Ligges __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Condition indexes and variance inflation factors
Dear John An interesting discussion! I would be the last to suggest ignoring such diagnostics as Cook's D; as you point out, it diagnoses a problem which condition indices do not: Whether a point is influential. OTOH, condition indices diagnose a problem which Cook's D does not: Would shifting the data slightly change the results. Consider the data given in Belsley (1991) on p. 5 y - c( 3.3979, 1.6094, 3.7131, 1.6767, 0.0419, 3.3768, 1.1661, 0.4701) x2a - c(-3.138, -0.297, -4.582, 0.301, 2.729, -4.836, 0.065, 4.102) x2b - c(-3.136, -0.296, -4.581, 0.300, 2.730, -4.834, 0.064, 4.103) x3a - c(1.286, 0.250, 1.247, 0.498, -0.280, 0.350, 0.208, 1.069) x3b - c(1.288, 0.251, 1.246, 0.498, -0.281, 0.349, 0.206, 1.069) x4a - c(0.169, 0.044, 0.109, 0.117, 0.035, -0.094, 0.047, 0.375) x4b - c(0.170, 0.043, 0.108, 0.118, 0.036, -0.093, 0.048, 0.376) where x2a , x3a and x4a are very similar to x2b, x3b and x4b, respecttively, and where both are generated from y = 1.2I - 0.4 x2 + 0.6x3 + 0.9x4 + e e ~ N(0, 0.01) Then modela - lm(y~ x2a + x3a + x4a) and modelb - lm(y~x2b + x3b + x4b) give radically different results, with neither having any significant parameters other than the intercept. Admittedly, the standard errors for a couple of the parameters are large. But why are they large? I have certainly dealt with models with large standard errors that have nothing to do with collinearity. here, the function PI.lm (supplied by Juergen Gross) gives huge condition indices, and indicates that the nature of the problem is that all three of the x variables are highly collinear. Variance-Decomposition Proportions for Scaled Condition Indexes: (Intercept) x2b x3b x4b 10.0494 0 0 0 10.0009 0 0 0 30.8101 0 0 0 464 0.1396 1 1 1 Regards Peter Peter L. Flom, PhD Assistant Director, Statistics and Data Analysis Core Center for Drug Use and HIV Research National Development and Research Institutes 71 W. 23rd St www.peterflom.com New York, NY 10010 (212) 845-4485 (voice) (917) 438-0894 (fax) __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Problem w/ source
Peter Muhlberger wrote: I'm trying to use the source command to run commands from a file. For instance: source(do.R), where do.R is a file in the same directory in which I am running R. The contents of do.R are: ls() print(hello) sum(y1) mean(y1) After source(do.R), all I see is: source(do.R) [1] hello I'm using the X11 version of R for Mac OS X (downloadable binary). Does anyone know how to get source to work? Thanks! Peter Works perfectly, but no automatic print() is executed as the default when sourcing. When you want the other objects to be printed either use print(ls()) print(sum(y1)) ... or use source(do.R, print.eval = TRUE) as described in ?source. Uwe Ligges __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] data.frame subsetting
Dirk Repsilber wrote: Hi, is there advice how to subset a data.frame, where until R version 1.7.0 it was possible to write data[[subset]] which meant the same as data$subset (data is a data.frame). From 1.7.1 on this does not seem to work longer. Could there be a bug in downwards compatibility? sincerely yours Works for me. Are you sure the object data is a data.frame in your case? Uwe Ligges __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Plotting math functions
I was wondering whether it is possible to plot math functions, for example sin ... also maybe: ?plot snipped from the help page Examples: ... plot(sin, -pi, 2*pi) unsnip bob -Original Message- From: Uwe Ligges [mailto:[EMAIL PROTECTED] Sent: Thursday, July 24, 2003 11:04 AM To: Jonck van der Kogel Cc: [EMAIL PROTECTED] Subject: Re: [R] Plotting math functions Jonck van der Kogel wrote: Hi all, I was wondering whether it is possible to plot math functions, for example sin, cos or a Gaussian type function, in R, and if so, how to do it. I have been searching through the archives and the R manual but had no luck in finding any hints on how to go about this. Any help is much appreciated! Thanks, Jonck __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help See ?curve Uwe Ligges __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] data.frame subsetting
On Thu, 24 Jul 2003 16:27:28 +0200, Dirk Repsilber [EMAIL PROTECTED] wrote : Hi, is there advice how to subset a data.frame, where until R version 1.7.0 it was possible to write data[[subset]] which meant the same as data$subset (data is a data.frame). From 1.7.1 on this does not seem to work longer. Could there be a bug in downwards compatibility? What are you trying to achieve? The notation you give still works if subset is one of the column names; I don't think it ever worked if subset was a logical vector, like its name would suggest. Duncan Murdoch __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Problem w/ source
On Thu, 24 Jul 2003 10:42:05 -0400, Peter Muhlberger [EMAIL PROTECTED] wrote : I'm trying to use the source command to run commands from a file. For instance: source(do.R), where do.R is a file in the same directory in which I am running R. The contents of do.R are: ls() print(hello) sum(y1) mean(y1) After source(do.R), all I see is: source(do.R) [1] hello I'm using the X11 version of R for Mac OS X (downloadable binary). Does anyone know how to get source to work? You probably want source(do.R, echo = TRUE) or maybe not; see ?source. Duncan Murdoch __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] trellis plot question
Suzette Blanchard [EMAIL PROTECTED] writes: Does anyone know how to get an id number in the little header above each individual plot within a trellis plot? The default seems to be to print the word id and add a line indicating on a linear scale where the current id sits. See the description of the strip argument in ?xyplot and the documentation for strip.default. There are currently 6 different styles available or you can add your own. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Problem w/ source
Well, we feel that source does work the way that it is documented to work. Please read the documentation and notice that the entire file is evaluated as one step in the read-eval-print loop. If you want to print the results of individual function calls you will need to change your script to print(ls()) print(hello) print(sum(y1)) print(mean(y1)) Peter Muhlberger [EMAIL PROTECTED] writes: I'm trying to use the source command to run commands from a file. For instance: source(do.R), where do.R is a file in the same directory in which I am running R. The contents of do.R are: ls() print(hello) sum(y1) mean(y1) After source(do.R), all I see is: source(do.R) [1] hello I'm using the X11 version of R for Mac OS X (downloadable binary). Does anyone know how to get source to work? Thanks! Peter __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- Douglas Bates[EMAIL PROTECTED] Statistics Department608/262-2598 University of Wisconsin - Madisonhttp://www.stat.wisc.edu/~bates/ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Plotting math functions
On Thu, 24 Jul 2003 16:55:55 +0200, Jonck van der Kogel [EMAIL PROTECTED] wrote : Hi all, I was wondering whether it is possible to plot math functions, for example sin, cos or a Gaussian type function, in R, and if so, how to do it. I have been searching through the archives and the R manual but had no luck in finding any hints on how to go about this. Any help is much appreciated! You can use plot(sin, from=-10, to=10) This calls plot.function; look at ?plot.function to read about the options. Duncan Murdoch __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Integer programming in R
Dear all, I am a relative newcomer to the R language, and am sussing out the possibilities of using R to do integer programming (which I am also new to). Is there something along the lines of the NUOPT S- PLUS package that is available, or on the way, for R? Are there other options for integer programming in R? Thanks very much in advance, Robin Naidoo University of Alberta Edmonton, AB, Canada __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Problem w/ source
Thanks to everyone for their suggestions on getting source to print! It seems not everyone was aware of a couple options that gets source to print out everything. I'm now using the following command: source(do.R, print.eval=TRUE, echo=TRUE) __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] scatterplot smoothing using gam
All: I am trying to use gam in a scatterplot smoothing problem. The data being smoothed have greater 1000 observation and have multiple humps. I can smooth the data fine using a function something like: out - ksmooth(x,y,normal,bandwidth=0.25) plot(x,out$y,type=l) The problem is when I try to fit the same data using gam out - predict.gam(gam(y~s(x)),se=TRUE) plot(x,out$fit,type=l) I only seem to get fits that would correspond to big bandwidths using ksmooth, and straight lines are always fit to the data. I do not appear to appreciate how to control bandwidth using gam. As even if I apply something like the gam model above to the smoothed out$y generated using ksmooth it tends to flatten out the smoothing curve. -- ### Tony Long Ecology and Evolutionary Biology Steinhaus Hall University of California at Irvine Irvine, CA 92697-2525 Tel: (949) 824-2562 (office) Tel: (949) 824-5994 (lab) Fax: (949) 824-2181 email: [EMAIL PROTECTED] http://hjmuller.bio.uci.edu/~labhome/ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] wireframe: how to remove the frame around my plot?
Hi, I've got a wireframe 3D surface plot, but I don't want a frame around it. Is there any way to remove the frame, or (worst case) change the color of the frame to the background color (which looks like grey). I'm using ver 1.7.1 I've tried frame.plot = F, but that doesn't seem to work for 'wireframe'. Many thanks, Alexis Diamond __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] data.frame subsetting
Still works for me in both 1.7.1 and 1.8.0 (in development) data(women) women[[height]] [1] 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 women$height [1] 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 Can you provide an example of how it is failing for you? Dirk Repsilber [EMAIL PROTECTED] writes: Hi, is there advice how to subset a data.frame, where until R version 1.7.0 it was possible to write data[[subset]] which meant the same as data$subset (data is a data.frame). From 1.7.1 on this does not seem to work longer. Could there be a bug in downwards compatibility? sincerely yours Dirk __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help -- Douglas Bates[EMAIL PROTECTED] Statistics Department608/262-2598 University of Wisconsin - Madisonhttp://www.stat.wisc.edu/~bates/ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] wireframe: how to remove the frame around my plot?
You can specify some options in par.box. Use col=NA to make the frame transparent. See example below (which was modified from the help file). See also the scales parameter if you want to remove the arrows as well. Cheers, Jerome library(lattice) x - seq(-pi, pi, len = 20) y - seq(-pi, pi, len = 20) g - expand.grid(x = x, y = y) g$z - sin(sqrt(g$x^2 + g$y^2)) wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=NA)) On July 24, 2003 12:44 pm, Alexis J. Diamond wrote: Hi, I've got a wireframe 3D surface plot, but I don't want a frame around it. Is there any way to remove the frame, or (worst case) change the color of the frame to the background color (which looks like grey). I'm using ver 1.7.1 I've tried frame.plot = F, but that doesn't seem to work for 'wireframe'. Many thanks, Alexis Diamond __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] wireframe: how to remove the frame around my plot?
Jerome Asselin wrote: You can specify some options in par.box. Use col=NA to make the frame transparent. See example below (which was modified from the help file). See also the scales parameter if you want to remove the arrows as well. Cheers, Jerome library(lattice) x - seq(-pi, pi, len = 20) y - seq(-pi, pi, len = 20) g - expand.grid(x = x, y = y) g$z - sin(sqrt(g$x^2 + g$y^2)) wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=NA)) Which doesn't work on device trellis.device(windows) but works well on, e.g., trellis.device(postscript) Deepayan, since I don't have the time to check that right now: Is this bug known? Is this a bug in R, grid or lattice? Uwe On July 24, 2003 12:44 pm, Alexis J. Diamond wrote: Hi, I've got a wireframe 3D surface plot, but I don't want a frame around it. Is there any way to remove the frame, or (worst case) change the color of the frame to the background color (which looks like grey). I'm using ver 1.7.1 I've tried frame.plot = F, but that doesn't seem to work for 'wireframe'. Many thanks, Alexis Diamond __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] wireframe: how to remove the frame around my plot?
hi jerome, thank you for your quick reply. your advice removes the 3D box in which the 3D plot is generated, but i like THAT box. (sorry for being unclear earlier) what i want to do is remove the 2D frame (a box of thin black lines) that circumscribes all of my plot area. any ideas? thanks again, alexis On Thu, 24 Jul 2003, Jerome Asselin wrote: You can specify some options in par.box. Use col=NA to make the frame transparent. See example below (which was modified from the help file). See also the scales parameter if you want to remove the arrows as well. Cheers, Jerome library(lattice) x - seq(-pi, pi, len = 20) y - seq(-pi, pi, len = 20) g - expand.grid(x = x, y = y) g$z - sin(sqrt(g$x^2 + g$y^2)) wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=NA)) On July 24, 2003 12:44 pm, Alexis J. Diamond wrote: Hi, I've got a wireframe 3D surface plot, but I don't want a frame around it. Is there any way to remove the frame, or (worst case) change the color of the frame to the background color (which looks like grey). I'm using ver 1.7.1 I've tried frame.plot = F, but that doesn't seem to work for 'wireframe'. Many thanks, Alexis Diamond __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] wireframe: how to remove the frame around my plot?
Hi Y'all, Alexis, I think Jerome's example works except for one change: This one has a box but no wires: wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=1),col=NA) versus with no box and no wires: wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=NA),col=NA) Hope this helps! Jake - Jake Bowers Dept of Political Science University of Michigan On Thu, 24 Jul 2003, Alexis J. Diamond wrote: hi jerome, thank you for your quick reply. your advice removes the 3D box in which the 3D plot is generated, but i like THAT box. (sorry for being unclear earlier) what i want to do is remove the 2D frame (a box of thin black lines) that circumscribes all of my plot area. any ideas? thanks again, alexis On Thu, 24 Jul 2003, Jerome Asselin wrote: You can specify some options in par.box. Use col=NA to make the frame transparent. See example below (which was modified from the help file). See also the scales parameter if you want to remove the arrows as well. Cheers, Jerome library(lattice) x - seq(-pi, pi, len = 20) y - seq(-pi, pi, len = 20) g - expand.grid(x = x, y = y) g$z - sin(sqrt(g$x^2 + g$y^2)) wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=NA)) On July 24, 2003 12:44 pm, Alexis J. Diamond wrote: Hi, I've got a wireframe 3D surface plot, but I don't want a frame around it. Is there any way to remove the frame, or (worst case) change the color of the frame to the background color (which looks like grey). I'm using ver 1.7.1 I've tried frame.plot = F, but that doesn't seem to work for 'wireframe'. Many thanks, Alexis Diamond __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] postscript device: pch='.' size?
hi ladies and gents: I am using R 1.7.0. Alas, the pch=. in the postscript device is too large for me, but apparently not scaleable via cex. Maybe a bug, though I read a complaing about this in the list archives from 1999, and given that it is still around, maybe it is a feature. Now, I would be happy with pch=o and cex=0.05, but for some strange reason, the resulting .eps file then becomes 3 times the size (200K, rather than 70K). For web-downloadable files, this can become painful. Are there any remedies? Regards, /iaw __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] wireframe: how to remove the frame around my plot?
hi jake, thanks for your e-mail. what i am actually trying to do is remove the 'picture frame' 2D box (the super-frame) that circumscribes the entire figure-- for example, the left vertical side of this box i'm concerned about is to the left of the 'z' axis label. i hope i'm finally making my query clear. both of your examples below retain this pictureframe box (jerome's examples do too), when the output is viewed as a .eps file in ghostcript viewer, so i am still stuck, unfortunately. there's got to be a parameter that modifies this picureframe box's color and style, right? thanks again, alexis On Thu, 24 Jul 2003, Jake Bowers wrote: Hi Y'all, Alexis, I think Jerome's example works except for one change: This one has a box but no wires: wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=1),col=NA) versus with no box and no wires: wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=NA),col=NA) Hope this helps! Jake - Jake Bowers Dept of Political Science University of Michigan On Thu, 24 Jul 2003, Alexis J. Diamond wrote: hi jerome, thank you for your quick reply. your advice removes the 3D box in which the 3D plot is generated, but i like THAT box. (sorry for being unclear earlier) what i want to do is remove the 2D frame (a box of thin black lines) that circumscribes all of my plot area. any ideas? thanks again, alexis On Thu, 24 Jul 2003, Jerome Asselin wrote: You can specify some options in par.box. Use col=NA to make the frame transparent. See example below (which was modified from the help file). See also the scales parameter if you want to remove the arrows as well. Cheers, Jerome library(lattice) x - seq(-pi, pi, len = 20) y - seq(-pi, pi, len = 20) g - expand.grid(x = x, y = y) g$z - sin(sqrt(g$x^2 + g$y^2)) wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=NA)) On July 24, 2003 12:44 pm, Alexis J. Diamond wrote: Hi, I've got a wireframe 3D surface plot, but I don't want a frame around it. Is there any way to remove the frame, or (worst case) change the color of the frame to the background color (which looks like grey). I'm using ver 1.7.1 I've tried frame.plot = F, but that doesn't seem to work for 'wireframe'. Many thanks, Alexis Diamond __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] wireframe: how to remove the frame around my plot?
Finally, I think this works. The function trellis.par.get() contains all default values of the parameters. I had to reset the option axis.line as below. The bounding (square) box is made transparent. See trellis.par.get() for the list of all default parameter values. Cheers, Jerome library(lattice) x - seq(-pi, pi, len = 20) y - seq(-pi, pi, len = 20) g - expand.grid(x = x, y = y) g$z - sin(sqrt(g$x^2 + g$y^2)) trellis.par.set(axis.line,list(col=NA,lty=1,lwd=1)) wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE) On July 24, 2003 02:41 pm, Alexis J. Diamond wrote: hi jake, thanks for your e-mail. what i am actually trying to do is remove the 'picture frame' 2D box (the super-frame) that circumscribes the entire figure-- for example, the left vertical side of this box i'm concerned about is to the left of the 'z' axis label. i hope i'm finally making my query clear. both of your examples below retain this pictureframe box (jerome's examples do too), when the output is viewed as a .eps file in ghostcript viewer, so i am still stuck, unfortunately. there's got to be a parameter that modifies this picureframe box's color and style, right? thanks again, alexis On Thu, 24 Jul 2003, Jake Bowers wrote: Hi Y'all, Alexis, I think Jerome's example works except for one change: This one has a box but no wires: wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=1),col=NA) versus with no box and no wires: wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=NA),col=NA) Hope this helps! Jake - Jake Bowers Dept of Political Science University of Michigan On Thu, 24 Jul 2003, Alexis J. Diamond wrote: hi jerome, thank you for your quick reply. your advice removes the 3D box in which the 3D plot is generated, but i like THAT box. (sorry for being unclear earlier) what i want to do is remove the 2D frame (a box of thin black lines) that circumscribes all of my plot area. any ideas? thanks again, alexis On Thu, 24 Jul 2003, Jerome Asselin wrote: You can specify some options in par.box. Use col=NA to make the frame transparent. See example below (which was modified from the help file). See also the scales parameter if you want to remove the arrows as well. Cheers, Jerome library(lattice) x - seq(-pi, pi, len = 20) y - seq(-pi, pi, len = 20) g - expand.grid(x = x, y = y) g$z - sin(sqrt(g$x^2 + g$y^2)) wireframe(z ~ x * y, g, drape = TRUE, perspective = FALSE, aspect = c(3,1), colorkey = FALSE, par.box = list(col=NA)) On July 24, 2003 12:44 pm, Alexis J. Diamond wrote: Hi, I've got a wireframe 3D surface plot, but I don't want a frame around it. Is there any way to remove the frame, or (worst case) change the color of the frame to the background color (which looks like grey). I'm using ver 1.7.1 I've tried frame.plot = F, but that doesn't seem to work for 'wireframe'. Many thanks, Alexis Diamond __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
RE: [R] Confidence Band for empirical distribution function
On Thu, 24 Jul 2003, Hotz, T. wrote: Dear Kjetil, As I already mentioned, it appears that there isn't a function available calculating the quantiles directly (at least, it doesn't appear in the C source of ctest). So as I already suggested, uniroot (or a similar C routine which calls the corresponding C code directly) is probably the best you can do (apart from writing it completely yourself). I didn't program this using uniroot, but I'd certainly try the following for speed-up: - For symmetry reasons, you only need to compute half of the quantiles. - The quantiles depend smoothly on the probabilities (of your reference distribution). Therefore, calculating only a few for probabilities between 0 and 0.5, and using (e.g. linear) interpolation should be satisfying. There is an example of this in my package HyperbolicDist which has just appeared on CRAN. It is a little more sophisticated in that it fits a spline in preference to linear interpolation, before using uniroot. Look at qhyperb if this is of interest. David Scott _ David Scott Department of Statistics, Tamaki Campus The University of Auckland, PB 92019 AucklandNEW ZEALAND Phone: +64 9 373 7599 ext 86830 Fax: +64 9 373 7000 Email: [EMAIL PROTECTED] Graduate Officer, Department of Statistics Webmaster, New Zealand Statistical Association: http://www.stat.auckland.ac.nz/nzsa/ __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] inverse prediction and Poisson regression
Hello to all, I'm a biologist trying to tackle a fish (Poisson Regression) which is just too big for my modest understanding of stats!!! Here goes... I want to find good literature or proper mathematical procedure to calculate a confidence interval for an inverse prediction of a Poisson regression using R. I'm currently trying to analyse a dose-response experiment. I want to calculate the dose (X) for 50% inhibition of a biological response (Y). My response is a count data that fits a Poisson distribution perfectly. I could make my life easy and calculate: dose response/control response = % of total response... and then use logistic regression, but somehow, that doesn't sound right. Should I just stick to logistic regression and go on with my life? Can I be cured of this paranoia? ;-) I thought a Poisson regression would be more appropriate, but I don't know how to properly calculate the dose equivalent to 50% inhibition. i/e confidence intervals, etc on the X = dose. Basically an inverse prediction problem. By the way, my data is graphically linear for Log(Y) = log(X) where Y is counts and X is dose. I use a Poisson regression to fit my dose-response experiment by EXCLUDING the response for dose = 0, because of log(0) Under R = glm.dose - glm(response[-1] ~ log(dose[-1]),family=poisson()) (that's why you see the dose[-1] term. The first dose in the dose vector is 0. This is really a nice fit. I can obtain a nice slope (B) and intercept (A): log(Y) = B log(x) + A I do have a biological value for dose = 0 from my control. i/e Ymax = some number with a Poisson error again So, what I want is EC50x : Y/Ymax = 0.5 = exp(B log(EC50x) + A) / Ymax exp((log(0.5) + Log(Ymax)) - A)/B) = EC50x That's all fine, except I don't have a clue on how to calculate the confidence intervals of EC50x or even if I can model this inverse prediction with a Poisson regression. In OLS linear regression, fitting X based on Y is not a good idea because of the way OLS calculates the slope and intercept. Is the same problem found in GLM/Poisson regression? Moreover, I also have a Poisson error on Ymax that I would have to consider, right? Help -- Vincent Philion, M.Sc. agr. Phytopathologiste Institut de Recherche et de Développement en Agroenvironnement (IRDA) 3300 Sicotte, St-Hyacinthe Québec J2S 7B8 téléphone: 450-778-6522 poste 233 courriel: [EMAIL PROTECTED] Site internet : www.irda.qc.ca __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] R-WinEdt problems
Hello, I've done most all the various steps outlined in a recent posting to the mailing list archives (and in the help files) to load and run R-WinEdt. I can get it to run fine but I am not successful in getting it to interface with RGui (1.6.2, not minimized). I try to use the R-line, R-source, R-paste buttons with no success. [All the necessary *.edt files appear to be in the proper directories.] I've done everything except modify the rprofile or the options(). Is that necessary, and if so how exactly do I do that? And, if I do modify them how do I restore them back to their default settings? Any help would be much appreciated. Thanks, Jeff Jorgensen [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] contourplot: how to ensure all (or certain) 'cuts' get labelled
hi all, contourplot generates a nice plot, but there are cut lines that don't have labels in my figure. i think this is because there's a high ridge stretching north-west/south-east entirely across the plot, diagonal-to-diagonal, and contourplot only wants to label each elevation-level only once, which means only lines in the upper-right quadrant got labelled. is there some way to *identify* and label a cutpoint that has no label, or to force contourplot to label each (or a particular) cutpoint? if i knew what the label should be on a given label-less cutpoint, i could probably insert it in as text myself. but i don't even know what it should be. by the way, does anyone know why i got only 5 cuts, when i specified cuts = 10? thanks, alexis ps-- i am open to using *contour* instead of *contourplot* if this gives me more control over the output. __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] R-WinEdt problems
Right click on your WinEdt icon and go to Properties menu-item. In the Target box look for something like, [PathToWinEdt]\WinEdt\WinEdt.exe -C=R-WinEdt -e=r.ini If you don't find something like this, make the change and try it. * It may be better to have a new icon for R-WinEdt on your desktop with the above modified Target, and use the existing icon for other work. Not sure how to solve the switching problem, given that there is a R specific initialization file being used. ---Anupam. In a message dated 7/24/03 6:15:46 PM Eastern Daylight Time, [EMAIL PROTECTED] writes: Hello, I've done most all the various steps outlined in a recent posting to the mailing list archives (and in the help files) to load and run R-WinEdt. I can get it to run fine but I am not successful in getting it to interface with RGui (1.6.2, not minimized). I try to use the R-line, R-source, R-paste buttons with no success. [All the necessary *.edt files appear to be in the proper directories.] I've done everything except modify the rprofile or the options(). Is that necessary, and if so how exactly do I do that? And, if I do modify them how do I restore them back to their default settings? Any help would be much appreciated. Thanks, Jeff Jorgensen [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Plotting Vector Fields
Dear R List,
Is there a function to plot vector fields in R?
I'm looking for something similar to PlotVectorField in Mathematica
(see below), e.g. given functions f(x) and f(y), I would like to plot
the resulting field of vectors (f(x),f(y)) over some range of x and y.
PlotVectorField[{f(x), f(y)}, {x, xmin, xmax}, {y, ymin, ymax}]
Thanks for your help,
Brian
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Re: [R] inverse prediction and Poisson regression
1. If you provide a toy data set with, e.g., 5 observations, to accompany your example, it would be much easier for people to try out ideas and then give you a more solid response. 2. Have you tried something like log(dose+0.5) or I(log(dose+0.5)) in your model statement in conjunction with predict or predict.glm on the output from glm? hope this helps. spencer graves Vincent Philion wrote: Hello to all, I'm a biologist trying to tackle a fish (Poisson Regression) which is just too big for my modest understanding of stats!!! Here goes... I want to find good literature or proper mathematical procedure to calculate a confidence interval for an inverse prediction of a Poisson regression using R. I'm currently trying to analyse a dose-response experiment. I want to calculate the dose (X) for 50% inhibition of a biological response (Y). My response is a count data that fits a Poisson distribution perfectly. I could make my life easy and calculate: dose response/control response = % of total response... and then use logistic regression, but somehow, that doesn't sound right. Should I just stick to logistic regression and go on with my life? Can I be cured of this paranoia? ;-) I thought a Poisson regression would be more appropriate, but I don't know how to properly calculate the dose equivalent to 50% inhibition. i/e confidence intervals, etc on the X = dose. Basically an inverse prediction problem. By the way, my data is graphically linear for Log(Y) = log(X) where Y is counts and X is dose. I use a Poisson regression to fit my dose-response experiment by EXCLUDING the response for dose = 0, because of log(0) Under R = glm.dose - glm(response[-1] ~ log(dose[-1]),family=poisson()) (that's why you see the dose[-1] term. The first dose in the dose vector is 0. This is really a nice fit. I can obtain a nice slope (B) and intercept (A): log(Y) = B log(x) + A I do have a biological value for dose = 0 from my control. i/e Ymax = some number with a Poisson error again So, what I want is EC50x : Y/Ymax = 0.5 = exp(B log(EC50x) + A) / Ymax exp((log(0.5) + Log(Ymax)) - A)/B) = EC50x That's all fine, except I don't have a clue on how to calculate the confidence intervals of EC50x or even if I can model this inverse prediction with a Poisson regression. In OLS linear regression, fitting X based on Y is not a good idea because of the way OLS calculates the slope and intercept. Is the same problem found in GLM/Poisson regression? Moreover, I also have a Poisson error on Ymax that I would have to consider, right? Help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] animal model in R
If your datasets are small, then it is not difficult to roll your own using optim(). If you look at http://www.qimr.edu.au/davidD/sib-pair.R you will find such a routine at lines 1511-1562. This calls kinship.rel() that produces NRM etc. -- | David Duffy (MBBS PhD) ,-_|\ | email: [EMAIL PROTECTED] ph: INT+61+7+3362-0217 fax: -0101 / * | Epidemiology Unit, Queensland Institute of Medical Research \_,-._/ | 300 Herston Rd, Brisbane, Queensland 4029, Australia GPG 4D0B994A v __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Memory explosion, plotting nmle grouped data object
Hi I am using R 1.7.1 on RH linux 9.0 sum(unlist(lapply(ls(),function(x)object.size(get(x)/1024^2 [1] 2.424263 so I am not using much memory (I have a gig of ram on my machine) now in nlme gtest-groupedData(log(X8)~Time|sub,all[,c(names(all)[1:9],X8)],outer=~A*B) object.size(gtest)/1024 [1] 59.98438 plot(gtest,outer=~Dose*chem,key=FALSE,asp=.5) Plotting takes forever and from top while it is running 5663 wf00223 15 0 1151M 546M 1780 R 6.1 54.2 3:19 0 R.bin Any Ideas why R is using so much memory? Thanks Nicholas __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Integer programming in R
Robin, This is not a direct answer to your question, but there exists a pretty good linear, mixted and integer programming package called lp_solve. I have used it successfully on a couple of projects. As far as I know it is in the public domain and there is source code available. There is also a precompiled version available for Windows Visual Basic environment containing a precompiled DLL and an interface module defining VB function calls to the DLL. I think, that having all that it should be possible to get something going in R, at least in the Windows environment. Hope this helps, Andy __ Andy Jaworski Engineering Systems Technology Center 3M Center, 518-1-01 St. Paul, MN 55144-1000 - E-mail: [EMAIL PROTECTED] Tel: (651) 733-6092 Fax: (651) 736-3122 |-+ | | Robin Naidoo | | | [EMAIL PROTECTED]| | | Sent by: | | | [EMAIL PROTECTED]| | | ath.ethz.ch | | || | || | | 07/24/2003 12:18 | | || |-+ -| | | | To: [EMAIL PROTECTED] | | cc: | | Subject: [R] Integer programming in R | -| Dear all, I am a relative newcomer to the R language, and am sussing out the possibilities of using R to do integer programming (which I am also new to). Is there something along the lines of the NUOPT S- PLUS package that is available, or on the way, for R? Are there other options for integer programming in R? Thanks very much in advance, Robin Naidoo University of Alberta Edmonton, AB, Canada __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
[R] Multiple expressions in system.time()?
Hi All, Is it possible for system.time() to measure the time it takes for a machine to evaluate more than one R expression? For example, # This I can do: system.time(x - rnorm(10)) [1] 0.07 0.00 0.13 0.00 0.00 # But this I can't: system.time(x - rnorm(10); new - sample(x, 10, replace=T)) Error: syntax error # Nor this: system.time(x - rnorm(10) + new - sample(x, 10, replace=T) Error: syntax error I'm trying to compare two fairly large code chunks, so executing system.time() individually for each expression and then summing the results seems suboptimal. Have I missed something simple? Thanks in advance for any suggestions- Rob version: platform i386-pc-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major1 minor7.1 year 2003 month06 day 16 language R _ Rob Keefe Lab: (208) 885-5165 M.S. student Home: (208) 882-9749 University of Idaho __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
Re: [R] Multiple expressions in system.time()?
On Thu, Jul 24, 2003 at 07:10:27PM -0700, Robert Keefe wrote:
# This I can do:
system.time(x - rnorm(10))
[1] 0.07 0.00 0.13 0.00 0.00
# But this I can't:
system.time(x - rnorm(10); new - sample(x, 10, replace=T))
Error: syntax error
Just use curly braces:
system.time({x - rnorm(10); new - sample(x, 10, replace=T)})
[1] 0.11 0.00 0.11 0.00 0.00
Have I missed something simple?
See above. Another related way is to define a function; you could even use a
'throw-away anonymous' function [1].
Lastly, you probably also want to learn about profiling your code. There is
an introductory article in one of the R News issues.
Regards, Dirk
--
Those are my principles, and if you don't like them... well, I have others.
-- Groucho Marx
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