[R] R-2.6.2 installation (64bit) falling over with the grid package
It's a long time since I had a problem compiling R, but I've now
encountered one with R-2.6.2 on Fedora Core 6 (64bit).
The problem arises when the installation process gets to the grid
package, the unit.c part in particular. This is the output I get:
gcc -std=gnu99 -I../../../../include -I../../../../include
-I/usr/local/include -fvisibility=hidden -fpic -g -O2 -c unit.c -o unit.o
unit.c: In function ��pureNullUnit��:
unit.c:332: error: stray ��#�� in program
unit.c:332: error: �?R_fcall�� undeclared (first use in this function)
unit.c:332: error: (Each undeclared identifier is reported only once
unit.c:332: error: for each function it appears in.)
make[5]: *** [unit.o] Error 1
make[5]: Leaving directory `/usr/local/R-2.6.2/src/library/grid/src'
make[4]: *** [all] Error 2
make[4]: Leaving directory `/usr/local/R-2.6.2/src/library/grid/src'
make[3]: *** [all] Error 1
make[3]: Leaving directory `/usr/local/R-2.6.2/src/library/grid'
make[2]: *** [R] Error 1
make[2]: Leaving directory `/usr/local/R-2.6.2/src/library'
make[1]: *** [R] Error 1
make[1]: Leaving directory `/usr/local/R-2.6.2/src'
make: *** [R] Error 1
I installed the same tgz file on a 32-bit Mepis machine and on a
CentOS5 machine without such any problem . All previous 64bit
compilations have also been without a problem.
There has been a problem with one PCI slot on the machine in question
after an electrical misadventure. If nobody else has had a problem on
64 bit machines, could it be that I have a hardware problem? Seems a
bit unlikely it would get as far as it did were that the case, but I
could be convinced otherwise.
Ideas? Other experience?
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_Middle minds discuss events
(:_~*~_:)Small minds discuss people
(_)-(_) . Anon
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Re: [R] R-2.6.2 installation (64bit) falling over with the grid package
It's your system: try unpacking the tarball again. That line is eval(R_fcall3, R_gridEvalEnv); and has no # there. I suspect FC6 systems are thin on the ground (it is end-of-life, I believe), but R-2.6.2.tar.gz has been installed on many, many 64-bit Linux systems (e.g. CRAN tests on one), including an FC5 system (which we still, just, have for testing compatibility with an old OS). People seem to fail to appreciate the effort that goes into checking that the release tarballs are complete and uncorrupted. On Sun, 24 Feb 2008, Patrick Connolly wrote: It's a long time since I had a problem compiling R, but I've now encountered one with R-2.6.2 on Fedora Core 6 (64bit). The problem arises when the installation process gets to the grid package, the unit.c part in particular. This is the output I get: gcc -std=gnu99 -I../../../../include -I../../../../include -I/usr/local/include -fvisibility=hidden -fpic -g -O2 -c unit.c -o unit.o unit.c: In function pureNullUnit: unit.c:332: error: stray # in program unit.c:332: error: ÿÿ?R_fcall undeclared (first use in this function) unit.c:332: error: (Each undeclared identifier is reported only once unit.c:332: error: for each function it appears in.) make[5]: *** [unit.o] Error 1 make[5]: Leaving directory `/usr/local/R-2.6.2/src/library/grid/src' make[4]: *** [all] Error 2 make[4]: Leaving directory `/usr/local/R-2.6.2/src/library/grid/src' make[3]: *** [all] Error 1 make[3]: Leaving directory `/usr/local/R-2.6.2/src/library/grid' make[2]: *** [R] Error 1 make[2]: Leaving directory `/usr/local/R-2.6.2/src/library' make[1]: *** [R] Error 1 make[1]: Leaving directory `/usr/local/R-2.6.2/src' make: *** [R] Error 1 I installed the same tgz file on a 32-bit Mepis machine and on a CentOS5 machine without such any problem . All previous 64bit compilations have also been without a problem. There has been a problem with one PCI slot on the machine in question after an electrical misadventure. If nobody else has had a problem on 64 bit machines, could it be that I have a hardware problem? Seems a bit unlikely it would get as far as it did were that the case, but I could be convinced otherwise. Ideas? Other experience? -- 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__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem with ML estimation
dear list,
as a part my problem. I have to estimate some parameters using ML
estimation. The form of the likelihood function
is not straight forward and I had to use a for loop to define the function.
I used optim to maximise the result but
was not sure of the programme.
To validate my results, I tried to write a function to obtain the MLE of a
bivariate normal in the same manner.
On generating a simulated data and running the script, I am getting results
which are very inaccurate.
I am sending my programme. It would be of great help if someone can point
me, where I am going wrong.
library(MASS)
Sigma - matrix(c(4,2.8,2.8,4),2,2)
y-mvrnorm(n=1000, rep(0, 2), Sigma)
mlebinorm- function(startval, y)# startval = Initial Values to be passed ,
y= data
{
startval-as.numeric(startval)
oneside=matrix(c(0,0,0,0,-1,0,0,0,0,1),5,2)
otherside = c(-0.999,-0.99)
n- ncol(y)
nf- function(x)
{
mu1-x[1]
mu2-x[2]
sig1-x[3]
sig2-x[4]
rho-x[5]
nf-(-n/2)*(log(sig1*sig2*(1-rho^2)))-(0.5/(1-rho^2)*
(((y[1,1]-mu1)^2/sig1)-
(2*rho*(y[1,1]-mu1)*(y[1,2]-mu2))/sqrt(sig1*sig2)+
(y[1,2]-mu2)^2/sig2))
for(i in 2:n)
{
nf- nf - (0.5*(1-rho^2)*
(((y[i,1]-mu1)^2/sig1)-
(2*rho*(y[i,1]-mu1)*(y[i,2]-mu2))/sqrt(sig1*sig2)+
(y[i,2]-mu2)^2/sig2))
}
return(nf)
}
constrOptim(startval,nf,NULL,ui=t(oneside),
ci=otherside,control=list(fnscale=-1))
}
mlebinorm(c(1,1,2,2,0.3),y)
Souvik Banerjee
Lecturer
Department of statistics
Memari College
Burdwan
India
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[R] Generic Functions
Hi
I have some problems in defining new generic functions and classes. Just
have a look at the following example:
require(fPortfolio)
setClass(PROBECLASS,
representation(
type=character
)
)
isGeneric(setType-)
#Returns
TRUE
#I would like to define a specific function for class PROBECLASS with other
arguments than for the generic function setType of fPortfolio.
setGeneric(setType-, function(object, value)
standardGeneric(setType-))
#Returns
Fehler in makeGeneric(name, fdef, fdeflt, group = group, valueClass =
valueClass, :
the formal arguments of the generic function for setType- (object,
value) differ from those of the non-generic to be used as the default (spec,
value)
setReplaceMethod(setType, PROBECLASS, function(object, value){
[EMAIL PROTECTED] - value
object
})
#Example
obj = new(PROBECLASS)
setType(obj) = test
obj
##
If I don't require fPortfolio it works fine. However, is it not possible to
create two generic functions with the same name but different arguments?
setType for fPortfolio may be differ completely from setType of
PROBECLASS...
What's the best way to have functions which belongs to an object of a
specific class? I had a look at the paper S4 Classes in 15 pages, more or
less (feb12, 2003), however, I could not found what I did wrong...
Any help is highly appreciated.
Thanks
Dominik
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Re: [R] Newbie: Measuring distance between clusters.
It works! I was banging my head for a week. thank you so much. Bill.Venables wrote: One way to do it is to find the distances between ther centers (= centres in English) of the clusters. dist(kc$centers) It rather depends on how you define distances between clusters, though. There are many possibilities. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Keizer_71 Sent: Saturday, 23 February 2008 7:40 PM To: r-help@r-project.org Subject: [R] Newbie: Measuring distance between clusters. Hi, I had 1 genes, and I clustered them using K-means clustering in R. kc-kmeans(data.sub,7) kc n cluster sum of squares by cluster: [1] 60631.76 135886.19 159049.71 101783.27 90040.72 183335.60 158867.81 Available components: [1] cluster centers withinss size I am very new to R. How do i measure the distance between those cluster? I tried I am trying to do a complete linkage z-hclust(kc,method=complete) Error in if (n 2) stop(must have n = 2 objects to cluster) : argument is of length zero thanks. -- View this message in context: http://www.nabble.com/Newbie%3A-Measuring-distance-between-clusters.-tp1 5650066p15650066.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Newbie%3A-Measuring-distance-between-clusters.-tp15650066p15665075.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] Announcement: obsSens Package
The new package obsSens is now on the CRAN mirrors. This package has tools for doing senstitivity analysis for observational studies. The common criticism of observational studies is that there is the possibility of an unmeasured variable that is related to both the response and the predictor of interst that could explain the observed relationship. The sensitivity analysis is a What If? game where you specify potential relationships between the unmeasured variable and the response and predictor variable and see how that affects the relationship between predictor and response. [[alternative HTML version deleted]] ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] mixed model nested ANOVA (part two)
First of all thank you for the responses. I appreciate the suggestions i have received thus far. Just to reiterate I am trying to analyze a data set that has been collected from a hierarchical sampling design. The model should be a mixed model nested ANOVA. The purpose of my study is to analyze the variability at each spatial scale in my design (random factors, variance components), and say something about the variability between regions (fixed factor, contrast of means). The data is as follows; region (fixed) Location (random) Site(random) site nested in location nested in region. Also i have read in Quinn and Keough 2002, design and analysis of experiments for biologists, that a variance component analysis should only be conducted after a rejection of the null hypothesis of no variance at that level. I have tried to implement mod1-lmer(density ~ 1 + (1|site) + (1|location) + (1|region)) However, as i understand it, this treats all my factors as random. Plus I do not know how to extract SS or MS from this model. anova(mod1) gives me Analysis of Variance Table Df Sum Sq Mean Sq and summary(mod1) gives me Linear mixed-effects model fit by REML Formula: density ~ 1 + (1 | site) + (1 | location) + (1 | region) AIC BIC logLik MLdeviance REMLdeviance 15658 15678 -7825 1566215650 Random effects: Groups NameVariance Std.Dev. site (Intercept) 22191 148.97 location (Intercept) 33544 183.15 region (Intercept) 41412 203.50 Residual 696189 834.38 number of obs: 960, groups: site, 4; location, 4; region, 3 Fixed effects: Estimate Std. Error t value (Intercept)261.3 168.7 1.549 from what i understand the variance in the penultimate column are my variance components. But how do i conduct my significance test? I have also tried mod1-lmer(density ~ region + (1|site) + (1|location)) Which i think is the correct mixed model for my design. However once again i do not know how to evaluate significance for the random factors. Thank-you again for any additional advice i receive Stephen Cole __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] what missed ----- CART
Hi all, Can anyone who is familar with CART tell me what I missed in my tree code? library (MASS) myfit - tree (y ~ x1 + x2 + x3 + x4 ) # tree.screens () # useless plot(myfit); text (myfit, all= TRUE, cex=0.5, pretty=0) # tile.tree (myfit, fgl$type) # useless # close.screen (all= TRUE) # useless My current tree plot resulted from above code shows as: 1. overlapped #s caused by unsuitable length of branch. 2. no misclassification rates: 'misclass.tree' only brings up the error of ' misclassification error rate is appropriate for factor responses only', but my response y is 0/1 data. 3. Unsuitable location of notations: there are not two notation of splitting criteria on the two branches when a node is split, instead only one notation of splitting criteria is on the node location. thanks, xiao yue - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Standard method for S4 object
/I think your question should be more relevant on Rdev./ ok, I will Personnally I would find stuff like names, $, $-, or [ useful as these are usual operation with S3 objects. Is it possible in S4 to define $- ? If there is a slot name 'a' in object 'B', I find (in S4 in 15 pages more or less) setGeneric(a-, function(x, value) standardGeneric(a-)) setReplaceMethod(a, . Then we can use obj - new(B) a(obj)- 3 But I did not find how to define obj$a - 3 Is it possible ? Is it the 'usual' way ? Christophe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Standard method for S4 object
Christophe Genolini wrote:
/I think your question should be more relevant on Rdev./
ok, I will
Personnally I would find stuff like names, $, $-, or [
useful as these are usual operation with S3 objects.
Is it possible in S4 to define $- ? If there is a slot name 'a' in
Possible yes, see below. Usual? It implies to the user that this is a
list- or data.frame-like object (where $ is used most commonly). Maybe
your object is not like that? It also seems like a short step from
direct slot access (although it doesn't have to be), which might be
breaking the abstraction layer that object orientation provides. And it
shifts the responsibility for confirming that 'name' is a slot, and
dispatching on different values of 'name' (e.g. for some slots perhaps
one doesn't want to allow access, and then the code inside the '$'
method has to check that), back to the developer (instead of allowing
the method definition and dispatch to do its work). Generally these seem
like backward steps to me.
setClass(A, representation(x=numeric))
[1] A
setReplaceMethod($,
+ signature=signature(
+ x=A,
+ name=ANY,
+ value=ANY),
+ function(x, name, value) {
+ slot(x, name) - value
+ x
+ })
[1] $-
a - new(A)
a$x - 1:5
a
An object of class A
Slot x:
[1] 1 2 3 4 5
object 'B', I find (in S4 in 15 pages more or less)
setGeneric(a-, function(x, value) standardGeneric(a-))
setReplaceMethod(a, .
Then we can use
obj - new(B)
a(obj)- 3
But I did not find how to define
obj$a - 3
Is it possible ? Is it the 'usual' way ?
Christophe
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Re: [R] Generic Functions
See the 'useAsDefault' argument to setGeneric.
As an aside, if 'setType-' is meant to be a 'setter' to change the
value of a slot 'type', then I find the syntax a little redundant --
it's use
setType(x) - foo
implies that it is already a 'setter' without 'set' at the front. Why
not just
type(x) - foo
(though perhaps 'type' is not such a good name, either)?
As a second aside, if you're writing code that you expect to be used
with fPortfolio, then having two functions with the same name but
different signatures or overall goals will confuse your user -- with
fPortfolio, setType- works fine, but then for mysterious reasons (i.e.,
when your package is loaded, with a different definition of setType-)
code that worked before no longer works! So I'd either use setType in a
way consistent with it's use in fPortfolio, or define a new generic for
your own purposes (setType- is not a generic in my version of fPortfolio,
packageDescription('fPortfolio')$Version
[1] 260.72
).
As a third aside, I think questions about S4 probably belong on R-devel,
as they seem to fall in the realm of 'questions likely to prompt
discussion unintelligible to non-programmers' (from the R-devel mailing
list description).
Martin
Dominik Locher wrote:
Hi
I have some problems in defining new generic functions and classes. Just
have a look at the following example:
require(fPortfolio)
setClass(PROBECLASS,
representation(
type=character
)
)
isGeneric(setType-)
#Returns
TRUE
#I would like to define a specific function for class PROBECLASS with other
arguments than for the generic function setType of fPortfolio.
setGeneric(setType-, function(object, value)
standardGeneric(setType-))
#Returns
Fehler in makeGeneric(name, fdef, fdeflt, group = group, valueClass =
valueClass, :
the formal arguments of the generic function for setType- (object,
value) differ from those of the non-generic to be used as the default (spec,
value)
setReplaceMethod(setType, PROBECLASS, function(object, value){
[EMAIL PROTECTED] - value
object
})
#Example
obj = new(PROBECLASS)
setType(obj) = test
obj
##
If I don't require fPortfolio it works fine. However, is it not possible to
create two generic functions with the same name but different arguments?
setType for fPortfolio may be differ completely from setType of
PROBECLASS...
What's the best way to have functions which belongs to an object of a
specific class? I had a look at the paper S4 Classes in 15 pages, more or
less (feb12, 2003), however, I could not found what I did wrong...
Any help is highly appreciated.
Thanks
Dominik
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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[R] Minor tick marks
Hello, Is there a way to add minor tick marks to the Y-axis of a lattice plots? Thank you Saptarshi Saptarshi Guha | [EMAIL PROTECTED] | http://www.stat.purdue.edu/~sguha [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Standard method for S4 object
Martin Morgan wrote: Christophe Genolini wrote: Personnally I would find stuff like names, $, $-, or [ useful as these are usual operation with S3 objects. Is it possible in S4 to define $- ? If there is a slot name 'a' in Possible yes, see below. Usual? It implies to the user that this is a list- or data.frame-like object (where $ is used most commonly). Maybe your object is not like that? It also seems like a short step from direct slot access (although it doesn't have to be), which might be breaking the abstraction layer that object orientation provides. And it shifts the responsibility for confirming that 'name' is a slot, and dispatching on different values of 'name' (e.g. for some slots perhaps one doesn't want to allow access, and then the code inside the '$' method has to check that), back to the developer (instead of allowing the method definition and dispatch to do its work). Generally these seem like backward steps to me. Yes, to me too. But it is nonetheless useful when you convert an 'old' S3 class to S4 and want to re-use your old code as much as possible (or want users to be able to do so). I don't know if this is the case here. Plus, as $ and $- are just wrappers for @ and @-, the procedure checking the validity of a value assigned to a slot is still called when using $-. Another, maybe interesting option is to use accessors to change the values of the slots. This might be more object-oriented, and would allow for more control on the content of the object. Cheers, Thibaut. -- ## Thibaut JOMBART CNRS UMR 5558 - Laboratoire de Biométrie et Biologie Evolutive Universite Lyon 1 43 bd du 11 novembre 1918 69622 Villeurbanne Cedex Tél. : 04.72.43.29.35 Fax : 04.72.43.13.88 [EMAIL PROTECTED] http://lbbe.univ-lyon1.fr/-Jombart-Thibaut-.html?lang=en http://pbil.univ-lyon1.fr/software/adegenet/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] running R as a web service
Hi all, As R is gaining space in our research group as language and environment for statistical computing and graphics it will be really helpful to have the option of using R as a web service. We are currently using a collaborative edition wiki tool (XWiki) to develop R code. And we can call this code with source() from any R installation. But it will be great to be able to send this code to a web service and get back just the result (textual or graphics). Rserver seems to fit into this requirement. Please, could anybody using any of this or other option share with us his/her experience? Is it feasible to run such a service with the current R release? Thanks! Best regards, Ricardo -- Ricardo Rodríguez Your XEN ICT Team __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] Bayesian Prediction with High-order Interactions
Hi Everybody, A new package called ``Bayesian Prediction with High-order Interactions'' is available from CRAN. The description of this package is as follows This R package is used in two situations. The first is to predict the next outcome based on the previous states of a discrete sequence. The second is to classify a discrete response based on a number of discreate covariates. In both situations, we use Bayesian logistic regression models that consider the high-order interactions. The time arising from using high-order interactions is reduced greatly by our compression technique that represents a group of original parameters as a single one in MCMC step. In this version, we use log-normal prior for the hyperparameters. When it is used for the second situation --- classification, we consider the full set of interaction patterns up to a specified order. The website of this package is http://fisher.utstat.toronto.edu/~longhai/software/BPHO/release.html -- Longhai Li, PhD Assistant Professor Department of Mathematics and Statistics University of Saskatchewan Homepage: http://math.usask.ca/~longhai ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] what missed ----- CART
o ha wang wrote: Hi all, Can anyone who is familar with CART tell me what I missed in my tree code? library (MASS) myfit - tree (y ~ x1 + x2 + x3 + x4 ) There is not function tree() in MASS. I guess you have loaded package tree? Note that the author and maintainer of the tree package suggests to use package rpart instead. # tree.screens () # useless plot(myfit); text (myfit, all= TRUE, cex=0.5, pretty=0) # tile.tree (myfit, fgl$type) # useless # close.screen (all= TRUE) # useless My current tree plot resulted from above code shows as: 1. overlapped #s caused by unsuitable length of branch. 2. no misclassification rates: 'misclass.tree' only brings up the error of ' misclassification error rate is appropriate for factor responses only', but my response y is 0/1 data. If you data is 0/1, you should make it a factor otherwise the cose will assume you want a regression rather than a classification tree. 3. Unsuitable location of notations: there are not two notation of splitting criteria on the two branches when a node is split, instead only one notation of splitting criteria is on the node location. It indicates which items go into the left branch. Uwe Ligges thanks, xiao yue - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Split a matrix for mixture distribution
Evgenia wrote: Hello, I have a matrix with 4000 rows and 8 columns with 1 and zeros Every 1000 rows comes for a different rule, f1,f2,f3,f4. 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 ... 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 ... 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 How can I extract 4 matrices each one containes 1000 elements from a certain rule,since i want to a) choose between f1, f2, f3 and f4 at random according to the probabilities p1, p2, p3 ,p4 (it is assumed p1+p2+p3=1). b) sample 1 value from whichever of f1, f2, f3 ,f4 was chosen I guess we need some better explanations or examples - at least I do not understand your aims. Uwe Ligges Thanks a lot Evgenia __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ginv and matlab's pinv give different results
Pedro Mardones wrote: Dear all; I'm kind of confused with the results obtained using the ginv function from package MASS and pinv function from Matlab. Accroding to the documentation both functions performs a Moore-Penrose generalized inverse of a matrix X. The problem is when I change the tolerance value, say to 1E-3. Here is some output from ginv 195.2674402 235.6758714 335.0830253 8.977515484 -291.7798965 428.8266383 165.9809056 284.0561017 -74.35007018-210.3373981 440.6537582 42.44715935 290.4538488 -58.74366626-86.71457799 246.4469887 -36.24936825397.6034028 -107.757444810.7545897 -17.07138404 30.43896492 399.3831596 -88.4968195912.33524917 ad here from pinv 1.988794963 4.931306729 -12.72706642-3.903229019-4.516082822 3.017403231 4.799144574 -14.43317599-5.388596501-5.872449411 4.951830097 4.684716091 -17.13075023-6.584430072-7.142673623 7.457154317 2.896685565 -21.59318352-11.38286985-10.58210901 7.841116927 0.836247791 -23.12560946-13.99460333-11.8333927 I'm sure it must be some explanation that I'm not aware of. If anyone can (want) to run the test I can email the data set Without the original data we cannot say if one or both of them are wrong or correct. Hence, please make the data and the code you have used available - as the posting guide suggests. Uwe Ligges Thanks for any idea PM __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] projection.matrix() {popbio} for more than one matrix
Michelle DePrenger-Levin wrote: Hello, I am trying to use the projection.matrix( ) function and am following the example given. I have my data formatted very similar to the test.census example. str(AsMi05mat) `data.frame': 1854 obs. of 6 variables: $ Tag : num 501 502 503 504 505 506 507 508 509 510 ... $ Year : int 1995 1995 1995 1995 1995 1995 1995 1995 1995 1995 ... $ Length : num 34 37 11 24 7 44 4 7 12 20 ... $ Flowering: int 1 1 0 1 0 1 0 0 0 1 ... $ Fruits : int 22 22 0 89 0 15 0 0 0 0 ... $ Stage: Factor w/ 5 levels ,Dead,Dormant,..: 4 4 5 4 5 4 5 5 5 4 ... The example data includes three years but only shows how to create a matrix from year 1 to 2. I have 13 years of data and would like to automate creating all matrices for each pair of years. I tried a for( ) loop but don't know how to assign new names for each matrix so end up with only the final comparison (2005 to 2006). I assume an apply( ) function is the way to go but can't see how to do it. Well, we cannot help to improve your code if we have neither, code nor data. Please read the posting guide and specify your code. If something is wrong in your code, we might be able to help. Uwe Ligges Thanks for any help! Michelle [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Combining Data Frames
stephen sefick wrote: I have two dataframes in R that were tab seperated .txt files y-read.table(foo.txt, header=T) x-read.table(foo.txt, header=T) these are set up like this: Datetime Temp 01/01/07 00:01 11.5 01/01/07 00:16 11.6 etc etc to 66000 rows in y and 33000 rows in x The two files overlap with the same data for a period of time but contain different values outside of these. Is there a way to merge these two data sets based on the shared date time column into one large dataset of ~90,000 lines. I want to eventually make this into a time series with frequency= 1/15. Have you tried ?merge. Uwe Ligges Thanks Stephen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R-2.6.2 installation (64bit) falling over with the grid package
On Sun, 24-Feb-2008 at 10:00AM +, Prof Brian Ripley wrote:
| It's your system: try unpacking the tarball again.
No surprises there: Brian is spot on again. Thank you Brian.
I'd not have thought of trying unpacking again. I thought there would
be no stochastic element to such a process. Perhaps the hardware
could introduce a stochastic element. (This machine is only 3 years
old, but already it's impossible to buy a replacement mobo.)
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_Middle minds discuss events
(:_~*~_:)Small minds discuss people
(_)-(_) . Anon
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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Re: [R] Error in ma.svd(X, 0, 0) : 0 extent dimensions
I guess you don't have any degrees of freedom left ... How much observations versus how much estimated parameters do you have for estimating your model parameters? Uwe Ligges Marcelo Luiz de Laia wrote: Hi, I run a maanova analysis and found this message error: Error in ma.svd(X, 0, 0) : 0 extent dimensions I did a google search and found this: \item ma.svd: function to compute the sigular-value decomposition of a rectangular matrix by using LAPACK routines DEGSVD AND ZGESVD. \item fdr: function to calculate the adjusted P values for FDR control. I did a search for LAPACK and not found a package. Could you help me on how I could solve this problem? I am try to do this: library(maanova) read the data file and design # Make the full model based on the design anova.full.mix - fitmaanova(data, formula=~ Var+ind+Trat+Time+ Var:ind+Var:Trat+Var:Time+ ind:Trat+ind:Time+ Trat:Time+ Var:ind:Trat+ Var:ind:Time+ Sample,random=~Sample) If I remove the Var:ind:Trat and Var:ind:Time from formula the script runs very well. Did is not possible to do that interactions (Var:ind:Trat and Var:ind:Time)? What you suggest me to do that? Thank you very much! Marcelo __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] rapply on a data.frame
Is it possible to get something like rapply for a data.frame? For example, if I do this: x - list(a=mtcars, b=mtcars) str(rapply(x, nrow, classes=data.frame, how=replace)) List of 2 $ a:List of 11 ..$ mpg : num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ... ..$ cyl : num [1:32] 6 6 4 6 8 6 8 4 4 6 ... ..$ disp: num [1:32] 160 160 108 258 360 ... ..$ hp : num [1:32] 110 110 93 110 175 105 245 62 95 123 ... ..$ drat: num [1:32] 3.9 3.9 3.85 3.08 3.15 2.76 3.21 3.69 3.92 3.92 ... ..$ wt : num [1:32] 2.62 2.88 2.32 3.21 3.44 ... ..$ qsec: num [1:32] 16.5 17.0 18.6 19.4 17.0 ... ..$ vs : num [1:32] 0 0 1 1 0 1 0 1 1 1 ... ..$ am : num [1:32] 1 1 1 0 0 0 0 0 0 0 ... ..$ gear: num [1:32] 4 4 4 3 3 3 3 4 4 4 ... ..$ carb: num [1:32] 4 4 1 1 2 1 4 2 2 4 ... $ b:List of 11 ..$ mpg : num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ... ..$ cyl : num [1:32] 6 6 4 6 8 6 8 4 4 6 ... ..$ disp: num [1:32] 160 160 108 258 360 ... ..$ hp : num [1:32] 110 110 93 110 175 105 245 62 95 123 ... ..$ drat: num [1:32] 3.9 3.9 3.85 3.08 3.15 2.76 3.21 3.69 3.92 3.92 ... ..$ wt : num [1:32] 2.62 2.88 2.32 3.21 3.44 ... ..$ qsec: num [1:32] 16.5 17.0 18.6 19.4 17.0 ... ..$ vs : num [1:32] 0 0 1 1 0 1 0 1 1 1 ... ..$ am : num [1:32] 1 1 1 0 0 0 0 0 0 0 ... ..$ gear: num [1:32] 4 4 4 3 3 3 3 4 4 4 ... ..$ carb: num [1:32] 4 4 1 1 2 1 4 2 2 4 ... I was hoping to get something like list(a=11, b=11). I suspect this is because the data.frame is actually implemented as a list. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Bootstrapping data with a regression model
Hello all: Just wondering if I can get advice on what kind of bootstrapping I should use when using a regression model to estimate juvenile fish passage data. I use rotary screw traps to do fish mark-recapture trials and the efficiency of every trial is added to the graph generating a different R-square and equation everytime. I plot river flows along the X axis and % trap efficiency along the Y axis. I have read about the different bootstrapping approaches but I am not sure how to do it..any help is appreciated. Felipe D. Carrillo Fishery Biologist US Fish Wildlife Service California, USA Never miss a thing. Make Yahoo your home page. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Bootstrapping data with a regression model
On 2/24/2008 4:03 PM, Felipe Carrillo wrote: Hello all: Just wondering if I can get advice on what kind of bootstrapping I should use when using a regression model to estimate juvenile fish passage data. I use rotary screw traps to do fish mark-recapture trials and the efficiency of every trial is added to the graph generating a different R-square and equation everytime. I plot river flows along the X axis and % trap efficiency along the Y axis. I have read about the different bootstrapping approaches but I am not sure how to do it..any help is appreciated. You might check out the the relevant web appendix by John Fox in the Contributed Documentation section of CRAN: http://cran.r-project.org/doc/contrib/Fox-Companion/appendix.html http://cran.r-project.org/doc/contrib/Fox-Companion/appendix-bootstrapping.pdf Felipe D. Carrillo Fishery Biologist US Fish Wildlife Service California, USA Never miss a thing. Make Yahoo your home page. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie: Where is lmFit function?
It's in the 'limma' Bioconductor package. Next time you can try help.search(lmFit) RSiteSearch(lmFit) Gabor On Sun, Feb 24, 2008 at 01:02:41PM -0800, Keizer_71 wrote: Hi Everyone, I am trying to use lmFit function; however, i cannot find it function anywhere. I have been trying to find the function in Bioconductor and elsewhere. I re-install bioconductor source, update package and update R as well. no luck Is there a command in R where i can just type, and it will download it for me? -- View this message in context: http://www.nabble.com/Newbie%3A-Where-is-lmFit-function--tp15669332p15669332.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Csardi Gabor [EMAIL PROTECTED]UNIL DGM __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] rapply on a data.frame
Michael Hoffman wrote:
Is it possible to get something like rapply for a data.frame?
On second thought, I can do something like this:
happly - function(x, fun) {
if (class(x) == list) {
lapply(x, happly, fun)
} else {
fun(x)
}
}
happly(list(a=list(c=mtcars, d=Titanic), x=mtcars), nrow)
$a
$a$c
[1] 32
$a$d
[1] 4
$x
[1] 32
Better suggestions welcome.
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Re: [R] Newbie: Where is lmFit function?
thank you. Chuck Cleland wrote: On 2/24/2008 4:02 PM, Keizer_71 wrote: Hi Everyone, I am trying to use lmFit function; however, i cannot find it function anywhere. I have been trying to find the function in Bioconductor and elsewhere. I re-install bioconductor source, update package and update R as well. no luck Is there a command in R where i can just type, and it will download it for me? RSiteSearch(lmFit) shows there is a function with that name in the limma package. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Newbie%3A-Where-is-lmFit-function--tp15669332p15669414.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie: Where is lmFit function?
On 2/24/2008 4:02 PM, Keizer_71 wrote: Hi Everyone, I am trying to use lmFit function; however, i cannot find it function anywhere. I have been trying to find the function in Bioconductor and elsewhere. I re-install bioconductor source, update package and update R as well. no luck Is there a command in R where i can just type, and it will download it for me? RSiteSearch(lmFit) shows there is a function with that name in the limma package. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mixed model nested ANOVA (part two)
Hi Stephen, Also i have read in Quinn and Keough 2002, design and analysis of experiments for biologists, that a variance component analysis should only be conducted after a rejection of the null hypothesis of no variance at that level. Once again the caveat: there are experts on this list who really know about this stuff, and I am not one of them. Your general strategy would be to set up two models with the same fixed effects, one of which doesn't have random effects. You then test the two models using anova(mod.withRandom, modWithoutRandom). I haven't tried this using lmer/2(), but with lme() you do this by fitting your fixed+random effects model using lme() and your fixed-only effects model using lm(). If you are using weights to model heteroskedasticity, then it's better to use gls(), as this will accept the same weights argument as the call to lme(). Then you simply do anova(lme.model, lm/gls.model). This tells you about the significance of your random effects, i.e. whether you need a random-effects component. To recap: mod.rand - lme(fixed=y ~ x, random=~x|Site, data=...) mod,fix - lm(fixed=y ~ x, data=...) anova(mod.rand, mod.fix) HTH, Mark. Stephen Cole-2 wrote: First of all thank you for the responses. I appreciate the suggestions i have received thus far. Just to reiterate I am trying to analyze a data set that has been collected from a hierarchical sampling design. The model should be a mixed model nested ANOVA. The purpose of my study is to analyze the variability at each spatial scale in my design (random factors, variance components), and say something about the variability between regions (fixed factor, contrast of means). The data is as follows; region (fixed) Location (random) Site(random) site nested in location nested in region. Also i have read in Quinn and Keough 2002, design and analysis of experiments for biologists, that a variance component analysis should only be conducted after a rejection of the null hypothesis of no variance at that level. I have tried to implement mod1-lmer(density ~ 1 + (1|site) + (1|location) + (1|region)) However, as i understand it, this treats all my factors as random. Plus I do not know how to extract SS or MS from this model. anova(mod1) gives me Analysis of Variance Table Df Sum Sq Mean Sq and summary(mod1) gives me Linear mixed-effects model fit by REML Formula: density ~ 1 + (1 | site) + (1 | location) + (1 | region) AIC BIC logLik MLdeviance REMLdeviance 15658 15678 -7825 1566215650 Random effects: Groups NameVariance Std.Dev. site (Intercept) 22191 148.97 location (Intercept) 33544 183.15 region (Intercept) 41412 203.50 Residual 696189 834.38 number of obs: 960, groups: site, 4; location, 4; region, 3 Fixed effects: Estimate Std. Error t value (Intercept)261.3 168.7 1.549 from what i understand the variance in the penultimate column are my variance components. But how do i conduct my significance test? I have also tried mod1-lmer(density ~ region + (1|site) + (1|location)) Which i think is the correct mixed model for my design. However once again i do not know how to evaluate significance for the random factors. Thank-you again for any additional advice i receive Stephen Cole __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/mixed-model-nested-ANOVA-%28part-two%29-tp15665478p15669384.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Newbie: Where is lmFit function?
Hi Everyone, I am trying to use lmFit function; however, i cannot find it function anywhere. I have been trying to find the function in Bioconductor and elsewhere. I re-install bioconductor source, update package and update R as well. no luck Is there a command in R where i can just type, and it will download it for me? -- View this message in context: http://www.nabble.com/Newbie%3A-Where-is-lmFit-function--tp15669332p15669332.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to put n and yprob on the tree plot ----- Re: what missed ----- CART
o ha wang wrote: Thanks! Yes, you were right. I loaded package 'tree'. I tried 'tree' and 'rpart' respectively, it looks like results for regression tree are same, and very similar for classification tree. (more biref using 'rpart' than using 'tree'). But my problems are still not been solved using rpart. 1. how to put number (n, ie. class counts) and probability (yprob) on the plot for classification tree. There are only 0/1 on the plot. n and yprob can be seen using 'myfit' but not on the plot. 2. The notation of Numbers are overlapped for regression tree. please see my rpart code: library (rpart) myfit - rpart (y ~ x1 + x2 + x3 + x4 ) plot (myfit)); text (myfit, all= TRUE, cex=0.5, pretty=0) myfit summary(myfit) I still cannot reproduce without y, x1, x2, x3, x4 Uwe Ligges thanks, xiao yue Uwe Ligges [EMAIL PROTECTED] wrote: o ha wang wrote: Hi all, Can anyone who is familar with CART tell me what I missed in my tree code? library (MASS) myfit - tree (y ~ x1 + x2 + x3 + x4 ) There is not function tree() in MASS. I guess you have loaded package tree? Note that the author and maintainer of the tree package suggests to use package rpart instead. # tree.screens () # useless plot(myfit); text (myfit, all= TRUE, cex=0.5, pretty=0) # tile.tree (myfit, fgl$type) # useless # close.screen (all= TRUE) # useless My current tree plot resulted from above code shows as: 1. overlapped #s caused by unsuitable length of branch. 2. no misclassification rates: 'misclass.tree' only brings up the error of ' misclassification error rate is appropriate for factor responses only', but my response y is 0/1 data. If you data is 0/1, you should make it a factor otherwise the cose will assume you want a regression rather than a classification tree. 3. Unsuitable location of notations: there are not two notation of splitting criteria on the two branches when a node is split, instead only one notation of splitting criteria is on the node location. It indicates which items go into the left branch. Uwe Ligges thanks, xiao yue - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mixed model nested ANOVA (part two)
Hi Stephen, Slip of the dactylus: lm() does not, of course, take a fixed=arg. So you need To recap: mod.rand - lme(fixed=y ~ x, random=~x|Site, data=...) mod,fix - lm(y ~ x, data=...) ## or ##mod,fix - lm(formula=y ~ x, data=...) Bye. Mark Difford wrote: Hi Stephen, Also i have read in Quinn and Keough 2002, design and analysis of experiments for biologists, that a variance component analysis should only be conducted after a rejection of the null hypothesis of no variance at that level. Once again the caveat: there are experts on this list who really know about this stuff, and I am not one of them. Your general strategy would be to set up two models with the same fixed effects, one of which doesn't have random effects. You then test the two models using anova(mod.withRandom, modWithoutRandom). I haven't tried this using lmer/2(), but with lme() you do this by fitting your fixed+random effects model using lme() and your fixed-only effects model using lm(). If you are using weights to model heteroskedasticity, then it's better to use gls(), as this will accept the same weights argument as the call to lme(). Then you simply do anova(lme.model, lm/gls.model). This tells you about the significance of your random effects, i.e. whether you need a random-effects component. To recap: mod.rand - lme(fixed=y ~ x, random=~x|Site, data=...) mod,fix - lm(fixed=y ~ x, data=...) anova(mod.rand, mod.fix) HTH, Mark. Stephen Cole-2 wrote: First of all thank you for the responses. I appreciate the suggestions i have received thus far. Just to reiterate I am trying to analyze a data set that has been collected from a hierarchical sampling design. The model should be a mixed model nested ANOVA. The purpose of my study is to analyze the variability at each spatial scale in my design (random factors, variance components), and say something about the variability between regions (fixed factor, contrast of means). The data is as follows; region (fixed) Location (random) Site(random) site nested in location nested in region. Also i have read in Quinn and Keough 2002, design and analysis of experiments for biologists, that a variance component analysis should only be conducted after a rejection of the null hypothesis of no variance at that level. I have tried to implement mod1-lmer(density ~ 1 + (1|site) + (1|location) + (1|region)) However, as i understand it, this treats all my factors as random. Plus I do not know how to extract SS or MS from this model. anova(mod1) gives me Analysis of Variance Table Df Sum Sq Mean Sq and summary(mod1) gives me Linear mixed-effects model fit by REML Formula: density ~ 1 + (1 | site) + (1 | location) + (1 | region) AIC BIC logLik MLdeviance REMLdeviance 15658 15678 -7825 1566215650 Random effects: Groups NameVariance Std.Dev. site (Intercept) 22191 148.97 location (Intercept) 33544 183.15 region (Intercept) 41412 203.50 Residual 696189 834.38 number of obs: 960, groups: site, 4; location, 4; region, 3 Fixed effects: Estimate Std. Error t value (Intercept)261.3 168.7 1.549 from what i understand the variance in the penultimate column are my variance components. But how do i conduct my significance test? I have also tried mod1-lmer(density ~ region + (1|site) + (1|location)) Which i think is the correct mixed model for my design. However once again i do not know how to evaluate significance for the random factors. Thank-you again for any additional advice i receive Stephen Cole __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/mixed-model-nested-ANOVA-%28part-two%29-tp15665478p15669608.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to put n and yprob on the tree plot ----- Re: what missed ----- CART
Thanks! Yes, you were right. I loaded package 'tree'. I tried 'tree' and 'rpart' respectively, it looks like results for regression tree are same, and very similar for classification tree. (more biref using 'rpart' than using 'tree'). But my problems are still not been solved using rpart. 1. how to put number (n, ie. class counts) and probability (yprob) on the plot for classification tree. There are only 0/1 on the plot. n and yprob can be seen using 'myfit' but not on the plot. 2. The notation of Numbers are overlapped for regression tree. please see my rpart code: library (rpart) myfit - rpart (y ~ x1 + x2 + x3 + x4 ) plot (myfit)); text (myfit, all= TRUE, cex=0.5, pretty=0) myfit summary(myfit) thanks, xiao yue Uwe Ligges [EMAIL PROTECTED] wrote: o ha wang wrote: Hi all, Can anyone who is familar with CART tell me what I missed in my tree code? library (MASS) myfit - tree (y ~ x1 + x2 + x3 + x4 ) There is not function tree() in MASS. I guess you have loaded package tree? Note that the author and maintainer of the tree package suggests to use package rpart instead. # tree.screens () # useless plot(myfit); text (myfit, all= TRUE, cex=0.5, pretty=0) # tile.tree (myfit, fgl$type) # useless # close.screen (all= TRUE) # useless My current tree plot resulted from above code shows as: 1. overlapped #s caused by unsuitable length of branch. 2. no misclassification rates: 'misclass.tree' only brings up the error of ' misclassification error rate is appropriate for factor responses only', but my response y is 0/1 data. If you data is 0/1, you should make it a factor otherwise the cose will assume you want a regression rather than a classification tree. 3. Unsuitable location of notations: there are not two notation of splitting criteria on the two branches when a node is split, instead only one notation of splitting criteria is on the node location. It indicates which items go into the left branch. Uwe Ligges thanks, xiao yue - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Color of CI bars in Hmisc::Dotplot()
Dear R-helpers,
(1) I can't figure out how to tell Dotplot that I want the colors of
the CI bars to be the same as the colors of the dots.
For example:
t2a.ci3 - data.frame(est = c(7, 20, 75), lower = c(-9, 0.5, 42),
upper = c(22, 39, 109))
mypal - c('skyblue3', 'mistyrose3')
condNames - c('noise', 'horn', 'marimba')
Dotplot(1:2 ~ Cbind(est, lower, upper), data = t2a.ci3,
scales = list(y = list(at = 1:2, labels = condNames)),
ylab = '',
xlab = 'rated duration', aspect = 0.2,
ylim = c(0.5, 2.5), panel = function(x, y){
panel.Dotplot(
x, y, col = mypal[c(1, 2)],
col.segments = mypal[c(1, 2)])
panel.abline(v = 50, col.line = 'red', lty = 2)}
)
My guess was col.segments but that doesn't do the trick.
(2) How do I insure that as I change the number of lines and ylim I
just change the the height of the y-axis, keep the distance between
horizontal lines constant, and do not add white space above and below
the figure? This shows how white space changes:
Dotplot(1:3 ~ Cbind(est, lower, upper), data = t2a.ci3,
scales = list(y = list(at = 1:3, labels = condNames)),
ylab = '',
xlab = 'rated duration', aspect = 0.3,
ylim = c(0.5, 3.5), panel = function(x, y){
panel.Dotplot(
x, y, col = mypal[c(1, 2, 1)],
col.segments = mypal[c(1, 2, 1)])
panel.abline(v = 50, col.line = 'red', lty = 2)}
)
_
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Box 400400Charlottesville, VA 22904-4400
Parcels:Room 102Gilmer Hall
McCormick RoadCharlottesville, VA 22903
Office:B011+1-434-982-4729
Lab:B019+1-434-982-4751
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Re: [R] mixed model nested ANOVA (part two)
Also i have read in Quinn and Keough 2002, design and analysis of
experiments for
biologists, that a variance component analysis should only be conducted
after a rejection
of the null hypothesis of no variance at that level.
Hmmm...
This does rather assume that 'no significant result' means 'near-zero variance
contribution'.
These are not identical statements if the anova has low power; an
'insignificant' term can conceal practically important sizes of effect. So if
you have a smallish number of groups (say, ten or less) you might want to find
out what that estimated between-group variance could have been before you throw
it away. That's especially important if you're expecting to say something about
standard errors or confidence intervals of fixed effects.
I may well be biased, here, though. In the kinds of nested design I get
involved in (often inter-laboratory or homogeneity studies in chemistry), there
is nearly always a between-group effect; the only question is its size. Under
those circumstances, the null hypothesis is not a particularly compelling
starting point. I'd rather have a variance component estimate and know how
vague it was than assume it wasn't there at all.
But if you have good power and a good reason for believing there's no case to
answer, sure; assume zero unless proven otherwise.
S
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[R] Jitter in correlation matrix?
Hi,
I am just starting to use R for a graduate course, and I like how the
correlation matrix at
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=137
I did something similar by copying from the examples(pairs), but it
seems that I need to jitter the bottom panel... and I have no idea how
to do that, and I mean no idea at all. I'd appreciate any help...
Here are the graphs:
http://socy602.pbwiki.com/f/dep_correlationmatrix.jpeg
http://socy602.pbwiki.com/f/indep_correlationmatrix.jpeg
And this is the code I used to produce the graphs (not that I
understand it, but):
##
# Primary Component Analyses
##
##
# Dependent variables
##
# List of these variables:
# langfamr , ethdifxx , catness , gc7 , gc8r , gc12 , culdifxx , poldifxx ,
ecdifxx
##
# Look at correlation matrix for these
cor(MAR[,c(langfamr , ethdifxx , catness , gc7 , gc8r , gc12 ,
culdifxx , poldifxx , ecdifxx )], use=complete.obs)
# Big time correlation matrix
panel.cor - function(x, y, digits=2, prefix=, cex.cor)
{
usr - par(usr); on.exit(par(usr))
par(usr = c(0, 1, 0, 1))
r - abs(cor(x, y))
txt - format(c(r, 0.123456789), digits=digits)[1]
txt - paste(prefix, txt, sep=)
if(missing(cex.cor)) cex - 0.8/strwidth(txt)
text(0.5, 0.5, txt, cex = cex * r)
}
MAR.omitdep - na.omit(MAR[,c(langfamr , ethdifxx , catness , gc7 ,
gc8r , gc12 , culdifxx , poldifxx , ecdifxx )])
pairs(MAR.omitdep[,c(langfamr , ethdifxx , catness , gc7 , gc8r ,
gc12 , culdifxx , poldifxx , ecdifxx )], lower.panel=panel.smooth,
upper.panel=panel.cor)
# Save it
dev.copy(jpeg,filename=dep_correlationmatrix.jpeg,height=600,
width=800,bg=white)
dev.off()
##
# Independent variables
##
# List of variables
# poldis , polres , ecdis , culres , gcc1
##
# Look at the correlation matrix
cor(MAR[,c(poldis , polres , ecdis , culres , gcc1 )],
use=complete.obs)
# Big time correlation matrix
panel.cor - function(x, y, digits=2, prefix=, cex.cor)
{
usr - par(usr); on.exit(par(usr))
par(usr = c(0, 1, 0, 1))
r - abs(cor(x, y))
txt - format(c(r, 0.123456789), digits=digits)[1]
txt - paste(prefix, txt, sep=)
if(missing(cex.cor)) cex - 0.8/strwidth(txt)
text(0.5, 0.5, txt, cex = cex * r)
}
MAR.omitindep - na.omit(MAR[,c(poldis , polres , ecdis , culres ,
gcc1 )])
pairs(MAR.omitindep[,c(poldis , polres , ecdis , culres , gcc1 )],
lower.panel=panel.smooth, upper.panel=panel.cor)
# Save it
dev.copy(jpeg,filename=indep_correlationmatrix.jpeg,height=600,
width=800,bg=white)
dev.off()
Thanks a lot in advance,
Sincerely,
Mehmet.
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[R] S4 : Signature
Hi the list Is it possible to define a method with a signature that will have more argument than the generic method? For exemple, print has only one argument, plot has two, can I do something like : setMethod(print,signature=c(x=numeric,y=character),function(x,y,...) .. setMethod(plot,signature=c(x=numeric,y=character,z=numeric),function(x,y,z,...) .. Thanks Christophe __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] color area between two time-series via polygon()?
Hi all, I would like to color the area between two time-series. I tried it by using the polygon() function but I keeps drawing lines between beginning and end points. Is there another more appropriate function or how could I close the polygon at the end en the beginning of the time series (e.g., drawing a straight line)? The following doesn't plot a polygon between the two time-series: z - ts(matrix(rnorm(200), 100), start=c(1961, 1), frequency=12) plot(z, plot.type=single, lty=1:2) xx - cbind(time(z[,1]),rev(z[,2])) yy - cbind(as.vector(z[,1]),rev(as.vector(z[,2]))) polygon(xx,yy, col=gray, border = red) I would like to make it look like this (but then for time series) n - 100 xx - c(0:n, n:0) yy - c(c(0,cumsum(stats::rnorm(n))), rev(c(0,cumsum(stats::rnorm(n) plot (xx, yy, type=n, xlab=Time, ylab=Distance) polygon(xx, yy, col=gray, border = red) Thanks for your help, Jan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] A more idiomatic way to write this
Hello,
I have a vector of 1,000,000 numbers and another vector of 1,000
divisors. What I'd like to do is to divide the first 1,000 numbers of
the first vector by the first divisor, then the next 1,000 by the second
divisor and so on. I came up with this, but I was wondering if there is
a more idiomatic, R-like way to write it:
x - ...
divs - ...
for (i in seq(from = 1, to = 100, by = 1000)) {
x[i:(i - 1 + 1000)] - x[i:(i - 1 + 1000)] / divs[i %/% 1000 + 1]
}
Any suggestions are welcome.
Thanks in advance,
Andre
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Re: [R] A more idiomatic way to write this
x/rep(divs,each=1000)
cheers,
Rolf Turner
On 25/02/2008, at 2:36 PM, Andre Nathan wrote:
Hello,
I have a vector of 1,000,000 numbers and another vector of 1,000
divisors. What I'd like to do is to divide the first 1,000 numbers of
the first vector by the first divisor, then the next 1,000 by the
second
divisor and so on. I came up with this, but I was wondering if
there is
a more idiomatic, R-like way to write it:
x - ...
divs - ...
for (i in seq(from = 1, to = 100, by = 1000)) {
x[i:(i - 1 + 1000)] - x[i:(i - 1 + 1000)] / divs[i %/% 1000 + 1]
}
Any suggestions are welcome.
Thanks in advance,
Andre
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##
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Re: [R] A more idiomatic way to write this
x - x/rep(divs, each = 1000)
Bill Venables
CSIRO Laboratories
PO Box 120, Cleveland, 4163
AUSTRALIA
Office Phone (email preferred): +61 7 3826 7251
Fax (if absolutely necessary): +61 7 3826 7304
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-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Andre Nathan
Sent: Monday, 25 February 2008 11:36 AM
To: r-help@r-project.org
Subject: [R] A more idiomatic way to write this
Hello,
I have a vector of 1,000,000 numbers and another vector of 1,000
divisors. What I'd like to do is to divide the first 1,000 numbers of
the first vector by the first divisor, then the next 1,000 by the second
divisor and so on. I came up with this, but I was wondering if there is
a more idiomatic, R-like way to write it:
x - ...
divs - ...
for (i in seq(from = 1, to = 100, by = 1000)) {
x[i:(i - 1 + 1000)] - x[i:(i - 1 + 1000)] / divs[i %/% 1000 + 1]
}
Any suggestions are welcome.
Thanks in advance,
Andre
__
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Re: [R] A more idiomatic way to write this
How about
x - x / rep(divs, rep(1000, 1000))
?
Cheers,
Andrew
On Sun, Feb 24, 2008 at 10:36:23PM -0300, Andre Nathan wrote:
Hello,
I have a vector of 1,000,000 numbers and another vector of 1,000
divisors. What I'd like to do is to divide the first 1,000 numbers of
the first vector by the first divisor, then the next 1,000 by the second
divisor and so on. I came up with this, but I was wondering if there is
a more idiomatic, R-like way to write it:
x - ...
divs - ...
for (i in seq(from = 1, to = 100, by = 1000)) {
x[i:(i - 1 + 1000)] - x[i:(i - 1 + 1000)] / divs[i %/% 1000 + 1]
}
Any suggestions are welcome.
Thanks in advance,
Andre
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Department of Mathematics and StatisticsTel: +61-3-8344-6410
University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599
http://www.ms.unimelb.edu.au/~andrewpr
http://blogs.mbs.edu/fishing-in-the-bay/
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Re: [R] color area between two time-series via polygon()?
I think you have to change one statement in your program: xx - cbind(time(z[,1]),rev(time(z[,2]))) On Sun, Feb 24, 2008 at 7:44 PM, [EMAIL PROTECTED] wrote: Hi all, I would like to color the area between two time-series. I tried it by using the polygon() function but I keeps drawing lines between beginning and end points. Is there another more appropriate function or how could I close the polygon at the end en the beginning of the time series (e.g., drawing a straight line)? The following doesn't plot a polygon between the two time-series: z - ts(matrix(rnorm(200), 100), start=c(1961, 1), frequency=12) plot(z, plot.type=single, lty=1:2) xx - cbind(time(z[,1]),rev(z[,2])) yy - cbind(as.vector(z[,1]),rev(as.vector(z[,2]))) polygon(xx,yy, col=gray, border = red) I would like to make it look like this (but then for time series) n - 100 xx - c(0:n, n:0) yy - c(c(0,cumsum(stats::rnorm(n))), rev(c(0,cumsum(stats::rnorm(n) plot (xx, yy, type=n, xlab=Time, ylab=Distance) polygon(xx, yy, col=gray, border = red) Thanks for your help, Jan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? Tell me what you want to do, not how you want to do it. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How do I use as.Date when day values are missing?
I have a data frame which contains some valuable date information. But for a few of the dates, the day information missing . Viz: interesting.data$date [1] 1/22/93 1/22/93 1/23/93 1/00/93 1/28/93 1/31/93 1/12/93 i.e. for dates where the day info is missing, the %d part of the %m/%d/%yy format is simply represented as 00. When I apply as.Date to the date information, the dates which don't contain exact day information are converted to NA. Viz: as.Date(interesting.data$date) [1] 1993-01-22 1993-01-22 1993-01-23 NA 1993-01-28 1993-01-31 1993-01-12 Is there a way of using the as.Date function when I only have partial dates (eg missing day information which is represented as 00, as above) such that the date isn't represented as NA? Thanks, Anupa Never miss a thing. Make Yahoo your home page. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A more idiomatic way to write this
On Mon, 2008-02-25 at 11:52 +1000, [EMAIL PROTECTED] wrote: x - x/rep(divs, each = 1000) Fantastic :) Thanks Bill, Andrew and Rolf Andre __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Geometric progression
On Sun, 2008-02-24 at 23:26 -0300, Andre Nathan wrote: gp(1, 2, 10) [1]1248 16 32 64 128 256 512 1024 Actually, [1]1248 16 32 64 128 256 512 Andre __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Geometric progression
On Feb 24, 2008, at 9:36 PM, Andre Nathan wrote: On Sun, 2008-02-24 at 23:26 -0300, Andre Nathan wrote: gp(1, 2, 10) [1]1248 16 32 64 128 256 512 1024 Actually, [1]1248 16 32 64 128 256 512 2^(0:9) Andre Haris Skiadas Department of Mathematics and Computer Science Hanover College __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How do I use as.Date when day values are missing?
It really depends on what you want to do with them but one possibility might be to represent them as chron dates and use a time of 0 for true dates and noon for missing dates replacing the missing day with 01 or 15 or some other day: library(chron) x - c(01/00/05, 01/22/06) no.day - regexpr(/00/, x) 0 as.chron(ifelse(no.day, sub(/00/, /15/, x), x )) + no.day/2 [1] (01/15/05 12:00:00) (01/22/06 00:00:00) We can then tell which have the incomplete dates with as.numeric(xx) %% 1 0 On Sun, Feb 24, 2008 at 9:00 PM, Anupa Fabian [EMAIL PROTECTED] wrote: I have a data frame which contains some valuable date information. But for a few of the dates, the day information missing . Viz: interesting.data$date [1] 1/22/93 1/22/93 1/23/93 1/00/93 1/28/93 1/31/93 1/12/93 i.e. for dates where the day info is missing, the %d part of the %m/%d/%yy format is simply represented as 00. When I apply as.Date to the date information, the dates which don't contain exact day information are converted to NA. Viz: as.Date(interesting.data$date) [1] 1993-01-22 1993-01-22 1993-01-23 NA 1993-01-28 1993-01-31 1993-01-12 Is there a way of using the as.Date function when I only have partial dates (eg missing day information which is represented as 00, as above) such that the date isn't represented as NA? Thanks, Anupa Never miss a thing. Make Yahoo your home page. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Geometric progression
On Sun, 2008-02-24 at 21:39 -0500, Charilaos Skiadas wrote: 2^(0:9) I guess it's so simple that I'd never think of that... Thanks! Andre __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Geometric progression
Hi
I'm pretty sure there's some built-in function to do the equivalent of
this, but I couldn't find one anywhere:
gp - function(init, mult, n)
{
if (n == 1)
init
else
pg(c(init, init[length(init)] * mult), mult, n-1)
}
gp(1, 2, 10)
[1]1248 16 32 64 128 256 512 1024
Any pointers?
Thanks,
Andre
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Re: [R] How do I use as.Date when day values are missing?
In looking at this again here is a slight simplification. Its now only one line: library(chron) x - c(01/00/05, 01/22/06) as.chron(sub(/00/, /15/, x)) + (regexpr(/00/, x) 0) / 2 [1] (01/15/05 12:00:00) (01/22/06 00:00:00) On Sun, Feb 24, 2008 at 9:45 PM, Gabor Grothendieck [EMAIL PROTECTED] wrote: It really depends on what you want to do with them but one possibility might be to represent them as chron dates and use a time of 0 for true dates and noon for missing dates replacing the missing day with 01 or 15 or some other day: library(chron) x - c(01/00/05, 01/22/06) no.day - regexpr(/00/, x) 0 as.chron(ifelse(no.day, sub(/00/, /15/, x), x )) + no.day/2 [1] (01/15/05 12:00:00) (01/22/06 00:00:00) We can then tell which have the incomplete dates with as.numeric(xx) %% 1 0 On Sun, Feb 24, 2008 at 9:00 PM, Anupa Fabian [EMAIL PROTECTED] wrote: I have a data frame which contains some valuable date information. But for a few of the dates, the day information missing . Viz: interesting.data$date [1] 1/22/93 1/22/93 1/23/93 1/00/93 1/28/93 1/31/93 1/12/93 i.e. for dates where the day info is missing, the %d part of the %m/%d/%yy format is simply represented as 00. When I apply as.Date to the date information, the dates which don't contain exact day information are converted to NA. Viz: as.Date(interesting.data$date) [1] 1993-01-22 1993-01-22 1993-01-23 NA 1993-01-28 1993-01-31 1993-01-12 Is there a way of using the as.Date function when I only have partial dates (eg missing day information which is represented as 00, as above) such that the date isn't represented as NA? Thanks, Anupa Never miss a thing. Make Yahoo your home page. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] projection.matrix() {popbio} for more than one matrix
Michelle,
I would probably run a loop as well and save all matrices to a single list
(see hudsonia and calathea on working with lists of matrices).
First, run the example(test.census) to get the stage-fate data frame trans
and then run this code to save the matrices into a list all.
years-2001:2002 #or unique(trans$year)
all-vector(list, length(years) )
names(all)-years
## loop through years
for (s.year in years)
{
## Add individual fertilities
test.trans - subset(trans, year==s.year)
seedlings-nrow(subset(test.census, year==s.year+1 stage==seedling))
test.trans$seedling-test.trans$fruits/sum(test.trans$fruits) * seedlings
## save test.trans to another list if needed for bootstrapping,
calculating pooled matrix etc.
## test.trans
## either one will work to get list index
all[[as.character(s.year)]] -projection.matrix(test.trans)
##all[[ s.year- (years[1]-1) ]] -projection.matrix(test.trans)
}
all
$`2001`
seedling vegetative reproductive
seedling 0.00.0 1.6667
vegetative0.50.5 0.
reproductive 0.00.5 0.6667
$`2002`
seedling vegetative reproductive
seedling 0.00.0 0.6667
vegetative0.20.5 0.
reproductive 0.00.0 0.
There's also a loop in the demo(fillmore) example that uses the aq.census
data to create matrices for seven years.
Chris
Michelle DePrenger-Levin wrote:
Hello,
I am trying to use the projection.matrix( ) function and am following the
example given. I have my data formatted very similar to the test.census
example.
str(AsMi05mat)
`data.frame': 1854 obs. of 6 variables:
$ Tag : num 501 502 503 504 505 506 507 508 509 510 ...
$ Year : int 1995 1995 1995 1995 1995 1995 1995 1995 1995 1995 ...
$ Length : num 34 37 11 24 7 44 4 7 12 20 ...
$ Flowering: int 1 1 0 1 0 1 0 0 0 1 ...
$ Fruits : int 22 22 0 89 0 15 0 0 0 0 ...
$ Stage: Factor w/ 5 levels ,Dead,Dormant,..: 4 4 5 4 5 4 5 5 5
4
...
The example data includes three years but only shows how to create a
matrix
from year 1 to 2. I have 13 years of data and would like to automate
creating
all matrices for each pair of years.
I tried a for( ) loop but don't know how to assign new names for each
matrix
so end up with only the final comparison (2005 to 2006). I assume an
apply( )
function is the way to go but can't see how to do it.
Thanks for any help!
Michelle
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[R] including data frames in R packages
useR's, Does any one know if there is a size limitation on the data frames that can be included in R packages. I have a data set in a text file that I would like to include in a package I am building and it is 8.5 MB in size. Will this be problematic? Is the process for including data sets in packages documented in WRE? Thanks, dxc -- View this message in context: http://www.nabble.com/including-data-frames-in-R-packages-tp15672705p15672705.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Plotting series marked with a symbol on every nth data point, preferably in ggplot...
Hello! I am working with signals and a plot of several signals on the same axes can get quite messy. With lines that are very fractured, distinction by only the linestyle is not very clear. If I add symbols to the plot however, there are so many symbols, that they overplot and the whole plot is unreadable once again. I am looking for advice on how to make a plot with continuous lines and symbols appearing at every nth point. An example of this problem and a solution in SAS can be found here: www2.sas.com/proceedings/sugi26/p072-26.pdf The obvious solution would be to extract the n-th point from the dataset and overplot as a new line with symbols, but this does not change the legend, does it? How can I then have a line+symbol in the legend represent my curve? Now, ideally I would prefer a solution in ggplot, because of the ease of plotting and changing the plot labels and other properties. However, I remember, Hadley mentioned that he is still working on a new version to have combined line+symbol plots, and probably he hasn't finished that yet. So any advice is welcome. Regards, TL PS. I did RTFM and I am not posting code, because at this time I expect a general outline how to do it and what commands to lookup, not code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plotting series marked with a symbol on every nth data point, preferably in ggplot...
Try this: pch - c(+, ); lwd - 1 plot(1:10, type = o, pch = pch, lwd = lwd) legend(topleft, legend = data, pch = pch, lwd = lwd) On Sun, Feb 24, 2008 at 11:01 PM, Tribo Laboy [EMAIL PROTECTED] wrote: Hello! I am working with signals and a plot of several signals on the same axes can get quite messy. With lines that are very fractured, distinction by only the linestyle is not very clear. If I add symbols to the plot however, there are so many symbols, that they overplot and the whole plot is unreadable once again. I am looking for advice on how to make a plot with continuous lines and symbols appearing at every nth point. An example of this problem and a solution in SAS can be found here: www2.sas.com/proceedings/sugi26/p072-26.pdf The obvious solution would be to extract the n-th point from the dataset and overplot as a new line with symbols, but this does not change the legend, does it? How can I then have a line+symbol in the legend represent my curve? Now, ideally I would prefer a solution in ggplot, because of the ease of plotting and changing the plot labels and other properties. However, I remember, Hadley mentioned that he is still working on a new version to have combined line+symbol plots, and probably he hasn't finished that yet. So any advice is welcome. Regards, TL PS. I did RTFM and I am not posting code, because at this time I expect a general outline how to do it and what commands to lookup, not code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] To get more digits in precision of predict function of randomForests
Hi, I am using randomForests for a classification problem. The predict function in the randomForest library, when asked to return the probabilities, has precision of two digits after the decimal. I need at least four digits of precision for the predicted probabilities. How do I achieve this? Thank you, Nagu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] version 2.5.1 and version 2.3.1
Dear R users. I have written my R code. It runs in the version 2.3.1, but not in the version 2.5.1. The error message is shown below. Error in .Fortran(DEPTH, u, v, as.integer(size), as.single(x), as.single(y), : Fortran symbol name depth not in load table Could any one advise me what I could do to run my code in 2.5.1? Sincerely, SUNGSU KIM [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: version 2.5.1 and version 2.3.1
Hi I can not help you to run your code in R 2.5.1 but I would recommend to use even more recent R version (2.6.1) or even maybe R 2.7.0. Regards Petr [EMAIL PROTECTED] [EMAIL PROTECTED] napsal dne 25.02.2008 06:18:10: Dear R users. I have written my R code. It runs in the version 2.3.1, but not in the version 2.5.1. The error message is shown below. Error in .Fortran(DEPTH, u, v, as.integer(size), as.single(x), as.single(y), : Fortran symbol name depth not in load table Could any one advise me what I could do to run my code in 2.5.1? Sincerely, SUNGSU KIM [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plotting series marked with a symbol on every nth data point, preferably in ggplot...
Hello Gabor and Charilaos Thanks for the help. I did not realize that the legend argument can take both the symbol and the line. Following your suggestions I have come with the code that follows. I realized that I can use the 'legend' function to draw over ggplot2 produced plot, so that is a good start, but I am not quite sure how to prepare the data and plot it in ggplot. If I use print with option pretty = FALSE on the object produced by ggplot, it displays the graph without legend, so I can overlay the legend from traditional graphics. This option (pretty = FALSE), however, also removes the X-label and Y-label from the plot. Thanks again. Maybe someone can help with the ggplot version of this thing? Regards, TL. #---START CODE SNIPPET--- #- #Try traditional graphics first #- # 1 . Generate signal-like data # x1 - c(26:300) y1- sin(0.1*x1) + rnorm(275,mean = 0,sd = 0.1); x2 - c(1:350) y2-sin(0.1*x2+10) + rnorm(350,mean = 0,sd = 0.1); x3 - c(41:280) y3-sin(0.1*x3+30) + rnorm(240,mean = 0,sd = 0.1); # 2 . Plot the three signals (and symbols) # # How many points to skip: N = 15; # Draw empty axes with limits and labels plot(NA, NA,type=n, xlim = c(0,400),ylim = c(-1.5,1.5),xlab =Time,ylab =Amplitude ) # Draw the lines pch1 - c(15,rep(NA,N)); pch2 - c(16,rep(NA,N)); pch3 - c(17,rep(NA,N)); lines(x1, y1, type = o, col = red, pch=pch1, cex=1.5, lwd=1.5, lty=1); lines(x2, y2, type = o, col = green, pch = pch2, cex=1.5, lwd=1.5, lty=2) lines(x3, y3, type = o, col = blue, pch = pch3, cex=1.5, lwd=1.5, lty=3) # # 4 . Add the legend # legend(topright,col=c(red,green,blue), pch=c(15,16,17), lty=c(1,2,3), legend=c(series 1, series 2, series 3), inset=0.05, bg='white') #- #Try the same thing with GGPLOT #- dataset - rbind(data.frame(Test = Test 1, x = x1, y = y1), data.frame(Test = Test 2, x = x2, y = y2), data.frame(Test = Test 3, x = x3, y = y3)); # 2 . Plot the three signals (and every point as a symbol) # ggp - ggplot(data = dataset, aes(x = x, y = y, colour = Test),) + geom_line() + geom_point() print(ggp,pretty = FALSE) #Do not plot the legend, but this also removes the x and y labels # # 4 . Add the legend (Traditional graphics) over the GGPLOT # legend(topright,col=c(red,green,blue), pch=c(15,16,17), lty=c(1,2,3), legend=c(Test 1, Test 2, Test 3), inset=0.05, bg='white') #--END CODE SNIPPET--- On Mon, Feb 25, 2008 at 2:05 PM, Gabor Grothendieck [EMAIL PROTECTED] wrote: Try this: pch - c(+, ); lwd - 1 plot(1:10, type = o, pch = pch, lwd = lwd) legend(topleft, legend = data, pch = pch, lwd = lwd) On Sun, Feb 24, 2008 at 11:01 PM, Tribo Laboy [EMAIL PROTECTED] wrote: Hello! I am working with signals and a plot of several signals on the same axes can get quite messy. With lines that are very fractured, distinction by only the linestyle is not very clear. If I add symbols to the plot however, there are so many symbols, that they overplot and the whole plot is unreadable once again. I am looking for advice on how to make a plot with continuous lines and symbols appearing at every nth point. An example of this problem and a solution in SAS can be found here: www2.sas.com/proceedings/sugi26/p072-26.pdf The obvious solution would be to extract the n-th point from the dataset and overplot as a new line with symbols, but this does not change the legend, does it? How can I then have a line+symbol in the legend represent my curve? Now, ideally I would prefer a solution in ggplot, because of the ease of plotting and changing the plot labels and other properties. However, I remember, Hadley mentioned that he is still working on a new version to have combined line+symbol plots, and probably he hasn't finished that yet. So any advice is welcome. Regards, TL PS. I did RTFM and I am not posting code, because at this time I expect a general outline how to do it and what commands to lookup, not code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Course: STATISTICAL PRACTICE IN EPIDEMIOLOGY USING R
Course in STATISTICAL PRACTICE IN EPIDEMIOLOGY USING R Tartu, Estonia, Wednesday 28 May to Monday 2 June 2008. Aimed at young statisticians and epidemiologists wishing to broaden their epidemiological skills, in particular with respect to practical statistical analysis. Participants will gain access to the versatile analysis tool R which has good analytical in particular graphical capabilities. The Epi-package for R will be extensively used. Participants are required to have a fairly good understanding of statistical principles and some familiarity with epidemiological concepts. The course will be mainly practically oriented with more than half the time at the computer. Price: Commercial: 1000 EUR, Academic: 750 EUR / 500 EUR. The 750 EUR are for particpants from the EU pre-2004, North America and the like. 500 for particants from other countries. Application deadline: 15 April 2004. Send an application by mail to the organizers, briefly stating your qualifications in epidemiology and statistics and a brief reason to participate: [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED] Further information at the course homepage: www.biostat.ku.dk/~bxc/SPE -- Krista Fischer, University of Tartu, Estonia Esa Läärä, University of Oulu, Finland Bendix Carstensen, Steno Diabetes Center, Denmark (Organizers) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
