Huang-Wen == Huang-Wen Chen [EMAIL PROTECTED]
on Thu, 24 Nov 2005 02:51:18 -0500 writes:
Huang-Wen Hi,
Huang-Wen In R, qnorm is the inversion function of pnorm. (c.d.f)
Huang-Wen But there is no inversion function for dnorm. (p.d.f).
Huang-Wen Is there any easy (and quick)
I am analysing binomial data using a generalised mixed effects model. I
understand that if I use glmmPQL it is not appropriate to compare AIC
values to obtain a minimum adequate model.
I am assuming that this means it is also inappropriate to use AIC values
from lmer since, when analysing
Hi,
I cannot find source code of package stats. Do you know where
I can find it?
Regards
Jakub Kalarus
Andrzejki - wróżby tylko dla pań!
Katarzynki - wróżby tylko dla kawalerów!
Sprawdź - zanim będzie za późno ;)
Hello everyone,
does somebody know how can I save an object list as a txt-file or as a
csv-file? The problem is, that the elements of this list have different row
length. So I cannot convert the list into a matrix or something else and then
use write-matrix or so as usual.
Thank you
Hi All,
I have conducted the following survival analysis which appears to be OK
(thanks BRipley for solving my earlier problem).
surv.mod1 - survreg( Surv(timep1, relall6)~randgrpc, data=Dataset,
dist=weibull, scale = 1)
summary(surv.mod1)
Call:
survreg(formula = Surv(timep1,
note that although PQL is the default method in lmer() for GLMMs, the
recent version of the function allow also for Laplace or adaptive
Gauss-Hermite approximations. In these cases it might be reasonable to
compute AIC values depending on how good the approximation to the
likelihood is;
https://svn.r-project.org/R/trunk/src/library/stats/
Kuba wrote:
Hi,
I cannot find source code of package stats. Do you know where
I can find it?
Regards
Jakub Kalarus
Andrzejki - wróżby tylko dla pań!
Katarzynki - wróżby tylko
Hi,
Does anyone know an R function to impute hamming distance?
Thanks
Ana
O@ nb @@@O@@O@ Centro de Genómica
@OO@ Instituto Valenciano de Investigaciones Agrarias (IVIA)
@@@O Carretera Moncada - Naquera, Km. 4,5
O@46113 Moncada
Hello,
I'm trying to calculate a chi-squared test to see if my data are
different from the theoretical distribution or not:
chisq.test(rbind(c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49,
60),c(80,80,80, 80, 80, 80, 80, 80, 80, 80, 80, 80)))
Pearson's Chi-squared test
data: rbind(c(79,
Claus Atzenbeck writes:
On Tue, 22 Nov 2005, Marc Schwartz wrote:
The disadvantage with pdfcrop is that it takes an additional step:
First, I have to produce the diagrams, then, I have to call pdfcrop on
them to crop the white space. pdfcrop does a perfect job here: It crops
the
Folks,
1. Is there a more appropriate list (r-devel?) for posting such
suggestions? I am a newbie to R, and doubtless will have some
suggestions for the documentation -- some good, others not quite so. I
would actually like to help give back to the community (I was
motivated by Prof. Ripley's
In this case the environment contains the variables local to the function, and
its enclosure is the environment of the enclosing function.(R-lang:p11)
I want to know if the enclosing function means the closure of the function?
for example ,if I call function mean(),and the create an
Bianca Vieru- Dimulescu wrote:
Hello,
I'm trying to calculate a chi-squared test to see if my data are
different from the theoretical distribution or not:
chisq.test(rbind(c(79, 52, 69, 71, 82, 87, 95, 74, 55, 78, 49,
60),c(80,80,80, 80, 80, 80, 80, 80, 80, 80, 80, 80)))
Hello,
I have prepared an svm on some training data and would like to use the svm
model for predicting binary outcome from new data.
The input data frame contains several numeric and factor variables. Usually I
construct the input matrix of the entities to be predicted with a perl script
that
Firstly, I assume that your variable is a numeric one. For seperat values p
and lambda for diferent categories, you should convert it to factor.
However, this has no effect in your case, since you have only 2 categories.
You can have only one p and lambda for a variable with only 2 values. The
Hi all,
I'm looking for to interpolate hourly temperature date collected from more
than 140 automatic weather station (irregularly spaced) using 4 independent
variable:
1-2) geografic coordinates (x,y) (from DEM - 40m)
3) altitude (z) (from DEM - 40m)
4) solar radiation (from a model calculated
Hi Marc
I use this function for finding maxima in some spectral
data (eg. from Xray diffraction) and it satisfied my
needs. The function itself was modified probably due to
some reasons for ploting my data so it dropped values
from the end rather than from both sides.
Peaks in those cases
ronggui [EMAIL PROTECTED] writes:
In this case the environment contains the variables local to the
function, and its enclosure is the environment of the enclosing
function.(R-lang:p11)
Page number are of dubious value for documents in multiple formats. It
is page 5 in the PDF version on CRAN!
To construct a nonparametric (1-alpha) confidence set for an arbitrary
CDF F, you can use the Dvoretzky-Kiefer-Wolfowitz (DKW) inequality
(e.g., see Wasserman, L. (2005). All of Statistics. 2nd, corr. printing.
NY: Springer, p. 99).
With n=sample size and eps=sqrt(log(2/alpha)/(2*n)),
the lower
Hi Bianca,
you could see my contribute Fitting distribution with
R, pagg. 16-18:
http://cran.r-project.org/doc/contrib/Ricci-distributions-en.pdf
Hoping it could help you.
Regards,
Vito
From: Bianca Vieru- Dimulescu bianca.vieru at
free.fr
Subject: [R] Chi-squared test
Date: 2005-11-24
Hi Ales,
Thanks I had tried converting randgrpc previously to a factor
- still didn't get the effect for the different groups.
Weird - do you think not getting different effects for each
group might have to do with tidying up tmp files and the like?
I have had tmp file issues with well known
Gabor == Gabor Grothendieck [EMAIL PROTECTED]
on Wed, 23 Nov 2005 16:51:17 -0500 writes:
Gabor One idea might be, rather than have a peaks function, enhance
Gabor rle so that it optionally produces a third component with the
Gabor peak information, perhaps 1, 0, -1 for peak,
Hi Ales,
Sorry Mis-read your posting - I blindly saw 'no effect' and was thrown
- so its like a dummy regression you loose 1 category
I can conclude that p of randgrp is 1.47e-04 or 0.000147 which is a
significant difference between the groups?
Sorry this is the first time I have done this.
Thanks very much.
ronggui [EMAIL PROTECTED] writes:
In this case the environment contains the variables local to the
function, and its enclosure is the environment of the enclosing
function.(R-lang:p11)
Page number are of dubious value for documents in multiple formats. It
is page 5 in the
You could install.packages(e1071)
and see help(hamming.distance)
JeeBee.
hamming.distancepackage:e1071R Documentation
Hamming Distances of Vectors
Description:
If both 'x' and 'y' are vectors, 'hamming.distance' returns the
Hamming distance (number of different
Hy all
I use barplot to draw frequencies by dates.
On the x axis it shows only 1 date for 2 bars, i've understand why, because R
cant put so much date on the same axis, it will get out of the graph.
that's why i tought about reducing the size of the font used to stamp the x
axis.
anyone
Dearest All,
Ok so I've had a couple of glasses of wine over lunch today... This is
likely to be trivial but I'm struggling to find a more elegant way to
obtain the following matrix rotations:
M - matrix(c(1,0,0,0), ncol=2)
M
[,1] [,2]
[1,]10
[2,]00
N -
Hi everyone,
Can someone explain me how to calculate SAS type III sums of squares in
R? Not that I would like to use them, I know they are problematic. I
would like to know how to calculate them in order to demonstrate that
strange things happen when you use them (for a course for example). I
Ok I warned you that I'd been drinking! What I really meant was
something to go from:
[,1] [,2]
[1,]12
[2,]43
to
[,1] [,2]
[1,]41
[2,]32
to
[,1] [,2]
[1,]34
[2,]21
to
[,1] [,2]
[1,]23
[2,]14
Sorry for
On 11/24/05, Martin Maechler [EMAIL PROTECTED] wrote:
NA's have the big drawback of producing a logical vector that
can NOT be used for subsetting -- and subsetting was exactly a main
reason for the padding...
Since 'which' throws away NAs, one extra 'which' can solve
that. In the
True, but if
x - c(1, 0, 0)
y - c(1, 0, 1),
then you can just define
hamming.distance - function(x,y){
sum(x != y)
}
The problem is just a bit harder when x and y are, for instance, strings.
Carlos J. Gil Bellosta
http://www.datanalytics.com
Quoting JeeBee [EMAIL PROTECTED]:
[EMAIL PROTECTED] wrote:
Hy all
I use barplot to draw frequencies by dates.
On the x axis it shows only 1 date for 2 bars, i've understand why, because R
cant put so much date on the same axis, it will get out of the graph.
that's why i tought about reducing the size of the font used
On 11/24/05, Vivek Satsangi [EMAIL PROTECTED] wrote:
motivated by Prof. Ripley's 2001 talk in which he had commented that
open source software users rarely give back anything.) -- but I know
With 600+ addon packages this does not seem to be the case for R.
Stefanie von Felten, IPWIfU [EMAIL PROTECTED] writes:
Hi everyone,
Can someone explain me how to calculate SAS type III sums of squares in
R? Not that I would like to use them, I know they are problematic. I
would like to know how to calculate them in order to demonstrate that
strange
If you use the Anova function in the car package and specify contr.sum or
contr.SAS for the contrasts for your categorical factors, you will get the
same results as outputted by SAS. I've tried this with a variety of data sets
and it works.
On Thu November 24 2005 08:27, Stefanie von Felten,
[Apologies if this posting is appears twice -- I think I was
unsuccessful in posting it previously.]
I've read through many postings about principle component analysis in
the R-help archives, but haven't been able to piece together the
information I need. I'd like to recreate an SPSS-like
Dear Stefanie and Mike,
To elaborate slightly on Mike's points, the Anova() function in car
calculates Type-III (and Type-II) tests differently from SAS. (The
difference originates in the fact that SAS uses a deficient-rank
parametrization of the model while R uses a full-rank parametrization; it
Hi all,
I need to write and read a list in R. I did r.site.search, found there is a
package rmutil doing this, unfortunately it is not on the list of package.
In another words, I can't install it from any CRAN mirror.
Anybody has idea about this? or any suggestion about the list? Thanks!
Try this:
# cyclically rotate a vector to the right
rot - function(x) c(x[length(x)], x[-length(x)])
# create a vector from a 2x2 matrix moving left to right along
# first row and then right to left along second row
m2v - function(m) m[c(1,3,4,2)]
# inverse of m2v. Note that c(1,4,2,3) equals
In the logistic regression model, there is no residual
log (pi/(1-pi)) = beta_0 + beta_1*X_1 + .
But glm model will return
residuals
What is that?
How to understand this? Can we put some residual in the logistic regression
model by replacing pi with pi' (the estimated pi)?
log
Dr. Herwig Meschke wrote:
To construct a nonparametric (1-alpha) confidence set for an arbitrary
CDF F, you can use the Dvoretzky-Kiefer-Wolfowitz (DKW) inequality
(e.g., see Wasserman, L. (2005). All of Statistics. 2nd, corr. printing.
NY: Springer, p. 99).
With n=sample size and
Mango Solutions, providers of R and S-PLUS consulting, development and
training Services, are looking for 2 consultants to join their UK-based
technical team. We are looking for highly motivated individuals to work in
a customer-focused environment.
R/S Pharmaceutical Consultant (Reading, UK)
Dear Urania,
The residuals method for glm objects can compute several kinds of residuals;
the default is deviance residuals. See ?residuals.glm for details and
references.
I hope this helps.
John
John Fox
Department of Sociology
McMaster University
Hamilton,
Hello,
I have a doubt in using the function step (step wise) to select glm models.
Usually I apply the gamma distribution to analyze fishery data. To select the
terms I use a routine where I first compare single term models to the null model
(eg. U~1 vs. U~depth; U~1 vs. U~latitude; etc. where
On 24-Nov-05 P Ehlers wrote:
Bianca Vieru- Dimulescu wrote:
Hello,
I'm trying to calculate a chi-squared test to see if my data are
different from the theoretical distribution or not:
chisq.test(rbind(c(79,52,69,71,82,87,95,74,55,78,49,60),
In thinking about this a bit more it can be simplified to
the following where f takes a 2x2 matrix as its first
argument and permutes it according to the second
argument returning the permuted 2x2 matrix:
f - function(m = c(1,4,2,3), idx = c(2,4,1,3)) matrix(m[idx],2)
print(m - f(,1:4))
for(i in
Antonio Olinto wrote:
Hello,
I have a doubt in using the function step (step wise) to select glm models.
Usually I apply the gamma distribution to analyze fishery data. To select the
terms I use a routine where I first compare single term models to the null
model
(eg. U~1 vs. U~depth;
Hi, R-Help,
I am a newbie.
what I concern most recently is the analysis of the time series,
But there are a lot of package in my eyes.
All I want to try is as follow:
How to test whether a time series fit the Poisson or other process in R?
Thank you very much in advance.
[EMAIL PROTECTED]
On Thu, 2005-11-24 at 21:55 +, Ted Harding wrote:
On 24-Nov-05 P Ehlers wrote:
Bianca Vieru- Dimulescu wrote:
Hello,
I'm trying to calculate a chi-squared test to see if my data are
different from the theoretical distribution or not:
(Ted Harding) [EMAIL PROTECTED] writes:
On 24-Nov-05 P Ehlers wrote:
Bianca Vieru- Dimulescu wrote:
Hello,
I'm trying to calculate a chi-squared test to see if my data are
different from the theoretical distribution or not:
chisq.test(rbind(c(79,52,69,71,82,87,95,74,55,78,49,60),
Marc Schwartz wrote:
On Thu, 2005-11-24 at 21:55 +, Ted Harding wrote:
On 24-Nov-05 P Ehlers wrote:
Bianca Vieru- Dimulescu wrote:
Hello,
I'm trying to calculate a chi-squared test to see if my data are
different from the theoretical distribution or not:
Thanks, Professor.
But is it ok to write residuals in the right hand side of the logistic
regression formula? Some people said I cannot since the generalized linear
model is to use a function to link the expectation to a linear model. So
there should not be residuals in the right hand side.
My
On Wed, 2005-11-23 at 16:13 -0700, nhepburn wrote:
I have a couple of data sets that I want to combine into one data frame.
One set contains a number of records on individual observations and includes
a geographic descriptor called dacode. The dacode is not unique in that
table. The other
Dear Urania,
-Original Message-
From: Urania Sun [mailto:[EMAIL PROTECTED]
Sent: Thursday, November 24, 2005 8:52 PM
To: John Fox
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] residuals in logistic regression model
Thanks, Professor.
But is it ok to write residuals in the
Thanks a lot, Professor.
Now I know if I put some additive residuals in the right handside of the
logistic regression equation, they are different with any glm returned
residuals.
But is it ever ok or legal to put some additive residuals in the right-hand
side of the logistic equation
On Thu, 2005-11-24 at 18:50 -0700, P Ehlers wrote:
Marc Schwartz wrote:
On Thu, 2005-11-24 at 21:55 +, Ted Harding wrote:
On 24-Nov-05 P Ehlers wrote:
Bianca Vieru- Dimulescu wrote:
Hello,
I'm trying to calculate a chi-squared test to see if my data are
different from the
Hi all,
does anyone know if there exists some library that implements the
sensitivity tests for matched pairs and unmatched groups proposed in
Rosenbaum's Observational Studies (2002: ch.4)?
Cheers,
Holger
__
R-help@stat.math.ethz.ch mailing list
Dear all,
I have n objects and I want to select k of these with replacement. Do you know
of code which would generate all the possible arrangements? Note that this is
different from the selection of k of n objects without replacement and wanting
to generate all the possible permutations.
Any
I think you just need a sample with replacement
sample(1:n,k,replace=TRUE)
Best regards,
Kristel
Globe Trotter wrote:
Dear all,
I have n objects and I want to select k of these with replacement. Do you know
of code which would generate all the possible arrangements? Note that this is
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