Neat piece of code, Jim, but it still uses a nested loop. If you order
the matrix first, you only need one passage through the whole matrix
to find the information you need.
Off course I don't take into account the ordering. If the ordering
algorithm doesn't work in linear time, then it doesn't re
Quite a strange error. The function installed.packages should be in
the utils package.
Try ?installed.packages and see if you get a help file. If it doesn't
find that function, there is a problem with your R installation.
It might be something went wrong with downloading the package itself.
DAAG i
First, you should define your function as :
test <- function(cand2,phi,lambda,
whatever-arguments-you-want-to-use-further){... insert code here ...}
All variables you use inside a function only exist within that
function. Your parameters/arguments is the interface between the
function and the outs
Actually, it's a bit more complex than just plot(). First you have to
make Days an ordered factor, otherwise you get indeed that error.
days <- c("Mon", "Tue", "Wed", "Thu", "Fri", "Sat","Sun")
Days <- factor(days,levels=days,ordered=T)
Then you want to make a plot with 2 axes, you should check :
Hi Lauren,
from the error message it looks like you have a problem with the
package "Hmisc". doBy is dependent on that one. Can you check whether
Hmisc is installed in your R version? For me, doBy loads without
problems.
On Tue, Oct 6, 2009 at 12:18 PM, Lauren Szathmary wrote:
> Hi all,
>
> I am
Hi all,
this one left me a bit puzzled, as I don't seem to find a function to
perform this easily. I must have overlooked the obvious, so sorry in
advance.
I have a list of dates in numerical format (i.e. 34576), defined as
the number of days that passed since january 1st 1900. So I apply the
fun
Confint doesn't work if you have a multi-dimensional dependent variable.
Kind regards
Joris
On Fri, Oct 2, 2009 at 4:29 AM, smith_cc wrote:
>
> I am unable to calculate confidence intervals for the slope estimate in a
> quasibinomial glm using confint(). Below is the output and the package info
Hi Daniel,
I presume you mean coefficients by values. Try ?gamObject and read the
info in the help file. A GAM object (returned by the function gam() )
contains an element coefficients that is easily accessible. That one
contains all estimated coefficients, including those for the smooth
functions
r: package 'psych' does not have a name space
>
> is there a way to clean up and reinstall the packages?
>
> thanks,
> Kenny
>
> On Thu, Oct 1, 2009 at 4:02 PM, joris meys wrote:
>>
>> Hi,
>>
>> It installed perfectly for me, both packages
The minimum of a character vector is returned according to the unicode
or ascii code values, but I guess that's not what Chris is looking
for.
Chris, please always try to provide a minimal working example of code
to see what is wrong.
Something goes wrong with the input in your example. As Peter
Hi Ben,
other members of the list also pointed me to this possibility. I'll
certainly add the manual.
Kind regards
Joris
On Thu, Oct 1, 2009 at 4:14 AM, Ben Bolker wrote:
> Tal Galili gmail.com> writes:
>
>>
>> Hi Joris,
>>
>> Good luck with your work.
>>
>> I know how to set up a wiki. But I w
ne number: 972-52-7275845
> FaceBook: Tal Galili
> My Blogs:
> http://www.talgalili.com (Web and general, Hebrew)
> http://www.biostatistics.co.il (Statistics, Hebrew)
> http://www.r-statistics.com/ (Statistics,R, English)
>
>
>
>
> On Wed, Sep 30, 2009 at 2:50 PM, jor
n R. Off course this
mailing list is an extensive source of all kind or errors and
examples, but I was wondering if there is a somewhat more structured
overview. All help appreciated.
Kind regards
Joris Meys
__
R-help@r-project.org mailing list
https
Dear Jacy,
As Jorge said.
To clarify : what you ask, is actually to compare the first element of
genotype with the first element of FREQ, the second of genotype with the
second of FREQ and so on. If the lengths differ, the shorter vector is
recycled. This toy example illustrates that principle :
e plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body of
> data.
> ~ John Tukey
>
> -Oorspronkelijk bericht-
> Van: r-help-bou
t; More seriously, you could transform the proportions to logits
> logit <- log(p/(1-p))
> and fit a linear regression.
>
> Kjetil
>
> On Tue, Mar 24, 2009 at 3:30 PM, joris meys wrote:
>
>> Dear all,
>>
>> I have a dataset where I reduced the dimensionality, a
Dear all,
I have a dataset where I reduced the dimensionality, and now I have a
response variable with probabilities/proportions between 0 and 1. I wanted
to do a logistic regression on those, but the function glm refuses to do
that with non-integer values in the response. I also tried lrm, but th
Just a side remark : "any" clustering method in R is pretty broad. Actually,
k-means clustering is a completely different animal than distance-based
neighbour-joining methods. You should be a bit more specific about your
research
I wouldn't worry too much about the scientific practice, as cluster
Dear all,
I want to get the likelihood (or AIC or BIC) of a ridge regression model
using lm.ridge from the MASS library. Yet, I can't really find it. As
lm.ridge does not return a standard fit object, it doesn't work with
functions like e.g. BIC (nlme package). Is there a way around it? I would
ca
Dear all,
I would like to perform a sparse PCA, but I didn't find any library offering
me this in R. Is there one available, or do I have to write the functions
myself?
Kind regards
Joris Meys
[[alternative HTML version deleted]]
__
R-h
d Harding wrote:
> On 07-Mar-09 10:57:17, Thomas Lumley wrote:
> > On Fri, 6 Mar 2009, joris meys wrote:
> >> Dear all,
> >> I have a dataset where the interaction is more than obvious,
> >> but I was asked to give a p-value, so I ran a logistic regression
> &g
One thing I notice immediately is a number of NA values for your
coefficients. If I were you, I would try a model with less parameters, and
use the anova() function to compare models, to see if the extra terms really
improve the model.
e.g.
fit1 <- glm(Y~X1+X2+X3,...)
fit2 <- glm(Y~X1+X2+X3+X1:X2,.
Dear all,
I have a dataset where the interaction is more than obvious, but I was asked
to give a p-value, so I ran a logistic regression using glm. Very funny, in
the outcome the interaction term is NOT significant, although that's
completely counterintuitive. There are 3 variables : spot (binary
On 3/6/09, kenji_aoyagi wrote:
>
> Hi,
>
> I am using glm().
> I'd like to know what the command means.
>
> For example,
> glm(family=binomial(link=logit))
> means logit model.
Means : binomial response variable transformed with the logit
Then,
> glm(family=gaussian(link=logit)),
> does this me
tion? It's contained in the package
e1071.
Thank you in advance
On Thu, Feb 19, 2009 at 2:16 PM, joris meys wrote:
> Dear all,
>
> I tried a simple naive Bayes classification on an artificial dataset, but I
> have troubles getting the predict function to work with the type=&qu
type="class")
Tried it also with pred.bayes <-predict(Bayes.res,
mixture.test,type="class"), but that gives the same effect. Is this a bug or
am I missing something?
Kind regards
Joris Meys
University Ghent
[[alternative HTML version deleted]]
___
To clarify : I am aware of the interpretation problems. Thank you for
the tip! (it's getting late here...)
On Sat, Dec 13, 2008 at 9:56 PM, joris meys wrote:
> Thanks for the answers.
>
> @David : I am aware of that, but this is far from the last model actually.
>
> @ Fran
I missing
something?
Kind regards
Joris
On Sat, Dec 13, 2008 at 8:44 PM, David Winsemius wrote:
>
> On Dec 13, 2008, at 1:12 PM, joris meys wrote:
>
>> Sent this mail in rich text format before. Excuse me for this.
>>
>>
>> Dear all,
>&g
Sent this mail in rich text format before. Excuse me for this.
Dear all,
I'm using the lrm function from the package "Design", and I want to
extract the p-values from the results of that function. Given an lrm
object constructed as follows :
fit <- lrm(Y~(X1+X2+X3+X4+X5+
Dear all,
I'm using the lrm function from the package "Design", and I want to extract
the p-values from the results of that function. Given an lrm object
constructed as follows :
fit <- lrm(Y~(X1+X2+X3+X4+X5+X6+X7)^2, data=dataset)
I need the p-values for the coefficients printed by calling "fit
On Sun, Nov 30, 2008 at 9:59 PM, Stefan Evert <[EMAIL PROTECTED]> wrote:
> Still, I think it's better to write a few lines of R code than to abuse
> regular expressions to do something they were never intended to do. How do
> other people on this list feel about that issue?
>
Honestly, I believe
I reread the question, and I don't actually see the problem too much.
If you plot the transformed dataset from most of your models, the
problem is quite obvious : if you transform your predictor to the
logscale, the result of a a linear regression on those outcomes will
be naturally an exponential
I'm not sure on what kind of dataset would be most appropriate, but
following code I used to create a dataset with a linear response and
an increasing variance (the megaphone type, common in ecological
datasets if I'm right) :
beta0 <- 10
beta1 <- 1
x <- c(1:40)
y <- beta0 + x*beta1 +rnorm(40,0,1
Just for clarification : this is an extension, where you can take the
probability of switching to any possible value. Take into account that
the p-value in the function rbinom is 1-P for the switching of the
sign.
On Tue, Nov 18, 2008 at 11:21 PM, joris meys <[EMAIL PROTECTED]> wrote:
The function rbinom might be a solution.
Try following simple program :
vec <- c(-1,1,-1,1,1,-1,-1,1,1)
inv <-rbinom(length(vec),1,0.5)
inv <-ifelse(inv==0,-1,1)
vec2 <- vec*inv #switches sign with p=0.5
In this, inv is a random binomial vector, where the probability for
being 1 is 0.5 in all
m\;C:\Program Files\xpdf\
>
> Still didn't work i'm afraid, but cheers for the sugestion.
>
> Tony Breyal
>
>
> On 16 Nov, 20:00, "joris meys" <[EMAIL PROTECTED]> wrote:
>> Try putting "PATH" under name, and the directory path (not the fil
Try putting "PATH" under name, and the directory path (not the file)
under value. That looks more appropriate to me...
Kind regards
Joris Meys
On Sun, Nov 16, 2008 at 8:41 PM, Tony Breyal <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Uwe -- ahh, thank you kindly, I was able
> #this is my stab at - I am sure that I am missing something. If this
> doesn't make sense then please ask for more details. #This may show
> my low level of programing knowledge
>
> hester. <- c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4)
> value <- rnorm(16)
> x <- data.frame(value, hester.)
>
> z <-
> (
401 - 438 of 438 matches
Mail list logo
