Hi,
Is there any package for logistic model selection using BIC and Mallow's Cp
statistic? If not, then kindly suggest me some ways to deal with these
problems.
Thanks.
--
View this message in context:
http://www.nabble.com/Stepwise-logistic-model-selection-using-Cp-and-BIC-criteria-tf4464430.
Dear List,
Which is the most eficient method to perform a gage repeatability and
reproducibility using R?
Thanks and best regards.
Constant Depièreux
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
G*Power 3 is free software for Mac and PC, see
http://www.psycho.uni-duesseldorf.de/abteilungen/aap/gpower3/
Jay
On 9/16/07, MATTHEW BRIDGMAN <[EMAIL PROTECTED]> wrote:
> Is there a way to calculate power for repeated
> measures ANOVA (2 groups x 7 observations)? I have
> searched all over, but
Have you tried 'str'? What 'nature of the object' are you trying to identify?
On 9/16/07, Letticia Ramlal <[EMAIL PROTECTED]> wrote:
> Hello
> Given the following data for a data set called airquality. To identify the
> nature of the objects from the data set airquality example "Ozone" would it
It sounds like you are using 'integers'. Have you considered
'numeric' (floating point)? You can always look at the function
'samlmu' to see what it does in this case.
On 9/16/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> Hi everyone,
>
> In the package POT, there is a function that compute
Mind a book reference instead of a software reference?
Look for Bausell and Li's "Power Analysis for Experimental Research"
-- cookbook style power calculations, but has explicit RM ANOVA.
On 9/16/07, MATTHEW BRIDGMAN <[EMAIL PROTECTED]> wrote:
> Is there a way to calculate power for repeated
>
If I understand what you are trying to do is to find duplicated values
of rearrangements of words. If that is the case, this is probably
faster since your final loop is removed by using "duplicated". Most
of the time is in the sapply function.
> a <- c("superman", "xman", "spiderman", "wolfman",
Is there a way to calculate power for repeated
measures ANOVA (2 groups x 7 observations)? I have
searched all over, but all I can find is
power.anova.test, but that would not give accurate
results, right?
Thanks,
Matt Bridgman
__
R-help@r-project.org m
In the course of revising a paper I have had occasion to attempt to
maximize a rather
complicated log likelihood using the function nlm(). This is at the
demand of a referee
who claims that this will work better than my proposed use of a home-
grown implementation
of the Levenberg-Marquardt
Dear list,
I have a vector of numbers, let's say:
myvec <- c(2, 8, 24, 26, 51, 57, 58, 78, 219)
My task is to reduce this vector to non-reducible numbers; small numbers can
cross-out some of the larger ones, based on a function let's say called
reduce()
If I apply the function to the first e
I have been trying for hours now to perform an orthogonal contrast through an
ANOVA in R.
I have done a two-factor factorial experiment, each factor having three
levels. I converted this dataset to a dataframe with one factor with nine
treatments, as I couldn't work out what else to do.
I
Hi,
I am trying to create a set of scatter diagrams with coplot, and I
want to use different symbols for each class of points (depending on
where the data point came from).
I have the array with the required symbols, but I am not sure how to
pass that array to the panel function, and whether I
Dear Sir
In Kendall Package of R, when we use the function Kendall() or MannKendall()
then we get the value of Kendall's tau and a two-sided p-value. The question
to be asked is that is the two-sided p value for Z statistics defined by
z=(S-1)/sqrt(var(S)) where S is Kendall S statistics.
Regards
Hello,
Yes I agree but they are the same thing!
AIM
- Original Message -
From: "amna khan" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Sunday, September 16, 2007 2:02 PM
Subject: Kendall Test
> Dear Sir
>
> In Kendall Package of R, when we use the functio
On 9/16/07, Zheng Lu <[EMAIL PROTECTED]> wrote:
> I am curious how to generate high-quality plot and graph with R and
> input it into my word document. my plot was always generated in device
> 2, when I save it as PNG, the quality is poor. Thank you very much for
> your consideration and time.
To
Use windows metafile format. Its a vector graphic format so it will
display in full
resolution. png is bitmapped and so won't. Also you can edit a wmf
graphic in Word using Word's built in graphic editor so you could change
the labels, etc. even after you have imported it. Right click the graph
Dear all:
I am curious how to generate high-quality plot and graph with R and
input it into my word document. my plot was always generated in device
2, when I save it as PNG, the quality is poor. Thank you very much for
your consideration and time.
ZLu
___
Hi everyone,
In the package POT, there is a function that computes the L-moments of a given
sample (samlmu). However, to compute those L-moments, one needs to obtain the
total number of combinations between two numbers, which, by the way, requires
the use of a factorial. See, for example, Hoski
Hello
Given the following data for a data set called airquality. To identify the
nature of the objects from the data set airquality example "Ozone" would it be
best to use the command is. like is.character(airquality$Ozone) ... I tried
attributes(airquality$Ozone) but it came up null. Would
On Sun, 2007-09-16 at 08:46 -0700, kevinchang wrote:
> Hey everyone,
>
> The code I wrote executes correctly but is stalled seriously. Is there a
> way to hasten execution without coming up with a brand new algorithm
> ?please help. Thanks a lot for your time.
>
>
> #a simplified version of th
Hey everyone,
The code I wrote executes correctly but is stalled seriously. Is there a
way to hasten execution without coming up with a brand new algorithm
?please help. Thanks a lot for your time.
#a simplified version of the code
a<-c("superman" , "xman" , "spiderman" ,"wolfman" ,"mansuper
You will have to supply a lot more information than you have. Is it a
memory problem (are you paging), is it a function of your data
structure, is it your algorithm, etc.
Please follow the guidelines: "provide commented, minimal,
self-contained, reproducible code."
On 9/16/07, kevinchang <[EMAIL
Hi,
The loop I wrote executes correctly but is stalled seriously. Is there a
way to hasten execution without coming up with a brand new algorithm ?
please help. Thanks.
--
View this message in context:
http://www.nabble.com/stalled-loop-tf4451301.html#a12699524
Sent from the R help mailing
On Wed, 12 Sep 2007, Uwe Ligges wrote:
>
>
> Scott Hyde wrote:
>> I'd like to create a small "local" repository that would be used to
>> install a package for a class of students at their home. I don't want
>> to upload it to CRAN, as I don't think it should be disseminated at
>> that level.
>>
>
On Sat, 8 Sep 2007, shao ran wrote:
> Hi *,
>
> Firstly, thank you so much for your time to read my email.
>
> I am currently interested in how to use R to predict time series from
> models fitted by ARIMA. The package I used is basic stats package, and the
> method I used is predict.Arima.
>
> Wh
Thanks to your answer,
î'm trying to instaling it...
Armin Goralczyk wrote:
>
> On 9/15/07, christophe vuadens <[EMAIL PROTECTED]> wrote:
>>
>> Hello,
>>
>> Sorry for my english, in a R function, I want to read HTML files to
>> analyse
>> the text. Do somebody now, how can i read the text onl
Megh Dal schrieb:
> Dear all,
>
> I have following codes:
>
> colnames(data) = c("var", "var", "var")
> i = c(1,2,3)
>
> Now I want construct a "for" loop starting from 1 to 3 to give the new names
> of columns for dataframe "data" like below
>
> colnames(data)
>> c("var1", "var2", "var3")
>
Dear all,
I have following codes:
colnames(data) = c("var", "var", "var")
i = c(1,2,3)
Now I want construct a "for" loop starting from 1 to 3 to give the new names of
columns for dataframe "data" like below
colnames(data)
> c("var1", "var2", "var3")
Definitely I could do this manually, howev
Charles C. Berry wrote:
> Two corrections to my previous posting.
>
>> ..
>
> Not!
>
> res <- gregexpr( "\\<[[:upper:]]" , my.charvec )
>
> will do just fine.
>
Thanks for sorting this out, Chuck. Let me just highlight the fact that
[[:upper:]]
is much superior to [A-Z] for those of
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