in public on the r-help list!
F
>
>
>
> On Wed, Jun 30, 2021 at 1:03 PM Federico Calboli
> wrote:
> Hello All,
>
> I am playing with igraph (which seems to work for what I have used it).
> Nevetheless:
>
> demo('community',
_0.3.8
generics_0.1.0
[10] ellipsis_0.3.2 tools_4.1.0 glue_1.4.2 purrr_0.3.4
compiler_4.1.0 pkgconfig_2.0.3 tidyselect_1.1.1 tibble_3.1.2
--
Federico Calboli
LBEG - Laboratory of Biodiversity and Evolutionary Genomics
Charles Deberiotstraat 32 box
purrr_0.3.4 callr_3.6.0
fs_1.5.0 ps_1.6.0 curl_4.3 testthat_3.0.2
[25] memoise_2.0.0 glue_1.4.2cachem_1.0.4 compiler_4.0.4
desc_1.3.0prettyunits_1.1.1
--
Federico Calboli
LBEG - Laboratory of Biodiversity and Evolutionary Gen
by building data frames on the fly, but this is cumbresome, both as
code and very likely as performance, especially looping over lots of data
having to generate these dummy data frames.
If there is a trick I am missing I’d be grateful if anybody could set me
straight.
Best
F
--
Federico Calboli
LBEG
ls.
Cheers
F
>
> On October 12, 2018 8:00:11 AM PDT, Federico Calboli
> wrote:
>> Hi all,
>>
>> more and more people (sadly) are putting stuff on github, with either
>> no CRAN upload or an package. So I am stuck with using devtools and
>> ins
Hi all,
more and more people (sadly) are putting stuff on github, with either no CRAN
upload or an package. So I am stuck with using devtools and install_github.
I know how to keep stuff from CRAN updated — how do I do the same for github
stuff?
Cheers
F
--
Federico Calboli
LBEG
colourchart, but the visual differences between ’skyblue’ and
’slategrey’ elude me when plotted as dots on a plot).
Cheers
F
--
Federico Calboli
LBEG - Laboratory of Biodiversity and Evolutionary Genomics
Charles Deberiotstraat 32 box 2439
3000 Leuven
+32 16 32 87 67
that might be useful but I need
to sort out. I am looking at formal claims ‘we have developed a method to do X
and said method is available to the public as a R library’. If that is the
claim I expect it to be true.
Best
F
--
Federico Calboli
LBEG - Laboratory of Biodiversity
seems to do what I need.
Best
F
>
> Cheers,
> Bert
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" co
PLINK does, but I’d rather not reinvent
the wheel so it might be worth asking: is there any package/function that
would do (in R) what PLINK does with --genome?
Bets wishes
F
--
Federico Calboli
Ecological Genetics Research Unit
Department of Biosciences
PO Box 65 (Biocenter 3, Viikinkaari 1
this incredibly silly behaviour so that my table follows a
reasonable expectation that 9 comes before 10 (and so on and so forth)?
BW
F
--
Federico Calboli
Ecological Genetics Research Unit
Department of Biosciences
PO Box 65 (Biocenter 3, Viikinkaari 1)
FIN-00014 University of Helsinki
Finland
cserv.mcmaster.ca/jfox
>
>
>
>
>> -Original Message-
>> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Federico
>> Calboli
>> Sent: February 12, 2016 10:13 AM
>> To: R Help <r-help@r-project.org>
>> Subject: [R] why
that
distCayley() in Rankcluster does what you want. From the examples:
x=1:5
y=c(2,3,1,4,5)
distCayley(x,y)
8
Cheers,
Boris
On Aug 6, 2015, at 9:51 AM, Federico Calboli federico.calb...@helsinki.fi
wrote:
On 6 Aug 2015, at 15:40, Bert Gunter bgunter.4...@gmail.com wrote
in numbers and use a rank test, but I was left wondering
whether this is the only solution and whether there are more appropriate
solutions that are already implemented in R (I am not going to reinvent the
wheel if I can avoid it).
BW
F
--
Federico Calboli
Ecological Genetics Research Unit
On Thursday, August 6, 2015, Federico Calboli federico.calb...@helsinki.fi
wrote:
Hi All,
let’s assume I have a vector of letters drawn only once from the alphabet:
x = sample(letters, 15, replace = F)
x
[1] z t g l u d w x a q k j f n “v
y = x[c(1:7,9:8, 10:12, 14, 15, 13)]
I would
--
Federico Calboli
Ecological Genetics Research Unit
Department of Biosciences
PO Box 65 (Biocenter 3, Viikinkaari 1)
FIN-00014 University of Helsinki
Finland
federico.calb...@helsinki.fi
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https
, Federico Calboli federico.calb...@helsinki.fi
wrote:
On 3 Jul 2015, at 12:14, Hadley Wickham h.wick...@gmail.com wrote:
It might be a line break problem - I think you want:
Description: Functions designed to test for single gene/phenotype
association and
for pleiotropy on genetic
not help.
BW
F
Hadley
On Fri, Jul 3, 2015 at 10:09 AM, Federico Calboli
federico.calb...@helsinki.fi wrote:
Hi All,
I am upgrading a package for CRAN, and I get this note:
checking DESCRIPTION meta-information ... NOTE
Malformed Description field: should contain one or more
.
Cheers
Petr
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Hadley
Wickham
Sent: Friday, July 03, 2015 1:14 PM
To: Federico Calboli
Cc: R-help
Subject: Re: [R] what constitutes a 'complete sentence'?
In that case, you need to create
that
check?
Best
F
--
Federico Calboli
Ecological Genetics Research Unit
Department of Biosciences
PO Box 65 (Biocenter 3, Viikinkaari 1)
FIN-00014 University of Helsinki
Finland
federico.calb...@helsinki.fi
__
R-help@r-project.org mailing list
(heat.colors(100)))
# Your result is in the var1.pred-variable of res$krige_output
You will generally get more and quicker answers to questions about spatial
data and methods from the r-sig-geo list.
Best wishes,
Jon
On 8/20/2014 11:01 PM, Federico Calboli wrote:
Hi All,
I am trying
Hi All,
I am trying to do some kriging of a floor, based on a number of heat sensors.
My data looks like this:
sensortempxy
1 1 1.25437406 390 2960
2 2 0.64384594 830 2960
3 3 1.52067733 1420 2960
4 4 1.21441928 3127 2920
5 5 1.04227694
Hi All,
together with colleagues we are planning to submit a 2.0 version of a package
we have on CRAN. Because the package deals with high throughput genomic data
we though it would be nice to have some sort of guidance for the users. This
should ideally mean a 'vignette', but as the time of
Hi All,
I am trying to create an index that returns something like
1,2,1,2,3,4,3,4,5,6,5,6,7,8,7,8
and so on and so forth until a predetermined value (which is obviously even).
I am trying very hard to avoid for loops or for loops front ends.
I'd be obliged if anybody could offer a
, andrija djurovic djandr...@gmail.com wrote:
Hi. Here are two approaches:
c(mapply(function(x,y) rep(c(x,y), 2), (1:10)[c(T,F)], (1:10)[c(F,T)]))
c(tapply(1:10, rep(1:(10/2), each=2), rep, 2), recursive=T)
Andrija
On Mon, Nov 11, 2013 at 1:11 PM, Federico Calboli f.calb
Hi All,
if memory serves me well I recall some paper comparing the relative success in
getting mainstream acceptance (as mainstream as statistics can be) of both R
and Octave. I remember vaguely that the fact the development strategies (core
team vs one main developer) played a major role in
Dear All,
is there a simple way that covers all regression models to extract the number
of samples from a data frame/matrix actually used in a regression model?
For instance I might have a data of 100 rows and 4 colums (1 response + 3
explanatory variables). If 3 samples have one or more NAs
On Mon, Mar 18, 2013 at 8:39 AM, Marc Schwartz marc_schwa...@me.com wrote:
On Mar 18, 2013, at 7:36 AM, Federico Calboli f.calb...@imperial.ac.uk
wrote:
Dear All,
is there a simple way that covers all regression models to extract the
number of samples from a data frame/matrix actually
Hi All,
is there a reasonably simple way of using a black and white chequer/checker
board pattern as a colour:
barplot(mydata, col = c('red', 'blue' 'checkerboard'))
?
BW
F
--
Federico C. F. Calboli
Neuroepidemiology and Ageing Research
Imperial College, St. Mary's Campus
Norfolk Place,
something like this, too many patterned areas in a
plot can be more distracting and distorting than useful.
thanks, but that sounds way too much hassle than it's worth
BW
F
On Mon, Jan 7, 2013 at 8:00 AM, Federico Calboli f.calb...@imperial.ac.uk
wrote:
Hi All,
is there a reasonably
Hi,
I am subsetting a matrix thus:
test
[,1] [,2] [,3]
[1,]17 13
[2,]28 14
[3,]39 15
[4,]4 10 16
[5,]5 11 17
[6,]6 12 18
test[cbind(c(1,3,5), c(2,1,3))]
[1] 7 3 17
This works fine, and is the equivalent of c(test[1,2], test[3,1],
Hi,
according to the help file rtags does not support VI(M) yet. Is there any known
hack to ctags to get tags for R in VI(M)?
BW
F
--
Federico C. F. Calboli
Neuroepidemiology and Ageing Research
Imperial College, St. Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20 75941602 Fax
Hi All,
I am writing a function that reads a file in
myfile = file('myfile.raw', 'rb')
.
.
.
.
.
close(myfile)
No matter what, I get the warning
Warning message:
closing unused connection 3 (myfile.raw)
Since the whole thing is in a function, I'd like to avoid unecessary noise for
the user,
On 6 Mar 2012, at 18:01, Sarah Goslee wrote:
The help for warning offers some suggestions.
none that seem to work though.
F
Sarah
On Tue, Mar 6, 2012 at 12:53 PM, Federico Calboli
f.calb...@imperial.ac.uk wrote:
Hi All,
I am writing a function that reads a file in
myfile
Dear All,
a new version of MultiPhen (0.3) is available on CRAN, and will be available to
a mirror near you soon.
*Please upgrade* because this release is a bug fix release. A new version, with
improvements in the output and more useful error messages is basically ready,
but I will wait to
Hi,
I have a coxph model like
coxph(Surv(start, stop, censor) ~ x + y, mydata)
I would like to calculate the Schoenfeld residuals for the null, i.e the same
model where the beta hat vector (in practical terms, the coeff vector spat out
by summary()) is constrained to be all 0s --all lese
On 22 Feb 2012, at 14:01, Terry Therneau wrote:
--- begin included message ---
I have a left truncated, right censored cox model:
coxph(Surv(start, stop, censor) ~ x + y, mydata)
I would like to know how much of the observed variance (as a number
between 0 and 1) is explained by each
Hi All,
I have a left truncated, right censored cox model:
coxph(Surv(start, stop, censor) ~ x + y, mydata)
I would like to know how much of the observed variance (as a number between 0
and 1) is explained by each variable. How could I do that?
Adding terms sequentially and then using
Dear All,
just a quick example:
x = 1:25
x[12] = NA
x
[1] 1 2 3 4 5 6 7 8 9 10 11 NA 13 14 15 16 17 18 19 20 21 22 23 24 25
y = x[x10]
y
[1] 1 2 3 4 5 6 7 8 9 NA
Is there any way of NOT getting NA for y = x[x10]? Similarly
y = x[x15]
y
[1] 1 2 3 4 5 6 7
Hi All,
I have the following command:
cat(rbind(table(gen$sex, gen$ddn2)[1,],round(table(gen$sex,
gen$ddn2)[1,]/apply(table(gen$sex, gen$ddn2),2,sum) * 100)), sep = c('
(','%)\n'), file = '')
3 (20%)
17 (30%)
15 (31
i.e. it does NOT print the last separator, which is something that bugs me a
On 25 Jan 2012, at 11:54, Duncan Murdoch wrote:
I think the documentation for cat is a little ambiguous, but it is working as
documented if I read If any element of sep contains a newline character, to
mean that the element consists of a newline and nothing else. I'm not sure
if that was
Hi All,
I'm having a problem with barplot:
mydata
[1,] 2 108 0 0 0 1 3 0 0 0 0 0 7 18 3 4 8 20 26 20 19 7 1 1
mycol = c(rep('yellow', 2), rep('white', 3), rep('orange',2), rep('white', 5),
rep('orange',3), rep('red',9))
barplot(mydata, col = mycol)
gives me an uniformly
, using the 'x' object as you created the code works, and I can have a
workaround. I fail to see why it would not work using mydata.
Cheers
F
Michael
On Mon, Dec 5, 2011 at 10:44 AM, Federico Calboli
f.calb...@imperial.ac.uk wrote:
Hi All,
I'm having a problem with barplot:
mydata
Hi All,
I have a parameter that is bimodal, and I want to get some sort of linear model
done with it
results = some.linear.function(bimodal.param ~ factor1 + some other stuff,
mydata)
I want to see if factor 1 matters (it has 3 levels, of of which can be taken as
baseline), i.e:
Hi All,
is there a way of using strsplit with a forward slash '/' as the splitting
point?
For data such as:
1 T/TC/C 16/33
2 T/TC/C 33/36
3 T/TC/C 16/34
4 T/TC/C 16/31
5 C/CC/C 28/29
6 T/TC/C 16/34
strsplit(my.data[1,1], /) # and any
On 3 Aug 2011, at 17:41, Duncan Murdoch wrote:
It looks as though your my.data[1,1] value is a factor, not a character value.
strsplit(as.character(my.data[1,1]), /)
Thanks Duncan, this solved it.
Best
Federico
would work, or you could avoid getting factors in the first place,
strsplit(as.character(my.data[1,1]), /)
work?
yes!
Thanks
Federico
If you used read.table() to get your data in, you might want the
as.is=TRUE or the stringsAsFactors=FALSE argument.
Sarah
On Wed, Aug 3, 2011 at 12:37 PM, Federico Calboli
f.calb...@imperial.ac.uk wrote:
Hi All
Hi,
I want to apply a function to a matrix, taking the columns 3 by 3. I could use
a for loop:
for(i in 1:3){ # here I assume my data matrix has 9 columns
j = i*3
set = my.data[,c(j-2,j-1,j)]
my.function(set)
}
which looks cumbersome and possibly slow. I was hoping there is some function
in
-
De: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Em
nome de Federico Calboli
Enviada em: terça-feira, 12 de julho de 2011 10:07
Para: r-help
Assunto: [R] applying function to multiple columns of a matrix
Hi,
I want to apply a function to a matrix, taking
HI All,
I have written and succesfully uploaded a new package. The licence it is under
is 'GPL' --no version. My assumption is, since all the code is written in R the
licence R used for R would affect the code (hence my GPL stands for whatever
version of the GPL R is under)
I am happy with
On 8 Jul 2011, at 12:06, Duncan Murdoch wrote:
On 11-07-08 6:20 AM, Federico Calboli wrote:
HI All,
I have written and succesfully uploaded a new package. The licence it is
under is 'GPL' --no version. My assumption is, since all the code is written
in R the licence R used for R would
On 8 Jul 2011, at 15:56, Spencer Graves wrote:
Ok, thanks for that. I though that, since R in under GPL-v2, I can only
release my code under GPL-v2 because the code is written in R and probably
qualifies as a derivative work.
Did you include someone else's GPL-vx code (possibly
On 8 Jul 2011, at 16:12, Barry Rowlingson wrote:
On Fri, Jul 8, 2011 at 4:07 PM, Federico Calboli
f.calb...@imperial.ac.uk wrote:
On 8 Jul 2011, at 15:56, Spencer Graves wrote:
Ok, thanks for that. I though that, since R in under GPL-v2, I can only
release my code under GPL-v2 because
I am writing to get a better handle on a warning I am getting from a coxph
analysis I am doing.
I am analysing age of onset of dementia *after* the onset of parkinson disease.
My data looks like:
age.park age.dem age.death censor x1 x2 x3 x4
1 76 8788 0 16 33 E3
it helps.
Best,
Dimitris
On 6/3/2011 12:17 PM, Federico Calboli wrote:
I am writing to get a better handle on a warning I am getting from a coxph
analysis I am doing.
I am analysing age of onset of dementia *after* the onset of parkinson
disease. My data looks like:
age.park
Hi All,
I am trying to figure out how to get the position of the knots in a pspline
used in a cox model.
my.model = coxph(Surv(agein, ageout, status) ~ pspline(x), mydata) # x being
continuous
How do I find out where the knot of the spline are? I would like to know to
figure out how many
Hi All,
I'm running the now almost-to-be upgraded R 2.11.1 on a Intel Mac, and on a
Ubuntu machine, but the problem I see is the same. I noticed the following
behaviour:
407585.91 * 0.8
[1] 326068.7 -- the right asnwer is 326068.728
round(407585.91 * 0.8, 2)
[1] 326068.7 -- same issue
Hi everyone,
I'm running a cox ph model on a dataset with a number of variables. Each
variable has a different number of missing data, so that coxph() drops the
individuals who are missing data at one or more variables. Because of this
dropping (totally fine btw) I want to know how many events
On 15 Jun 2010, at 18:34, Federico Calboli wrote:
I'm running a cox ph model on a dataset with a number of variables. Each
variable has a different number of missing data, so that coxph() drops the
individuals who are missing data at one or more variables. Because of this
dropping (totally
Hi everyone,
I'm doing some coxph() analyses with a large and complex dataset. The data was
collected in different centers, so I am using strata(centers) to stratify the
analysis.
My main issue is, not all centers collected all the variables, so for a model
such as:
coxph(Surv(days, cancer)
Hi All,
I am in the annoying position of having to present some data to someone who
seems to be somewhat less than numerate. I need to label the y-axes of a
multhist with the y-axis labeled not as counts but as percentage of a
population. Plotting the standard histogram is in a way fine, all I
question were clearer, I might be able to help in more specific ways.
-tgs
On Fri, May 14, 2010 at 10:19 AM, Federico Calboli f.calb...@imperial.ac.uk
wrote:
Hi All,
I am in the annoying position of having to present some data to someone who
seems to be somewhat less than numerate. I need
, May 14, 2010 at 11:51 AM, Federico Calboli f.calb...@imperial.ac.uk
wrote:
On 14 May 2010, at 16:09, Thomas Stewart wrote:
Please be more specific with your question. Perhaps a simple subset of the
data you are trying to plot? Here is some non-specific advice:
Plotting histograms
Hi All,
I have a data frame where a couple of columns are factors, with long and
complex names. Everything works ok --in R, but I need to export the data so it
can be used on a dumber program (one with a three letters name starting with
S...). I know that those complex factor names are causing
Dear All,
I will soon be working with NIH and possibly FDA. Will I be able to
use R or will I be forced to use SAS?
Cheers,
Federico
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Norfolk Place, London W2 1PG
Tel +44 (0)20
On 13 Nov 2009, at 12:25, Barry Rowlingson wrote:
Working with is different to Working for. Assuming they want to
work with you then they want you for your abilities and skills, and if
those skills are with R then you go ahead and use R.
You don't employ a bricklayer to build a wall and then
Dear All,
I'm using apply to do some genetic association analysis along a chromosome, with
many thousands markers. For each marker the analysis is the same, so I was
planning to use apply(chrom, 2, somefunction)
In the specific case I do:
my.results = apply(chr, 2, function(x){anova(lrm(
than a measly script.
Cheers,
Fede
-thomas
On Tue, Nov 10, 2009 at 10:04 AM, Federico Calboli
f.calb...@imperial.ac.uk wrote:
Dear All,
I'm using apply to do some genetic association analysis along a
chromosome,
with many thousands markers. For each marker the analysis is the
same
Hi All,
I'm fitting an proportional odds model using vglm() from VGAM.
My response variable is the severity of diseases, going from 0 to 5 (the
severity is actually an ordered factor).
The independent variables are: 1 genetic marker, time of medical observation,
age, sex. What I *need* is a
On 4 Nov 2009, at 18:11, Gavin Simpson wrote:
Is there a particular reason for choosing a VGLM here? My reading of
your post suggests the response is an univariate, ordered factor and
VGLMs are especially for multivariate responses. In which case, can
you
not use polr() in package MASS that
On 4 Nov 2009, at 18:40, Gavin Simpson wrote:
Additionally, while extracting the t value is a piece of cake with
polr(), the p-value I get a nowhere close to a null distribution.
Yes - I see that polr() also doesn't produce p-values in the output
from
summary. You can use it to get a
Hi All,
I have some data where the dependent variable is a score, low (1:3) or
high (8:9), and the independent variables are 21 genotypic markers.
I'm fitting a logistic regression on the whole dataset after
transforming the score to 0/1 and normal linear regression on the high
and low
Actually, I tried doing
data2 = unique(data)
mod = lm(y ~ x1 + ... + xn, data2)
fitted(mod)
and I still get les fitted values than observations.
Federico
On 4 Aug 2009, at 12:18, Federico Calboli wrote:
Hi All,
I have some data where the dependent variable is a score, low (1:3) or
high (8
to the numbers involved? Even better
would be small *reproducible* example.
I'll have to cook that up, the data is more or less confidential. Not
very much but enough no to go on google ;)
F
--
David
On Aug 4, 2009, at 12:51 PM, Federico Calboli wrote:
Actually, I tried doing
data2 = unique
On Aug 4, 2009, at 12:51 PM, Federico Calboli wrote:
Actually, I tried doing
data2 = unique(data)
mod = lm(y ~ x1 + ... + xn, data2)
fitted(mod)
and I still get les fitted values than observations.
Federico
On 4 Aug 2009, at 12:18, Federico Calboli wrote:
Hi All,
I have some data where
On 17 Jun 2009, at 18:15, Dirk Eddelbuettel wrote:
On Wed, Jun 17, 2009 at 05:29:26PM +0100, Federico Calboli wrote:
Hello all,
a friend has a problem with tiff() which I was unable to help
about. I
searched the error messages to no avail. When he tries:
tiff(filename
I checked on the affected computer and both tiff and cairo
capabiliies
return FALSE. How that is I would not know, it's a bog standard R
installation on windows XP.
The 2.9.0 Windows install does not have TIFF capability. Try png() or
something else.
I don't know about that, but R is
Hello all,
a friend has a problem with tiff() which I was unable to help about. I
searched the error messages to no avail. When he tries:
tiff(filename = FedeWhyDoesntThisBloodyWork.tif, width = 5, height =
5, units = cm, bg = white, res = 1200)
Error in tiff(filename =
Hello,
I'm trying to use the function svyglm in the library survey.
I create a data survey object:
data_svy- svydesign(id=~PSU, strata=~sample_domain,
weights=~sample_weight, data=data, nest=TRUE)
and I try to use svyglm() with little success:
Dear All,
is there a package in R that implements parametric and non parametric
genetic linkage analysis (with both case/control or quantitative
phenos)? I tried using Merlin, which could not deal with pedigrees the
size I have, and SimWalk2 does not look like the answer either --for
On 11 May 2008, at 22:45, Andrew Robinson wrote:
lme(y ~ selection * males, random = ~1|replica/selection/males,
mydata)
forgive me, but I seem to see nesting in the random statement.
That is
what happens when we separate factors with a '/'; they are nested. We
would expect that
On 12 May 2008, at 01:05, Andrew Robinson wrote:
On Mon, May 12, 2008 at 10:34:40AM +1200, Rolf Turner wrote:
On 12/05/2008, at 9:45 AM, Andrew Robinson wrote:
On Sun, May 11, 2008 at 07:52:50PM +0100, Federico Calboli wrote:
The main point of my question is, having a 3 way anova
On 12 May 2008, at 09:29, Dieter Menne wrote:
Federico:
First, mixed models are different from standard 101 Anova, and
quite a lot
of the nesting stuff I used to ponder about 30 year ago when I started
teaching this is no longer relevant and works implicitely when you
code the
parameters
On 11 May 2008, at 23:34, Rolf Turner wrote:
It doesn't seem to me to be a complaint as such. It is a
request for insight. I too would like some insight as to
what on earth is going on. And why do you say Federico
shows no evidence of having searched the
On 12 May 2008, at 10:05, Ken Beath wrote:
There is only one random effect, so where does the crossing come
from ? The fixed effects vary across blocks, but they are fixed so
are just covariates. For this type of data the usual model in lme4
is y~fixed1+fixed2+1|group and for lme split into
On 12 May 2008, at 11:16, Andrew Robinson wrote:
Well. I have documentation relevant to nlme that goes back about 10
years. I don't know when it was first added to S-plus, but I assume
that it was about then. Now, do you think that if the thing that you
want to do was really bog standard,
On 12 May 2008, at 12:21, Nick Isaac wrote:
I *think* the syntax for the model Federico wants is this:
lmer(y~selection*males+ (selection|month) + (males|month))
I'll try and check against some back of the envelope calculations --
as I said, the model is, per se, nothing really new, and my
On 12 May 2008, at 14:37, Doran, Harold wrote:
I haven't followed this thread carefully, so apologies if I'm too off
base. But, in response to Rolf's questions/issues. First, SAS cannot
handle models with crossed random effects (at least well at all).
SAS is
horribly incapable of handling
On 12 May 2008, at 17:09, Douglas Bates wrote:
I'm entering this discussion late so I may be discussing issues that
have already been addressed.
As I understand it, Federico, you began by describing a model for data
in which two factors have a fixed set of levels and one factor has an
of the nesting, in any case I attach a toy dataset I created
especially to test how to specify the correct model (silly me).
Best,
Federico Calboli
[1] So much bog standard that the Zar, IV ed, gives a nice table of
how to compute the F-tests correctly, taking into account that one of
the 3
Note that random can be a list:
a one-sided formula of the form ~x1+...+xn, or a pdMat object with a formula
(i.e. a non-NULL value for formula(object)), or a list of such formulas or pdMat
objects.
If you can translate that into *informative* English I'd be grateful. I have the
Pinheiro
Hi everyone,
I am confused on how to specify some nesting and interaction terma with lme().
I have a dataset where some flies where selected for accessory gland size, made
to mate in presence/absence of another male and the level of some protein
measured. Now the complex stuff.
The
Hi All,
is there a way of predicting memory usage?
I need to build an array of 86000 by 2500 numbers (or I might create
a list of 2 by 2500 arrays 43000 long). How much memory should I
expect to use/need?
Cheers,
Fede
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Hi All,
I would like to output the results of a function into a text file,
legible as a such. The function produces a summary quite like:
summary(lm(x ~ y + w * z))
[for instance]
and I am not clear how to save this summary into a text file
'automagically', because I need to be able to do
, dn, lwd = lwd, lty = 1, col = colour)
segments(wid.lf, up, wid.rt, up, lwd = lwd, lty = 1, col = colour)
segments(wid.lf, dn, wid.rt, dn, lwd = lwd, lty = 1, col = colour)
}
invisible(data.frame(x, up, dn))
}
}
)
Regards,
Federico Calboli
--
Federico C. F. Calboli
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