Is there a way to estimate Robust standard errors when using a nls()
function? I'm trying to fit some data to a complicated model and everything
works fine with nls() but I also wanted to obtain a robust estimate of my
errors.
I tried "coeftest(m, vcov=sandwich)" and it seems to work, but so does
those functions can be found.
>
>
> Hope this helps,
>
> Rui Barradas
>
>
> Às 01:37 de 15/05/2022, De Simone escreveu:
> > Hi,
> > Excuse me for this silly question
> > how do I get a colour gradient like the one attached. Colours are a bit
> > limi
Hi all,
I'm using R remotely via ssh connection in linux. I need to save a large R
object from the remote server to my laptop. How can I specify the path in
the save() function?
Thanks much for your help,
Simone
[[alternative HTML version de
Dear R users,
I would like to announce a new package called "sparsevar" version 0.0.3:
https://cran.r-project.org/web/packages/sparsevar/
The package should be useful to estimate sparse VAR/VECM models.
The developing version can be found on Github:
https://github.com/svazzole/sparsevar
Best,
Dear Frede,
the initial question was to find probability proportional to the
values, so Rui, Dan and Boris are right...
I think I will go for Boris' solution...
best regards,
Simone
2014-04-11 7:29 GMT+02:00 Frede Aakmann Tøgersen :
> I think you have calculated the wrong proba
Hello, Rui,
it does, indeed!
thanks,
Simone
2014-04-10 20:55 GMT+02:00 Rui Barradas :
> Hello,
>
> Use ?sample.
>
> sample(x, 1, prob = x)
>
> Hope this helps,
>
> Rui Barradas
>
> Em 10-04-2014 19:49, Simone Gabbriellini escreveu:
>
>> Hello List,
>
proportional to
the element value, thus higher values should be picked up more often
than small values (i.e., picking up 38 should be more probable than
picking up 3)
Do you have any idea on how to code such a rich-get-richer mechanism?
Best regards,
Simone
To Whom It May Concern:
I am subscribed to this listserv, but I would like to switch my settings so
that I receive a daily batched email rather than individual messages. Is it
possible to do that, or do I need to unsubscribe and then resubscribe?
Best,
Simone Gill
[[alternative HTML
Hi all,
I have performed a binomial test to verify if the number of males in a study is
significantly different from a null hypothesis (say, H0:p of being a male= 0.5).
For instancee:
binom.test(10, 30, p=0.5, alternative="two.sided", conf.level=0.95)
Exact binomial test
data: 10 and 30
nu
it.
>
>
> On Wed, Jan 22, 2014 at 11:04 AM, Simone Gabbriellini
> wrote:
>> Dear List,
>>
>> I have a data.frame like this:
>>
>> name religion neighbor religion.neighbor
>> pippo a minnie a
>> pluto a mickey a
>> paperino b donald a
>
. Is
there anything I can do in order to keep the data.frame the way it is
but tell R to count values once if they are repeated?
the point is that each row represent a relation, thus I cannot simply
remove duplicates...
any help more than welcome!
Best regar
Thanks
Simone
> Il giorno 15/nov/2013, alle ore 15:30, Joshua Ulrich
> ha scritto:
>
>> On Fri, Nov 15, 2013 at 6:32 AM, Simone Medori wrote:
>> Hi,
>> I'm triyng to merge two xts time series objects, one of them is from
>> Return.portfolio (Performance
have different classes ("POSIXct" and
"Date"): might be this the reason? Why do I get different extra dates anyway?
Kind regards,
Simone
> require(PerformanceAnalytics)
> require(quantmod)
>
> benchmark<-c("^STOXX50E","^NDX")
> downloa
Hi all,
I have a data set made of 12 years each one with a number of males and a
number of females. I tested the relationship between the sex ratio
(proportion of males over the total) weighted for the number of
individuals of each year.
In R:
glm.1<-glm(cbind(males,females)~predictor,bino
Hi David,
your code works perfectly on my dataset, thanks for the detailed example
and for your time!
Best,
Simone
2013/6/26 David Carlson
> Your data is probably not arranged correctly. See if this works for
> your data:
>
> > # Creating a reproducible example
> >
plotmeans, but I
am unable to understand the syntax I should use. I tried:
times<-as.factor(seq(1,15))
plotmeans(data~times)
with no luck... what should I tell to plotmeans in order to evaluate my
dataset?
Best regards,
Simone
--
Simone Gabbriellini, PhD
PostDoc@DISI, University of Bolo
gt;
> Hope this helps,
>
> Rui Barradas
>
> Em 20-04-2013 10:58, Simone Gabbriellini escreveu:
>
> Hello,
>>
>> I am adding rows to a matrix using rbind() and I would like to also add a
>> row name to the added row in order to do some sorting after the add
Hello,
I am adding rows to a matrix using rbind() and I would like to also add a
row name to the added row in order to do some sorting after the adding rows
part is finished. is this possible?
Best,
Simone
--
Simone Gabbriellini, PhD
PostDoc@DISI, University of Bologna
mobile: +39 340 39 75
each random vectors.
I am not a statistician, and I think that this solution may already have
been implemented in R, but I do not know where to search for this, so I do
welcome every suggestion/advice you might give me.
Best regards,
Simone
--
Simone Gabbriellini, PhD
PostDoc@DISI, University
group 1, y
individuals of group 2, ... z individuals of group "n".
I need to bootstrap this data set to get a new data set of the same length
(resampling with replacement) and where the number of individuals of each
group is maintained the same.
Does anyone have an idea on how to do it?
Than
Hi,
Yes I know it works, but I'd like to assign the results like this:
V(g)$meanknn <- ONELINER
Where V(g) elencates all the nodes in my graph...
Thanks,
Simone
Sent from my iPhone. Please excuse brevity and odd typos
On 01/feb/2013, at 13:31, PIKAL Petr wrote:
> Hi
>
>
listOfLists){
print(mean(graph.knn(g, i)$knn))
}
Is there any way to convert this in a one-liner? I have tried to
figure it out with lapply() or mapply() but with no success.
Thank you in advance for your help.
Simone
--
Simone Gabbriellini, PhD
PostDoc@DISI, University of Bologna
mobile: +39
apply(aa, setdiff, bb)
but I simply get "aa" back as result. Could you please point me in the
right direction about the combination of functions that I need to
accomplish this task?
Best regards,
Simone
--
Simone Gabbriellini, PhD
PostDoc@DISI, University of Bologna
mobile: +39 340 39 75
])
plot(Pt)
Rt_init_cond<-0
Rt<-Rt_init_cond*0
Rt[2]<- -a*Rt[1]+a*Et[1]+e[2]
for(i in 2:(n)){
Rt[i]<- -a*Rt[i-1]+a*Et[i-1]+e[i]
}
Rt<-ts(Rt[(length(Rt)-n+1):length(Rt)])
plot(Rt)
I donât think the code above is correct, and I donât even know if this is
the approach
Took some googling, but it was worth it! :)
Best,
Simone
Sent from my iPhone. Please excuse brevity and odd typos
Il giorno 30/dic/2012, alle ore 19:44, Bert Gunter ha
scritto:
> Are you "learning the ropes" or "on the ropes"?
agged values of Y_t+1 from Y_t to Y_t-n. Epsilon_t is, as You
said a vector of random errors.
Any suggestion is very appreciated.
thanks and best regards,
Simone
From: Rui Barradas
Sent: Saturday, December 22, 2012 8:29 PM
To: Simone Gogna
Cc: r-help@r-project.org
Subject: Re: [R] creating a f
Dear R users,
Iâd like to create a function as:
Y.t+1 = Y.t + (1\p)*summation(x(Y.t,Y.t-1,...)) + epsilon.t
where x is a function of Y.t, Y.t-1 and so on, epsilon is a random error and p
is a parameter.
Do you think something of the following form might be appropriate?
Y<-function(Y,p,x,epsi
Thanks! I missed that argument... :)
2012/9/22 Bert Gunter :
> ?boxplot
> ... and note the "border" argument
>
> -- Bert
>
> On Fri, Sep 21, 2012 at 6:57 PM, Simone Gabbriellini
> wrote:
>> Hello,
>>
>> I would like to change the color of the bor
Hello,
I would like to change the color of the borders of my boxplot. Using col= I am
able only to change the inside background of the boxplot, while I would like to
have it transparent andchange the border instead.
Any hint aamore than welcome,
Best,
Simone
Inviato da iPhone
Thanks a lot for pointing me to that!
Best,
Simone
2012/5/7 David Winsemius :
>
> On May 7, 2012, at 7:15 AM, Simone Gabbriellini wrote:
>
>> Hello List,
>>
>> I have some plots with the wireframe() function, and I'd like to
>> display them in a single
ble(title, header=TRUE),
nrow=5,ncol=4,byrow=FALSE)
colnames(rank)<-c(.25,0.50,0.75,1.00)
wireframe(rank, scales=list(arrows=FALSE), drape=TRUE, colorkey=TRUE,
xlab="η", ylab="π",zlab="corr", screen = list(z = 45, x = -70),
zlim=c(0,1), main="Rank")
clear solution.
Any advice would be appreciated.
Simone Tenan
##
library(popbio)
fec<-seq(0.1,2,0.1)
delta <- seq(0,0.6,0.05)
lambda<-matrix(0,length(fec),
length(delta))
S0=0.7
S1=0.8
S2=0.85
S3=S2
for (j in 1:length(delta)){
for (i in 1:length(fec)){
F1=0
anybody help?
S
--
-------
Simone Salvadei
Faculty of Economics
Department of Financial and Economic Studies and Quantitative Methods
University of Rome Tor Vergata
e-mail: simone.salva...@uniroma2.it
url: http://www.economia.uniroma2.it/phd/econometricsempiricaleconomics/
<http://w
,,]=m_u
temp_u=kronecker(temp_u,matrix(rep(1,d),nr=d))
A special thank to David Winsemius,William Dunlap and Patrick Burns.
I hope I have been helpful.
---
Simone Salvadei
Faculty of Economics
Department of Financial a
g the remaining rows (full of
zeroes) of 'temp_u' with
copies of the corrensponding 1st row
--
-------
Simone Salvadei
Faculty of Economics
Department of Financial and Economic Studies and Quantitative Methods
University o
(previously defined) phi=array with dimensions c(c,d,T)*
*
*
*temp=array(0,dim=c(c,d,T))*
* for(i in 1:T)*
* {*
* temp[,,i]=t(phi[,,i])*
* }*
* phi=temp*
*
*
Thank you very much!
S
--
---
Simone Salvadei
Faculty of Economics
De
have
tried with the lme4 but after this I cannot figure out how to build this same
model without the random effect in order to make it comparable to the random
effect model.
Thanks for any help.
Simone
[[alternative HTML version deleted]]
__
empty cells with a 0... how to address those empty
cells?
thanks for your help!
best,
Simone
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and
Hi,
I have found quite a few posts on normality checking of response variables, but
I am still in doubt about that. As it is easy to understand I'm not a
statistician so be patient please.
I want to estimate the possible effects of some predictors on my response
variable that is nº of males an
Hi,
I have a data frame like this:
var1=years
var2=Sex ratio (0https://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.
Hi,
This is a mixed conceptual/methodological issue.
I have 3 years and 2 localities, I want to compare the Sex Ratio series between
the two localities.
I can do it year by year, for instance:
> SR2010<-data.frame(FAO=c(96,52),JUNC=c(60,42))
> SR2010
FAO JUNC
1 96 60
2 52 42
>
many thanks, this is really a great solution!
best,
Simone
Il giorno 02/mar/2011, alle ore 16.22, Scott Chamberlain ha scritto:
> see package plyr, especially the function ddply(), eg.., in your case:
>
> ddply(dataframe, .(columnA, columnB), summarise,
> columnC = le
data.frame, but how to count and store the
duplicates?
thanks in advance for any help.
best regards,
Simone
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xample of Zipf's law?
thanks in advance for any help.
best regards,
Simone
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provid
ualization of those data in the same R console window) and there are not
empty cells but, nevertheless, some columns have NA values but other ones have
values.
Why?
Could it possibly cause any problem to the analysis?
Thanks
Best wishes
Simone
thank you very much, I am now closer to the result I would like to achieve.
Still, I have to figure out how to "invert" the n axes, see figure:
http://www.digitaldust.it/materiali/mine2.pdf
is panel.3dwire the function I have to study for this last step?
thanks,
simone
Il giorno 0
I am trying to reproduce this graph:
http://www.digitaldust.it/materiali/their.png
the default axes orientation of wireframe gives me this:
http://www.digitaldust.it/materiali/mine.pdf
I am trying to understand if I can reproduce the axes orientation of the first
figure.
many thanks,
simone
Dear List,
I am using the wireframe function in the lattice package, and I am wondering if
it is possible to invert the default axes orientation for x and y axes... what
parameter should I look for?
Best regards,
Simone Gabbriellini
__
R-help@r
o force the tabled observations to
the same length, in order to put them into a data frame?
thanks in advance for any advice.
best regards,
Simone Gabbriellini
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PLEASE
many thanks, works perfectly!
best,
Simone
Il giorno 02/nov/2010, alle ore 17.17, David Winsemius ha scritto:
>
> On Nov 2, 2010, at 11:53 AM, Simone Gabbriellini wrote:
>
>> Hello List,
>>
>> this should be simple, but cannot figure it out. I am trying to subset
;
'2010-01-05 07:18:36', but I have no idea about the sintax to do that.
is something like this close to the correct way:
active<-data4['time'<= as.POSIXct("2010-01-05 07:18:36", origin="1970-01-01
00:00:00-00")]
thanks in
. For sure I can connect to remote server with RODBC, but since I
am on a Mac OSX 10.6, I have quite difficulties to find ODBC drivers and
compile them.
Any advice is more than welcome.
best regards,
Simone Gabbriellini
__
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78 4 2
is it possible? could someone point me to the right function to do this?
best regards,
Simone
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PLEASE do read the posting guide http://www.R
, the best fitted curve in comparison with my data.
Anyone could help me?
Best regards
Simone.
--
Simone R. Freitas
Universidade Federal do ABC (UFABC)
Centro de Ciências Naturais e Humanas (CCNH)
R. Catequese, 242
Bairro Jardim
09090-400 - Santo
1
1 5861
1 5861
1 5861
1 5861
1 5861
1 5861
1 5861
1 5861
After that, I'd like to join tables using KTOTAL500X column. How can I do
that?
Thank you very much.
Best regards
Simone.
-------
:
Mixed1 <- lme(density~year, data=mydata, random=~1|sector)
Mixed2 <- lme(density~year, data=mydata, random=~1|year/sector)
But I'm not sure I can fit the Mixed2 model.
Many thanks for the answers so far.
Any ideas?
Simone Vincenzi, PhD
Department of Environmental Sciences
Un
among streams from 2000 to
2007? Is it sufficient and appropriate a test of correlation (using rcorr of
the library Design, for example) to test if density trends in time are
similar among streams?
Any help would be greatly appreciated.
Simone
Simone Vincenzi, PhD
Department of
Hi,
Does anyone know a way to estimate the existence of a temporal trend (each unit
of the sample is a count) by resting the possible effect of a covariate (i.e.
climatic factor)?
I have periodical counts of several species of waterbirds during the last 13
years and I want to know if, resting
affecting it. I am already using a weight in my
model with the effort spent in each grid cell.
Thanks for your feedback, with my best regards,
Simone
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PLEASE do read the
Hi Paul,
I've tried your solutions and it works great... still my eps viewer
(CocoViewX) doesn't visualize them well, but the pdf converter (I use
mac, the Finder does this conversion automagically) visualize all the
images correctly
thank you very much for your help!
simone
-- Forwarded message --
From: Jonathan Baron
Date: 2009/9/4
Subject: Re: [R] eps file with embedded font
To: Simone Gabbriellini
On 09/04/09 17:16, Simone Gabbriellini wrote:
> thanks Jonathan,
>
> I was wondering about the difference between your second option and
thanks Jonathan,
I was wondering about the difference between your second option and
the Ted one: isn't it the same thing?
regards,
Simone
2009/9/4 Jonathan Baron :
> A couple of other ideas about embedding fonts and setting bounding
> boxes. These all work on Linux, so in theory
Thank for the help,
unfortunately it doesn't work... it looks impossible to embed fonts...
cairo should do this automatically but not for eps files at the
moment... really don't know how to solve it.
best regards,
Simone
2009/9/4 Ted Harding :
> On 04-Sep-09 14:01:44, Simone Gabbr
mbedFonts(file="indegdistr.eps", outfile="indegdistrEMB.eps",
fontpaths="System/Library/Fonts")
the problem is that the second file, with font embedded, is cutted
near the end, and the very last part of the plots and the border are
off the page...
I use R 2.8.
I change the block length?
2) In terms of block length and type of simulation (sim="fixed" or "geom"),
what is the best way of doing it?
Thanks in advance for any suggestion,
Best wishes
Simone
¿Quieres conocerte mejor? ¡Conoce lo que Windows Live tiene especialmente pa
I change the block length?
2) In terms of block length and type of simulation (sim="fixed" or "geom"),
what is the best way of doing it?
Thanks in advance for any suggestion,
Best wishes
Simone
_
[[elided Hotmail s
d
> if you are a beginner. You are right on the edge of having too little
> data for what you want to do (the rule of thumb is that you should
> have at least 10-20 responses per predictor), and stepwise regression is
> known
> to inflate p-values (see e.g. Whittingham e
Hi,
I'm a very new user of R and I hope not to be too "basic" (I tried to
find the answer to my questions by other ways but I was not able to).
I have 12 response variables (species growth rates) and two
environmental factors that I want to test to find out a possible
relation.
The sample s
estimate the density of both distribution? because the red
one looks like a pareto distribution, but I am not an expert...
many thanks,
Simone
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PLEASE do read the postin
yes it's THE solution!
thank you very much,
Simone
Il giorno 30/nov/08, alle ore 22:42, Kingsford Jones ha scritto:
It's generally easier to work with data frames, so read your data with
students <- read.spss(yourFile, to.data.frame=TRUE)
Then subset will work as expe
2 2 2 2 2 1 1
2 2
[171] 2 2 2 2 2 1 1 1 1 2 2 2 2 2 2 2 2 1 2 2 1 1 1 1 1 2 0 2 2 1 2 1
2 2
[205] 1 2 2 2 2 2 2 2 2 2 1 2 2 1 1 2 2 2 2 2
I would like to filter - or subset - the dataset for $Sex = 1 (in this
case means male...), for example...
thanks anyway,
Simone
Il giorno 30/
dear list,
I have read a spss file with read.spss()
now I have a list with all my variable stored as vectors.
is it possible to selec cases based on the value of one or more
variables?
thank you,
Simone
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https
t(g)-1)/
(vcount(g)-2
hope it helps,
Simone
Il giorno 30/nov/08, alle ore 17:05, Gábor Csárdi ha scritto:
Weijia, centralization was not included in igraph, because really, it
takes only writing a one line function to do it. But it is on our TODO
list and will be included soon.
E.g. for degree s
Dear ALL,
thank you for this sunday morning help! It works great!
all the best,
Simone
Il giorno 30/nov/08, alle ore 11:22, Jorge Ivan Velez ha scritto:
>
> Dear Simone,
>
> Try this:
>
> x=c(1,1,2,3,4,3,2,2,2,1,2,3,4,4,3)
> table(factor(x,levels=1:5))
> 1 2 3
Dear List,
is it possible to plot vertical labels under vertical bars? for
vertical labels I mean to rotate 90 degrees horizontal labels
I suppose yes, because if excel can do this, R should do it twice
better...
thank you,
Simone
__
R-help@r
,"2","3","4","5","6","7","8","9","10","11","12","13","14")
but all I have is:
error in table
all the arguments must have the same length
thank you,
Simone
___
great too), you can simply handle
it via RPy.
I've learned this strategy from Pietro Terna - http://web.econ.unito.it/terna/
hope it helps,
Simone Gabbriellini
Il giorno 07/nov/08, alle ore 13:06, Tom Backer Johnsen ha scritto:
Do anyone know anything about the use of R for agent-ba
ar.gz' had non-zero exit
status in: install.packages("quadprog_1.4-11.tar.gz", repos = NULL, type
= "/usr/local/free/gfortran.csh")
Thank you in advances!
Simone
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hello,
I have a last question on cohesive blocks: if there are multiple links
between some nodes in the graph, this is taken into account by
cohesive blocks? or the multiple links are simply ignored?
thank you,
Simone
__
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parts, if
I simply wrote
> plot(simone.m1, log="", col="red")
everything became red..
thank in advance,
Simone
Il giorno 06/dic/07, alle ore 09:00, Christian Ritz ha scritto:
> Dear Simone,
>
> you can use the package 'drc' to fit a logistic model, th
parts, if
I simply wrote
> plot(simone.m1, log="", col="red")
everything became red..
thank in advance,
Simone
Il giorno 06/dic/07, alle ore 09:00, Christian Ritz ha scritto:
> Dear Simone,
>
> you can use the package 'drc' to fit a logistic model, th
with
abline(lm(generalstats[,
1
]~
c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25)),
lty=3, col="red");
how to interpolate the data with a logistic curve? I cannot find the -
I suppose easy - solution..
thank you,
Simone
__
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20.83
33.33
21.428571
7.4074074
Thanks
Simone
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paramatric test right?
And how can do it on R software?
Thanks in advance
Simone
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