Hello everyone again,
I much appreciated the explanations.
On Wed, Sep 25, 2019 at 11:02:42AM +0200, Francesco Ariis wrote:
> Maybe the Introduction should link to it (or similar page) with text
> "In case you are interest in the difference between static and lexical
> sco
Dear R users/developers,
while ploughing through "An Introduction to R" [1], I found the
expression "static scope" (in contraposition to "lexical scope").
I was a bit puzzled by the difference (since e.g. Wikipedia conflates the
two) until I found this document [2].
Maybe the Introduction should
On Fri, Sep 20, 2019 at 03:14:44PM +1200, Richard O'Keefe wrote:
> The one thing "slave" does not mean in technology is any kind of human
> being.
At risk of repeating what someone else said, we are most likely
not dealing with a human but with a "supernatural being, often
represented as of diminu
lated comment to my
repository at:
https://github.com/contefranz/msmtools/issues
or contact me at francesco.grosse...@polimi.it
Thanks!
Francesco
--
Francesco Grossetti
Ph.D. Student
MOX - Modeling and Scientific Computing
Dipartimento di Matematica “F. Brioschi"
Politecnico di Milano
Via
Sema, E is just a computer way of saying 0. For the purpose of statistical
analysis, if you can't compute a calculation with E values (i.e. 0),
substitute all E values with a usable constant, say 50. I stumbled across a
few websites lately that did this.
Frank Romano Ph.D.
*Academia.edu*
https://
Dear all,I want to create a routine to generate an object for different value
of val:
z <- c(1,2,3,4,5,6,7,8,9)
for (val in z) {
neighbors.knn <- knn2nb(knearneigh(coord, val, longlat=F),
row.names=cod_pro,sym=F)
}
However, it seems it does not work.
How to store the neighbors.knn created
Thanks Boris for the detailed answer. I thought to make it backward
compatible because I did not know if gamma parameter would change the
result if absent. As far as I can see changes in my results are not
relevant, so I think to follow your advice to update the old code.
Thanks
~ Francesco
Hi Boris,
yes I tried this way and it worked. The fact is that I wanted to be
compliant with the old code, I did not want to change anything. So I wanted
to find a new way to rewrite the code.
Thanks
~ Francesco Brundu
On 19 April 2014 23:18, Boris Steipe wrote:
> Have you looked at ?rain
, gamma = 1.5)
?
It fails with:
Error in rainbow(100, s = 1, v = 0.75, start = 0, end = 0.75, gamma = 1.5)
:
unused argument (gamma = 1.5)
Calls: nmfconsensus ... matrix.abs.plot -> image -> image.default -> rainbow
Execution halted
Thanks
~ Francesco Brundu
[[alternative HTM
ho =0.5,mu = 0, sigma = 1),
trace = TRUE)
lines(x, predict(fit, newdata = x), lty = 2,col = "red")
deltasN[i, ] <- coef(fit)
}
deltasN
I am running R(3.03) on Mac OS 10.9 and quantreg(5.05)
Thank you!
Francesco
[[alternative HTML version deleted]]
Nice job Nicholas!thanks for the e-mail.
Regards,
Francesco
---
Francesco Nutini
CNR-IREA
Institute for Electromagnetic Sensing of the Environment
Via Bassini 15, 20133 Milano (Italy)
Tel: +39-02 23699 297
www.irea.cnr.it
nutin
How to calculate the probability P (xt https://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.
how can i calculate the probability of occurrence of an event and the
probability of an event upon the occurrence of another event.
P (A) and P (A | B) ..
[[alternative HTML version deleted]]
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R-help
;*",sep=" "),
> ifelse(lm2[,4]<.1,paste(lm2[,2],".",sep="
> "),lm2[,2]
>
> ## ONE OPTIONS ##
> lms <- as.data.frame(cbind(lm1[,1],lm2[,1],lm1[,2],lm2[,2]))
> rownames(lms) <- rownames(lm1)
> colnames(l
estimated parameters
>> from all imputation runs?
>>
>> Chris
>>
>>
>> On Sat, Aug 17, 2013 at 11:18 AM, Francesco Sarracino <
>> f.sarrac...@gmail.com> wrote:
>>
>>> Hi Christopher,
>>> thanks for your reply. Unfortunate
> }
>
> # xtable
> for(i in 1:m){
> print(xtable(summary(lm.imputed1[[i]])))
> print(xtable(summary(lm.imputed2[[i]])))
> }
>
>
> On Sat, Aug 17, 2013 at 6:37 AM, Francesco Sarracino <
> f.sarrac...@gmail.com> wrote:
>
>> Dear listers,
>>
>
Dear listers,
I am running some OLS on multiply imputed data using Amelia.
I first imputed the data with Amelia.
than I run a OLS using Zelig to obtain a table of results accounting for
the multiply imputed data-sets. And I'd like to do this for various models.
Finally, I want to output all the mo
What is the function to do the truncation to a certain decimal digit of a
number. And the function approximation?
Thanks.
Francesco M
[[alternative HTML version deleted
at you're looking for.
>
> Sarah
>
> On Mon, Jun 24, 2013 at 4:50 PM, Francesco Miranda
> wrote:
> > Hello,I was trying to enter commands input messages or error within a
> > routin.for example : msg (consistency error: at the time ( j : generic
> > ti
Hello,I was trying to enter commands input messages or error within a
routin.for example : msg (consistency error: at the time ( j : generic time )
the lower quantile is not less than the upper quantile)
thanks in advanceFrancesco Miranda
[[alterna
<>
what kind of error? how to solve it?
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PLEASE do read the posting guide http://www.
Hello,How do I extract only the value from the quantile
function?example:quantile (x, probs = 0.10) 10%-1.83442I want to add salt
only the number -1.83442
SincerelyFrancesco Miranda
[[alternative HTML version deleted]]
Thanks so much for your help, now I could solve the problem (using [.])!
2013/5/13 arun
> You may also try:
> grepl("[.]",c("ad.1","ads","ad.2"))
> #[1] TRUE FALSE TRUE
> A.K.
>
>
>
> - Original Message -
> From: Fra
t \. but it
does not work.
> grepl("\.",string)
Error: '\.' is an unrecognized escape in character string starting "\."
Thank you very much for your help
Francesco
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Dear Ellison,
thanks a lot for your reply. Your explanation makes things much clearer.
Sincerely,
f.
On 24 January 2013 05:58, S Ellison wrote:
>
>
> On 23 Jan 2013, at 21:36, "Francesco Sarracino"
> wrote:
>
> > what I meant refers to the fact that I&
urner wrote:
>
> Given that your labels are "no" and "yes", what do you expect R to
> do? To quote a well-known fortune, "R is lacking a mind_read() function!"
>
> cheers,
>
> Rolf Turner
>
>
> On 01/23/2013 10:58 PM, Frances
t; check also
>
> pp <- rep(0:1, 10)
> pp <- factor(pp, levels=(0:1), labels=c("no","yes"))
>
> unclass(pp)
> unclass(pp) - 1
>
>
> Best,
> Dimitris
>
>
> On 1/23/2013 10:48 AM, Francesco Sarracino wrote:
> > Dear Dimitris,
> &g
the output is a vector of NA.
Any ideas?
f.
On 23 January 2013 10:39, D. Rizopoulos wrote:
> Check R FAQ 7.10: How do I convert factors to numeric?
>
>
> I hope it helps.
>
> Best,
> Dimitris
>
>
> On 1/23/2013 10:33 AM, Francesco Sarracino wrote:
> > Dear R li
. Indeed, the result is 1.5 and not 0.5 as expected.
What am I doing wrong?
Thanks in advance for your kind support,
f.
--
Francesco Sarracino, Ph.D.
https://sites.google.com/site/fsarracino/
[[alternative HTML version deleted]]
__
R-help@r
r.com/4lKpw.png
>
> Regards,
> Yihui
> --
> Yihui Xie
> Phone: 515-294-2465 Web: http://yihui.name
> Department of Statistics, Iowa State University
> 2215 Snedecor Hall, Ames, IA
>
>
> On Tue, Jan 8, 2013 at 4:17 AM, Francesco Sarracino
> wrote:
> > Dear
Dear R-listers,
does anybody know of any package developed to implement the Oaxaca-Blinder
decomposition in R?
I've been googling around and my reserch has been unfruitful. The latest
news I've found were 1 year old. Does anybody know of any recent
development?
Has R ever been employed to run a Oa
Dear R helpers,
I am using knitr to run analysis with R and edit my document with Latex. I
am wondering whether there is a way to include 2 or more pictures per chunk
and being able to refer them in the text independently and eventually
whether it is possible to give them different captions. Let m
; plot <- tmpplot + geom_line()+scale_x_continuous(breaks=ii)
>
>
> Yao He
>
>
> 2013/1/8 Francesco Sarracino :
> > Dear R helpers,
> >
> > I am currently having hard time fixing the values on the x-axis of a plot
> > with ggplot: even though I have
Dear R helpers,
I am currently having hard time fixing the values on the x-axis of a plot
with ggplot: even though I have 12 years, ggplot plots only 3 of them.
Here is my example:
library(ggplot2)
ii <- 2000:2011
ss <- rnorm(12,0,1)
pm <- data.frame(ii,ss)
tmpplot <- ggplot(pm, aes(x = ii, y = s
nderstand what's going on. Does anybody have any ideas about
what is going on?
Thanks in advance for your help ,
f.
--
Francesco Sarracino, Ph.D.
https://sites.google.com/site/fsarracino/
[[alternative HTML version deleted]]
__
R-h
the level (the value label - in
Stata dictionary). Moreover, the little "hat" in "o.particip^4" drives
latex nuts. Any ideas on how to fix these things?
Thank you very much for your kind help,
f.
--
Francesco Sarracino, Ph.D.
https://sites.google.com/site/fsarracino/
ber of dimensions
Any ideas about what am I doing wrong?
Thanks in advance,
f.
--
Francesco Sarracino, Ph.D.
https://sites.google.com/site/fsarracino/
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
https://stat.ethz.ch
levels option I lost the initial information.
Can you help figuring out what I am doing wrong?
thanks in advance,
f.
--
Francesco Sarracino, Ph.D.
https://sites.google.com/site/fsarracino/
[[alternative HTML version deleted]]
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R-help@r-pro
Dear R-listers,
I am a newbie with R and I am struggling with something I consider very
basic. I wish to produce a table (to import in a latex file) of summary
statistics, but for as much as I've been looking around and trying various
alternatives (plyr, reporttools, pastecs and Hmisc) I haven't f
, length.out= 50)
data <- data.frame(var, dim1, dim2, dim3)
I am trying to build a fifth one (let's say: group_id) to uniquely identify
groups of observations as defined by dim1, dim2 and dim3, i.e. 30 groups.
can you please help me figuring out how to do it?
thanks in advance,
f.
--
F
conditions
on y must be true":
which(x==1 & (c(0,y[1:12])==-10 |c(0,0, y[1:11])==-10|
c(0,0,0, y[1:10])==-10))
thank in advance for you help,
Francesco
> Date: Wed, 17 Oct 2012 10:36:46 +1100
> Subject: Re: [R] vectors comparison
> From: sleepingw...@gmail.c
uestion does not seem to make sense - there is no value of -500
> in Y (did you mean -10?). Anyway, I think this might work:
>
> which(y==-10 & (x==1 | c(0, x[-length(x)]) == 1 | c(x[-1], 0) == 1))
>
> ... though one would think there is a more elegant way
>
>
> O
n x.i.e. in this
example only the -10 in position y[3] satisfies the criteria, because x has in
position x[2]the figure 1.
Thank in advance for you help,Francesco
[[alternative HTML version deleted]]
__
thank you very much David, I will have a look at your idea
Best,
On 8 October 2012 19:24, David Winsemius wrote:
>
> On Oct 8, 2012, at 10:12 AM, David Winsemius wrote:
>
>>
>> On Oct 8, 2012, at 9:18 AM, David Winsemius wrote:
>>
>>>
>>> On Oct 8,
I think I have my answer... ggplot2 uses ecdf which does NOT allow
weightings...
so there is no warning or error, but still the resulting plot do not
take into account the command weight=weight
Hope that helps someone, just in case ;-)
On 8 October 2012 15:40, Francesco wrote:
> Dear all,
&g
Dear all,
I am trying to draw a weighted cumulative distribution (as defined
here http://rss.acs.unt.edu/Rdoc/library/spatstat/html/ewcdf.html)
with ggplot2
however the syntax
temp<-qplot(X,weight=weight,data=data,stat = "ecdf", geom =
"step",colour=factor(year))
seems not to produce exactly th
Hi all,
I'm new to R and to this mailing list,
I'd need to display, in a 2x2 window (mfrow = c(2, 2)), a graph that is 2 rows
tall (so with 2 other graphs at his right).
Here is an example of what I mean
http://www.stefandahlen.com/pictures/rowspan.gif
in html there is an attribute called rowspa
---Original Message-
> > From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
> > Sent: Sunday, August 19, 2012 3:54 PM
> > To: William Dunlap; Francesco
> > Cc: r-help
> > Subject: Re: [R] merging and obtaining the nearest value
> >
> > Hello,
> >
>
combination of TYPE and DATE.
>
> Replace this
>
> x[which(min(a) == a), ]
>
> by this
>
> x[which.min(a), ]
>
> Rui Barradas
>
> Em 19-08-2012 12:00, Francesco escreveu:
>
>> Dear Riu, Many thanks for your suggestion
>>
>> However thes
me", row.names = c(NA, -5L))
>
> Now all we have to do is run the statement A <- structure(... etc...) to
> have an exact copy of the data example.
> Anyway, your example with input and the wanted result was very welcome.
>
> Hope this helps,
>
> Rui Barradas
>
&g
Dear R-help
Î would like to know if there is a short solution in R for this
merging problem...
Let say I have a dataset A as:
TYPE DATE
A2
A5
A20
B10
B2
(there can be duplicates for the same type and date)
and I have another dataset
Oh thank you Carlos!I wasted a lot of time formatting my xyplot by
powerpoint.Did you used a similar tips for ternaryplot (vcd)?
Many thanks.Regards,Francesco
Date: Wed, 6 Jun 2012 17:08:39 +0200
Subject: Re: [R] [r] par and complex graph
From: c...@qualityexcellence.es
To: nutini.france
Thank you Brian! So, that's why sometimes I can't use the par()
Now I'm using the ternaryplot in [vcd]. Then, I have to read the vcd help to
looking for a function similar to par().
Many thanks.
Francesco
> Date: Tue, 5 Jun 2012 19:01:25 +0100
> From: rip..
t).How can I print more than one graph per page, when I work with
this "elaborated" graph?Many thanks!Francesco
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Dear R-Users, I'd like to have some tips for a ternaryplot ("vcd").
I have this dataframe:
a<- c (0.1, 0.5, 0.5, 0.6, 0.2, 0, 0, 0.00417, 0.45) b<- c
(0.75,0.5,0,0.1,0.2,0.951612903,0.918103448,0.7875,0.45)c<- c
(0.15,0,0.5,0.3,0.6,0.048387097,0.081896552,0.20833,0.1)
Dear Peter,
You are absolutely right... I did not know that
;-)
What to say ? Many thanks for your right answer to my "silly" question ;-)
Best Regards
On 25 May 2012 15:19, peter dalgaard wrote:
>
> On May 25, 2012, at 14:46 , Francesco wrote:
>
> > Dear R help,
>
Dear R help,
I am using Stata, and I use a Stata ado file (Rsource) to run R in batch
mode within Stata
Everything works fine except for the fact that I cannot export the
graphics that I obtain with my computations written in my R source file
I believe this is related to the global Rterm_opt
Dear R listers,
I have a silly problem. I am trying to load a dta (Stata) file in R.
The dta is about 650 MB and contains the integrated World Values
Survey/ European Value Study data-set.
My problem is that I don't manage to load the file. After almost 3
hours after I issued the following command
Dear R listers,
I have a silly problem. I am trying to load a dta (Stata) file in R.
The dta is about 650 MB and contains the integrated World Values
Survey/ European Value Study data-set.
My problem is that I don't manage to load the file. After almost 1
hour I issued the following command:
data
Dear Giovanni,
I recalled the procedure and here is the output :
Erreur : impossible d'allouer un vecteur de taille 39.2 Mo
De plus : Messages d'avis :
1: In structure(list(message = as.character(message), call = call), :
Reached total allocation of 12279Mb: see help(memory.size)
2: In structu
Dear Andrew,
Thanks for your suggestion.
I will indeed have a look at Allison's booklet...
Best,
On 7 February 2012 23:39, Andrew Miles wrote:
> Based on Paul Allison's booklet "Fixed Effect Regression Models" (2009), the
> FE model can be estimated by person-mean centering all of your variable
Hi Petr,
thanks a lot for your reply. Unfortunately, your suggestion does not work
for me.
I even tried larger boxes such as 15,15 , but the result does not change.
Is there some setting that I am missing?
However, once more thanks a lot for your help.
f.
On 12 October 2011 15:58, Petr PIKAL wrot
Thanks a lot Andrés.
It was easier than I expected.
f.
2011/10/12 Andrés Aragón
> Francesco,
>
> Try cex.axis=0.6
>
> Regards,
>
> Andrés AM
>
> 2011/10/12, Francesco Sarracino :
> > Dear R-listers,
> >
> > I have a little problem with a boxplot and
Dear R-listers,
I have a little problem with a boxplot and I hope you can help me figuring
it out.
I'll try to make up some data to illustrate the issue. Sorry, if my
procedures look naive, but these are my first steps in R. Any comments
and/or suggestions are very welcome.
let's create a vector
That's cool!
it works :-)))
for me (as a stata user) these are quite basic things and I didn't find them
anywhere for what concerns R. I can't figure out why.
Really, thank you so much,
f.
On 27 September 2011 14:20, Petr PIKAL wrote:
> Hi Francesco
>
> > Dear Petr
Dear Petr,
thank you so much for your quick reply. I was sure that there were some
smart ways to address my issue. I went through it and took some time to look
at the help for lapply and mapply.
However, some doubts still remain. Following your example, I did:
lll <-vector(mode = "list", length =
Dear R listers,
I am trying to be a new R user, but life is not that easy.
My problem is the following one: let's assume to have 3 outcome variables
(y1, y2, y3) and 3 explanatory ones (x1, x2, x3).
How can I run the following three separate regressions without having to
repeat the lm command thre
eep track of.
>>
>> The two documents mentioned in my signature
>> may help you get up and running in R. Since
>> you are coming from Stata, you should also
>> look at http://r4stats.com
>>
>>
>> On 18/09/2011 17:05, Francesco Sarracino wrote:
>>
&g
Dear all,
I am a stata user and I am moving my first steps in R.
I am dealing with a silly issue concerning looping over variables. I read
previous posts on similar topics in the R help archive, but I did not find a
solution.
Here is my case:
I run a simple bivariate linear regression saving the
ound
we cannot understand what's wrong and where R searches for the mylib.so files.
Any suggestion?
Thanks,
Michela and Francesco
__
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PLEASE do read the postin
.23 859.0167 945.4912
1046.917 1147.703 1249.375 1359.794 1447.518
1553.838 1649.231 1735.217 1850.617 1957.85
2031.329 2156.8 2247.55 2340.6 2822.1
Thank you very much.
francesco
From: michael.weyla...@gmail.com
Date: Tue, 23 Aug 2011 11:27:13 -
Dear R-users,
I need to produce a histogram where for every breaks there are the mean of the
data.
I tried tu use the function >hist(x, break=20 ... ) but this return the
numerosity for every breaks, not the mean.
Any hint?
Thanks in advance,
france
In fact "temp.data$" is not necessary, but the command still not run.
Thanks for your response, I'm also tring other point of view (suggested by
r-helpers).
For example ?ddply and ?lmList.
If you are interested I will keep you updated.
Francesco
> Date: Fri, 20 May 2
y elements as there are
> levels in c) and extract the coefficients.
> Dimitri
>
>
> On Fri, May 20, 2011 at 9:57 AM, Francesco Nutini
> wrote:
> > Yes Dimitri that's what I mean!
> > Something like this?
> >
> > for(i in levels(c)) { lm(a ~ b * c
tors
> From: dimitri.liakhovit...@gmail.com
> To: nutini.france...@gmail.com
> CC: rb...@atsu.edu; r-help@r-project.org
>
> Francesco, do you just want a separate regression for each level of
> your factor c?
> You could write a loop - looping through levels of c:
>
> for(i in levels(c))
318801 y
So, I would like to know the r2 for a~b for every factors levels.
Off course I can made the regression separately for every factors, but my
dataset have 68 factors...
--
Francesco Nutini
PhD student
CNR-IREA (Institute for Electromagnetic Sensing of the Environment)
Milan
this information with excel, but the factor have 68
levels...maybe [r] have a useful command.
Thanks,
Francesco Nutini
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3, grid=T, par.settings =
list(strip.background = list(col = c("gray90"
enclose (?c.trellis):
c( z,double, x.same = TRUE, y.same = T, layout = c(13,2))
Maybe there is no solution but, anyway, thanks for help!
francesco
From: nutini.france...@gmail.com
To: baptiste.aug...@
yplot
Date: Thu, 14 Apr 2011 07:46:31 +
Did you mean this?
http://rgm2.lab.nig.ac.jp/RGM2/func.php?rd_id=latticeExtra:c.trellis
In fact I'm already using latticeExtra package because my xyplot is little bit
complicated...
So, I'm tring, thanks fro tricks baptiste!
Franc
Did you mean this?
http://rgm2.lab.nig.ac.jp/RGM2/func.php?rd_id=latticeExtra:c.trellis
In fact I'm already using latticeExtra package because my xyplot is little bit
complicated...
So, I'm tring, thanks fro tricks baptiste!
Francesco
> Date: Thu, 14 Apr 2011 08:52:45 +1200
>
: |_|_|_|
what i want: | | | |
|_|_|_|
I tried to use the command "par", but it's doesn't work with xyplot. The two
plot have, by default, the same x-axis scale.
I know it's just a "visual solution", but it could be nice for a paper!
Sorry, I mean "heading".
Thanks for the tip.
Francesco
> CC: djmu...@gmail.com; r-help@r-project.org
> From: dwinsem...@comcast.net
> To: ui...@hotmail.it
> Subject: Re: [R] [r] align xyplot
> Date: Tue, 22 Feb 2011 08:33:07 -0500
>
> I don't know the ter
default is
pale-pink, but light-gray is better for a paper.
Thanks,
Francesco
From: nutini.france...@gmail.com
To: djmu...@gmail.com
CC: r-help@r-project.org
Subject: RE: [R] [r] align xyplot
Date: Thu, 3 Feb 2011 10:58:35 +
Thanks Dennis,
I have used the codes as.table = TRUE and
Hi Dennis and [R]users!
as I said last week, I need more info about xyplot.
Is it possible to change the color of the intestation of xyplot? By default is
pale-pink, but light-gray is better for a paper.
Thanks,
Francesco
From: nutini.france...@gmail.com
To: djmu...@gmail.com
CC: r-help@r
Thanks Dennis,
I have used the codes as.table = TRUE and aspect = 1 to obtain what i needed.
Maybe I'll ask you more about xyplot, is very powerfull tool!
Cheers,
Francesco
Date: Mon, 31 Jan 2011 04:33:27 -0800
Subject: Re: [R] [r] align xyplot
From: djmu...@gmail.com
To: nutini.f
Hello everybody,
is there a way to tell Sweave to skip the R code and process only the
LaTeX part?
I know it sounds a little bit weird, but the R code of my document is
complete and _slow_ and I need to add the LaTeX part.
Thanks,
franZ
__
R-help@r-p
ce of graphs?
1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010
I have tried with "par" code but it's doesn't work.
Thanks for help,
Francesco Nutini
[[alterna
if (x*y>0) {...}
On Mon, Jan 24, 2011 at 1:27 PM, Rainer Schuermann
wrote:
> On Monday, January 24, 2011 07:18:03 pm Alaios wrote:
>> Hello :)
>> I wanted to right an expression to check when x and y have the same sign
>> and I wrote the following:
>>
>> if ((x<0 && y<0) || (x>0 && y>0))
>>
>> w
ommand "summary", and
generically about the finest post-hoc test for implemented model with the code
"lm()". I think that R-help is quite insufficient...
Which command did you use normally?
Thanks a lot,
Francesco Nutini
__
R-he
(from smaller to bigger) and
order the correspondent row to the new position.
Is it possible (apart from looping on the index) to do this with some
predefined R function?
Thanks,
Francesco
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R-help@r-project.org mailing list
https://stat.ethz.ch/ma
?
I wanna start with 2D set of points, but the real case scenario is
with a 5D set of points.
Thanks,
Francesco
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org
ce the correlation?
Any suggestions how to perform post-hoc comparions?
Thanks a lot!
Francesco Nutini
P.S. numbers have no significance, it's just an example
[[alternative HTML version deleted]]
> From: fe...@nfrac.org
> Date: Sun, 12 Dec 2010 11:47:55 +1100
> Subject: Re: [R] overlap different line in a xyplot (lattice)
> To: ehl...@ucalgary.ca
> CC: nutini.france...@gmail.com; r-help@r-project.org
>
> On 12 December 2010 00:08, Peter Ehlers wrote:
> > On
n the
dataset, but it's a factorial variables.
How can I use your method?
sorry for my ignorance!
Francesco Nutini
> Date: Fri, 10 Dec 2010 10:13:00 -0800
> From: ehl...@ucalgary.ca
> To: nutini.france...@gmail.com
> CC: r-help@r-project.org
> Subject: Re: [R] [r] overlap diffe
dear [R] users,
is there a way to plot different data (but with the same x-variables) in the
same xyplot window?
There are already a similar question, but the answer is not enought
explanatory...
Thanks a lot,
Francesco
[[alternative HTML
near model. But if I made a scatterplot the r2 is
very low.
How can I interpretate the information of command summary()?
Thank you in advance,
Francesco
#command for summary() of linear model
>summary(model_example)
Call:
lm(formula = dmp ~
Latitude + Longitude + Year + Tot.Prod + RFE.Cu
0.2339
---
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1
If I read the line "var_indipendent:factor" can I understand if the factor
influence significatvly the regression between dependent-indipendent variable?
Thanks a lot!
Francesco Nutini
P.S. numbers have
t the significativity of the variable.
I mean, variables that appear correlated with dependent variable in the matrix
result not correlated in the summary of linear model, and vice versa. Have I
made a mistake in the interpretation of the result, or not?
Thank you in advance,
Francesco
#command for m
Hi,
I was wondering if it would be possible to print the counts of common
items in the venneuler() diagrams.
Anybody knows a straightforward idea?
I can always count things from the data matrix, but I have vectors with
50 thousands elements.
thanks
Francesco
Sorry for the dumb question, but I couldn't figure this out myself.
Consider the following:
> str <- c("abc","def")
> array(str, c(2,1))
[,1]
[1,] "abc"
[2,] "def"
How can i obtain the outcome of the second instruction without
specifyin
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