n the underlying methods see:
https://arxiv.org/abs/1809.10632
The package is also available on github:
https://github.com/mfasiolo/mgcViz
so feel free to send comments, criticisms and feature requests via Github
issues.
Matteo Fasiolo
[[alternative HTML version de
://github.com/mfasiolo/qgam
Best,
Matteo Fasiolo
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Thanks Sarah (and all the others who replied) for your precious
suggestions! Matteo
On 22 November 2016 at 14:18, Sarah Goslee wrote:
> Here's how to get one:
>
> x <- c(9,9,1,3,2,7,6,10,5,6)
>> which.min(abs(x - quantile(x, .25)))
> [1] 4
>
> And here's one
what matters - in case the quantile is exactly half way
through).
This is an example.
> x <- c(9,9,1,3,2,7,6,10,5,6)
> quantile(x)
0% 25% 50% 75% 100%
1.0 3.5 6.0 8.5 10.0
What I need is a vector 'index' which looks like
> index
3 4 7 1 8
Many thanks
d.]
The trick works only for v[2]. I imagine this is because the vector is
not dynamically updated during the iteration, so all values v[2:10]
are retained as NA.
How can I solve the problem, without using a 'for' cycle?
Thanks for your help ! Matteo
___
nction
foo = function(i,j,h){
mycube[i,j,h] <<- i*j*h
}
# update
for(h in 1:H){
# males:
for(i in 1:I)
for(j in 1:J)
foo(i,j,h)
}
Thanks a lot for your help. Matteo
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Hi Daniel,
thanks for your answer. How can I populate the array with the
matrixes? Suppose I want to populate it with 10 matrixes
matrix(NA,5,5)
Matteo
On 15 January 2016 at 22:26, Dalthorp, Daniel wrote:
> How about: D<-array(dim=c(d1, d2, d3))?
>
>
>
> On Fri, Jan 1
but then D[i] returns a number, and not a matrix.
I imagine this is a dummy question, but I did search for an answer on
various R help sites, and found nothing straightforward. Being an
inexperienced R user, I prefer a simple solution, even at some efficiency
cost.
Many thanks for your help.
Mat
ot;"))
> M[i,j] = value
> assign(paste("M_",s,sep=""),M, envir = .GlobalEnv)
> }
>
> foo2("a",1,2,15)
>
> cheers
> Peter
>
> > On 23 Dec 2015, at 09:44, Matteo Richiardi
> wrote:
> >
> > I am following the ex
"),value, envir = .GlobalEnv)
}
but this produces the problem I described above (namely, a new
variable is created, rather than replacing the specified element of
the matrix).
I know that in this example I could simply pass the matrix as an
argument to the function, but I
Hi David,
that's a great answer, thanks so much. I imagined apply() was involved in
the solution, but I was unable to find how myself. Thanks again.
Matteo
On 17 December 2015 at 01:37, David Winsemius
wrote:
>
> > On Dec 16, 2015, at 5:34 PM, David Winsemius
> wrote:
>
;
> On December 16, 2015 4:18:56 PM PST, Matteo Richiardi <
> matteo.richia...@gmail.com> wrote:
>
>> I have to evolve each element of a matrix W
>>
>> W <- matrix(0,2,3)
>>
>> according to some function which uses the indices of the matrix [i,j] as
>&g
How can I do it, without recurring to a loop? Also, in my application I
need to pass the function another argument.
Thanks a lot for your suggestions.
Matteo
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Hi,
thanks a lot to everybody for your help. Very nice suggestions!
Matteo
On 16 December 2015 at 12:53, Matteo Richiardi
wrote:
> I apologise for this very basic question, but I found no answers on the
> web.
> I want to create a data.table with n columns, named i1 ... i'n
I apologise for this very basic question, but I found no answers on the web.
I want to create a data.table with n columns, named i1 ... i'n', with only
one row with value = 100 for every variable.
I can do it "by hand":
M.dt <- data.table(i1=100,i2=100,i3=100)
but I would like to make it in a lo
33708 0.5612197 0.6882252 0.8183746 0.7687452
This is the error I get:
Error in rank(x, ties.method = "min", na.last = "keep") :
unimplemented type 'list' in 'greater'
Matteo
On 3 October 2015 at 09:16, Matteo Richiardi
wrote:
> I have several est
I have several estimated time series, running from 2013 to 2050. 'y' values
are constrained between 0 and 1. I would like to plot them using shaded
colours of decreasing intensity, depending on an estimated density at each
point x in 2013-2050.
This is what I have done:
require(denstrip)
x <- 201
Hello R users,
I am trying to run the command rbind gathering the name of tables to bind
from a list argument.
Unfortunately I am not able to obtain it.
I would appreciate any suggestions. Below is a reproducible example with
the problem.
Thanks
Matteo
5d7893bf
And here is the code (the error occurs after executing the final line,
which is also a toy model to be estimated through the bsar function):
https://gist.github.com/lessermatter/0284be117a19620750aa
Any ideas?
Kind regards,
Matteo Villa
Università de
ine, which is also a toy model to be estimated through the bsar function):
https://gist.github.com/lessermatter/0284be117a19620750aa
Any ideas?
Kind regards,
Matteo Villa
Università degli Studi di Milano
Italy
[[alternative HTML version deleted]]
___
plot.new()
abline(0,1)
abline(1,1)
# or as Michael suggest
plot(1:10, 1:10, type = "n")
# then
abline(0,1)
abline(1,1)
Best
Matteo Murenu,
Cagliari, IT
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of David
Winsemius
Sent: 05 December 201
but
still clear and readable alternative?
Thanks a lot. Matteo
rm(list=ls())
library(plyr)
y0 = 0
lambda = 0.1
N = 20
T = 100
m_e = 0
sd_e = 1
# construct the data frame and initialize y
D = data.frame(
id = rep(1:N,T),
t = rep(1:T, each = N),
y = rep(y0,N*T)
)
# update y
for(t in 2:T
Dear all,
I am trying to use the nonlinear quantile regression which involves copula
functions. Following the Frank copula example provided in the "quantreg"
vignette I try do do the same using the Normal (Gaussian) copula. The
problem is that the "nlrq" algorithm stops by giving the following err
of different size classes, depending on the amount of available
food. Nevertheless the biomass does not reach steady state, and increases
exponentially even with parameters set to 0 (which in theory should result
in a biomass value of 0)
Thank you very much for your help!
I appreciate it
Matteo
11
x3138
Sorry if I'm not able to explain it better and as I said I'm very new to
R.
Thanks
Matteo
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PLEASE do read the posting gui
Hi,
result <- lapply(unique(dataset$ID), function(uid) dataset[dataset$ID
== uid, ])
Ok, I have the element result as a list
names(result) <- unique(dataset$ID)
Nothing happens. I don't have any submatrix...
Matteo
__
R-help@r-
every point of sampling.
I can do it manually, e.g. x1<-data.frame(dataset[dataset$ID=="x1",])
but is it possible to write a script and let do it to R?
So i got n submatrix of the n ID found in the original columns.
Cheers
Matteo
__
R-
I all,
I'm new on this list, so sorry if I do something wrong. Anyway, I have a
problem with the installation of the package StatDA.
Typing install.package("StatDA") I receive the error:
ERROR: configuration failed for package ‘MBA’
* removing ‘/home/matteo/R/i686-pc-linux-gnu-
c('mean'))
Error in Getlimi(name[i], Limval, need.all = TRUE) :
no limits defined by datadist for variable Age
--
can someone give me any advice about this problem and to create nice plot
with x axis a continuous predictor and y axis the predicted survival?
Thank you
Dr. Matteo Pettin
I also had a lot of problems with the
ftplistonly=T option so I decided to use the saver version FALSE and split it
by my own.
Matteo
MODIS:::filesUrl
filesUrl <- function(url)
{
require(RCurl)
if (substr(url,nchar(url),nchar(url))!="/")
{
url <- paste0(u
bject
is load()-ed on every need. That should be fairly fast as the object itself is
very small and it is not needed to often.
Thanks
Matteo
On 28/05/2013 08:20, Matteo Mattiuzzi wrote:
> Dear List,
>
>
> I'm writing the package MODIS (currently on R-forge), I lazy load s
rnal library gdal-bin. So
it takes only _once_ the time to produce that object that is from now on
available for any other function in the package environment.
I hope I was clear, thanks in advanced,
Matteo Mattiuzzi
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R-help@r-proje
77840-4352
>
>
>
>
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On
> Behalf Of Matteo Mura
> Sent: Wednesday, April 24, 2013 7:57 AM
> To: r-help@r-project.org
> Subject: [R] Sum up column values acc
are already ordered by id ascending and g descending using:
ipso <- ipso[with(ipso, order(ipso$id, -ipso$g)), ] # Order for id
ascending and g descending
I tried a lot of for loops and tapply() without results.
Can anyone help me to solve this?
Thanks for your attention
Best whishes
.DT
0.34447791 is the fifth value in B.DT, and it is left unassigned
because it is too big.
This is of course a simplified problem (and therefore may not make
much sense). Thanks for your time and input.
Matteo
[[alternative HTML version deleted]]
.DT
0.34447791 is the fifth value in B.DT, and it is left unassigned
because it is too big.
This is of course a simplified problem (and therefore may not make
much sense). Thanks for your time and input.
Matteo
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https
or:
> Error in if (sigma2$id < 0) stop(paste("the estimated variance of the", :
> missing value where TRUE/FALSE needed
Thanks
Matteo Ognibene
> Subject: Re: [R] plm random effect: the estimated variance of the individ
the same random effect model in Stata and it works... (and
for the within model the results are identical in R and Stata)
I'll be really gratefull if someone can help me.
Thanks,
Matteo
> CC: r-help@r-project.org
> From: dwinsem...@comcast.net
> To: ognibenemat...@hotmail.it
>
"))
I want to point the all the code works perfectly in a within model, and I tried
using all the random.method for the variance estimation.In addition I used the
same model in Stata and it works... (the results are identical for a within
mo
), model="within")
my output: Coefficients : Estimate Std. Error t-value Pr(>|t|)
L1.deltaS -0.063576 0.017950 -3.5419 0.0004031 ***L2.deltaS -0.068272
0.017966 -3.8001 0.0001474 ***---
I tried adding "-0", "+1", "intercept=TRUE"
Thanks
ften happens that I get a errorcode 550 an I'm not able to get
the information. If I put the very same adress to a browser I get te content,
so the connection is working.
Thanks in advance Matteo
library(RCurl)
filelisturl <- "ftp://e4ftl01.cr.usgs.gov/MOLT/MOD13Q1.005/200
t;
> On Fri, Jan 13, 2012 at 11:30 AM, matteo dossena wrote:
>> #Dear All,
>> #I'm having a bit of a trouble here, please help me...
>> #I have this data
>> set.seed(4)
>> mydata <- data.frame(var = rnorm(100),
>>
},
key = list(space = "right",
text = list(c("trt1", "trt2"), cex = 0.8),
points = list(pch = c(21), fill = c("black", "grey")),
rep = FALSE)
Thanks David and Patrick,
really appreciate tour help...
m.
Il giorno 14 Dec 2011, alle ore 18:33, David Winsemius ha scritto:
>
> On Dec 14, 2011, at 11:43 AM, matteo wrote:
>
>> set.seed(3)
>> mydata <- data.frame(var = rnorm(100,20,1),
>>
nd so on.
Any suggestions...
Many thanks
matteo
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View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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Hi,
I'm trying to use StatEt IDE for Eclipse as my R editor, but I'm completely
lost. I've read all I could find online, made apparently all I had to do
(installing rj, configuraing StatEt, etc.) but still cannot make R running.
Below is the error log file.
Thank you so much for ass
Hi,
I'm trying to use StatEt IDE for Eclipse as my R editor, but I'm completely
lost. I've read all I could find online, made apparently all I had to do
(installing rj, configuraing StatEt, etc.) but still cannot make R running.
Below is the error log file.
Thank you so much for ass
I don't want to go for
a matrix, because I'd loose the column names and everything will
become too much error-prone.
Any suggestions on how to do it?
Thanks in advance,
Matteo
--
Matteo Richiardi
University of Turin
Faculty of Law
Department of Economics "Cognetti De Mart
ata.frame in a ts or a xts object, but got
really lost there. I'd really apreciate help...
Thank you,
Matteo
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PLEASE do read the posting guide http://www.R-project
Thanks again,
really appreciate that.
Il giorno 18 Nov 2011, alle ore 23:06, Ben Bolker ha scritto:
> [cc'ing back to r-help]
>
> On Fri, Nov 18, 2011 at 4:39 PM, matteo dossena
> wrote:
>> Thanks a lot,
>>
>> just to make sure i got it right,
>>
write a 'fully functional'
script, though, because of quoting troubles and so on.
best regards,
--
Matteo Giani
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PLEASE do read the posting guide http://
Dear R team
I have a problem with the installation of the RCOR package; any time I try to
install it I get the error message: Error: package 'ROCR' was built before R
2.10.0: please re-install it.
What can I do to solve this problem?
Thank you
Kind regards
Dr. Matteo Pettin
model)
cat(forecast(air.model2, h=2)$mean, '\n')
}
But this approach, even if not fitting the model from scratch, is still very
slow for big timeseries.
Some hints to speed it up?
Thanks,
Matteo
[[alternative HTML version deleted]]
__
> Matteo Bertini schrieb:
>>
>> Hello everyone,
>>
>> I'm doing some benchmark comparing Arima [1] and SVR on time series data.
>>
>> I'm using an out-of-sample one-step-ahead prediction from Arima using
>> the "fitted" method
Hello everyone,
I'm doing some benchmark comparing Arima [1] and SVR on time series data.
I'm using an out-of-sample one-step-ahead prediction from Arima using
the "fitted" method [2].
Do someone know how to have a two-steps-ahead forecast timeseries from Arima?
Thanks
er "dates") but after
changing, "for" gave no more output. And I don't know why!!!
Why i use "for" twice? Because after the last file there is no "|"...
Thanks for help
Matteo
[[alternative HTML version deleted]]
_
e it takes is multiplied by this factor! (so getting this with
"rowSums(is.na(x))" it takes much to long)
I was trying to look inside the function maybe to get not only the
means but the valid "n" too, but I wasn't able.
Thanks for help Matteo
DI Matteo Mattiuzzi,
Inst
I'd like to fit a SARIMA model on a timeseries but the period I'd like to
use is too big (7 day in 15min samples = 672) for the algorithm used in R.
Some suggested alternatives?
Thanks,
Matteo Bertini
[[alternative HTML version deleted]]
__
Using the gretl.sf.net dialect:
Static forecasts are one step ahead, based on realized values from the previous
period, while dynamic forecasts employ the chain rule of forecasting.
What kind of forecast is doing forecast.Arima?
Thanks,
Matteo Bertini
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