Hi Micheal, Maybe there is a simple way but i wanted to get the lowest aicc ana i could not find a way to do so, that's why i created the table to store all possible outcomes and then i can easily get the minimum value and the values of (i,j and k) used for that minimum value. The first column in the table is AICc[,1] to store i and second column for j and so on. Maybe i am mistaken and this won't give me what i want, the code been running for 5 hours now. So i am waiting On Wed, Feb 6, 2019 at 4:59 PM Michael Dewey <lists at dewey.myzen.co.uk> wrote:> This is not an answer to your speed problem but are your assignments to > AICc[,1] and so on doing what you hope they are doing? > > Michael > > On 06/02/2019 12:03, salah maadawy wrote: > > i am a beginner regarding R but i am trying to do a simple thing, but it > is > > taking too much time and i am asking if there is any way to achieve what > i > > need, i have a time series data set with 730 data points, i detected 7, > 354 > > and 365 seasonality periods. i am trying to use Fourier terms for > > seasonality and for loop to get the K value for each while minimizing > AICc, > > my code is > > > > AICc<- data.table(matrix(nrow = 96642, ncol = 4))for (i in 1:3) { > > for (j in 1:177) { > > for (k in 182) { #i,j and k values are choosen > > with regad that K cannot exceed seasonality period/2 > > z1 <- fourier(ts(demand,frequency = 7), K=i) > > z2 <- fourier(ts(demand,frequency=354), K=j) > > z3 <- fourier(ts(demand,frequency = 365),K=k) > > fit <- auto.arima(demand, xreg =cbind(z1,z2,z3), > > seasonal = FALSE) > > fit$aicc > > AICc[,1]<-i > > AICc[,2]<-j > > AICc[,3]<-k > > AICc[,4]<-fit$aicc > > } > > > > } > > } > > AICc > > > > i have created a data table to store AICc values from all possible i,j,k > > combinations so that i can find later the minimum AICc value. the problem > > now is that it is taking forever to do so not only to iterate all > > combinations but also due to the large K values. > > > > , is there any possible solution for this? thank you in advance > > > > [[alternative HTML version deleted]] > > > > ______________________________________________ > > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > > 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. > > > > -- > Michael > http://www.dewey.myzen.co.uk/home.html >[[alternative HTML version deleted]]
Well I do not know about data.table but in standard R if you go AICc[,1] <- 3 it fills the whole column with 3 so you will end up with a table with the last value of AICc stored in every row which is almost certainly not what you want. Michael On 06/02/2019 14:15, salah maadawy wrote:> Hi Micheal, Maybe there is a simple way but i wanted to get the lowest > aicc ana i could not find a way to do so, that's why i created the > ?table to store all possible outcomes and then i can easily get the > minimum value and the values of (i,j and k) used for that minimum value. > The first column in the table is AICc[,1] to store i and second column > for j and so on. Maybe i am mistaken and this won't give me what i want, > the code been running for 5 hours now. So i am waiting > > On Wed, Feb 6, 2019 at 4:59 PM Michael Dewey <lists at dewey.myzen.co.uk > <mailto:lists at dewey.myzen.co.uk>> wrote: > > This is not an answer to your speed problem but are your assignments to > AICc[,1] and so on doing what you hope they are doing? > > Michael > > On 06/02/2019 12:03, salah maadawy wrote: > > i am a beginner regarding R but i am trying to do a simple thing, > but it is > > taking too much time and i am asking if there is any way to > achieve what i > > need, i have a time series data set with 730 data points, i > detected 7, 354 > > and 365 seasonality periods. i am trying to use Fourier terms for > > seasonality and for loop to get the K value for each while > minimizing AICc, > > my code is > > > >? ? ? AICc<- data.table(matrix(nrow = 96642, ncol = 4))for (i in > 1:3) { > >? ? for (j in 1:177) { > >? ? ? for (k in 182) {? ? ? ? ? ? ? ? ? ? ?#i,j and k values are > choosen > > with regad that K cannot exceed seasonality period/2 > >? ? ? ? z1 <- fourier(ts(demand,frequency = 7), K=i) > >? ? ? ? z2 <- fourier(ts(demand,frequency=354), K=j) > >? ? ? ? z3 <- fourier(ts(demand,frequency = 365),K=k) > >? ? ? ? fit <- auto.arima(demand, xreg =cbind(z1,z2,z3), > >? ? ? ? ? ?seasonal = FALSE) > >? ? ? ? fit$aicc > >? ? ? ? AICc[,1]<-i > >? ? ? ? AICc[,2]<-j > >? ? ? ? AICc[,3]<-k > >? ? ? ? AICc[,4]<-fit$aicc > >? ? ? } > > > >? ? } > > } > >? ? AICc > > > > i have created a data table to store AICc values from all > possible i,j,k > > combinations so that i can find later the minimum AICc value. the > problem > > now is that it is taking forever to do so not only to iterate all > > combinations but also due to the large K values. > > > > , is there any possible solution for this? thank you in advance > > > >? ? ? ?[[alternative HTML version deleted]] > > > > ______________________________________________ > > R-help at r-project.org <mailto:R-help at r-project.org> mailing list > -- To UNSUBSCRIBE and more, see > > 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. > > > > -- > Michael > http://www.dewey.myzen.co.uk/home.html >-- Michael http://www.dewey.myzen.co.uk/home.html
With 96k model fits it's going to be slow, so you might want to think
first about whether you need to do them all. Beyond that, I think this
is more in the R style, so might be quicker (I don't know how much the
loops are slowing you down), and even if not it should be easier to
adapt.
The other thing to think about is parallelising the code - the
parallel package should help.
FitModel <- function(K, data) {
z1 <- fourier(ts(data,frequency = 7), K=K["i"])
z2 <- fourier(ts(data,frequency=354), K=K["j"])
z3 <- fourier(ts(data,,frequency = 365),K=K["k"])
fit <- auto.arima(data,, xreg =cbind(z1,z2,z3), seasonal = FALSE)
fit$aicc
}
# smaller MaxOrders used so if you run it like this, it won't take 5 hours
MaxOrders <- expand.grid(i = 1:3, j=1=7, k=1:8)
AICc <- apply(MaxOrders, FitModel, data=demand)
Bob
AICc<- data.table(matrix(nrow = 96642, ncol = 4))for (i in 1:3)
{> > for (j in 1:177) {
> > for (k in 182) { #i,j and k values are
choosen
> > with regad that K cannot exceed seasonality period/2
> > z1 <- fourier(ts(demand,frequency = 7), K=i)
> > z2 <- fourier(ts(demand,frequency=354), K=j)
> > z3 <- fourier(ts(demand,frequency = 365),K=k)
> > fit <- auto.arima(demand, xreg =cbind(z1,z2,z3),
> > seasonal = FALSE)
> > fit$aicc
> > AICc[,1]<-i
> > AICc[,2]<-j
> > AICc[,3]<-k
> > AICc[,4]<-fit$aicc
> > }
> >change
> > }
> > }
> > AICc
On Thu, 7 Feb 2019 at 13:44, salah maadawy <salahmaadawy at gmail.com>
wrote:>
> Hi Micheal, Maybe there is a simple way but i wanted to get the lowest aicc
> ana i could not find a way to do so, that's why i created the table to
> store all possible outcomes and then i can easily get the minimum value and
> the values of (i,j and k) used for that minimum value. The first column in
> the table is AICc[,1] to store i and second column for j and so on. Maybe i
> am mistaken and this won't give me what i want, the code been running
for 5
> hours now. So i am waiting
>
> On Wed, Feb 6, 2019 at 4:59 PM Michael Dewey <lists at
dewey.myzen.co.uk>
> wrote:
>
> > This is not an answer to your speed problem but are your assignments
to
> > AICc[,1] and so on doing what you hope they are doing?
> >
> > Michael
> >
> > On 06/02/2019 12:03, salah maadawy wrote:
> > > i am a beginner regarding R but i am trying to do a simple thing,
but it
> > is
> > > taking too much time and i am asking if there is any way to
achieve what
> > i
> > > need, i have a time series data set with 730 data points, i
detected 7,
> > 354
> > > and 365 seasonality periods. i am trying to use Fourier terms for
> > > seasonality and for loop to get the K value for each while
minimizing
> > AICc,
> > > my code is
> > >
> > > AICc<- data.table(matrix(nrow = 96642, ncol = 4))for (i
in 1:3) {
> > > for (j in 1:177) {
> > > for (k in 182) { #i,j and k values are
choosen
> > > with regad that K cannot exceed seasonality period/2
> > > z1 <- fourier(ts(demand,frequency = 7), K=i)
> > > z2 <- fourier(ts(demand,frequency=354), K=j)
> > > z3 <- fourier(ts(demand,frequency = 365),K=k)
> > > fit <- auto.arima(demand, xreg =cbind(z1,z2,z3),
> > > seasonal = FALSE)
> > > fit$aicc
> > > AICc[,1]<-i
> > > AICc[,2]<-j
> > > AICc[,3]<-k
> > > AICc[,4]<-fit$aicc
> > > }
> > >
> > > }
> > > }
> > > AICc
> > >
> > > i have created a data table to store AICc values from all
possible i,j,k
> > > combinations so that i can find later the minimum AICc value. the
problem
> > > now is that it is taking forever to do so not only to iterate all
> > > combinations but also due to the large K values.
> > >
> > > , is there any possible solution for this? thank you in advance
> > >
> > > [[alternative HTML version deleted]]
> > >
> > > ______________________________________________
> > > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more,
see
> > > 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.
> > >
> >
> > --
> > Michael
> > http://www.dewey.myzen.co.uk/home.html
> >
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.
--
Bob O'Hara
Institutt for matematiske fag
NTNU
7491 Trondheim
Norway
Mobile: +47 915 54 416
Journal of Negative Results - EEB: www.jnr-eeb.org
thank you all for your suggestions and feedback, with further search and experimentation, the problem is with the auto.arima function with large k values, it takes 4 min to compute one model with k=30 and the time increases with K, so i used your suggestions for collecting the output but i limited my loop (i=3,j=25 and k=25) , it took around 17 hours to finish already (1875 models). thank you again. On Thu, Feb 7, 2019 at 4:10 PM Bob O'Hara <rni.boh at gmail.com> wrote:> With 96k model fits it's going to be slow, so you might want to think > first about whether you need to do them all. Beyond that, I think this > is more in the R style, so might be quicker (I don't know how much the > loops are slowing you down), and even if not it should be easier to > adapt. > > The other thing to think about is parallelising the code - the > parallel package should help. > > FitModel <- function(K, data) { > z1 <- fourier(ts(data,frequency = 7), K=K["i"]) > z2 <- fourier(ts(data,frequency=354), K=K["j"]) > z3 <- fourier(ts(data,,frequency = 365),K=K["k"]) > fit <- auto.arima(data,, xreg =cbind(z1,z2,z3), seasonal = FALSE) > fit$aicc > } > > # smaller MaxOrders used so if you run it like this, it won't take 5 hours > MaxOrders <- expand.grid(i = 1:3, j=1=7, k=1:8) > AICc <- apply(MaxOrders, FitModel, data=demand) > > Bob > > AICc<- data.table(matrix(nrow = 96642, ncol = 4))for (i in 1:3) { > > > for (j in 1:177) { > > > for (k in 182) { #i,j and k values are choosen > > > with regad that K cannot exceed seasonality period/2 > > > z1 <- fourier(ts(demand,frequency = 7), K=i) > > > z2 <- fourier(ts(demand,frequency=354), K=j) > > > z3 <- fourier(ts(demand,frequency = 365),K=k) > > > fit <- auto.arima(demand, xreg =cbind(z1,z2,z3), > > > seasonal = FALSE) > > > fit$aicc > > > AICc[,1]<-i > > > AICc[,2]<-j > > > AICc[,3]<-k > > > AICc[,4]<-fit$aicc > > > } > > >change > > > } > > > } > > > AICc > > On Thu, 7 Feb 2019 at 13:44, salah maadawy <salahmaadawy at gmail.com> wrote: > > > > Hi Micheal, Maybe there is a simple way but i wanted to get the lowest > aicc > > ana i could not find a way to do so, that's why i created the table to > > store all possible outcomes and then i can easily get the minimum value > and > > the values of (i,j and k) used for that minimum value. The first column > in > > the table is AICc[,1] to store i and second column for j and so on. > Maybe i > > am mistaken and this won't give me what i want, the code been running > for 5 > > hours now. So i am waiting > > > > On Wed, Feb 6, 2019 at 4:59 PM Michael Dewey <lists at dewey.myzen.co.uk> > > wrote: > > > > > This is not an answer to your speed problem but are your assignments to > > > AICc[,1] and so on doing what you hope they are doing? > > > > > > Michael > > > > > > On 06/02/2019 12:03, salah maadawy wrote: > > > > i am a beginner regarding R but i am trying to do a simple thing, > but it > > > is > > > > taking too much time and i am asking if there is any way to achieve > what > > > i > > > > need, i have a time series data set with 730 data points, i detected > 7, > > > 354 > > > > and 365 seasonality periods. i am trying to use Fourier terms for > > > > seasonality and for loop to get the K value for each while minimizing > > > AICc, > > > > my code is > > > > > > > > AICc<- data.table(matrix(nrow = 96642, ncol = 4))for (i in 1:3) > { > > > > for (j in 1:177) { > > > > for (k in 182) { #i,j and k values are > choosen > > > > with regad that K cannot exceed seasonality period/2 > > > > z1 <- fourier(ts(demand,frequency = 7), K=i) > > > > z2 <- fourier(ts(demand,frequency=354), K=j) > > > > z3 <- fourier(ts(demand,frequency = 365),K=k) > > > > fit <- auto.arima(demand, xreg =cbind(z1,z2,z3), > > > > seasonal = FALSE) > > > > fit$aicc > > > > AICc[,1]<-i > > > > AICc[,2]<-j > > > > AICc[,3]<-k > > > > AICc[,4]<-fit$aicc > > > > } > > > > > > > > } > > > > } > > > > AICc > > > > > > > > i have created a data table to store AICc values from all possible > i,j,k > > > > combinations so that i can find later the minimum AICc value. the > problem > > > > now is that it is taking forever to do so not only to iterate all > > > > combinations but also due to the large K values. > > > > > > > > , is there any possible solution for this? thank you in advance > > > > > > > > [[alternative HTML version deleted]] > > > > > > > > ______________________________________________ > > > > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > > > > 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. > > > > > > > > > > -- > > > Michael > > > http://www.dewey.myzen.co.uk/home.html > > > > > > > [[alternative HTML version deleted]] > > > > ______________________________________________ > > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > > 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. > > > > -- > Bob O'Hara > Institutt for matematiske fag > NTNU > 7491 Trondheim > Norway > > Mobile: +47 915 54 416 > Journal of Negative Results - EEB: www.jnr-eeb.org >[[alternative HTML version deleted]]