Hello,
Also, in the code
x <- data.matrix(Ozone_weekly)
[...omited...]
for(i in 1:nrow(x))
+ { for(j in 1:ncol(x))
+ {x[i,j] = 1}}
not only you rewrite x but the double for loop is equivalent to
x[] <- 1
courtesy R's vectorised behavior. (The square parenthesis are needed to
keep the dimensions, the matrix form.)
And, I'm not sure but isn't
head(gev.fit)[1:4]
equivalent to
head(gev.fit, n = 4)
?
Like Jim says, we need more information, can you post Ozone_weekly2 and
the code that produced gev.fit? But in the mean time you can revise your
code.
Hope this helps,
Rui Barradas
?s 11:08 de 08/07/21, Jim Lemon escreveu:> Hi Siti,
> I think we need a bit more information to respond helpfully. I have no
> idea what "Ozone_weekly2" is and Google is also ignorant.
"gev.fit" is
> also unknown. The name suggests that it is the output of some
> regression or similar. What function produced it, and from what
> library? "ti" is known as you have defined it. However, I
don't know
> what you want to do with it. Finally, as this is a text mailing list,
> we don't get any highlighting, so the text to which you refer cannot
> be identified. I can see you have a problem, but cannot offer any help
> right now.
>
> Jim
>
> On Thu, Jul 8, 2021 at 12:06 AM SITI AISYAH ZAKARIA
> <aisyahzakaria at unimap.edu.my> wrote:
>>
>> Dear all,
>>
>> Can I ask something about programming in marginal distribution for
spatial
>> extreme?
>> I really stuck on my coding to obtain the parameter estimation for
>> univariate or marginal distribution for new model in spatial extreme.
>>
>> I want to run my data in order to get the parameter estimation value
for 25
>> stations in one table. But I really didn't get the idea of the
correct
>> coding. Here I attached my coding
>>
>> x <- data.matrix(Ozone_weekly2)
>> x
>> head(gev.fit)[1:4]
>> ti = matrix(ncol = 3, nrow = 888)
>> ti[,1] = seq(1, 888, 1)
>> ti[,2]=sin(2*pi*(ti[,1])/52)
>> ti[,3]=cos(2*pi*(ti[,1])/52)
>> for(i in 1:nrow(x))
>> + { for(j in 1:ncol(x))
>> + {x[i,j] = 1}}
>>
>> My problem is highlighted in red color.
>> And if are not hesitate to all. Can someone share with me the
procedure,
>> how can I map my data using spatial extreme.
>> For example:
>> After I finish my marginal distribution, what the next procedure. It is
I
>> need to get the spatial independent value.
>>
>> That's all
>> Thank you.
>>
>> --
>>
>>
>>
>>
>>
>> "..Millions of trees are used to make papers, only to be thrown
away
>> after a couple of minutes reading from them. Our planet is at stake.
Please
>> be considerate. THINK TWICE BEFORE PRINTING THIS.."
>>
>> DISCLAIMER: This email \ and any files transmitte...{{dropped:24}}
>>
>> ______________________________________________
>> 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.
>
> ______________________________________________
> 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.
>
SITI AISYAH ZAKARIA
2021-Jul-08 23:59 UTC
[R] How to estimate the parameter for many variable?
Dear all,
Thank you very much for the feedback.
Sorry for the lack of information about this problem.
Here, I explain again.
I use this package to run my coding.
library(ismev)
library(mgcv)
library(nlme)
The purpose of this is I want to get the value of parameter estimation
using MLE by applying the GEV distribution.
x <- data.matrix(Ozone_weekly2) x refers to my data
that consists of 19 variables. I will attach the data together.
x
head(gev.fit)[1:4]
ti = matrix(ncol = 3, nrow = 888)
ti[,1] = seq(1, 888, 1)
ti[,2]=sin(2*pi*(ti[,1])/52)
ti[,3]=cos(2*pi*(ti[,1])/52)
*for(i in 1:nrow(x)) + { for(j in 1:ncol(x))
the problem in here, i don't no to create the coding. i target my
output will come out in matrix that + {x[i,j] = 1}}
show the parameter estimation for 19 variable which have
19 row and 3 column*
* row -- refer
to variable (station) ; column -- refer to parameter estimation for GEV
distribution*thank you.
On Thu, 8 Jul 2021 at 18:40, Rui Barradas <ruipbarradas at sapo.pt> wrote:
> Hello,
>
> Also, in the code
>
> x <- data.matrix(Ozone_weekly)
>
> [...omited...]
>
> for(i in 1:nrow(x))
> + { for(j in 1:ncol(x))
> + {x[i,j] = 1}}
>
> not only you rewrite x but the double for loop is equivalent to
>
>
> x[] <- 1
>
>
> courtesy R's vectorised behavior. (The square parenthesis are needed to
> keep the dimensions, the matrix form.)
> And, I'm not sure but isn't
>
> head(gev.fit)[1:4]
>
> equivalent to
>
> head(gev.fit, n = 4)
>
> ?
>
> Like Jim says, we need more information, can you post Ozone_weekly2 and
> the code that produced gev.fit? But in the mean time you can revise your
> code.
>
> Hope this helps,
>
> Rui Barradas
>
>
> ?s 11:08 de 08/07/21, Jim Lemon escreveu:
> > Hi Siti,
> > I think we need a bit more information to respond helpfully. I have no
> > idea what "Ozone_weekly2" is and Google is also ignorant.
"gev.fit" is
> > also unknown. The name suggests that it is the output of some
> > regression or similar. What function produced it, and from what
> > library? "ti" is known as you have defined it. However, I
don't know
> > what you want to do with it. Finally, as this is a text mailing list,
> > we don't get any highlighting, so the text to which you refer
cannot
> > be identified. I can see you have a problem, but cannot offer any help
> > right now.
> >
> > Jim
> >
> > On Thu, Jul 8, 2021 at 12:06 AM SITI AISYAH ZAKARIA
> > <aisyahzakaria at unimap.edu.my> wrote:
> >>
> >> Dear all,
> >>
> >> Can I ask something about programming in marginal distribution for
> spatial
> >> extreme?
> >> I really stuck on my coding to obtain the parameter estimation for
> >> univariate or marginal distribution for new model in spatial
extreme.
> >>
> >> I want to run my data in order to get the parameter estimation
value
> for 25
> >> stations in one table. But I really didn't get the idea of the
correct
> >> coding. Here I attached my coding
> >>
> >> x <- data.matrix(Ozone_weekly2)
> >> x
> >> head(gev.fit)[1:4]
> >> ti = matrix(ncol = 3, nrow = 888)
> >> ti[,1] = seq(1, 888, 1)
> >> ti[,2]=sin(2*pi*(ti[,1])/52)
> >> ti[,3]=cos(2*pi*(ti[,1])/52)
> >> for(i in 1:nrow(x))
> >> + { for(j in 1:ncol(x))
> >> + {x[i,j] = 1}}
> >>
> >> My problem is highlighted in red color.
> >> And if are not hesitate to all. Can someone share with me the
procedure,
> >> how can I map my data using spatial extreme.
> >> For example:
> >> After I finish my marginal distribution, what the next procedure.
It is
> I
> >> need to get the spatial independent value.
> >>
> >> That's all
> >> Thank you.
> >>
> >> --
> >>
> >>
> >>
> >>
> >>
> >> "..Millions of trees are used to make papers, only to be
thrown away
> >> after a couple of minutes reading from them. Our planet is at
stake.
> Please
> >> be considerate. THINK TWICE BEFORE PRINTING THIS.."
> >>
> >> DISCLAIMER: This email \ and any files transmitte...{{dropped:24}}
> >>
> >> ______________________________________________
> >> 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.
> >
> > ______________________________________________
> > 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.
> >
>
--
"..Millions of trees are used to make papers, only to be thrown away
after a couple of minutes reading from them. Our planet is at stake. Please
be considerate. THINK TWICE BEFORE PRINTING THIS.."
DISCLAIMER:?This email
and any files transmitted with it are confidential and intended solely for
the use of the individual orentity to whom they are addressed. If you have
received this email in error please notify the UniMAP's Email
Administrator. Please note that any views or opinions presented in this
email are solely those of the author and do not necessarily represent those
of the university. Finally, the recipient should check this email and any
attachments for the presence of viruses.The university accepts no liability
for any damage caused by any virus transmitted by this email.
Universiti
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MALAYSIA |?www.unimap.edu.my <http://www.unimap.edu.my/>
-------------- next part --------------
An embedded and charset-unspecified text was scrubbed...
Name: Ozone_weekly2.txt
URL:
<https://stat.ethz.ch/pipermail/r-help/attachments/20210709/1038141e/attachment.txt>
Hi Siti,
I think some progress has been made. You have a data set with 888 rows
and 19 columns:
ow2<-read.table("Ozone_weekly2.txt",
header=TRUE)
dim(ow2)
[1] 888 19
The values may be parts per million ozone in the atmosphere. The
columns may represent different measuring locations and my guess is
that rows are times of measurement, perhaps in weeks given the name.
If so, you might have run ismev::gev.fit() on all of the columns,
looking for estimated maxima at each location. Is this at all close to
what you are doing?
Jim
On Fri, Jul 9, 2021 at 9:59 AM SITI AISYAH ZAKARIA
<aisyahzakaria at unimap.edu.my> wrote:>
> Dear all,
>
> Thank you very much for the feedback.
>
> Sorry for the lack of information about this problem.
>
> Here, I explain again.
>
> I use this package to run my coding.
>
> library(ismev)
> library(mgcv)
> library(nlme)
>
> The purpose of this is I want to get the value of parameter estimation
using MLE by applying the GEV distribution.
>
> x <- data.matrix(Ozone_weekly2) x refers to my data
that consists of 19 variables. I will attach the data together.
> x
> head(gev.fit)[1:4]
> ti = matrix(ncol = 3, nrow = 888)
> ti[,1] = seq(1, 888, 1)
> ti[,2]=sin(2*pi*(ti[,1])/52)
> ti[,3]=cos(2*pi*(ti[,1])/52)
>
> for(i in 1:nrow(x))
> + { for(j in 1:ncol(x)) the problem in here, i
don't no to create the coding. i target my output will come out in matrix
that
> + {x[i,j] = 1}} show the
parameter estimation for 19 variable which have 19 row and 3 column
> row -- refer
to variable (station) ; column -- refer to parameter estimation for GEV
distribution
>
> thank you.
>
> On Thu, 8 Jul 2021 at 18:40, Rui Barradas <ruipbarradas at sapo.pt>
wrote:
>>
>> Hello,
>>
>> Also, in the code
>>
>> x <- data.matrix(Ozone_weekly)
>>
>> [...omited...]
>>
>> for(i in 1:nrow(x))
>> + { for(j in 1:ncol(x))
>> + {x[i,j] = 1}}
>>
>> not only you rewrite x but the double for loop is equivalent to
>>
>>
>> x[] <- 1
>>
>>
>> courtesy R's vectorised behavior. (The square parenthesis are
needed to
>> keep the dimensions, the matrix form.)
>> And, I'm not sure but isn't
>>
>> head(gev.fit)[1:4]
>>
>> equivalent to
>>
>> head(gev.fit, n = 4)
>>
>> ?
>>
>> Like Jim says, we need more information, can you post Ozone_weekly2 and
>> the code that produced gev.fit? But in the mean time you can revise
your
>> code.
>>
>> Hope this helps,
>>
>> Rui Barradas
>>
>>
>> ?s 11:08 de 08/07/21, Jim Lemon escreveu:
>> > Hi Siti,
>> > I think we need a bit more information to respond helpfully. I
have no
>> > idea what "Ozone_weekly2" is and Google is also
ignorant. "gev.fit" is
>> > also unknown. The name suggests that it is the output of some
>> > regression or similar. What function produced it, and from what
>> > library? "ti" is known as you have defined it. However,
I don't know
>> > what you want to do with it. Finally, as this is a text mailing
list,
>> > we don't get any highlighting, so the text to which you refer
cannot
>> > be identified. I can see you have a problem, but cannot offer any
help
>> > right now.
>> >
>> > Jim
>> >
>> > On Thu, Jul 8, 2021 at 12:06 AM SITI AISYAH ZAKARIA
>> > <aisyahzakaria at unimap.edu.my> wrote:
>> >>
>> >> Dear all,
>> >>
>> >> Can I ask something about programming in marginal distribution
for spatial
>> >> extreme?
>> >> I really stuck on my coding to obtain the parameter estimation
for
>> >> univariate or marginal distribution for new model in spatial
extreme.
>> >>
>> >> I want to run my data in order to get the parameter estimation
value for 25
>> >> stations in one table. But I really didn't get the idea of
the correct
>> >> coding. Here I attached my coding
>> >>
>> >> x <- data.matrix(Ozone_weekly2)
>> >> x
>> >> head(gev.fit)[1:4]
>> >> ti = matrix(ncol = 3, nrow = 888)
>> >> ti[,1] = seq(1, 888, 1)
>> >> ti[,2]=sin(2*pi*(ti[,1])/52)
>> >> ti[,3]=cos(2*pi*(ti[,1])/52)
>> >> for(i in 1:nrow(x))
>> >> + { for(j in 1:ncol(x))
>> >> + {x[i,j] = 1}}
>> >>
>> >> My problem is highlighted in red color.
>> >> And if are not hesitate to all. Can someone share with me the
procedure,
>> >> how can I map my data using spatial extreme.
>> >> For example:
>> >> After I finish my marginal distribution, what the next
procedure. It is I
>> >> need to get the spatial independent value.
>> >>
>> >> That's all
>> >> Thank you.
>> >>
>> >> --
>> >>
>> >>
>> >>
>> >>
>> >>
>> >> "..Millions of trees are used to make papers, only to be
thrown away
>> >> after a couple of minutes reading from them. Our planet is at
stake. Please
>> >> be considerate. THINK TWICE BEFORE PRINTING THIS.."
>> >>
>> >> DISCLAIMER: This email \ and any files
transmitte...{{dropped:24}}
>> >>
>> >> ______________________________________________
>> >> 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.
>> >
>> > ______________________________________________
>> > 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.
>> >
>
>
>
> "..Millions of trees are used to make papers, only to be thrown away
after a couple of minutes reading from them. Our planet is at stake. Please be
considerate. THINK TWICE BEFORE PRINTING THIS.."
>
> DISCLAIMER: This email and any files transmitted with it are confidential
and intended solely for the use of the individual orentity to whom they are
addressed. If you have received this email in error please notify the
UniMAP's Email Administrator. Please note that any views or opinions
presented in this email are solely those of the author and do not necessarily
represent those of the university. Finally, the recipient should check this
email and any attachments for the presence of viruses.The university accepts no
liability for any damage caused by any virus transmitted by this email.
>
> Universiti Malaysia Perlis (UniMAP) | Digital Management & Development
Centre (DMDC), Universiti Malaysia Perlis (UniMAP), Pauh Putra Campus, 02600
Arau, Perlis, MALAYSIA | www.unimap.edu.my