# [R] How to estimate the parameter for many variable?

SITI AISYAH ZAKARIA @|@y@hz@k@r|@ @end|ng |rom un|m@p@edu@my
Sat Jul 10 01:30:35 CEST 2021

```Dear Rui and Jim,

Thank you very much for your feedback.

Yes, now I get the output. And after this I will use this output as the
marginal distribution to continue the analysis on spatial extremes.

Thanks again. See you later.

On Fri, 9 Jul 2021 at 17:18, Rui Barradas <ruipbarradas using sapo.pt> wrote:

> Hello,
>
> With the condition for the location it can be estimated like the following.
>
>
> fit_list2 <- gev_fit_list <- lapply(Ozone_weekly2, gev.fit, ydat = ti,
> mul = c(2, 3), show = FALSE)
> mle_params2 <- t(sapply(fit_list2, '[[', 'mle'))
> # assign column names
> colnames(mle_params2) <- c("location", "scale", "shape", "mul2", "mul3")
>
>
> Hope this helps,
>
>
> Às 09:53 de 09/07/21, SITI AISYAH ZAKARIA escreveu:
> > Dear Rui ang Jim,
> >
> > Thank you very much.
> >
> > grateful I got the answer.
> >
> > ok now, I have some condition on the location parameter which is cyclic
> > condition.
> >
> > So, I will add another 2 variables for the column.
> >
> > and the condition for location is this one.
> >
> >   ti[,1] = seq(1, 888, 1)
> >    ti[,2]=sin(2*pi*(ti[,1])/52)
> >    ti[,3]=cos(2*pi*(ti[,1])/52)
> >    fit0<-gev.fit(x[,i], ydat = ti, mul=c(2, 3))
> >
> > Thank you again.
> >
> > On Fri, 9 Jul 2021 at 14:38, Rui Barradas <ruipbarradas using sapo.pt
> > <mailto:ruipbarradas using sapo.pt>> wrote:
> >
> >     Hello,
> >
> >     The following lapply one-liner fits a GEV to each column vector,
> there
> >     is no need for the double for loop. There's also no need to create a
> >     data set x.
> >
> >
> >     library(ismev)
> >     library(mgcv)
> >     library(EnvStats)
> >
> >
> >     # fit a GEV to each column
> >     gev_fit_list <- lapply(Ozone_weekly2, gev.fit, show = FALSE)
> >
> >     # extract the parameters MLE estimates
> >     mle_params <- t(sapply(gev_fit_list, '[[', 'mle'))
> >
> >     # assign column names
> >     colnames(mle_params) <- c("location", "scale", "shape")
> >
> >     # see first few rows
> >
> >
> >
> >     The OP doesn't ask for plots but, here they go.
> >
> >
> >     y_vals <- function(x, params){
> >         loc <- params[1]
> >         scale <- params[2]
> >         shape <- params[3]
> >         EnvStats::dgevd(x, loc, scale, shape)
> >     }
> >     plot_fit <- function(data, vec, verbose = FALSE){
> >         fit <- gev.fit(data[[vec]], show = verbose)
> >         x <- sort(data[[vec]])
> >         hist(x, freq = FALSE)
> >         lines(x, y_vals(x, params = fit\$mle))
> >     }
> >
> >     # seems a good fit
> >     plot_fit(Ozone_weekly2, 1)       # column number
> >     plot_fit(Ozone_weekly2, "CA01")  # col name, equivalent
> >
> >     # the data seems gaussian, not a good fit
> >     plot_fit(Ozone_weekly2, 4)       # column number
> >     plot_fit(Ozone_weekly2, "CA08")  # col name, equivalent
> >
> >
> >
> >     Hope this helps,
> >
> >
> >
> >     Às 00:59 de 09/07/21, SITI AISYAH ZAKARIA escreveu:
> >      > Dear all,
> >      >
> >      > Thank you very much for the feedback.
> >      >
> >      >
> >      > 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
> >      > 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 using 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
> >      >
> >      >
> >      >     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,
> >      >
> >      >
> >      >
> >      >     Às 11:08 de 08/07/21, Jim Lemon escreveu:
> >      >      > Hi Siti,
> >      >      > I think we need a bit more information to respond
> >      >     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 using unimap.edu.my
> >     <mailto:aisyahzakaria using unimap.edu.my>
> >      >     <mailto:aisyahzakaria using unimap.edu.my
> >     <mailto:aisyahzakaria using 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
> >      >      >> 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
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> is at
> >      >      >> be considerate. THINK TWICE BEFORE PRINTING THIS.."
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--

"..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}}

```