How to estimate the parameter for many variable?

Dear Rui ang Jim,

Thank you very much.

Thank you Rui Barradas, I already tried using your coding and I'm 
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 at sapo.pt 
<mailto:ruipbarradas at 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)

    Ozone_weekly2 <- read.table("~/tmp/Ozone_weekly2.txt", header = TRUE)

    # 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
    head(mle_params)



    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,

    Rui Barradas


    ?s 00:59 de 09/07/21, SITI AISYAH ZAKARIA escreveu:

     > 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

    <mailto:ruipbarradas at sapo.pt>

     > <mailto:ruipbarradas at sapo.pt <mailto: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

    <mailto:aisyahzakaria at unimap.edu.my>

     >? ? ?<mailto:aisyahzakaria at unimap.edu.my

    <mailto: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

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     >? ? ? >>
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Please be considerate. THINK TWICE BEFORE PRINTING THIS.."

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