[R] How to estimate the parameter for many variable?
Rui Barradas
ru|pb@rr@d@@ @end|ng |rom @@po@pt
Fri Jul 9 11:18:29 CEST 2021
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")
head(mle_params2)
Hope this helps,
Rui Barradas
Às 09:53 de 09/07/21, SITI AISYAH ZAKARIA escreveu:
> 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 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)
>
> 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
>
>
> Às 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 using sapo.pt
> <mailto:ruipbarradas using sapo.pt>
> > <mailto:ruipbarradas using sapo.pt <mailto: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
> >
> > 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
> >
> >
> > Às 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 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
> > >> 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}}
> > >>
> > >> ______________________________________________
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> > > 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.."
> >
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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 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
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> recipient should check this email and any attachments for the presence
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