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> 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>> 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>> 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.."
>>
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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}}