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

[Rui Barradas]

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> 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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>>
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> 
> ______________________________________________
> R-help using 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.
>