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
<aisyahzaka...@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.
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