If I understand what you?re trying to do, then I believe this will do the
same as your little loop:
opt.fc <- fc
opt.fc[rmax < 0.1] <- fc1[rmax < 0.1]
(this is an example of vectorization, and it?s fundamental to how R works
and the power of R)
Then to extend it, use the same method
opt.fx[ {some expression} ] <- fc2[ {some expression} ]
But I have no idea what {some expression} should be. Perhaps it has
something to do with the maximized r squared values, but I can?t tell from
what you?ve said. In any case, the {some expression} has to result in a
logical vector of the same length as the others (373 in your example).
Hope this helps.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 10/16/14, 8:51 AM, "moeby" <fabian_schlegel at hotmail.com>
wrote:
>I try to run an if-else command line where the else argument should be the
>corresponding value of the pmax command.
>
>#####################
>#opt.fc is the optimal forecast
>#rmax is the vector of the maximized r squared from pmax-command of 2 data
>sets containing r squares
>#fc1 are the estimates of a specific forecast model (e.g. AR1)
>#fc are the estimates of a specific forecast model, namely the Benchmark
>(e.g. mean)
>
>opt.fc <- as.numeric(373) #data length is 373
>for (x in 1:373)
>{
>if (rmax[x] < 0.1)
>opt.fc[x] <- fc[x] #if rmax smaller than the value 0.1 the forecast of
the
>benchmark should be used.
>else
>opt.fc[x] <- fc1[x] #else the forecast of fc model should be used.
>}
>So far, so easy.
>
>How can I compute the opt.fc when I have more than one alternative
>forecast
>model, let's say fc2 and fc3.
>In Addition, I want to use the values from the forecast model where the
>values of the corresponding r squared are maxed (pmax(fcr1,fcr2,fcr3...)
>..... :(
>
>any ideas?
>
>Thank you very much in advance.
>
>
>
>
>
>--
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