Avoiding a loop

2011/4/8 Juan Carlos Borr?s <jcborras at gmail.com>:
#Use the indexes of S in a sapply function.

N <- 10
S <- sample(c(0,1), size=N, replace=TRUE)
v1 <- sapply(c(1:N-1), function(i) S[i]&&S[i+1])
You can achieve the same v1 using

v1.2 <-  S[2:N-1] & S[2:N]

.. or if you insist on having NA as the first element, -- c(NA, v1.2)

Vectorization is more efficient than loops but this need not be true
for the *apply functions.
# Then
v2 <- (P > m)

# And I guess you can fill up the rest. Beware of the boundary
condition (the NA in v1)

Cheers,
jcb!
_______________________
http://twitter.com/jcborras

On Fri, Apr 8, 2011 at 5:30 AM, Worik R <worikr at gmail.com> wrote:
Friends.

I cannot simplify this much, and I think the loop is unavoidable. ?As a
recovering C programmer I want to avoid loops and in cases like this I
almost allways can by using an apply function. ?But I suspect in this case
there is nothing I can do.

It is a finance example where a price series is compared to a moving
average. ?If the price goes above the average, plus a bit, buy the
security. ?If we are holding the security and the price dips below the
moving average sell it.

P is the series of prices

m is the moving average series

S <- P>(m*1.005)
S[S]<-1

Now S is my signal it is 1 when P > m plus a margin of 0.005 x m

But now I need to control when S goes from 1 to 0. ?As far as I can tell
this is the only way...

for(i in 2:length(S)){
?if(S[i]==0 && S[i-1] == 1){
? ?## Was long, now short. ?SHould I be short? ?Is P>m still?
? ?if(P[i] > m[i]){
? ? ?## Stay long
? ? ?S[i] <- 1
? ?}
?}
}

As I mentioned I am a recovering C programmer, so I have a buit in loop
reflex, but I am struggling to adapt. ?But this one has me beat! ?Can anyone
help?

cheers
W

? ? ? ?[[alternative HTML version deleted]]

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Avoiding a loop

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