I believe the solutions proposed ignore the recursive nature of the
original problem and hence produce wrong solutions.
P <- c(5, 7, 6.01, 6.01, 7)
m <- rep(6, 5)
S0 <- as.numeric(P>(m*1.005))
Then the original loop from Worik gives
S <- S0
for(i in 2:length(S)){
if(S[i]==0 && S[i-1] == 1){
if(P[i] > m[i]){
S[i] <- 1
}
}
}
> S
[1] 0 1 1 1 1
The other solutions I have seen on the list suggest looking
upfront on differences in S, and I am pretty sure the intentions
were to produce
S <- S0
v <- c(FALSE, (S[-1] == 0) & (S[-5] == 1)) & (P > m)
where v equals c(FALSE, FALSE, TRUE, FALSE, FALSE) and then
S[v] <- 1
giving
> S
[1] 0 1 1 0 1
When you recursively update a vector like this I don't know any
general vectorization fix.
- Niels
On 08/04/11 00.58, Juan Carlos Borrás wrote:
Kenn,
I find your solution more elegant.
2011/4/8 Kenn Konstabel<lebats...@gmail.com>:
2011/4/8 Juan Carlos Borrás<jcbor...@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)
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