# [R] Avoiding a loop

Niels Richard Hansen Niels.R.Hansen+lists at math.ku.dk
Fri Apr 8 20:05:32 CEST 2011

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<lebatsnok at gmail.com>:
>> 2011/4/8 Juan Carlos Borrás<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)
>>>
>
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--
Niels Richard Hansen                     Web:   www.math.ku.dk/~richard
Associate Professor                      Email: Niels.R.Hansen at math.ku.dk
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