[R] updating elements of a vector sequentially - is there a faster way?

[Petr Savicky]

On Fri, Aug 24, 2012 at 10:34:14AM +0200, Petr Savicky wrote:
> On Thu, Aug 23, 2012 at 09:49:33PM -0700, Gopi Goteti wrote:
> > I would like to know whether there is a faster way to do the below
> > operation (updating vec1).
> > 
> > My objective is to update the elements of a vector (vec1), where a
> > particular element i is dependent on the previous one. I need to do this on
> > vectors that are 1 million or longer and need to repeat that process
> > several hundred times. The for loop works but is slow. If there is a faster
> > way, please let me know.
> > 
> > probs <- c(.1, .3, .2, .4, .7, .9, .3, .4, .5, .6)
> > p10 <- 0.6
> > p00 <- 0.4
> > vec1 <- rep(0, 10)
> > for (i in 2:10) {
> >   vec1[i] <- ifelse(vec1[i-1] == 0,
> >                     ifelse(probs[i]<p10, 0, 1),
> >                     ifelse(probs[i]<p00, 0, 1))
> > }
> 
> Hi.
> 
> If p10 is always more than p00, then try the following.
[...]
>   # modification 
>   a10 <- ifelse(probs<p10, 0, 1)
>   a00 <- ifelse(probs<p00, 0, 1)
>   vec2 <- ifelse(a10 == a00, a10, NA)
>   vec2[1] <- 0
>   n <- length(vec2)
>   while (any(is.na(vec2))) {
>       shift <- c(NA, vec2[-n])
>       vec2 <- ifelse(is.na(vec2), shift, vec2)
>   }

Hi.

Let me suggest a variant of this, which can be more efficient
in some cases.

  probs <- c(.1, .3, .2, .4, .7, .9, .3, .4, .5, .6)

  p10 <- 0.6
  p00 <- 0.4

  # original code  
  vec1 <- rep(0, 10)
  for (i in 2:10) {
    vec1[i] <- ifelse(vec1[i-1] == 0,
                      ifelse(probs[i]<p10, 0, 1),
                      ifelse(probs[i]<p00, 0, 1))
  }

  # modification 
  a10 <- ifelse(probs<p10, 0, 1)
  a00 <- ifelse(probs<p00, 0, 1)
  vec2 <- ifelse(a10 == a00, a10, NA)
  vec2[1] <- 0
  while (1) {
      i <- which(is.na(vec2))
      if (length(i) == 0) break
      vec2[i] <- vec2[i-1]
  }

  all(vec1 == vec2)

  [1] TRUE

Hope this helps.

Petr Savicky.