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

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.

There are already several solutions, which use the fact that each
output value either does not depend on the previous value or is its
copy. The following implements this using rle() function.

  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, -1)
  vec2[1] <- 0
  r <- rle(vec2)
  rlen <- r$lengths
  rval <- r$values
  i <- which(rval == -1)
  rval[i] <- rval[i-1]
  vec2 <- rep(rval, times=rlen)

  all(vec1 == vec2)

  [1] TRUE

Hope this helps.

Petr Savicky.