The beauty of code vectorisation

I came across this problem in Twitter:

The basic pieces of the problem are:

• We need to generate pseudorandom numbers with an identical distribution, add them up until they go over 1 and report back how many numbers we needed to achieve that condition.
• Well, do the above “quite a few times” and take the average, which should converge to the number e (2.7182…).

In most traditional languages, the first step would look more or less like this:

counter <- 0
while(total <= 1) {
  total <- total + runif(1)
  counter <- counter + 1
}

or by setting a “big enough” predefined number of iterations:

total <- 0 
for(counter in 1:100) { 
  total <- total + runif(1) 
  if(total > 1) return(counter)
}

We would need to define another loop to run the above code “quite a few times”, which for ten million times (10^7) could look like:

results <- 0
average <- 0
for(iteration in 1:10^7) {
  total <- 0
  counter <- 0
  while(total < 1) {
    total <- total + runif(1)
    counter <- counter + 1
  }
results <- results + counter
average <- results/iteration
}

However, we are working with R, which happens to be an array-based (or matrix-based, or vector-based) language and it has functionality for data processing. A first, rough approach could be:

mean(vapply(1:10^7, 
     FUN.VALUE = 1, FUN = function(x) {
  sum <- 0
  i <- 1
  while(TRUE) {
    sum <- sum + runif(1)
    if(sum > 1) return(i) else {
      i <- i+1
    next
    }
  }
}))

It can be much shorter if, rather than thinking of loops, we think of vectors:

mean(replicate(10^7, 
     min(which(cumsum(runif(100)) > 1, TRUE))))

The code works the following way:

1. Generate 100 uniformly distributed random numbers (0, 1): runif(100)
2. Calculate the cumulative sum of the 100 numbers: cumsum()
3. Check if the cumulative sum is greater than 1: which(cumsum() > 1, TRUE)
4. Find the minimum position that meets the criterion: min()
5. Replicate that process ten million times: replicate(10^7, )
6. Take the average of all the positions: mean()

Half a tweet to solve the problem. Look, ma. No explicit loops!