Function with two arguments that returns when arguments are equal

Question

I'm trying to create a function that takes two arguments. It will repeatedly call the first argument, in my case

f()

until f returns that same value 3x in a row. It will then call the second argument

g()

and if g returns the same value as f returned earlier, then the function must return this value. Otherwise it will go back and call the first argument f() again and repeat the cycle.

This is what I have so far.

call_until = function(f, g) {
    while(1) {
        n = f %>% chunk(3) %>% filter(f, function(v) length(unique(v)) == 1)
        x = n()[1]
        if (x == g()) {
            return(x)
        }
    }
}

For example, if f returns 4 three times in a row, then go to g. If g equals what f returned three times in a row (which in this example is 4); so if g == 4, then call_until should return 4.

Answer 1

Here's another example of how you might do this

call_until = function(f, g) {
  while(TRUE) {
    value <- f()
    seen <- 1
    while(seen < 3) {
      next_value <- f()
      if (next_value == value) {
        seen <- seen + 1
      } else {
        value <- next_value
        seen <- 1
      }
    }
    if (value == g()) {
        return(value)
    }
  }
}

Though it kind of matters if you are supposed to draw three new f values if g doesn't match or just one new value.

Here are some useful testing functions. They just return values from a vector in order, repeating as necessary.

cycler <- function(vals) {
  i <- 1
  function() {
    i<<-i+1
    vals[(i-2) %% length(vals)+1]
  }
}

f <- cycler(c(1,2,2,2,3,4,5,5,5,1,1))
g <- cycler(c(5,4))

With this we get

call_until(f, g)
# [1] 5