How to incrementally sample without replacement?

Question

Python has my_sample = random.sample(range(100), 10) to randomly sample without replacement from [0, 100).

Suppose I have sampled n

Answer 1

Note to readers from OP: Please consider looking at the originally accepted answer to understand the logic, and then understand this answer.

Aaaaaand for completeness sake: This is the concept of necromancer’s answer, but adapted so it takes a list of forbidden numbers as input. This is just the same code as in my previous answer, but we build a state from forbid, before we generate numbers.

• This is time O(f+k) and memory O(f+k). Obviously this is the fastest thing possible without requirements towards the format of forbid (sorted/set). I think this makes this a winner in some way ^^.
• If forbid is a set, the repeated guessing method is faster with O(k⋅n/(n-(f+k))), which is very close to O(k) for f+k not very close to n.
• If forbid is sorted, my ridiculous algorithm is faster with:
  

import random
def sample_gen(n, forbid):
    state = dict()
    track = dict()
    for (i, o) in enumerate(forbid):
        x = track.get(o, o)
        t = state.get(n-i-1, n-i-1)
        state[x] = t
        track[t] = x
        state.pop(n-i-1, None)
        track.pop(o, None)
    del track
    for remaining in xrange(n-len(forbid), 0, -1):
        i = random.randrange(remaining)
        yield state.get(i, i)
        state[i] = state.get(remaining - 1, remaining - 1)
        state.pop(remaining - 1, None)

usage:

gen = sample_gen(10, [1, 2, 4, 8])
print gen.next()
print gen.next()
print gen.next()
print gen.next()