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

This is a rewritten version of @necromancer's cool solution. Wraps it in a class to make it much easier to use correctly, and uses more dict methods to cut the lines of code.

from random import randrange

class Sampler:
    def __init__(self, n):
        self.n = n # number remaining from original range(n)
        # i is a key iff i < n and i already returned;
        # in that case, state[i] is a value to return
        # instead of i.
        self.state = dict()

    def get(self):
        n = self.n
        if n <= 0:
            raise ValueError("range exhausted")
        result = i = randrange(n)
        state = self.state
        # Most of the fiddling here is just to get
        # rid of state[n-1] (if it exists).  It's a
        # space optimization.
        if i == n - 1:
            if i in state:
                result = state.pop(i)
        elif i in state:
            result = state[i]
            if n - 1 in state:
                state[i] = state.pop(n - 1)
            else:
                state[i] = n - 1
        elif n - 1 in state:
            state[i] = state.pop(n - 1)
        else:
            state[i] = n - 1
        self.n = n-1
        return result

Here's a basic driver:

s = Sampler(100)
allx = [s.get() for _ in range(100)]
assert sorted(allx) == list(range(100))

from collections import Counter
c = Counter()
for i in range(6000):
    s = Sampler(3)
    one = tuple(s.get() for _ in range(3))
    c[one] += 1
for k, v in sorted(c.items()):
    print(k, v)

and sample output:

(0, 1, 2) 1001
(0, 2, 1) 991
(1, 0, 2) 995
(1, 2, 0) 1044
(2, 0, 1) 950
(2, 1, 0) 1019

By eyeball, that distribution is fine (run a chi-squared test if you're skeptical). Some of the solutions here don't give each permutation with equal probability (even though they return each k-subset of n with equal probability), so are unlike random.sample() in that respect.