以概率添加元素到数组

Question

So I am building a list in Python, for example, let us say the first 100 integers, but I do need all the 100 integers but only a sample lets say 3.

import random 

def f():
    list_ = []
    for i in range(100):
        list_.append(i)
    return list_

def g(list_,k):
     return random.sample(list_, k)

print(g(f(),3))

>>>[50, 92, 6]

Now can I get away with not building the whole list in the first place, but directly build the sample, maybe by adding a probability with which elements get added to the list in f()

Because if I am building a huge list which does not integers numbers but some other objects, this approach could be costly, in terms of memory and computation.

Answer 1

def random_no_dups_k_of_n(k, n):
    res = list(range(k))
    for i in range(k, n):
        v = random.randint(0, i) # this is 0-i inclusive
        if v == i:
            ir = random.randint(0,k-1)
            res[ir] = i
    return res

What's happening here: it's a telescoping product. Each element from 0 to k-1 starts out having a k/k chance of being selected. After 1st iteration k has 1/(k+1) chance of getting selected, while all others (not just remaining, but all) have a (k-1)/k * k/(k+1) = (k-1)/(k+1) chance of getting selected. After 2nd iteration, k+1 has a 1/(k+2) chance of getting selected, while all the others have a (k-1)/(k+1) * (k+1)/(k+2) = (k-1)/(k+2) chance of getting selected. And so on. In the end, each number will have a k/n chance of getting selected.

Actually, I just saw that you can just do random.sample(range(n), k) . I just assumed it wasn't available.

EDIT : I got the probabilities reversed above. The correct version should be:

def random_no_dups_k_of_n(k, n):
    res = list(range(k))
    for i in range(k, n):
        v = random.randint(0, i) # this is 0-i inclusive
        if v < k:
            ir = random.randint(0,k-1)
            res[ir] = i
    return res

Each element from 0 to k-1 starts out having a k/k chance of being selected. After 1st iteration k has k/(k+1) chance of getting selected, while all others (not just remaining, but all) have a k/k*((k-1)/k * k/(k+1) + 1(k+1) = k/(k+1) chance of getting selected. After 2nd iteration, k+1 has a k/(k+2) chance of getting selected, while all the others have a k/(k+1)*((k-1)/k * k/(k+2) + 2/(k+2))= k/(k+2) chance of getting selected.

And this actually does collapse all the calculations to give each element a k/(k+m) chance after m th step.