从列表中随机选择至少一次

Question

I have 10 elements in a Python list. I want to select an item randomly from this list - multiple times (more than 1000 times...). And, I want to ensure that all 10 are elements chosen at least once. Is there a way to do it in Python?

random.choice is not helping me...(or, at least, I do not know how to use it).

Answer 1

As I mentioned in a comment, if it's truly random, and it iterates a set number of times, it is not 100% certain all items are chosen at least once, though if you are doing it over 1000 times on a list of 10 items, it's of course highly probably they will be. To make 100% sure all items are chosen once, you'd have to impose some kind of rule that would preclude complete randomness. BUT if you didn't want to do that, you could set your minimum number of random selections, and if for some reason all elements had not been selected, it would keep going until they are.

import random
original_list = [1,2,3,4,5,6,7,8,9,10]
list = original_list[:]
minimum_iterations = 1000

while minimum_iterations > 0:  
    choice = random.choice(list)
    print(choice)
    if choice in original_list:
        original_list.remove(choice)
    minimum_iterations = minimum_iterations - 1

while len(original_list) > 0: # only executes if items remain unchosen
    choice = random.choice(list)
    print(choice)
    if choice in original_list:
        original_list.remove(choice)

This again is not completely random, since it's continuing until every item is chosen ; but it's the most random way I can think of to accomplish your original question.

Another solution, if you were willing to "distribute" the randomness a little more, would be to have it randomly select each list value once until all were selected, then continue doing so for the total number of iterations. That would guarantee each value is chosen an equal number of times, but in an order that is random.

Answer 2

You could start by cloning the list, ensuring each is in the new list at least once, which simulates choosing with probability 1 for at least one each.

Then you use random.choice in a range loop to randomly choose your_number of times from the list.

import random

your_number = 1000
old_list = [1,2,3,4,5,6,7,8,9,10]
new_list = list(old_list)

for x in range(your_number):
    new_list.append(random.choice(old_list))

if you don't like the numbers being in predictable locations, increase the entropy by using random.shuffle like so:

random.shuffle(new_list)

Answer 3

Moving my answer from the comments to a post.

statistically, if you iterate over an array with ten elements a thousand times, each element will get chosen once. So why not just do something like this? with a correction after the expensive operation to ensure presence.

import random

data = range(10) # or really any iterable sequence
occurences = {}

for i in range(1000):
    choice = random.choice(data)

    # increment the occurence of the random choice by one
    # if the random choice was not found assume 0
    occurences[choice] = occurences.get(choice, 0) + 1

# error correction to ensure each element is counted once
# keep in mind this is extremely unlikely.

# iterate over each element in data that is not a key in occurences
for forgotten in (item for item in data if item not in occurences):
    choice = random.choice(tuple(data.keys()))

    # ensure each data element has been counted once
    # so simply borrow from another occurence
    occurences[choice] -= 1
    occurences[forgotten] = 1

Answer 4

Use a set to keep track of what items have been used; re-roll if not all are used in the set.

The amortized cost is negligible (as the odds to failing are very low) and the added tracking is O(1), so it doesn't change the O(n) time complexity.

It is not truly random, but the internal randomness of the set is maintained and it guarantees as random a result as you can get while guaranteeing all ten terms.

opts = range(10)  # or your list
result = []
while len(set(result)) != len(opts):
    for _ in range(1000):
        choice = random.choice(opts)
        result.append(choice)