如何计算随机列表中出现的数字列表 Python

Question

I run 1000 times of random.choice()from a list from 0-11. how to track of the number of random selections necessary before all 12 number have been selected at least once. (Many will be selected more than once.) For instance, suppose the simulation yields the following sequence of random choices for a single trial: 2 5 6 8 2 9 11 10 6 3 1 9 7 10 0 7 0 7 4, where all 12 numbers have been selected at least once. The count for this example trial is 19. Collect the count for each trial of the simulation in a single list (ultimately consisting of 1,000 counts).

Answer 1

Here is a solution using collections.Counter as a container:

from collections import Counter
import random

nums = list(range(12))
n = 1000
counts = [0]*n
for trial in range(n):
    c = Counter()
    while len(c)<len(nums):
        c[random.choice(nums)]+=1
    counts[trial] = sum(c.values()) # c.total() in python ≥ 3.10

counts

Output:

[28, 24, 39, 27, 40, 36, ...] # 1000 elements

Distribution of the counts:

Answer 2

One simple (but inefficient) way to do this would be with

check = range(11)
if all(elem in randlist for elem in check): # check after each choice
    # do something

As others have said, tell us what you have tried so far so we can help further.

Answer 3

Maybe you can try using a set to store your results in a non-redundant way, while checking to see if all numbers have been used:

import random

guesses = set()
count = 0
for i in range(1000):
    count += 1
    set.add(random.randrange(0, 12))
    if len(guesses) == 12:
        break
print(count)