python中的简单monte carlo模拟

通过使用 set 而不是一个列表,还可以更改迭代逻辑:

import random

def roll_till_all_sides_appeared():
    sides_seen = set()
    n = 0

    while len(sides_seen) < 6:
        side = random.randint(1, 6)
        sides_seen.add(side)         # will only be added if it isn't present
        n += 1

    return n

def test(repetitions):
    max_n = float('-inf')
    min_n = float('inf')
    sum_n = 0

    for _ in range(repetitions):
        n = roll_till_all_sides_appeared()
        max_n = max(max_n, n)
        min_n = min(min_n, n)
        sum_n += n

    print('max:', max_n)
    print('min:', min_n)
    print('avg:', sum_n / repetitions)

此代码可以如下使用:

>>> test(10)
max: 32
min: 8
avg: 14.3
>>> test(100)
max: 45
min: 6
avg: 14.13
>>> test(1000)
max: 56
min: 6
avg: 14.749
>>> test(10000)
max: 62
min: 6
avg: 14.6422

你的 while 条件不能表达你所期望的。你可能想用列表理解

while any([x not in collection for x in [1, 2, 3, 4, 5, 6])

另外,你不需要三层循环,只需要两层:一层 for 每次审判 虽然 审判尚未完成。一个有效的例子,接近你原来的职位

 import random

 def roll(n=1000):
     trials = []
     sides = 6
     start = 1
     possible_sides = [1, 2, 3, 4, 5, 6]

     for i in range (n):
         collection = [random.randint(1,sides)]
         while any([side not in collection for side in possible_sides]):
             collection.append(random.randint(1,6))
         trials.append(len(collection))

     return sum(trials)/ len(trials)

而更有效的解决方案 set 有效地执行与以前的解决方案相同的操作 any([side not in collection for side in possible_sides]) :

import random

 def roll(n=1000):
     trials = []
     sides = 6
     start = 1
     possible_sides = set([1, 2, 3, 4, 5, 6])

     for i in range (n):
         n_rolls = 0
         sides_rolled = set()
         while not sides_rolled == possible_sides:
             sides_rolled.add(random.randint(1, sides))
             n_rolls += 1

         trials.append(n_rolls)

     return sum(trials)/ len(trials)

或者,更有效的是,检查一下 len(sides_rolled) < 6 ,正如帕特里克·阿特纳在回答中指出的那样。

你可能不是 印刷 不管函数返回什么-这就是为什么它什么都不显示。

使用 print(roll()) 而不是 roll() 打印得到的结果。

循环太多,解决方案占用的内存空间太大。

考虑一下运气不好,必须掷1.000.000.000.000次才能得到前6个-你将在列表中保存1.000.000.000个其他数字。那是很多记忆。

你可以用 set 要记住看到的数字和一个计数器来计算查找所有数据所需的时间:

def roll(sides=6, n=1000):
    """Tests 'n' times to roll all numbers from 1 to 'sides' randomly.
    Returns average tries needed to see all numbers over 'n' tries."""
    trials = []   # collects all sinly tried counters

    for _ in range(n):
        seen = set()   # empty, will only ever store 6 elements at most
        tried = 0      # how long did it take to find all 6?
        while len(seen) < sides:  # use sides here as well
            seen.add(random.randint(1,sides))
            tried += 1
        trials.append(tried)

    return sum(trials)/n


print(roll())  

输出(4次启动):

14.878

14.694

14.732

14.516