如何在 Python 中打乱列表

Question

def make_sorted_deck():
     ''' Return a sorted deck of cards. Each card is
         represented by a string as follows:
         "queen of hearts". The cards are ordered by rank and then suit within rank.
     :return: The sorted deck of cards, as a list of strings
     EXAMPLE: make_sorted_deck() == ['2 of spades', '2 of hearts', '2 of    clubs', ..., 'ace of clubs', 'ace of diamonds'] '''
     #Hint: Use the previous functions and two nested for loops.
     sorted_deck = []
     for i in get_ranks():
         for j in get_suits():
             sorted_deck.append("{0} of {1}".format(i,j))
         return sorted_deck
     print(make_sorted_deck())

def shuffle(deck):
    ''' Randomly shuffle the cards in deck into a newly created deck of cards (list).
    :param: deck: A deck of cards
    :return: A new list, consisting of a random shuffle of deck.
    EXAMPLE: shuffle(['2 of hearts', '3 of diamonds', 'jack of spades', '2 of   clubs']) could return ['jack of spades', '3 of diamonds', '2 of hearts', '2 of clubs'] 
     #REQUIREMENTS: Please implement the following algorithm: Use a while loop to repeatedly pick a random card from deck, remove it, and add it to a newly created list. '''

如何随机播放make_sorted_deck() 创建的列表？

我知道我可以导入一个函数来洗牌，但我需要这样做的方法是取出 1 张随机牌并将其附加到一个新列表中以获得一个洗牌列表。

Answer 1

我不会解决你的作业，但让我给你一些提示：

• while x: 将循环，只要x 是真实的。非空列表是真实的。

• 您可以通过x = random.randrange(n) (docs) 选择一个随机数x 其中0 <= x < n

• 您可以使用l.pop(i) (docs) 从列表l（即l[i]）中删除索引为i 的项目

Answer 2

另一个没有回答 OP 问题的答案...

要随机播放长度为n 的列表，您需要一个索引列表，从0 到n-1，以随机顺序排列...

我们开始从random 模块导入randrange 函数，

from random import randrange

像这样调用的randrange(n) 返回一个随机整数i、0 <= i <= n-1。

当我们选择第一个随机索引时，包括在0 和n-1 之间，我们会在更窄的区间内选择下一个索引，依此类推...

l = [randrange(n-i) for i in range(n)]

当然l 中的最后一个数字是0，因为i==n-1 和randrange(1) 必须返回0。

l中的数字不能直接用来寻址要洗牌的列表， 因为它们指的是 available 元素列表在洗牌过程的某个点上的位置，所以对于 n 中的每个数字，我们必须查看有多少元素已经被洗牌，以及它们的相对位置对于当前元素，假设我们要将 _real_indices 存储在一个列表中，最初为空

indices = []

我们必须小心...

for i in l:                       # the randomized, partial indices
    j = 0                         # aux variable
    while j <= i:                 # we will increment j later
        if j in indices:          # if this number j, smaller than i, is in the
             i += 1               # list of used indices, i must be incremented
        j += 1
    indices.append(i)             # the (possibly) incremented i is stored

这就是洗牌的全部内容。

我在这里报告一个简短的 IPython 会话，以证明这种方法的正确性：

In [1]: from random import randrange

In [2]: def shuffle(n):
    l = [randrange(n-i) for i in range(n)]
    indices = []
    for i in l:
        j = 0
        while j <= i:
            if j in indices: i = i+1
            j = j+1
        indices.append(i)
    return indices
   ...: 

In [3]: sh = shuffle(10) ; print sh ; print sorted(sh)
[7, 6, 4, 9, 1, 5, 0, 2, 8, 3]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [4]: sh = shuffle(10) ; print sh ; print sorted(sh)
[6, 9, 5, 1, 4, 3, 0, 2, 8, 7]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [5]: sh = shuffle(10) ; print sh ; print sorted(sh)
[3, 6, 4, 9, 0, 7, 8, 1, 2, 5]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

In [6]: