如何在Python中隨機播放列表

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. '''

我如何洗牌make_sorted_deck()創建的列表？

我知道我可以導入一個隨機播放卡組的功能，但是我需要這樣做的方法是取出1張隨機卡並將其附加到新列表中以產生隨機播放的列表。

Answer 1

我不會解決您的作業，但讓我給您一些提示：

• while x:只要x為真，就會循環。 非空列表為真。

• 您可以通過執行x = random.randrange(n) （ docs ）選擇一個隨機數x ，其中0 <= x < n

• 您可以使用l.pop(i) （ docs ）從列表l （即l[i] ）中刪除索引為i的項目。

Answer 2

另一個不能回答OP問題的答案...

要隨機排列長度為n的列表，您需要一個索引列表，從0到n-1 ，以隨機順序...

我們開始從random模塊導入randrange函數，

from random import randrange

這樣調用randrange(n)返回一個隨機整數i ， 0 <= i <= n-1 。

當我們選擇第一個隨機索引（包括0到n-1 ，我們將以較窄的間隔選擇下一個索引，依此類推……

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

當然， l的最后一個數字將為0因為i==n-1並且randrange(1)必須返回0 。

l的數字不能直接用於尋址要改組的列表，因為它們指的是改組過程中特定點在可用元素列表中的位置，因此對於n中的每個數字，我們必須查看有多少個元素具有已經改組，它們相對於當前元素的位置，假設我們要將_real_indices存儲在一個列表中，最初為空

indices = []

我們必須要小心...

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

這就是洗牌。

在這里，我報告了一個簡短的IPython會話，演示了這種方法的正確性：

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]: