在 python 列表中移动项目？

Question

这是最终产品。如果其他人有任何削减它的技巧，请告诉我！非常感谢您的帮助！

def triple_cut(deck):
    ''' (list of int) -> NoneType

    Modify deck by finding the first joker and putting all the cards above it
    to the bottom of deck, and all the cards below the second joker to the top
    of deck. 
    >>> deck = [2, 7, 3, 27, 11, 23, 28, 1, 6, 9, 13, 4]
    >>> triple_cut(deck) 
    >>> deck 
    [1, 6, 9, 13, 4, 27, 11, 23, 28, 2, 7, 3]
    '''     
    joker1 = deck.index(JOKER1)
    joker2 = deck.index(JOKER2)
    first = min(joker1, joker2)
    first_cards = []

    for cards in range(len(deck[:first])):
        cards = 0 
        pop = deck.pop(cards)
        first_cards.append(pop)

    joker1 = deck.index(JOKER1)
    joker2 = deck.index(JOKER2)
    second = max(joker1, joker2)    
    second_cards = []

    for cards in deck[second + 1:]:
        pop = deck.pop(deck.index(cards))
        second_cards.append(pop)

    second_cards.reverse()

    for card in second_cards: 
        deck.insert(0, card)

    deck.extend(first_cards)   

我需要输入更多，因为我的帖子主要是代码：请添加更多详细信息 sss ss

Answer 1

提示：

p = list('abcdefghijkl')
pivot = p.index('g')
q = p[pivot:] + p[:pivot]

列表切片是你的朋友。

Answer 2

回答第二次修订

函数完成后，不会改变牌组。

问题是你没有给出完整的代码。我猜它看起来像这样：

def triple_cut(deck)
    …
    deck = q

根据您的文档字符串，您将其称为

deck = […]
triple_cut(deck)

并且对分配deck = q 不会传播到triple_cut() 感到困惑。您不能修改方法的形式参数，因此赋值对triple_cut 保持本地化并且不影响模块级变量deck

正确的写法是

def triple_cut(cuttable)
    …
    return cuttable[first:] + cuttable[:first]

deck = […]
deck = triple_cut(deck)

为了便于解释，我将参数名称更改为 cuttable 。您可以将参数名称保留为deck，但我想表明，当您认为您正在分配到甲板时，您实际上是在分配给可切割的并且该分配不会执行triple_cut()。