如何在while循环中从列表中删除元组

Question

我有一个函数，它可以确定填充容量为200的堆栈的最佳方式。 要求如下：

• 仍然合适的小物件应该放在第一位
• 如果两件物品大小相同，则价格最高的物品应放在第一位

capacity = 200

dairy_items=[('p1', 10, 3), ('p2', 13, 5),
             ('p3', 15, 2), ('p4', 26, 2),
             ('p5', 18, 6), ('p6', 25, 3),
             ('p7', 20, 4), ('p8', 10, 5),
             ('p9', 15, 4), ('p10', 12, 7),
             ('p11', 19, 3), ('p12', 27, 6),
             ('p13', 16, 4), ('p14', 23, 5),
             ('p15', 14, 2), ('p16', 23, 5), 
             ('p17', 12, 7), ('p18', 11, 3),
             ('p19', 16, 5), ('p20', 11, 4)]

第一个元组元素是唯一的产品id，第二个是产品大小，第三个是价格

所以我创建了这个函数。我只有一个问题。堆叠的第一件东西是产品p8，因此需要从dairy_项中删除此项，以便函数不会再次选择该值，因为它已被使用。我用dairy_items.remove（）尝试了这个方法，但不起作用。是否有其他方法可以使用，或者我只是没有正确使用

def shelving(dairy_items):
    filled = 0
    worth = 0
    shelves = []
    while filled <= capacity:
        smallest_product = math.inf
        most_expensive_product = 0
        
        for productNr in range(len(dairy_items)):
            if dairy_items[productNr][1] < smallest_product:
                smallest_product = dairy_items[productNr][1]
                most_expensive_product = dairy_items[productNr][2]
                productID = dairy_items[productNr][0]
  
            elif smallest_product == dairy_items[productNr][1]:
                if dairy_items[productNr][2] > most_expensive_product:
                    smallest_product = dairy_items[productNr][1]
                    productID = dairy_items[productNr][0]
        filled = filled + smallest_product
        worth = worth + most_expensive_product
        shelves.append(productID)
        dairy_items.remove((productID, smallest_product, most_expensive_product))
       
    return (shelves,filled,worth)

Answer 1

问题是elif块没有分配most_expensive_product。因此，您可以在if块中分配most_expensive_product，在elif块中分配smallest_product和productID。这将导致在remove调用中构造一个元组，该元组一开始就不存在，因此无法删除

您可以通过在elif中添加赋值来解决这个问题，假设这是您的期望行为

most_expensive_product = dairy_items[productNr][2]

我对您的变量名有点困惑，因为most_expensive_product似乎不是最昂贵的产品，而是最小产品的价格（或者如果有多个最小的项目，则价格更高）

如果我对您尝试执行的操作的假设是正确的，那么使用for循环会更好，这样您就避免了在索引和元组重建方面的麻烦。例如，下面的内容似乎与您的意图相符：

def shelving(dairy_items):
    filled = 0
    worth = 0
    shelves = []
    while filled <= capacity:
        chosen_product = dairy_items[0]

        for product in dairy_items:
            if product[1] < chosen_product[1]:
                chosen_product = product

            elif chosen_product[1] == product[1]:
                if product[2] > chosen_product[2]:
                    chosen_product = product
        filled = filled + chosen_product[1]
        worth = worth + chosen_product[2]
        shelves.append(chosen_product[0])
        dairy_items.remove(chosen_product)

    return (shelves, filled, worth)

然而，这仍然有一个问题，您的原始代码有；因为它只在循环开始时检查容量，所以它会将货架填满。这可以通过在末尾添加支票来解决，如下所示：

def shelving(dairy_items):
    filled = 0
    worth = 0
    shelves = []
    while filled <= capacity:
        chosen_product = dairy_items[0]

        for product in dairy_items:
            if product[1] < chosen_product[1]:
                chosen_product = product

            elif chosen_product[1] == product[1]:
                if product[2] > chosen_product[2]:
                    chosen_product = product
        if filled + chosen_product[1] <= capacity:
            filled = filled + chosen_product[1]
            worth = worth + chosen_product[2]
            shelves.append(chosen_product[0])
            dairy_items.remove(chosen_product)
        else:
            break

    return (shelves, filled, worth)

Answer 2

我的代码中有一些愚蠢的错误。我改变了一些事情，下面的事情对我来说是可行的

def shelving(dairy_items):
    filled = 0
    worth = 0
    shelves = []
    while filled <= capacity:
        smallest_product = math.inf
        most_expensive_product = 0
        for productNr in range(len(dairy_items)):
            if dairy_items[productNr][1] < smallest_product:
                smallest_product = dairy_items[productNr][1]
                most_expensive_product = dairy_items[productNr][2]
                productID = dairy_items[productNr][0]
            elif smallest_product == dairy_items[productNr][1]:
                if dairy_items[productNr][2] > most_expensive_product:
                    smallest_product = dairy_items[productNr][1]
                    productID = dairy_items[productNr][0]
                    most_expensive_product = dairy_items[productNr][2]
        dairy_items.remove((productID, smallest_product, most_expensive_product))
        if filled + smallest_product >= 200:
            break
        filled = filled + smallest_product
        worth = worth + most_expensive_product
        shelves.append(productID)
        
       
    return (shelves, filled , worth)