python代码在一个文件上工作，但在另一个文件上失败

Question

大家好，我有这个代码，它打印出商品的最低成本和餐厅 ID。客户不想访问多家餐厅。因此，例如，如果他要求“A，B”，那么代码应该打印同时提供它们的商店，而不是将用户需求分散到不同的餐厅（即使某些餐厅提供的价格便宜）。

另外，如果假设用户要汉堡。那么如果某个餐厅“X”以 4 美元的价格提供“汉堡”，而另一家餐厅“Y”以 3 美元的价格提供“汉堡+金枪鱼+豆腐”，那么我们会告诉用户去 RESTAURANT 'Y'，即使它除了用户要求的 'burger' 之外还有额外的物品，但我们很乐意给他们额外的物品，只要它便宜。

一切都很好，但是代码在两个格式相同的输入文件（在 input.csv 上失败但在 input-2.csv 上运行）上的行为奇怪地不同，它为一个文件提供正确的输出，而对另一个文件则失败。这是我需要您帮助修复的唯一微小错误。请帮帮我，我想我已经撞墙了，无法超越这一切。

def build_shops(shop_text):
    shops = {}
    for item_info in shop_text:
        shop_id,cost,items = item_info.replace('\n', '').split(',')
        cost = float(cost)
        items = items.split('+')

        if shop_id not in shops:
            shops[shop_id] = {}
        shop_dict = shops[shop_id]

        for item in items:
            if item not in shop_dict:
                shop_dict[item] = []
            shop_dict[item].append([cost,items])
    return shops


def solve_one_shop(shop, items):
    if len(items) == 0:
        return [0.0, []]
    all_possible = []
    first_item = items[0]
    if first_item in shop:
        print "SHOP",shop.get(first_item)
    for (price,combo) in shop[first_item]:
        #print "items,combo=",items,combo
        sub_set = [x for x in items if x not in combo]
        #print "sub_set=",sub_set
        price_sub_set,solution = solve_one_shop(shop, sub_set)
        solution.append([price,combo])
        all_possible.append([price+price_sub_set, solution])

    cheapest = min(all_possible, key=(lambda x: x[0]))
    return cheapest


def solver(input_data, required_items):
    shops = build_shops(input_data)
    #print shops
    result_all_shops = []
    for shop_id,shop_info in shops.iteritems():
        (price, solution) = solve_one_shop(shop_info, required_items)
        result_all_shops.append([shop_id, price, solution])

    shop_id,total_price,solution = min(result_all_shops, key=(lambda x: x[1]))
    print('SHOP_ID=%s' % shop_id)
    sln_str = [','.join(items)+'(%0.2f)'%price for (price,items) in solution]
    sln_str = '+'.join(sln_str)
    print(sln_str + ' = %0.2f' % total_price)



shop_text = open('input-1.csv','rb')    
solver(shop_text,['burger'])

=====input-1.csv=====restaurant_id、价格、商品

1,2.00,burger
1,1.25,tofulog
1,2.00,tofulog
1,1.00,chef_salad
1,1.00,A+B
1,1.50,A+CCC
1,2.50,A
2,3.00,A
2,1.00,B
2,1.20,CCC
2,1.25,D

=====输出和错误====：

{'1': {'A': [[1.0, ['A', 'B']], [1.5, ['A', 'CCC']], [2.5, ['A', 'D']]], 'B': [[1.0, ['A', 'B']]], 'D': [[2.5, ['A', 'D']]], 'chef_salad': [[1.0, ['chef_salad']]], 'burger': [[2.0, ['burger']]], 'tofulog': [[1.25, ['tofulog']], [2.0, ['tofulog']]], 'CCC': [[1.5, ['A', 'CCC']]]}, '2': {'A': [[3.0, ['A']]], 'B': [[1.0, ['B']]], 'D': [[1.25, ['D']]], 'CCC': [[1.2, ['CCC']]]}}
SHOP [[2.0, ['burger']]]
Traceback (most recent call last):
  File "work.py", line 55, in <module>
    solver(shop_text,['burger'])
  File "work.py", line 43, in solver
    (price, solution) = solve_one_shop(shop_info, required_items)
  File "work.py", line 26, in solve_one_shop
    for (price,combo) in shop[first_item]:
KeyError: 'burger'

而如果我在 input-2.csv 上运行相同的代码，并查询求解器（shop_text，['A'，'CCC']），我会得到正确的结果

=====input-2.csv======

1,2.00,A
1,1.25,B
1,2.00,B
1,1.00,A
1,1.00,A+B
1,1.50,A+CCC
1,2.50,A+D
2,3.00,A
2,1.00,B
2,1.20,CCC
2,1.25,D

=========输出====

{'1': {'A': [[2.0, ['A']], [1.0, ['A']], [1.0, ['A', 'B']], [1.5, ['A', 'CCC']], [2.5, ['A', 'D']]], 'B': [[1.25, ['B']], [2.0, ['B']], [1.0, ['A', 'B']]], 'D': [[2.5, ['A', 'D']]], 'CCC': [[1.5, ['A', 'CCC']]]}, '2': {'A': [[3.0, ['A']]], 'B': [[1.0, ['B']]], 'D': [[1.25, ['D']]], 'CCC': [[1.2, ['CCC']]]}}
SHOP [[2.0, ['A']], [1.0, ['A']], [1.0, ['A', 'B']], [1.5, ['A', 'CCC']], [2.5, ['A', 'D']]]
SHOP [[1.5, ['A', 'CCC']]]
SHOP [[1.5, ['A', 'CCC']]]
SHOP [[1.5, ['A', 'CCC']]]
SHOP [[1.5, ['A', 'CCC']]]
SHOP [[3.0, ['A']]]
SHOP [[1.2, ['CCC']]]
SHOP_ID=1
A,CCC(1.50) = 1.50

Answer 1

如果你这样做，你可以找出错误：

在您的solve_one_shop 方法中，在first_item = items[0] 行之后打印字典shop。这样做会打印出来：

{'A': [[3.0, ['A']]], 'B': [[1.0, ['B']]], 'D': [[1.25, ['D']]], 'CCC': [[1.2, ['CCC']]]}

所以，burger 不是它的键之一，因此它会抛出一个 KeyError

添加这一行： 2,1.25,burger 到您的 input.csv 文件的末尾，您的代码可以正常工作。

在 try except 块中读取 shop 字典中的值，以处理可能不存在项目的情况。

注意：
在您的方法build_shops 中：

shop_id,cost,items = item_info.replace('\n', '').split(',')

虽然去掉了换行符，但并没有去掉回车符。要解决此问题，请执行以下操作：

shop_id,cost,items = item_info.replace('\n', '').replace('\r', '').split(',')

希望这会有所帮助。

Answer 2

我想我已经解决了...

solve_one_shop

for 循环应该只发生在if 内，否则你会得到一个KeyError。此外，我已对其进行了更改，使其仅在 all_possible 包含任何内容时返回（空的 list 评估为 False。

编辑为了防止 TypeError 我已经分配了一个临时值 this_subset 并且循环的其余部分只发生它不是 None。

def solve_one_shop(shop, items):
    if len(items) == 0:
        return [0.0, []]
    all_possible = []
    first_item = items[0]
    if first_item in shop:
        for (price,combo) in shop[first_item]:
            sub_set = [x for x in items if x not in combo]
            this_subset = solve_one_shop(shop, sub_set)
            if this_subset is not None:
                price_sub_set,solution = this_subset
                solution.append([price,combo])
                all_possible.append([price+price_sub_set, solution])

    if all_possible:
        cheapest = min(all_possible, key=(lambda x: x[0]))
        return cheapest

求解器

我已将solve_one_shop 的返回值分配给一个中间变量。如果是None，则该店铺未添加到result_all_shops。

编辑如果result_all_shops为空，则打印一条消息而不是尝试查找min。

def solver(input_data, required_items):
    shops = build_shops(input_data)
    result_all_shops = []
    for shop_id,shop_info in shops.iteritems():
        this_shop = solve_one_shop(shop_info, required_items)
        if this_shop is not None:
            (price, solution) = this_shop
            result_all_shops.append([shop_id, price, solution])

    if result_all_shops:
        shop_id,total_price,solution = min(result_all_shops, key=(lambda x: x[1]))
        print('SHOP_ID=%s' % shop_id)
        sln_str = [','.join(items)+'(%0.2f)'%price for (price,items) in solution]
        sln_str = '+'.join(sln_str)
        print(sln_str + ' = %0.2f' % total_price)
    else:
        print "Item not available"