当使用不同的csv文件作为输入时的KeyError

Question

我有这个代码基本上为csv输入文件，打印出项目/项目的最低成本和餐馆ID。但它在一个输入文件上抛出“KeyError”，但在另一个输入文件上完美地工作（两者都是相同的样式）。有人能指出它错在哪里吗？由于

有一些重要事项:: 挑战::

1. 我的客户不想要访问多个餐馆。例如，如果他要求“extreme_fajita，jalapeno_poppers，extra_salsa”，那么代码应该打印商店6（它可以作为组合项目使用），而不是将用户需求分散在不同的餐馆周围（即使某些餐馆提供便宜）。

2. 其次&最重要的是：假设用户要求汉堡包。那么如果某个餐厅'X'给4个汉堡“汉堡”，而另一个餐厅'Y'给予“汉堡+金枪鱼+豆腐”如果价格为3美元，那么我们会告诉用户购买餐厅'Y'，即使除了用户要求的'汉堡'外还有其他物品，但只要价格便宜，我们很乐意为他们提供额外物品。

---

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] 
    for (price,combo) in shop[first_item]: 
     sub_set = [x for x in items if x not in combo] 
     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.csv','rb')  
#shops = build_shops(shop_text) 
#cheapest=solve_one_shop(shops,items) 
solver(shop_text,['A']) 

input.csv

1,4.00,tuna 
1,8.00,tofu 
2,5.00,tuna 
2,6.50,tofu 
3,4.00,chef_salad 
3,8.00,steak__sandwich 
4,5.00,steak__sandwich 
4,2.50,wine_spritzer 
5,4.00,extreme_fajita 
5,8.00,fancy_eu_water 
6,5.00,fancy_eu_water 
6,6.00,extreme_fajita+jalapeno_poppers+extra_salsa 

但我得到这个错误： -

Traceback (most recent call last): 
    File "working.py", line 56, in <module> 
    solver(shop_text,['extra_salsa']) 
    File "working.py", line 42, in solver 
    (price, solution) = solve_one_shop(shop_info, required_items) 
    File "working.py", line 27, in solve_one_shop 
    for (price,combo) in shop[first_item]: 
KeyError: 'extra_salsa' 

而如果我在另一个输入文件运行它，我得到正确的答案，并不克等任何错误。

input.csv

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

========输出=========

{'1': {'A': [[2.0, ['A']], [1.0, ['A', 'B']], [1.5, ['A', 'C']], [2.5, ['A', 'D']]], 'C': [[2.0, ['C']], [1.5, ['A', 'C']]], 'B': [[1.25, ['B']], [1.0, ['A', 'B']]], 'D': [[1.0, ['D']], [2.5, ['A', 'D']]]}, '2': {'A': [[3.0, ['A']]], 'C': [[1.2, ['C']]], 'B': [[1.0, ['B']]], 'D': [[1.25, ['D']]]}} 

SHOP_ID=1 
A,B(1.00) = 1.00 

Answer 1

会发生什么地方是没有extra_salsa在你的店里？

除了愤怒的萨尔萨爱好客户，您的脚本不能正常工作，因为密钥不存在。

就像你正在检查是否items是空的，你需要检查请求的项目实际上是在商店或不：

def solve_one_shop(shop, items): 
    if len(items) == 0: 
     return [0.0, []] 
    all_possible = [] 
    # first_item = items[0] 
    for item in items: 
     price,combo = shop.get(item, (0.0,[])) # This will return 
               # default values when 
               # the key doesn't exist  

---

让您通过优化加载代码开始：

import csv 

from collections import defaultdict 

def build_shops(shop_file_name): 
    shops = defaultdict(list) 
    with open(shop_file_name, 'r') as f: 
     reader = csv.reader(f, delimiter=',') 
     for row in reader: 
      id, cost, items = row 
      cost = float(cost) 
      items = items.split('+') 
      shops[id].append((cost, items,)) 
    return shops 

现在我们有一个返回字典的函数，每个键都是代表成本和项目列表的元组列表。

接下来，让优化求解：

def solver(shops, required_items): 
    result_all_shops = [] 
    shops_with_items = [] 

    for i in required_items: 
     for shop, inventory in shops.iteritems(): 
      for price, items in inventory: 
       if i in items: 
        shops_with_items.append((shop, price, i)) 

    if not shops_with_items: 
     return [] # No shops contained the items 

    for i in required_items: 
     result_all_shops.append(min(filter(lambda x: x[2] == i, shops_with_items), 
               key=lambda x: x[1])) 

    return result_all_shops 

最后，加载了一切：

if __name__ == '__main__': 
    shops = build_shops('input.csv') 
    items = ['extra_salsa','tofu'] 
    result = solver(shops, items) 
    if not result: 
     print('Sorry, no shops contained {}'.format(','.join(items))) 
    else: 
     for shop, item, price in result: 
      print('Shop {} had the item {} for {}'.format(shop,price,item)))