在Python代碼中處理和繞過「TypeError：'NoneType'」

所以我有這個代碼打印出項目/項目的最低成本和餐館ID。客戶不想訪問多家餐館。例如，如果他要求「A，B」，那麼代碼應該打印提供這兩者的商店，而不是將用戶需求分散在不同的餐館周圍（即使一些餐館提供便宜）。在Python代碼中處理和繞過「TypeError：'NoneType'」

的誤差基本進入，因爲無論是項（「漢堡」和「d」）不提供「一起」在任何兩個餐館（1 & 2）的。但是，我不想扔這麼長的錯誤，我只想打印「其中一個項目不可用於恢復」等。

其他此類錯誤引發的協議是solver(shop_text,['tofulog', 'D'])，因爲'tofulog'僅在restaurant_1，wheras 'D'只能在restaurant_2上使用。

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: 
     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]) 

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


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]) 

    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', 'D'])

=====輸入1.csv ===== restaurant_id，價格，項目

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

======輸出====== =

Traceback (most recent call last): 
    File "26mar_cheap.py", line 106, in <module> 
    final_out(restaurant_read,sys.argv[2:]) 
    File "26mar_cheap.py", line 92, in final_out 
    this_resto = requirement_one_restaurant(shop_info, required_items) 
    File "26mar_cheap.py", line 77, in requirement_one_restaurant 
    cost_sub_set,solution = requirement_one_restaurant(shop, sub_set) 
TypeError: 'NoneType' object is not iterable

來源

2014-03-26 sunita

回溯跟蹤似乎沒有要切合你已經發布的代碼：該錯誤是在requirement_one_restaurant但你只發布solve_one_shop。

但是，假設代碼的模式是相似的，你顯然正在做某種遞歸調用。當你這樣做時，你需要確保所有可能的路徑返回一個值。例如，在solve_one_shop中，如果all_possible在該函數結束時仍然爲空，則不會返回任何內容，這將導致出現NoneType錯誤。

來源

2014-03-26 09:32:46

在Python代碼中處理和繞過「TypeError：'NoneType'」

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