我有這個代碼基本上爲csv輸入文件,打印出項目/項目的最低成本和餐館ID。但它在一個輸入文件上拋出「KeyError」,但在另一個輸入文件上完美地工作(兩者都是相同的樣式)。有人能指出它錯在哪裏嗎?由於當使用不同的csv文件作爲輸入時的KeyError
有一些重要事項:: 挑戰::
我的客戶不想要訪問多個餐館。例如,如果他要求「extreme_fajita,jalapeno_poppers,extra_salsa」,那麼代碼應該打印商店6(它可以作爲組合項目使用),而不是將用戶需求分散在不同的餐館周圍(即使某些餐館提供便宜)。
其次&最重要的是:假設用戶要求漢堡包。那麼如果某個餐廳'X'給4個漢堡「漢堡」,而另一個餐廳'Y'給予「漢堡+金槍魚+豆腐」如果價格爲3美元,那麼我們會告訴用戶購買餐廳'Y',即使除了用戶要求的'漢堡'外還有其他物品,但只要價格便宜,我們很樂意爲他們提供額外物品。
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
但我得到這個錯誤: -
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'
而如果我在另一個輸入文件運行它,我得到正確的答案,並不克等任何錯誤。
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
========輸出=========
{'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
感謝BURHAN。但extra_salsa在餐廳6。其次,如果我查詢求解器(shop_text,['tuna']),我會得到同樣的錯誤。第三,如果我只用字母替換這些項目(例如,'A'代表'金槍魚','B'代表'豆腐'等),如果我解決了這個問題(shop_text,['A']),我會得到根據需要修正結果。爲什麼在兩個csv文件上這種不同的行爲? – sunita
Burhan Ive編輯我的帖子以包含正確的輸出,以及我的代碼正在工作的文件。請看看並幫我修復它。謝謝 !!!! – sunita
Burkhan,但如果客戶想要:-'extreme_fajita,jalapeno_poppers,extra_salsa',那麼你的代碼給5,6個獨立餐廳。但客戶只希望在一家餐廳(如果可以的話)提供所需的餐點,在這種情況下,餐廳是6.請參閱 - 我編輯了我的帖子並添加了「重要考慮因素」 – sunita