如何計算總和許多對象

-2

ids = [] 
menu = {} 
for offer in customer_order.split(';'): 
    id, count = offer.split(':') 
    ids.append(id) 
    menu[int(id)] = int(count) 
mList = Goods.objects.filter(id__in=ids, 
          kind_cost=1).values('id', 'cost') 
for item in mList: 
    id = item['id'] 
    cost = item['cost'] 
    menu[id] = menu[id] * cost 
return sum(menu.values())

customer_order是包括字符串：'32：5; 45：2; 555：23' 和等

我的問題：我確定有最好的解決方案來實現結果。任何人都可以幫我尋找解決方案嗎？請分享鏈接閱讀如何改進代碼

Tnx！

UPD：我需要總結的所有商品

來源

2015-02-05 Abbasov Alexander

'customer_order'是字典嗎？ 'dict'沒有'split'方法....你在做什麼 - 你用文字而不是代碼來描述它？ – 2015-02-05 23:15:15

'Goods.objects.'應該做什麼？ – Marcin 2015-02-05 23:26:15

@Marcin'Goods'是一個django模型。 'objects.filter'是一個數據庫查詢，它返回該模型中所有在該過濾器後面的對象（在這種情況下，'id in ids AND kind_cost == 1'） – 2015-02-05 23:31:29

看起來成本像customer_order是一個字符串，它看起來像一個字典（可能是JSON？）如果是JSON你應該使用json模塊解析它，但我會繼續工作，假設它實際上是一個本地Python字典對象作爲一個字符串。

import ast 

customer_order = ast.literal_eval(customer_order) 
# makes this into an ACTUAL dict... 

mList = Goods.objects.filter(id__in=customer_order.keys(), 
           kind_cost=1).values('id', 'cost') 
# mList is now a list of dicts, each dict has keys `id` and `cost` 

mList = dict([k.values() for k in mList]) 
# mList is now a single dict with id:cost pairings 

# As a lengthy aside, you might be better off just doing: 
# 
# # mList = Goods.objects.filter(id__in=customer_order.keys(),kind_cost=1) 
# # mList = {item.id:item.cost for item in mList} 
# 
# I find that must easier to read. If your items have a huge number of 
# columns, of course, this will vastly increase your query time. YMMV. 

result = [qty * mList[id] for id,qty in customer_order.items()]

如果customer_order實際上只是鍵值配對，看起來像key1:value1;key2:value2那麼你就必須做一些預處理。

customer_order = {keyvalue.split(":") for keyvalue in customer_order.split(";")}

來源

2015-02-05 23:37:58

'customer_order'不是JSON它更類似於CommaSeparatedField，但使用分號。 – 2015-02-06 01:10:54

如何計算總和許多對象

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