我有以下代码创建属于某个价格范围的水果,蔬菜和饮料的组合：

fruits = [('apple', 1), ('banana', 2), ('orange', 3)]
veggies = [('tomato', 1), ('cucumber', 2), ('onion', 3)]
drinks = [('water', 1), ('juice', 2), ('soda', 3)]

for fruit1 in fruits:
    for fruit2 in fruits:
        for veggie1 in veggies:
            for veggie2 in veggies:
                for drink1 in drinks:
                    for drink2 in drinks:
                        if 12 <= fruit1[1]+fruit2[1]+veggie1[1]+veggie2[1]+drink1[1]+drink2[1] <= 14: 
                            if fruit1 != fruit2: 
                                if veggie1 != veggie2: 
                                    if drink1 != drink2:
                                        combos.append((fruit1[0], fruit2[0], veggie1[0], veggie2[0], drink1[0], drink2[0], fruit1[1]+fruit2[1]+veggie1[1]+veggie2[1]+drink1[1]+drink2[1])) 

它会检查以确保不会两次采摘相同的水果/蔬菜/饮料,但我需要确保组合是唯一的.

例如,如果我最终得到

('apple', 'banana', 'tomato', 'cucumber', 'water', 'soda', 10) 

我也不要

('banana', 'apple', 'tomato', 'cucumber', 'water', 'soda', 10)

等等.这给我带来了应有的麻烦,因此我们将不胜感激.

解决方法:

您可以像这样处理数据

prices = {k:v for items in [fruits, veggies, drinks] for k, v in items}
fru,veg,dri=[i[0] for i in fruits],[i[0] for i in veggies],[i[0] for i in drinks]

from itertools import combinations, product, chain
for items in product(*(combinations(i, r = 2) for i in (fru, veg, dri))):
    total = sum(prices[i] for item in items for i in item)
    if 12 <= total <= 14:
        print tuple(chain.from_iterable(items)) + (total,)

输出量

('apple', 'banana', 'tomato', 'onion', 'juice', 'soda', 12)
('apple', 'banana', 'cucumber', 'onion', 'water', 'soda', 12)
('apple', 'banana', 'cucumber', 'onion', 'juice', 'soda', 13)
('apple', 'orange', 'tomato', 'cucumber', 'juice', 'soda', 12)
('apple', 'orange', 'tomato', 'onion', 'water', 'soda', 12)
('apple', 'orange', 'tomato', 'onion', 'juice', 'soda', 13)
('apple', 'orange', 'cucumber', 'onion', 'water', 'juice', 12)
('apple', 'orange', 'cucumber', 'onion', 'water', 'soda', 13)
('apple', 'orange', 'cucumber', 'onion', 'juice', 'soda', 14)
('banana', 'orange', 'tomato', 'cucumber', 'water', 'soda', 12)
('banana', 'orange', 'tomato', 'cucumber', 'juice', 'soda', 13)
('banana', 'orange', 'tomato', 'onion', 'water', 'juice', 12)
('banana', 'orange', 'tomato', 'onion', 'water', 'soda', 13)
('banana', 'orange', 'tomato', 'onion', 'juice', 'soda', 14)
('banana', 'orange', 'cucumber', 'onion', 'water', 'juice', 13)
('banana', 'orange', 'cucumber', 'onion', 'water', 'soda', 14)

如果您只想从饮料中选择一种元素,则可以像这样更改组合部分

d = {0: 2, 1: 2, 2: 1}
for items in product(*(combinations(j, r=d.get(i)) for i, j in enumerate((fru,veg,dri)))):

更改后,输出变为

('apple', 'orange', 'cucumber', 'onion', 'soda', 12)
('banana', 'orange', 'tomato', 'onion', 'soda', 12)
('banana', 'orange', 'cucumber', 'onion', 'juice', 12)
('banana', 'orange', 'cucumber', 'onion', 'soda', 13)