假设我有三个产品(A,B C)的清单。每个产品都有价格。考虑到总成本,我希望找到所有可能的产品组合,使其与该成本完全相同。
到目前为止,我尝试了类似的东西:
for price in product:
ret = []
for i in range(int(totalCost / price), -1, -1):
ret.append(i)
for c in range(1, len(products)+1, 1):
ret.append(int(products[c-1][1]/products[c][1]))
这就是我被卡住的地方。这将为我提供一系列可能性,但它只包含列表中较晚(比当前位置)的产品。它不会包含开头,因此,给我一切可能性。
我需要做些什么才能获得所有可能性?
答案 0 :(得分:6)
def possibilities(available_products, target_price):
if target_price == 0 or not available_products:
return []
this_price = available_products[0]
remaining_products = available_products[1:]
results = []
for qty in range(1 + target_price / this_price):
remaining_price = target_price - qty*this_price
if remaining_price == 0:
results.append([qty] + [0] * len(remaining_products))
else:
for option in possibilities(remaining_products, remaining_price):
results.append([qty] + option)
return results
这会给你:
pprint.pprint(possibilities([1, 2, 5], 10))
[[0, 0, 2],
[0, 5, 0],
[1, 2, 1],
[2, 4, 0],
[3, 1, 1],
[4, 3, 0],
[5, 0, 1],
[6, 2, 0],
[8, 1, 0],
[10, 0, 0]]
答案 1 :(得分:3)
itertools 模块提供组合生成器来帮助解决以下问题:
>>> from itertools import *
>>> prices = dict(a=10, b=15, c=8, d=2, e=5)
>>> total_cost = 20
>>> for r in range(1, 30):
for t in combinations_with_replacement(prices, r):
cost = sum(prices[p] for p in t)
if cost == total_cost:
print t