我在将返回的解决方案正确装箱为 min_cals <= sum(calories) <= max_cals 时遇到一些困难。有一些组合可以产生 sum(cals) < min_cals 的解决方案。除了保持在热量范围内之外，解决方案还应该生成一个列表，其中总成本尽可能接近预算而不超过预算限制。这是一些重新上下文化的代码。我真的可以帮忙：menu = [ {'name':'Cheese Pizza Slice', 'calories': 700, 'cost': 4}, {'name':'House Salad', 'calories': 100, 'cost': 8.5}, {'name':'Grilled Shrimp', 'calories': 400, 'cost': 15}, {'name':'Beef Brisket', 'calories': 400, 'cost': 12}, {'name':'Soda', 'calories': 100, 'cost': 1}, {'name':'Cake', 'calories': 300, 'cost': 3},]def menu_recommendation(menu, min_cal, max_cal, budget): menu = [item for item in menu if item['calories'] <= max_cal and item['cost'] <= budget] if len(menu) == 0: return [] return min(( [item] + menu_recommendation(menu, min_cal - item['calories'], max_cal - item['calories'], budget - item['cost']) for item in menu ), key= lambda recommendations: [budget - sum(item['cost'] for item in recommendations) and min_cal <= sum(item['calories'] for item in recommendations) <= max_cal, -sum(item['calories'] for item in recommendations)] )recommendation = menu_recommendation(menu, 1000, 1200, 15)total_cost = sum(item['cost'] for item in recommendation)total_cals = sum(item['calories'] for item in recommendation)print(f'recommendation: {recommendation}')print(f'total cost: {total_cost}')print(f'total calories: {total_cals}')例如，以下命令返回总卡路里数为 700 的解决方案，低于最小值 1000。推荐 = menu_recommendation(菜单, 1000, 1200, 15)
4 回答
慕工程0101907
TA贡献1887条经验 获得超5个赞
我想出了这个，它基本上是一个背包，但如果不适合推荐，它会从菜单中递归删除菜肴：
menu = [
{'name':'Cheese Pizza Slice', 'calories': 700, 'cost': 4},
{'name':'House Salad', 'calories': 100, 'cost': 8.5},
{'name':'Grilled Shrimp', 'calories': 400, 'cost': 15},
{'name':'Beef Brisket', 'calories': 400, 'cost': 12},
{'name':'Soda', 'calories': 100, 'cost': 1},
{'name':'Cake', 'calories': 300, 'cost': 3},
]
def get_price(recommendation):
return sum(dish["cost"] for dish in recommendation)
def get_calories(recommendation):
return sum(dish["calories"] for dish in recommendation)
def menu_recommendation(menu, min_cal, max_cal, budget):
sorted_menu = sorted(menu, key=lambda dish: dish["cost"], reverse=True)
recommendation = []
for dish in sorted_menu:
if dish["cost"] + get_price(recommendation) <= budget:
recommendation.append(dish)
if recommendation:
if get_calories(recommendation) < min_cal:
sorted_menu.remove(min(recommendation, key=lambda dish: dish["calories"]/dish["cost"]))
return menu_recommendation(sorted_menu, min_cal, max_cal, budget)
if get_calories(recommendation) > max_cal:
sorted_menu.remove(max(recommendation, key=lambda dish: dish["calories"]/dish["cost"]))
return menu_recommendation(sorted_menu, min_cal, max_cal, budget)
return recommendation
recommendation = menu_recommendation(menu, 500, 800, 15)
total_cost = sum(item['cost'] for item in recommendation)
total_cals = sum(item['calories'] for item in recommendation)
print(f'recommendation: {recommendation}')
print(f'total cost: {total_cost}')
它根据卡路里/成本率删除元素，因为这是应用背包的成本。
如果您有任何疑问，请告诉我。
ibeautiful
TA贡献1993条经验 获得超5个赞
(calorie, cost)这是一个动态编程解决方案，它构建了一个数据结构，显示我们可以最终得到的所有选项以及每个选项。我们寻找符合标准的最佳方案，然后找到推荐方案。
def menu_recommendation(menu, min_cal, max_cal, budget):
# This finds the best possible solution in pseudo-polynomial time.
recommendation_tree = {(0, 0.0): None}
for item in menu:
# This tree will wind up being the old plus new entries from adding this item.
new_recommendation_tree = {}
for key in recommendation_tree.keys():
calories, cost = key
new_recommendation_tree[key] = recommendation_tree[key]
new_key = (calories + item['calories'], cost + item['cost'])
if new_key not in recommendation_tree and new_key[0] <= max_cal:
# This is a new calorie/cost combination to look at.
new_recommendation_tree[new_key] = item
# And now save the work.
recommendation_tree = new_recommendation_tree
# Now we look for the best combination.
best = None
for key in recommendation_tree:
calories, cost = key
# By construction, we know that calories <= max_cal
if min_cal <= calories:
if best is None or abs(budget - cost) < abs(budget - best[1]):
# We improved!
best = key
if best is None:
return None
else:
# We need to follow the tree back to the root to find the recommendation
calories, cost = best
item = recommendation_tree[best]
answer = []
while item is not None:
# This item is part of the menu.
answer.append(item)
# And now find where we were before adding this item.
calories = calories - item['calories']
cost = cost - item['cost']
best = (calories, cost)
item = recommendation_tree[best]
return answer
GCT1015
TA贡献1827条经验 获得超4个赞
也许我们可以做一些递归的事情。
def smallest_combo(lst, m, n, z):
# filter list to remove elements we can't use next without breaking the rules
lst = [dct for dct in lst if m <= dct['x'] <= n and dct['y'] <= z]
# recursive base case
if len(lst) == 0:
return []
# go through our list of eligibles
# simulate 'what would the best possibility be if we picked that one to go with next'
# then of those results select the one with the sum('y') closest to z
# (breaking ties with the largest sum('x'))
return min((
[dct] + smallest_combo(lst, m - dct['x'], n - dct['x'], z - dct['y'])
for dct in lst
), key=
lambda com: [z - sum(d['y'] for d in com), -sum(d['x'] for d in com)]
)
inp = [{'name': 'item1', 'x': 600, 'y': 5},
{'name': 'item2', 'x': 200, 'y': 8},
{'name': 'item3', 'x': 500, 'y': 12.5},
{'name': 'item4', 'x': 0, 'y': 1.5},
{'name': 'item5', 'x': 100, 'y': 1}]
print(smallest_combo(inp, 500, 1500, 25))
# [{'name': 'item3', 'x': 500, 'y': 12.5}, {'name': 'item3', 'x': 500, 'y': 12.5}]
有多种方法可以加快这一速度。首先通过制作递归缓存，然后采用动态编程方法（即从底部开始而不是从顶部开始）。
烙印99
TA贡献1829条经验 获得超13个赞
我相信我已经解决了提出超出范围 min_cal / max_cal 边界的建议的问题，但我仍然觉得可能有一个更接近预算的解决方案。
这是我更新的代码：
menu = [
{'name':'Cheese Pizza Slice', 'calories': 700, 'cost': 4},
{'name':'House Salad', 'calories': 100, 'cost': 8.5},
{'name':'Grilled Shrimp', 'calories': 400, 'cost': 15},
{'name':'Beef Brisket', 'calories': 400, 'cost': 12},
{'name':'Soda', 'calories': 100, 'cost': 1},
{'name':'Cake', 'calories': 300, 'cost': 3},
]
def menu_recommendation(menu, min_cal, max_cal, budget):
menu = [item for item in menu if item['calories'] <= max_cal and item['cost'] <= budget]
if len(menu) == 0: return []
return min(([item] + menu_recommendation(menu, min_cal - item['calories'], max_cal - item['calories'], budget - item['cost'])
for item in menu
), key=
lambda recommendations: [budget - sum(item['cost'] for item in recommendations)
and min_cal - sum(item['calories'] for item in recommendations) >= 0
and max_cal - sum(item['calories'] for item in recommendations) >= 0,
-sum(item['calories'] for item in recommendations)]
)
recommendation = menu_recommendation(menu, 1000, 1200, 15)
total_cost = sum(item['cost'] for item in recommendation)
total_cals = sum(item['calories'] for item in recommendation)
print(f'recommendation: {recommendation}')
print(f'total cost: {total_cost}')
print(f'total calories: {total_cals}')
如果有人有任何改进，我很乐意听到他们的声音！
添加回答
举报
0/150
提交
取消