Python返回适合设置参数的项目列表

Question

我在将返回的解决方案正确装箱到 min_cals

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，以下返回总卡路里数为 700 的解决方案，低于 1000 的最小值。 推荐 = menu_recommendation(menu, 1000, 1200, 15)

Answer 1

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}]

有很多方法可以加快速度。首先是制作递归缓存，其次是采用动态编程方法（即从底部开始而不是从顶部开始）。

Answer 2

这是一个动态编程解决方案，它构建了一个数据结构，显示了我们可以使用的所有 (calorie, cost) 选项以及每个选项。我们会寻找符合标准的最佳方案，然后找出是什么建议。

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

Answer 3

我想出了这个，它基本上是一个背包，但是如果它们不适合推荐，它会递归地从菜单中删除它们：

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/成本比率去除元素，因为它是应用背包的成本。

如果您有任何问题，请告诉我。

Answer 4

我相信我已经解决了提出超出范围 min_cal / max_cal 界限的建议的问题，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！