pulp solve 函数给出相同的输出

Question

我已经为不同的日子编写了以下代餐代码，但我每天都吃同样的餐。我想隔天 "meat" 和 "Vegetarian" food_groups。

my dataframe is as follows:

id      name               energy   sugar   Food_Groups
1       4-Grain Flakes      140     58.8    Breakfast
2       Beef Mince, Fried   1443    8.0     Meat
3       Pork                1000    3.0     Meat
4       cake                1200    150     Sweet
5       cheese              1100    140     Sweet
6       Juice               700     85      Drink
7       cabbage             60      13      vegetarian
8       cucumber            10      10      vegetarian
9       eggs                45      30      Breakfast

我正在使用 PuLP 来减少糖分，同时限制卡路里摄入量。

# Create the 'prob' variable to contain the problem data
prob = LpProblem("Simple Diet Problem",LpMinimize)
#create data variables and dictionary
food_items = list(df['name'])
calories = dict(zip(food_items,df['energy']))
sugars = dict(zip(food_items,df['sugar']))

food_vars =LpVariable.dicts("Food",food_items,lowBound=0,cat='Integer')

#Building the LP problem by adding the main objective function.
prob += lpSum([sugars[i]*food_vars[i] for i in food_items])

#adding calorie constraint
prob += lpSum([calories[f] * food_vars[f] for f in food_items]) >= 
1800.0, "CalorieMinimum"
prob += lpSum([calories[f] * food_vars[f] for f in food_items]) <= 
2200.0, "CalorieMaximum"

我循环 prob.solve() 以生成不同日期的菜单

prob.writeLP("SimpleDietProblem.lp")
days = ['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday']
for i in days:
    print(i)

    prob.solve(PULP_CBC_CMD())
#    print("Status:", LpStatus[prob.status])
    print("Therefore, the optimal balanced diet consists of\n"+"-")
    for v in prob.variables():
        if v.varValue:
            print(v.name , "=", v.varValue)
    print("The total sugar of this balanced diet is: {}\n\n".format(round(value(prob.objective),2)))

我的问题是每天都重复输出。如何隔天获得 "meat" 和 "vegetarian"？

Answer 1

@Khaned，做你想做的最简单的方法是设置两个 Problem 实例。一个有肉类选择，另一个有素食选择。在不同的日子使用每一个。您可以根据需要运行每周交替开始问题，以获得为期两周的膳食计划。

您可以像这样设置求解器：

prob1 = LpProblem("Simple Diet Problem Meat Day",LpMinimize)
prob2 = LpProblem("Simple Diet Problem Vegetarian Day",LpMinimize)
#create data variables and dictionary
day1_df = df[df['Food_Groups'] != 'vegetarian']
day1_items = list(day1_df['name'])
day1_calories = dict(zip(day1_items,day1_df['energy']))
day1_sugars = dict(zip(day1_items,day1_df['sugar']))
day2_df = df[df['Food_Groups'] != 'Meat']
day2_items = list(day2_df['name'])
day2_calories = dict(zip(day2_items,day2_df['energy']))
day2_sugars = dict(zip(day2_items,day2_df['sugar']))
# variables
day1_vars =LpVariable.dicts("Food",day1_items,lowBound=0,cat='Integer')
day2_vars =LpVariable.dicts("Food",day2_items,lowBound=0,cat='Integer')

#Building the LP problem by adding the main objective function.
prob1 += lpSum([day1_sugars[i]*day1_vars[i] for i in day1_items])
prob2 += lpSum([day2_sugars[i]*day2_vars[i] for i in day2_items])

如果您仍想显示并非全天都在肉类和素食之间选择的选项，则需要创建一个更复杂的模型，并为这些项目指定 food_vars 为零的约束。

两个问题各做一次。

接下来，在一周中的每一天分配一个列表中的问题，例如：

days = [('Monday', prob1), ('Tuesday', prob2), ...]

然后像你已经做的那样循环这些天并打印变量。

for day, prob in days:
    print(day)
    print("Therefore, the optimal balanced diet consists of\n"+"-")
    for v in prob.variables():
        if v.varValue:
            print(v.name , "=", v.varValue)
    print("The total sugar of this balanced diet is: {}\n\n".format(round(value(prob.objective),2)))

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