我必须用纸浆解决整数线性优化问题。我解决了问题并获得了等于42的优化值。但是当我编写更通用的代码时，比如在循环内声明变量，在循环内定义约束以及使用lpSum函数定义优化，我没有解决方案。 我认为我的问题在于定义下一个约束。
for a in itemset_dict.keys():
for b in itemset_dict[a][0]:
my_lp_program +=b >= a, "2Constraint"
我收到了下一个警告：
C:\Program Files\Python36\lib\site-packages\pulp\pulp.py:1353: UserWarning: Overwriting previously set objective.
warnings.warn("Overwriting previously set objective.")
Status: Optimal
Total Optimum= None
__dummy = None
products_beer = 0.0
products_cheese = 0.0
products_cola = 0.0
products_peanuts = 0.0
The thread 'MainThread' (0xb30) has exited with code 0 (0x0).
The program '[10140] python.exe' has exited with code 0 (0x0).
谢谢。
from pulp import *
# defining list of products
products = ['cola','peanuts', 'cheese', 'beer']
itemsets = ['x1','x2', 'x3']
#disctionary of the costs of each of the products is created
costs = {'cola' : 5, 'peanuts' : 3, 'cheese' : 1, 'beer' : 4 }
# dictionary of frequent itemsets
itemset_dict = { "x1" : (("cola", "peanuts"),10),
"x2" : (("peanuts","cheese"),20),
"x3" : (("peanuts","beer"),30)
}
products_var=LpVariable.dicts("Products", products, 0)
itemsets_var=LpVariable.dicts("Itemsets", itemsets, 0)
# defining itemsets variables
'''
for x in itemsets:
x = LpVariable('x', lowBound=0, upBound=1, cat='Binary')
'''
#x = LpVariable.dicts("x", itemsets, lowBound=0, upBound=1, cat='Binary')
#option2
i1=LpVariable.dict("itemsets", itemsets, lowBound=0, upBound=1, cat='Binary')
#print(i1);
#print(i1['x1'])
#print(type(i1['x1']))
'''
x1 = LpVariable('x1', lowBound=0, upBound=1, cat='Binary')
x2 = LpVariable('x2', lowBound=0, upBound=1, cat='Binary')
x3 = LpVariable('x3', lowBound=0, upBound=1, cat='Binary')
'''
# defining products variables
'''
for p in products:
p = LpVariable(p, lowBound=0, upBound=1, cat='Binary')
'''
#p = [LpVariable.dicts("p", i, lowBound=0, upBound=1, cat='Binary') for i in products]
'''
option1
p=[LpVariable(i, lowBound=0, upBound=1, cat='Binary') for i in products]
print(p[2])
print(type(p[2]))
'''
#option2
p1=LpVariable.dict("products", products, lowBound=0, upBound=1, cat='Binary')
#print(p1);
#print(p1['cola'])
#print(type(p1['cola']))
'''
cola = LpVariable('cola', lowBound=0, upBound=1, cat='Binary')
peanuts = LpVariable('peanuts', lowBound=0, upBound=1, cat='Binary')
cheese = LpVariable('cheese', lowBound=0, upBound=1, cat='Binary')
beer = LpVariable('beer', lowBound=0, upBound=1, cat='Binary')
'''
my_lp_program = LpProblem('My LP Problem', LpMaximize)
#my_lp_program += 10*x1+20*x2+30*x3-5*p1-3*p2-1*p3-4*p4 , "Maximization"
#my_lp_program += 10*x1+20*x2+30*x3 - lpSum([costs[i] * p1[i] for i in products]) , "Maximization"
my_lp_program += lpSum([itemset_dict[i][1] * i1[i] for i in itemsets]) - lpSum([costs[i] * p1[i] for i in products]) , "Maximization"
#my_lp_program += lpSum([itemset_dict[i][1] * itemsets_var[i] for i in itemsets]) - lpSum([costs[i] * products_var[i] for i in products]) , "Maximization"
#my_lp_program +=cola+peanuts+cheese+beer<=3, "1Constained"
my_lp_program +=lpSum([p1[i] for i in products]) <= 3, "1Constaint"
'''
my_lp_program +=cola>=x1, "2Constained"
my_lp_program +=peanuts>=x1, "3Constained"
my_lp_program +=peanuts>=x2, "4Constained"
my_lp_program +=cheese>=x2, "5Constained"
my_lp_program +=peanuts>=x3, "6Constained"
my_lp_program +=beer>=x3, "7Constained"
'''
for a in itemset_dict.keys():
for b in itemset_dict[a][0]:
my_lp_program +=b >= a, "2Constraint"
my_lp_program.writeLP("CheckLpProgram.lp")
my_lp_program.solve()
print("Status:", LpStatus[my_lp_program.status])
print("Total Optimum=", value(my_lp_program.objective))
for v in my_lp_program.variables():
print(v.name, "=", v.varValue)
答案 0 :(得分：4)
我在这里重写了你的代码，提供了一个可行的解决方案。请查看注释，这些注释提供了一些关于调试此类模型有用的提示。
from pulp import *
# defining list of products
products = ['cola','peanuts', 'cheese', 'beer']
itemsets = ['x1','x2', 'x3']
#disctionary of the costs of each of the products is created
costs = {'cola' : 5, 'peanuts' : 3, 'cheese' : 1, 'beer' : 4 }
# dictionary of frequent itemsets
# ~~> This is hard to maintain - I would select a different data structure
# it gets really complicated below as you will see!
itemset_dict = { "x1" : (("cola", "peanuts"),10),
"x2" : (("peanuts","cheese"),20),
"x3" : (("peanuts","beer"),30)
}
# Good practice to first define your problem
my_lp_program = LpProblem('My LP Problem', LpMaximize)
# ~~>You do not need bounds for binary variables, they are automatically 0/1
products_var=LpVariable.dicts("Products", products, cat='Binary')
itemsets_var=LpVariable.dicts("Itemsets", itemsets, cat='Binary')
# ~~> Not necessary - commended out
# defining itemsets variables
# for x in itemsets:
# x = LpVariable(x, lowBound=0, upBound=1, cat='Binary')
'''
x1 = LpVariable('x1', lowBound=0, upBound=1, cat='Binary')
x2 = LpVariable('x2', lowBound=0, upBound=1, cat='Binary')
x3 = LpVariable('x3', lowBound=0, upBound=1, cat='Binary')
'''
# ~~> Not necessary - commended out
# defining products variables
# for p in products:
# p = LpVariable(p, lowBound=0, upBound=1, cat='Binary')
'''
cola = LpVariable('cola', lowBound=0, upBound=1, cat='Binary')
peanuts = LpVariable('peanuts', lowBound=0, upBound=1, cat='Binary')
cheese = LpVariable('cheese', lowBound=0, upBound=1, cat='Binary')
beer = LpVariable('beer', lowBound=0, upBound=1, cat='Binary')
'''
# ~~> Not necessary - commended out, see below
#my_lp_program += 10*x1+20*x2+30*x3-5*p1-3*p2-1*p3-4*p4 , "Maximization"
# my_lp_program += lpSum([itemset_dict[i][1] * itemsets_var[i] for i in itemsets]) - lpSum([costs[i] * products_var[i] for i in products]) , "Maximization"
# ~~> Use an affine expression to define your objective.
# ~~> Even better, define two objects as LpAffineExpression and add them,
# ~~> it keeps the code cleaner
my_lp_program += LpAffineExpression([(
itemsets_var[x], itemset_dict[x][1]) for x in itemsets_var]) + \
LpAffineExpression([(
products_var[x], -costs[x]) for x in products_var])
# ~~> Not necessary - commended out
#my_lp_program +=cola+peanuts+cheese+beer<=3, "1Constained"
# my_lp_program +=lpSum([products_var[i] for i in products]) <= 3, "1Constaint"
# ~~> This is the right way to enter this constraint.
# ~~> I do not like the naming though..
my_lp_program += lpSum(products_var) <= 3, '1Constraint'
'''
my_lp_program +=cola>=x1, "2Constained"
my_lp_program +=peanuts>=x1, "3Constained"
my_lp_program +=peanuts>=x2, "4Constained"
my_lp_program +=cheese>=x2, "5Constained"
my_lp_program +=peanuts>=x3, "6Constained"
my_lp_program +=beer>=x3, "7Constained"
'''
# ~~> Here are your constraints
counter = 1
for a in itemset_dict.keys():
item = itemsets_var[a]
for b in itemset_dict[a][0]:
product = products_var[b]
counter +=1
my_lp_program += product >= item, "{}Constraint".format(counter)
# ~~> Great that you export the lp! If you look at the file you can
# ~~> spot a lot of issues with the model during debugging
my_lp_program.writeLP("CheckLpProgram.lp")
my_lp_program.solve()
print("Status:", LpStatus[my_lp_program.status])
print("Total Optimum=", value(my_lp_program.objective))
for v in my_lp_program.variables():
print(v.name, "=", v.varValue)
输出：
('Status:', 'Optimal')
('Total Optimum=', 42.0)
('Itemsets_x1', '=', 0.0)
('Itemsets_x2', '=', 1.0)
('Itemsets_x3', '=', 1.0)
('Products_beer', '=', 1.0)
('Products_cheese', '=', 1.0)
('Products_cola', '=', 0.0)
('Products_peanuts', '=', 1.0)