如何修复 'gurobipy.LinExpr' 对象不可迭代
how to fix 'gurobipy.LinExpr' object is not iterable
我在 gurobi 中使用线性规划模型时遇到一些问题。我对编程很陌生,并不真正理解错误是什么。它出现是因为 'gurobipy.LinExpr' 对象不能用我的 Objective 函数迭代。这是我目前的代码
from gurobipy import *
IDtoLVC = [
[40.4,61.2,69.3,60.9,86.8,13.1,70.5,69.7],
[59.5,5.6,45.8,85.6,62.6,46.9,77.8,29.7],
[27.2,62.7,26.9,30.1,30.3,62.6,11.7,45.0]
]
CCDPop = [2180,4887,2174,4244,2444,2156,2433,2096,4148,3636,4935,3160,4153,2058,4614,3862,4588,4264,2559,2770,3964,2333,3681,4786,3256]
CCDtoLVC = [
[48.8,0,0,0,0,19.2,0,0],
[46.3,0,0,0,0,8.3,0,0],
[0,37.2,0,0,0,17.7,0,0],
[0,14.0,0,0,0,0,0,36.1],
[0,18.4,0,0,0,0,0,42.2],
[35.5,0,0,0,0,20.1,0,0],
[33.3,0,0,0,0,6.8,0,0],
[24.9,0,0,0,0,18.7,0,0],
[0,17.4,0,0,0,30.8,0,0],
[0,12.4,0,0,0,0,0,20.1],
[23.5,0,0,31.6,0,0,0,0],
[6.9,0,0,28.3,0,0,0,0],
[17.9,0,27.7,0,0,0,0,0],
[0,24.4,20.3,0,0,0,0,16.4],
[0,28.9,26.1,0,0,0,0,11.5],
[28.3,0,0,12.0,0,0,0,0],
[13.0,0,0,14.4,0,0,24.9,0],
[13.1,0,26.7,31.0,0,0,24.0,0],
[0,0,6.5,0,17.1,0,0,25.4],
[0,0,20.3,0,16.4,0,0,23.3],
[0,0,0,15.1,0,0,35.3,0],
[27.2,0,0,14.1,0,0,7.8,0],
[0,0,25.6,0,25.0,0,14.7,0],
[0,0,26.7,0,12.0,0,0,0],
[0,0,30.2,0,15.5,0,0,0]
]
LVCCost = [194, 105, 184]
# Sets
I = range(3) # no. of IDs
J = range(8) # no. of LVCs
K = range(25) # no. of CCDs
T = range(6) # length of time
# Model
m = Model('Vaccine Distribution')
X = { (i,j,t): m.addVar() for i in I for j in J for t in T }
Y = { (j,k,t): m.addVar() for j in J for k in K for t in T }
m.setObjective((0.20 * quicksum(quicksum(quicksum(X[i,j,t] * IDtoLVC[i][j] for i in I for j in J for t in T)))) +
(quicksum(quicksum(quicksum(Y[j,k,t] * CCDtoLVC[k][j] for j in J for k in K for t in T)))) +
(quicksum(quicksum(quicksum(X[i,j,t] * LVCCost[i] for i in I for j in J for t in T)))),
GRB.MINIMIZE)
m.optimize()
您不需要对 quicksum
进行三次嵌套调用来表示三重和。相反,您只需调用 quicksum
一次:
sum1 = quicksum(X[i,j,t] * IDtoLVC[i][j] for i in I for j in J for t in T)
sum2 = quicksum(Y[j,k,t] * CCDtoLVC[k][j] for j in J for k in K for t in T)
sum3 = quicksum(X[i,j,t] * LVCCost[i] for i in I for j in J for t in T)
m.setObjective(0.20 * sum1 + sum2 + sum3, GRB.MINIMIZE)
我在 gurobi 中使用线性规划模型时遇到一些问题。我对编程很陌生,并不真正理解错误是什么。它出现是因为 'gurobipy.LinExpr' 对象不能用我的 Objective 函数迭代。这是我目前的代码
from gurobipy import *
IDtoLVC = [
[40.4,61.2,69.3,60.9,86.8,13.1,70.5,69.7],
[59.5,5.6,45.8,85.6,62.6,46.9,77.8,29.7],
[27.2,62.7,26.9,30.1,30.3,62.6,11.7,45.0]
]
CCDPop = [2180,4887,2174,4244,2444,2156,2433,2096,4148,3636,4935,3160,4153,2058,4614,3862,4588,4264,2559,2770,3964,2333,3681,4786,3256]
CCDtoLVC = [
[48.8,0,0,0,0,19.2,0,0],
[46.3,0,0,0,0,8.3,0,0],
[0,37.2,0,0,0,17.7,0,0],
[0,14.0,0,0,0,0,0,36.1],
[0,18.4,0,0,0,0,0,42.2],
[35.5,0,0,0,0,20.1,0,0],
[33.3,0,0,0,0,6.8,0,0],
[24.9,0,0,0,0,18.7,0,0],
[0,17.4,0,0,0,30.8,0,0],
[0,12.4,0,0,0,0,0,20.1],
[23.5,0,0,31.6,0,0,0,0],
[6.9,0,0,28.3,0,0,0,0],
[17.9,0,27.7,0,0,0,0,0],
[0,24.4,20.3,0,0,0,0,16.4],
[0,28.9,26.1,0,0,0,0,11.5],
[28.3,0,0,12.0,0,0,0,0],
[13.0,0,0,14.4,0,0,24.9,0],
[13.1,0,26.7,31.0,0,0,24.0,0],
[0,0,6.5,0,17.1,0,0,25.4],
[0,0,20.3,0,16.4,0,0,23.3],
[0,0,0,15.1,0,0,35.3,0],
[27.2,0,0,14.1,0,0,7.8,0],
[0,0,25.6,0,25.0,0,14.7,0],
[0,0,26.7,0,12.0,0,0,0],
[0,0,30.2,0,15.5,0,0,0]
]
LVCCost = [194, 105, 184]
# Sets
I = range(3) # no. of IDs
J = range(8) # no. of LVCs
K = range(25) # no. of CCDs
T = range(6) # length of time
# Model
m = Model('Vaccine Distribution')
X = { (i,j,t): m.addVar() for i in I for j in J for t in T }
Y = { (j,k,t): m.addVar() for j in J for k in K for t in T }
m.setObjective((0.20 * quicksum(quicksum(quicksum(X[i,j,t] * IDtoLVC[i][j] for i in I for j in J for t in T)))) +
(quicksum(quicksum(quicksum(Y[j,k,t] * CCDtoLVC[k][j] for j in J for k in K for t in T)))) +
(quicksum(quicksum(quicksum(X[i,j,t] * LVCCost[i] for i in I for j in J for t in T)))),
GRB.MINIMIZE)
m.optimize()
您不需要对 quicksum
进行三次嵌套调用来表示三重和。相反,您只需调用 quicksum
一次:
sum1 = quicksum(X[i,j,t] * IDtoLVC[i][j] for i in I for j in J for t in T)
sum2 = quicksum(Y[j,k,t] * CCDtoLVC[k][j] for j in J for k in K for t in T)
sum3 = quicksum(X[i,j,t] * LVCCost[i] for i in I for j in J for t in T)
m.setObjective(0.20 * sum1 + sum2 + sum3, GRB.MINIMIZE)