我是Pyomo/Python的新用户，我只是想知道如何在类类型的Pyomo模型中显示最佳变量值。

我刚刚尝试了Pyomo示例库中的标准示例“最小成本流模型”。该代码在中提供https://github.com/Pyomo/PyomoGallery/wiki/Min-cost-flow

在代码的底部，它说：

sp = MinCostFlow('nodes.csv', 'arcs.csv') 
sp.solve()
print('\n\n---------------------------')
print('Cost: ', sp.m.OBJ())

输出是

Academic license - for non-commercial use only
Read LP format model from file.    
Reading time = 0.00 seconds
x8: 7 rows, 8 columns, 16 nonzeros
No parameters matching 'mip_tolerances_integrality' found
No parameters matching 'mip_tolerances_mipgap' found
Optimize a model with 7 rows, 8 columns and 16 nonzeros
Coefficient statistics:
  Matrix range     [1e+00, 1e+00]
  Objective range  [1e+00, 5e+00]
  Bounds range     [0e+00, 0e+00]
  RHS range        [1e+00, 1e+00]
Presolve removed 7 rows and 8 columns
Presolve time: 0.00s
Presolve: All rows and columns removed
Iteration    Objective       Primal Inf.    Dual Inf.      Time
       0    5.0000000e+00   0.000000e+00   0.000000e+00      0s

Solved in 0 iterations and 0.01 seconds
Optimal objective  5.000000000e+00

我只能得到最优目标，但最优变量值呢？我还搜索了文档，其中告诉我使用以下内容：

print("x[2]=",pyo.value(model.x[2])).  

但它不适用于类类型模型，如最小成本流模型。

我还试图修改类中的函数定义：

def solve(self):
        """Solve the model."""
        solver = pyomo.opt.SolverFactory('gurobi')
        results = solver.solve(self.m, tee=True, keepfiles=False, options_string="mip_tolerances_integrality=1e-9, mip_tolerances_mipgap=0")
        print('\n\n---------------------------')
        print('First Variable: ', self.m.Y[0])

但效果并不理想。输出为：

KeyError: "Index '0' is not valid for indexed component 'Y'"

你能帮我吗？谢谢！

加布里埃尔