我目前正在 Python GEKKO 中实现 MINLP 优化问题，以确定三代能源系统的最佳运行策略。由于我将不同代表性日子的所有时期的能源需求视为输入数据，因此基本上我所有的决策变量、中间体等都是二维数组。
我怀疑 2D 中间体的声明是我的问题。现在我使用列表理解来声明 2D 中间体，但似乎 python 不能使用这些中间体。此外，误差这个稳态 IMODE 只允许标量值。发生。
每当我像这样使用 GEKKO m.Array 函数时:e_GT = m.Array(m.Intermediate(E_GT[z][p]/E_max_GT) for z in range(Z) for p in range(P), (Z,P))它说，不能调用 GEKKO 对象 m.Intermediate。
如果有人能给我提示，我将不胜感激。
这是完整的代码:
"""
Created on Fri Nov 22 10:18:33 2019
@author: julia
"""
# __Get GEKKO & numpy___
from gekko import GEKKO
import numpy as np
# ___Initialize model___
m = GEKKO()
# ___Global options_____
m.options.SOLVER = 1 # APOPT is MINLP Solver
# ______Constants_______
i = m.Const(value=0.05)
n = m.Const(value=10)
C_GT = m.Const(value=100000)
C_RB = m.Const(value=10000)
C_HB = m.Const(value=10000)
C_RS = m.Const(value=10000)
C_RE = m.Const(value=10000)
Z = 12
P = 24
E_min_GT = m.Const(value=1000)
E_max_GT = m.Const(value=50000)
F_max_GT = m.Const(value=100000)
Q_max_GT = m.Const(value=100000)
a = m.Const(value=1)
b = m.Const(value=1)
c = m.Const(value=1)
d = m.Const(value=1)
eta_RB = m.Const(value=0.01)
eta_HB = m.Const(value=0.01)
eta_RS = m.Const(value=0.01)
eta_RE = m.Const(value=0.01)
alpha = m.Const(value=0.01)
# ______Parameters______
T_z = m.Param([31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31])
C_p_Gas = m.Param(np.ones([P]))
C_p_Elec = m.Param(np.ones([P]))
E_d = np.ones([Z,P])
H_d = np.ones([Z,P])
K_d = np.ones([Z,P])
# _______Variables______
E_purch = m.Array(m.Var, (Z,P), lb=0)
E_GT = m.Array(m.Var, (Z,P), lb=0)
F_GT = m.Array(m.Var, (Z,P), lb=0)
Q_GT = m.Array(m.Var, (Z,P), lb=0)
Q_GT_RB = m.Array(m.Var, (Z,P), lb=0)
Q_disp = m.Array(m.Var, (Z,P), lb=0)
Q_HB = m.Array(m.Var, (Z,P), lb=0)
K_RS = m.Array(m.Var, (Z,P), lb=0)
K_RE = m.Array(m.Var, (Z,P), lb=0)
delta_GT = m.Array(m.Var, (Z,P), lb=0, ub=1, integer=True)
delta_RB = m.Array(m.Var, (Z,P), lb=0, ub=1, integer=True)
delta_HB = m.Array(m.Var, (Z,P), lb=0, ub=1, integer=True)
delta_RS = m.Array(m.Var, (Z,P), lb=0, ub=1, integer=True)
delta_RE = m.Array(m.Var, (Z,P), lb=0, ub=1, integer=True)
# ____Intermediates_____
R = m.Intermediate((i*(1+i)**n)/((1+i)**n-1))
e_min_GT = m.Intermediate(E_min_GT/E_max_GT)
e_GT = [m.Intermediate(E_GT[z][p]/E_max_GT) for z in range(Z) for p in range(P)]
f_GT = [m.Intermediate(F_GT[z][p]/F_max_GT) for z in range(Z) for p in range(P)]
q_GT = [m.Intermediate(Q_GT[z][p]/Q_max_GT) for z in range(Z) for p in range(P)]
Q_RB = [m.Intermediate(eta_RB*Q_GT_RB[z][p]*delta_RB[z][p]) for z in range(Z) for p in range(P)]
F_HB = [m.Intermediate(eta_HB*Q_HB[z][p]*delta_HB[z][p]) for z in range(Z) for p in range(P)]
Q_RS = [m.Intermediate(eta_RS*K_RS[z][p]*delta_RS[z][p]) for z in range(Z) for p in range(P)]
E_RE = [m.Intermediate(eta_RE*K_RE[z][p]*delta_RE[z][p]) for z in range(Z) for p in range(P)]
F_Gas = [m.Intermediate(F_GT[z][p] + eta_HB*Q_HB[z][p]*delta_HB[z][p]) for z in range(Z) for p in range(P)]
Cc = m.Intermediate(R*(C_GT + C_RB + C_HB + C_RS + C_RE))
Cr_z = m.Intermediate((sum(C_p_Gas[p]*F_Gas[z][p] + C_p_Elec[p]*E_purch[z][p]) for p in range(P)) for z in range(Z))
Cr = m.Intermediate(sum(Cr_z[z]*T_z[z]) for z in range(Z))
# ______Equations_______
m.Equation(e_min_GT[z][p]*delta_GT[z][p] <= e_GT[z][p] for z in range(Z) for p in range(P))
m.Equation(e_GT[z][p] <= 1*delta_GT[z][p] for z in range(Z) for p in range(P))
m.Equation(f_GT [z][p]== a*delta_GT[z][p] + b*e_GT[z][p] for z in range(Z) for p in range(P))
m.Equation(q_GT [z][p]== c*delta_GT[z][p] + d*e_GT[z][p] for z in range(Z) for p in range(P))
m.Equation(E_purch[z][p] + E_GT[z][p] == E_RE[z][p] + E_d[z][p] for z in range(Z) for p in range(P))
m.Equation(Q_GT[z][p] == Q_disp[z][p] + Q_GT_RB[z][p] for z in range(Z) for p in range(P))
m.Equation(Q_RB[z][p] + Q_HB[z][p] == Q_RS[z][p] + H_d[z][p] for z in range(Z) for p in range(P))
m.Equation(K_RS[z][p] + K_RE[z][p] == K_d[z][p] for z in range(Z) for p in range(P))
m.Equation(Q_disp[z][p] <= alpha*Q_GT[z][p] for z in range(Z) for p in range(P))
# ______Objective_______
m.Obj(Cc + Cr)
#_____Solve Problem_____
m.solve()
最佳答案
二维列表定义需要额外的方括号。这给出了一个 3 行 4 列的 2D 列表。
[[p+10*z for p in range(3)] for z in range(4)]
# Result: [[0, 1, 2], [10, 11, 12], [20, 21, 22], [30, 31, 32]]
如果省略内括号，则它是长度为 12 的一维列表。
[p+10*z for p in range(3) for z in range(4)]
# Result: [0, 10, 20, 30, 1, 11, 21, 31, 2, 12, 22, 32]
当列表的每个元素都是 Gekko
Intermediate 时，它也有效.[[m.Intermediate(p+10*z) for p in range(3)] for z in range(4)]
关于能源系统: How to build intermediates that are 2D arrays的Python GEKKO MINLP优化，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/59103401/