关于 GEKKO 中的条件语句（'m.if3'）的问题

Question

我想在 GEKKO 代码中添加一些条件语句。我用“m.if3”添加了以下语句。但是，它返回了没有条件语句我没有的语法错误。'R1_1 = m.if3(R1<0, 0, R1)'#%% GEKKOm = GEKKO(remote=False)#print(m.version)#m.time = np.linspace(0, 3600, 100)m.time = np.array([0,tstep])cH = m.CV(value=cs0[0])cM = m.CV(value=cs0[1])cW = m.CV(value=cs0[2])cF = m.CV(value=cs0[3])R1_1 = m.Var()r3 = m.Intermediate(r0*(1-cF/cFeMax)**(1/3))r2 = m.Intermediate(r0*((2*cH + 3*cM)/cFeMax)**(1/3))r1 = m.Intermediate(r0*(2*cH/cFeMax)**(1/3))x = m.Intermediate(r1/r0)y = m.Intermediate(r2/r0)z = m.Intermediate(r3/r0)A1 = m.Intermediate(1/x**2/(kd[0]*(1+1/Keq[0])))A2 = m.Intermediate(1/y**2/(kd[1]*(1+1/Keq[1])))A3 = m.Intermediate(1/z**2/(kd[2]*(1+1/Keq[2])))B1 = m.Intermediate((y-x)/x/y*r0/Dif[1])B2 = m.Intermediate((z-y)/y/z*r0/Dif[2])B3 = m.Intermediate((1-z)/z*r0/Dif[3])F = 0W = m.Intermediate((A1+B1)*(A3*(A2+B2+B3+F)+(A2+B2)*(B3+F))+A2*(A3*(B2+B3+F))+B2*(B3+F))ceq1 = m.Intermediate((cg0[0]+cg0[1])/(1+Keq[0]))ceq2 = m.Intermediate((cg0[0]+cg0[1])/(1+Keq[1]))ceq3 = m.Intermediate((cg0[0]+cg0[1])/(1+Keq[2]))R1 = m.Intermediate(3/r0/W*((A3*(A2+B2+B3+F)+(A2+B2)*(B3+F))*(cg0[0]-ceq1) \                     -(A3*(B2+B3+F)+B2*(B3+F))*(cg0[0]-ceq2) \                     -A2*(B3+F)*(cg0[0]-ceq3)))R2 = m.Intermediate(3/r0/W*(-(B2*(A3+B3+F)+A3*(B3+F))*(cg0[0]-ceq1) \                     +((A1+B1+B2)*(A3+B3+F)+A3*(B3+F))*(cg0[0]-ceq2) \                     -(A1+B1)*(B3+F)*(cg0[0]-ceq3)))R3 = m.Intermediate(3/r0/W*(-A2*(B3+F)*(cg0[0]-ceq1) \                    -(A1+B1)*(B3+F)*(cg0[0]-ceq2) \                    +((A1+B1)*(A2+B2+B3+F)+A2*(B2+B1+F))*(cg0[0]-ceq3)))R1_1 = m.if3(R1<0, 0, R1)m.Equation(cH.dt() == nus[0].dot([R1_1, R2, R3]))m.Equation(cM.dt() == nus[1].dot([R1_1, R2, R3]))m.Equation(cW.dt() == nus[2].dot([R1_1, R2, R3]))m.Equation(cF.dt() == nus[3].dot([R1_1, R2, R3]))m.options.IMODE = 4m.options.SOLVER = 3m.options.nodes = 2异常：@error：模型表达式 *** 函数字符串语法错误：括号不匹配位置：4(0)))-((((1-int_v5))*(i35)-slk_1 ?

Answer 1

您只需进行一次更正（删除<0）即可使用if3Gekko 中的功能。

R1_1 = m.if3(R1, 0, R1)

该m.if3函数使用一个条件，根据条件是小于零还是大于或等于零来切换使用的内容（参数 2 或 3）。这是帮助 (m.if3) 的结果以及一些额外的解释：

if3(condition, x1, x2) method of gekko.gekko.GEKKO instance

IF conditional with a binary switch variable.

The traditional method for IF statements is not continuously

differentiable and can cause a gradient-based optimizer to fail

to converge.

Usage: y = m.if3(condition,x1,x2)

Inputs:

   condition: GEKKO variable, parameter, or expression

   x1 and x2: GEKKO variable, parameter, or expression

Output: GEKKO variable y = x1 when condition<0

                       y = x2 when condition>=0

要考虑的另一件事是该if3函数使用二进制变量，您需要使用混合整数求解器来找到整数解。您可以删除将求解器切换到 IPOPT 的行（if3默认情况下切换到 APOPT 求解器），也可以手动将选项切换到 APOPT。

m.options.SOLVER = 1

您的脚本中缺少一些常量。我添加了一些虚拟变量只是为了让它运行。

from gekko import GEKKO

import numpy as np

m = GEKKO()

help(m.if3)

tstep = 1.0

cs0=[1,1,1,1]

r0 = 1.0

cFeMax = 1.0

kd = [1,1,1]

Keq = [1,1,1]

Dif = [1,1,1,1]

cg0 = [1,1,1]

nus = np.array([[1,1,1],[1,1,1],[1,1,1],[1,1,1]])

m.time = np.array([0,tstep])

cH = m.CV(value=cs0[0])

cM = m.CV(value=cs0[1])

cW = m.CV(value=cs0[2])

cF = m.CV(value=cs0[3])

R1_1 = m.Var()

r3 = m.Intermediate(r0*(1-cF/cFeMax)**(1/3))

r2 = m.Intermediate(r0*((2*cH + 3*cM)/cFeMax)**(1/3))

r1 = m.Intermediate(r0*(2*cH/cFeMax)**(1/3))

x = m.Intermediate(r1/r0)

y = m.Intermediate(r2/r0)

z = m.Intermediate(r3/r0)

A1 = m.Intermediate(1/x**2/(kd[0]*(1+1/Keq[0])))

A2 = m.Intermediate(1/y**2/(kd[1]*(1+1/Keq[1])))

A3 = m.Intermediate(1/z**2/(kd[2]*(1+1/Keq[2])))

B1 = m.Intermediate((y-x)/x/y*r0/Dif[1])

B2 = m.Intermediate((z-y)/y/z*r0/Dif[2])

B3 = m.Intermediate((1-z)/z*r0/Dif[3])

F = 0

W = m.Intermediate((A1+B1)*(A3*(A2+B2+B3+F)+(A2+B2)*(B3+F))+A2*(A3*(B2+B3+F))+B2*(B3+F))

ceq1 = m.Intermediate((cg0[0]+cg0[1])/(1+Keq[0]))

ceq2 = m.Intermediate((cg0[0]+cg0[1])/(1+Keq[1]))

ceq3 = m.Intermediate((cg0[0]+cg0[1])/(1+Keq[2]))

R1 = m.Intermediate(3/r0/W*((A3*(A2+B2+B3+F)+(A2+B2)*(B3+F))*(cg0[0]-ceq1) \

                     -(A3*(B2+B3+F)+B2*(B3+F))*(cg0[0]-ceq2) \

                     -A2*(B3+F)*(cg0[0]-ceq3)))

R2 = m.Intermediate(3/r0/W*(-(B2*(A3+B3+F)+A3*(B3+F))*(cg0[0]-ceq1) \

                     +((A1+B1+B2)*(A3+B3+F)+A3*(B3+F))*(cg0[0]-ceq2) \

                     -(A1+B1)*(B3+F)*(cg0[0]-ceq3)))

R3 = m.Intermediate(3/r0/W*(-A2*(B3+F)*(cg0[0]-ceq1) \

                    -(A1+B1)*(B3+F)*(cg0[0]-ceq2) \

                    +((A1+B1)*(A2+B2+B3+F)+A2*(B2+B1+F))*(cg0[0]-ceq3)))

R1_1 = m.if3(R1, 0, R1)

m.Equation(cH.dt() == nus[0].dot([R1_1, R2, R3]))

m.Equation(cM.dt() == nus[1].dot([R1_1, R2, R3]))

m.Equation(cW.dt() == nus[2].dot([R1_1, R2, R3]))

m.Equation(cF.dt() == nus[3].dot([R1_1, R2, R3]))

m.options.IMODE = 4

m.options.SOLVER = 1

m.options.nodes = 2

m.solve()