我想在 GEKKO 代碼中添加一些條件語句。我用“m.if3”添加了以下語句。但是，它返回了沒有條件語句我沒有的語法錯誤。'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異常：@error：模型表達式 *** 函數字符串語法錯誤：括號不匹配位置：4(0)))-((((1-int_v5))*(i35)-slk_1 ?
1 回答
臨摹微笑
TA貢獻1982條經驗 獲得超2個贊
您只需進行一次更正（刪除<0）即可使用if3Gekko 中的功能。
R1_1 = m.if3(R1, 0, R1)
該m.if3函數使用一個條件，根據條件是小于零還是大于或等于零來切換使用的內容（參數 2 或 3）。這是幫助 (m.if3) 的結果以及一些額外的解釋：
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
要考慮的另一件事是該if3函數使用二進制變量，您需要使用混合整數求解器來找到整數解。您可以刪除將求解器切換到 IPOPT 的行（if3默認情況下切換到 APOPT 求解器），也可以手動將選項切換到 APOPT。
m.options.SOLVER = 1
您的腳本中缺少一些常量。我添加了一些虛擬變量只是為了讓它運行。
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()
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