Python 如何使用数组从Gekko应用m.connection？

我试图用数组“for cycle”并遵循来自

我不确定变量的初始化是否可以这样做。此外，我怀疑我是否正确使用了约束。先谢谢你

这是我正在编写的代码

from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt 
import math as math
import pandas as pd

####measured lab data - nh2cl decay
#data set 1 
t_data1 = [0,0.08333,0.5,1,4,22.6167] #nh2cl
x_data1 = [0,4.91e-5,4.57e-5,4.74e-5,4.17e-5,2.76e-5] #8.01e-5
x_data1mgl = [0, 3.48,3.24,3.36,2.96,1.96]#5.68

#data set 2
t_data2 = [0,0.08333,0.5,1,4,22.8167]
x_data2 = [0,5.92e-5,5.7e-5,5.64e-5,5.30e-5,4.6e-5] #8.01e-5
x_data2mgl = [0,4.2,4.04,4,3.76,2.88]#5.68


####Preparing lab data and time dataframe
#combine data and time in the same dataframe
data1 = pd.DataFrame({'time':t_data1,'x1':x_data1,'x1mgl': x_data1mgl})
data2 = pd.DataFrame({'time':t_data2,'x2':x_data2,'x2mgl': x_data2mgl})

data1.set_index('time', inplace=True)
data2.set_index('time', inplace=True)

# merge dataframes
data = data1.join(data2, how='outer')

# simulation time points
dftp = pd.DataFrame({'time':np.linspace(0,5,100)})
dftp.set_index('time', inplace=True)

# merge dataframes
dff = data.join(dftp,how='outer')

z1 = (dff['x1']==dff['x1']).astype(int)
z2 = (dff['x2']==dff['x2']).astype(int)

# replace NaN with zeros
df0 = dff.fillna(value=0)


####Estimator Model
m = GEKKO(remote=False)
#m = GEKKO() 

m.time = df0.index.values

# measurements - Gekko arrays to store measured data
xm = m.Array(m.Param,2)
xm[0].value = df0['x1'].values
xm[1].value = df0['x2'].values

####inital measured values
hocl_init_val=m.Array(m.Var,2)
hocl_init_val[0].value= 8.01e-5
hocl_init_val[1].value= 8.01e-5 

nh3_init_val=m.Array(m.Var,2)
nh3_init_val[0].value= 1.37e-3 
nh3_init_val[1].value= 6.82e-4 

h_init_val=m.Array(m.Var,2)
h_init_val[0].value= 6.31e-8 
h_init_val[1].value= 2e-8

h2co3_init_val=m.Array(m.Var,2)
h2co3_init_val[0].value= 5.39e-4
h2co3_init_val[1].value= 1.88e-4

hco3_init_val=m.Array(m.Var,2)
hco3_init_val[0].value= 3.46e-3
hco3_init_val[1].value= 3.80e-3

co32_init_val=m.Array(m.Var,2)
co32_init_val[0].value= 2.16e-6
co32_init_val[1].value= 7.5e-6

alk_init_val=m.Array(m.Var,2)
alk_init_val[0].value= 3.46e-3
alk_init_val[1].value= 3.82-3

hocl=m.Array(m.Var,2)
nh3=m.Array(m.Var,2)
nh2cl=m.Array(m.Var,2,lb=1e-10)
nh2cl_meas=m.Array(m.Param,2)
nhcl2=m.Array(m.Var,2,lb=1e-10)
h=m.Array(m.Var,2)
oh=m.Array(m.Var,2)
I=m.Array(m.Var,2,lb=1e-10)
ocl=m.Array(m.Var,2)
nh4=m.Array(m.Var,2)
h2co3=m.Array(m.Var,2)
hco3=m.Array(m.Var,2)
co32=m.Array(m.Var,2)
alk=m.Array(m.Var,2)
#DOC1=m.Array(m.Var,2)
#DOC2=m.Array(m.Var,2)
cnh3=m.Array(m.Var,2)
cnh2cl=m.Array(m.Var,2)

r1=m.Array(m.Var,2)
r2=m.Array(m.Var,2)
r3=m.Array(m.Var,2)
r4=m.Array(m.Var,2)
r5=m.Array(m.Var,2)
r6=m.Array(m.Var,2)
r7=m.Array(m.Var,2)
r8=m.Array(m.Var,2)
r9=m.Array(m.Var,2)
r10=m.Array(m.Var,2)

r11=m.Array(m.Var,2)
r12=m.Array(m.Var,2)
r13=m.Array(m.Var,2)
r14=m.Array(m.Var,2)

r15=m.Array(m.Var,2)
r16=m.Array(m.Var,2)



#Define GEKKO variables that determine if time point contains data to be used in regression
#index for objective (0=not measured, 1=measured)
zm = m.Array(m.Param,2)
zm[0].value=z1
zm[1].value=z2

# fit to measurement
x=m.Array(m.Var,2)                         
x[0].value=x_data1[0]
x[1].value=x_data2[0]

meas=m.Array(m.Param,2) 

#### adjustable parameters 

kdoc1 = m.FV(1.2334e6,lb=1,ub=100000000000) #m-1h-1
kdoc2 = m.FV(3.8809e9,lb=1,ub=10000000000) #m-1h-1

x10=3.1684e-5
x20=3.1308e-5
#DOC10 = m.FV(x10,lb=1e-6,ub=1)
#DOC20 = m.FV(x20,lb=1e-6,ub=1)

#constrains
DOC10=m.Array(m.FV,2)
DOC10[0].value=x10
DOC10[1].value=x10*0.5

DOC20=m.Array(m.FV,2)
DOC20[0].value=x20
DOC20[1].value=x20*0.5

#variables connected to the parameters DOC20 and DOC10: DOC10 is DOC1 when t=0 and DOC20 is DOC2 when t=0 
DOC1=m.Array(m.SV,2,fixed_initial=False)
#DOC1[0].value=x10
#DOC1[1].value=x10*0.5

DOC2=m.Array(m.SV,2,fixed_initial=False)
#DOC2[0].value=x20
#DOC2[1].value=x20*0.5

#t = m.Param(value=m.time)

####Rate constants
k1 = m.FV(1.5e10)
k2 = m.FV(7.6e-2)
k3 = m.FV(1e6)
k4 = m.FV(2.3e-3)
k6 = m.FV(2.2e8)
k7 = m.FV(4e5)
k8 = m.FV(1e8)
k9 = m.FV(3e7)
k10 = m.FV(55)
k11 = m.FV(3.16e-8*1e10)
k12 = m.FV(1e10)
k13 = m.FV(5.01e-10*1e10)
k14 = m.FV(1e10)

k5=m.Array(m.Var,2)

for i in range(2):
    m.Connection(DOC2[i],DOC20[i],pos1=1,pos2=1,node1=1,node2=1)
    m.Connection(DOC1[i],DOC10[i],pos1=1,pos2=1,node1=1,node2=1)
    
    #variables - initial values
    hocl[i] = m.Var(value=hocl_init_val[i])
    nh3[i] = m.Var(value=nh3_init_val[i])
    nh2cl[i] = m.Var(value=x[i])
    nh2cl_meas[i] = m.Param(xm[i])
    meas[i] = m.Param(zm[i])
    nhcl2[i] = m.Var(value=0)
    h[i] = m.Var(value=h_init_val[i])
    oh[i] = m.Var(value=1e-14/h_init_val[i])
    I[i] = m.Var(value=0)
    ocl[i]= m.Var(value=0)
    nh4[i] = m.Var(value=0)
    h2co3[i] = m.Var(value=h2co3_init_val[i])
    hco3[i] = m.Var(value=hco3_init_val[i])
    co32[i] = m.Var(value=co32_init_val[i])
    alk[i] = m.Var(value=alk_init_val[i])
    DOC1[i] = m.SV(value=DOC10[i])
    DOC2[i] = m.SV(value=DOC20[i])
    cnh3[i] = m.Var(value=0)
    cnh2cl[i] = m.Var(value=0)
    

###Equations 1   
    m.Equation(k5[i] == 2.5e7*h[i]+4e4*h2co3[i]+800*hco3[i])

###Equations 2 
    m.Equation(r1[i] == k1 * hocl[i] * nh3[i])
    m.Equation(r2[i] == k2 * nh2cl[i])
    m.Equation(r3[i] == k3 * hocl[i] * nh2cl[i])
    m.Equation(r4[i] == k4 * nhcl2[i])
    m.Equation(r5[i] == k5[i] * nh2cl[i] * nh2cl[i])
    m.Equation(r6[i] == k6 * nhcl2[i] * nh3[i]* h[i])
    m.Equation(r7[i] == k7 * nhcl2[i] * oh[i])
    m.Equation(r8[i] == k8 * I[i] * nhcl2[i])
    m.Equation(r9[i] == k9 * I[i] * nh2cl[i])
    m.Equation(r10[i] == k10 * nh2cl[i] * nhcl2[i])
    
    m.Equation(r11[i] == k11*hocl[i])
    m.Equation(r12[i] == k12*h[i]*ocl[i])
    m.Equation(r13[i] == k13*nh4[i])
    m.Equation(r14[i] == k14*h[i]*nh3[i])
    
    m.Equation(r15[i] == kdoc1*DOC1[i]*nh2cl[i])
    m.Equation(r16[i] == kdoc2*DOC2[i]*hocl[i])
    
####Equations 3

    m.Equation(hocl[i].dt() == -r1[i] + r2[i] - r3[i] + r4[i] + r8[i] - r16[i] - r11[i] + r12[i])
      
    m.Equation(nh3[i].dt() == -r1[i] + r2[i] + r5[i] - r6[i] + r13[i] - r14[i])
    m.Equation(nh2cl[i].dt() == r1[i] - r2[i] - r3[i] + r4[i] - r5[i] + r6[i] - r9[i] - r10[i] - r15[i])
    m.Equation(nhcl2[i].dt() == r3[i] - r4[i] + r5[i] - r6[i] - r7[i] - r8[i] - r10[i])
    m.Equation(h[i].dt() == 0)
    
    m.Equation(oh[i] == 1e-14/h[i])
    
    m.Equation(I[i].dt() == r7[i] - r8[i] - r9[i])  
    m.Equation(ocl[i].dt() == r11[i] - r12[i])
    m.Equation(nh4[i].dt() == -r13[i]  + r14[i])
    
    m.Equation(h2co3[i] == (hco3[i]*h[i])/5.01e-7)
    
    m.Equation(hco3[i] == alk[i] - 2*co32[i] - oh[i] + h[i])
    
    m.Equation(co32[i] == (5.01e-11*hco3[i])/h[i])
    
    m.Equation(alk[i].dt() == 0)
       
    m.Equation(DOC1[i].dt() == -r15[i])
    m.Equation(DOC2[i].dt() == -r16[i])
    
    m.Equation(cnh3[i] == 17000*nh3[i])
    m.Equation(cnh2cl[i] == 70900*nh2cl[i])
    
###obj function
    
    m.Minimize(meas[i]*((nh2cl[i]-nh2cl_meas[i]))**2)
        
#Application options
m.options.SOLVER = 1 #IPOPT solver=3 #APOPT solver=1
m.options.IMODE = 8 #Dynamic Simultaneous - estimation = 5 # Dynamic sequential - estimation = 8
m.options.RTOL = 1E-3
m.options.EV_TYPE = 2 #absolute error
m.options.NODES = 2 #collocation nodes (2 up to 6)
#m.options.MAX_ITER = 500
m.open_folder()

if True:
    kdoc1.STATUS=1
    kdoc2.STATUS=1
    DOC10[i].STATUS=1
    DOC20[i].STATUS=1

m.options.TIME_SHIFT = 0
try:
    m.solve(disp=True)
except:
    print("don't stop when not finding Doc10")


print('Final SSE Objective: ' + str(m.options.objfcnval))

print('Solution')
print('kdoc1 = ' + str(kdoc1.value[0]))
print('kdoc2 = ' + str(kdoc2.value[0]))
print('DOC10_1 = ' + str(DOC10[0].value[0]))
print('DOC20_1 = ' + str(DOC20[0].value[0]))
print('DOC10_2 = ' + str(DOC10[1].value[0]))
print('DOC20_2 = ' + str(DOC20[1].value[0]))

####Graphics

plt.figure(1,figsize=(8,5))
plt.subplot(2,1,1)
plt.plot(m.time,nh2cl[0],'m',label='Predicted 1')
plt.plot(m.time,nh2cl[1],'c',label='Predicted 2')

plt.plot(t_data1,x_data1,'mo',label='Meas 1')
plt.plot(t_data2,x_data2,'co',label='Meas 2')

plt.legend(loc='best')
plt.ylabel('mol/L')
plt.legend(loc='center right', bbox_to_anchor=(1.45, 0.5), ncol=2) #shadow=True,

plt.subplot(2,1,2)
plt.plot(m.time,cnh2cl[0].value,'m',label ='C2_M1')
plt.plot(m.time,cnh2cl[1].value,'c',label ='C2_M2')

plt.plot(t_data1,x_data1mgl,'mo',label='Meas 1')
plt.plot(t_data2,x_data2mgl,'co',label='Meas 2')

plt.legend(loc='best')
plt.ylabel('mg Cl2/L')
plt.xlabel('time (h)')
plt.legend(loc='right', bbox_to_anchor=(1.4, 0.5),  ncol=2) #shadow=True,

plt.figure(2,figsize=(12,8))
plt.subplot(4,3,1)
plt.plot(m.time,hocl[i].value,label ='hocl')
plt.legend()

plt.subplot(4,3,2)
plt.plot(m.time,nh3[i].value,label ='nh3')
plt.legend()

plt.subplot(4,3,3)
plt.plot(m.time,nhcl2[i].value,label ='nhcl2')
plt.legend()

plt.subplot(4,3,4)
plt.plot(m.time,h[i].value,label ='h')
plt.legend()

plt.subplot(4,3,5)
plt.plot(m.time,oh[i].value,label ='oh')
plt.legend()

plt.subplot(4,3,6)
plt.plot(m.time,I[i].value,label ='I')
plt.legend()

plt.subplot(4,3,7)
plt.plot(m.time,ocl[i].value,label ='ocl')
plt.legend()

plt.subplot(4,3,8)
plt.plot(m.time,nh4[i].value,label ='nh4')
plt.legend()

plt.subplot(4,3,9)
plt.plot(m.time,h2co3[i].value,label ='h2co3')
plt.legend()
plt.xlabel('time (h)')

plt.subplot(4,3,10)
plt.plot(m.time,hco3[i].value,label ='hco3')
plt.legend()
plt.xlabel('time (h)')

plt.subplot(4,3,11)
plt.plot(m.time,co32[i].value,label ='co32')
plt.legend()
plt.xlabel('time (h)')

plt.subplot(4,3,12)
plt.plot(m.time,alk[i].value,label ='alk')
plt.legend()
plt.xlabel('time (h)')

plt.show()

---

18546自由度的问题维度看起来不正确。似乎只有速率常数作为自由度。您可能需要检查每个变量是否都有一个方程，并且在所有速率常数固定的情况下，它最初是以零自由度（方程数等于变量数）求解的

 ----------------------------------------------------------------
 APMonitor, Version 1.0.0
 APMonitor Optimization Suite
 ----------------------------------------------------------------
 
 
 --------- APM Model Size ------------
 Each time step contains
   Objects      :  0
   Constants    :  0
   Variables    :  141
   Intermediates:  0
   Connections  :  8
   Equations    :  36
   Residuals    :  36
 
 Variable time shift OFF
 Number of state variables:    30082
 Number of total equations: -  11536
 Number of slack variables: -  0
 ---------------------------------------
 Degrees of freedom       :    18546
 
 ----------------------------------------------
 Dynamic Estimation with APOPT Solver
 ----------------------------------------------
 
 Iter    Objective  Convergence
    0  1.37152E+01  4.08093E-01
    1  2.83795E+01  9.57058E-05
    2  1.36177E-01  1.62209E-05

---

我对代码做了一些修改。我主要添加了m.options.RTOL=1E-3。它可以运行，但解决方案还不好。我还在努力。我会很感激其他的想法。此外，我遵循@JohnHedengren关于自由度的想法。我把它减少到6。。。将一些测量值改为Param而不是Var，将速率常数改为Const。我不确定我应该改变什么来获得零自由度。更新代码如下：

  247  3.06822E+28  5.05213E-04
  248  3.01561E+28  5.05213E-04
  249  3.04167E+28  5.05213E-04
 
 Iter    Objective  Convergence
  250  3.06822E+28  5.05213E-04
 Maximum iterations
 
 ---------------------------------------------------
 Solver         :  APOPT (v1.0)
 Solution time  :  264.2566 sec
 Objective      :  4.856644896504421E-8
 Unsuccessful with error code  0
 ---------------------------------------------------
 
 Creating file: infeasibilities.txt
 Use command apm_get(server,app,'infeasibilities.txt') to retrieve file
 @error: Solution Not Found