python - Pyomo 在时间序列缺失值的 ODE 系统中的参数估计

Question

我有一个包含 7 个方程的 ODE 系统，用于解释一组特定的微生物动力学，其形式为:

哪里是涉及的不同化学和微生物物种(甚至是化合物的子索引)，是产量系数，是伪 react :

我使用 Pyomo 来估计所有未知参数，这些参数基本上是所有屈服系数和动力学常数(总共 15 个)。

当与每个动态变量的完整实验时间序列一起使用时，以下代码可以完美运行:

from pyomo.environ import *
from pyomo.dae import *

m = AbstractModel()
m.t = ContinuousSet()   
m.MEAS_t = Set(within=m.t)  # Measurement times, must be subset of t
m.x1_meas = Param(m.MEAS_t)
m.x2_meas = Param(m.MEAS_t)
m.x3_meas = Param(m.MEAS_t)
m.x4_meas = Param(m.MEAS_t)
m.x5_meas = Param(m.MEAS_t)
m.x6_meas = Param(m.MEAS_t)
m.x7_meas = Param(m.MEAS_t)

m.x1 = Var(m.t,within=PositiveReals)
m.x2 = Var(m.t,within=PositiveReals)
m.x3 = Var(m.t,within=PositiveReals)
m.x4 = Var(m.t,within=PositiveReals)
m.x5 = Var(m.t,within=PositiveReals)
m.x6 = Var(m.t,within=PositiveReals)
m.x7 = Var(m.t,within=PositiveReals)

m.k1 = Var(within=PositiveReals)
m.k2 = Var(within=PositiveReals)
m.k3 = Var(within=PositiveReals)
m.k4 = Var(within=PositiveReals)
m.k5 = Var(within=PositiveReals)
m.k6 = Var(within=PositiveReals)
m.k7 = Var(within=PositiveReals)
m.k8 = Var(within=PositiveReals)
m.k9 = Var(within=PositiveReals)
m.y1 = Var(within=PositiveReals)
m.y2 = Var(within=PositiveReals)
m.y3 = Var(within=PositiveReals)
m.y4 = Var(within=PositiveReals)
m.y5 = Var(within=PositiveReals)
m.y6 = Var(within=PositiveReals)

m.x1dot = DerivativeVar(m.x1,wrt=m.t)
m.x2dot = DerivativeVar(m.x2,wrt=m.t)
m.x3dot = DerivativeVar(m.x3,wrt=m.t)
m.x4dot = DerivativeVar(m.x4,wrt=m.t)
m.x5dot = DerivativeVar(m.x5,wrt=m.t)
m.x6dot = DerivativeVar(m.x6,wrt=m.t)
m.x7dot = DerivativeVar(m.x7,wrt=m.t)

def _init_conditions(m):
    yield m.x1[0] == 51.963
    yield m.x2[0] == 6.289
    yield m.x3[0] == 0
    yield m.x4[0] == 6.799
    yield m.x5[0] == 0
    yield m.x6[0] == 4.08
    yield m.x7[0] == 0
m.init_conditions=ConstraintList(rule=_init_conditions)


def _x1dot(m,i):
    if i==0:
        return Constraint.Skip
    return m.x1dot[i] == - m.y1*m.k1*m.x1[i]*m.x2[i]/(m.k2+m.x1[i]) - m.y2*m.k3*m.x1[i]*m.x4[i]/(m.k4+m.x1[i])
m.x1dotcon = Constraint(m.t, rule=_x1dot)

def _x2dot(m,i):
    if i==0:
        return Constraint.Skip
    return m.x2dot[i] ==  m.k1*m.x1[i]*m.x2[i]/(m.k2+m.x1[i]) - m.k7*m.x2[i]*m.x3[i]
m.x2dotcon = Constraint(m.t, rule=_x2dot)

def _x3dot(m,i):
    if i==0:
        return Constraint.Skip
    return m.x3dot[i] ==  m.y3*m.k1*m.x1[i]*m.x2[i]/(m.k2+m.x1[i]) - m.y4*m.k5*m.x3[i]*m.x6[i]/(m.k6+m.x3[i])
m.x3dotcon = Constraint(m.t, rule=_x3dot)

def _x4dot(m,i):
    if i==0:
        return Constraint.Skip
    return m.x4dot[i] == m.k3*m.x1[i]*m.x4[i]/(m.k4+m.x1[i]) - m.k8*m.x4[i]*m.x3[i]
m.x4dotcon = Constraint(m.t, rule=_x4dot)

def _x5dot(m,i):
    if i==0:
        return Constraint.Skip
    return m.x5dot[i] == m.y5*m.k3*m.x1[i]*m.x4[i]/(m.k4+m.x1[i])
m.x5dotcon = Constraint(m.t, rule=_x5dot)

def _x6dot(m,i):
    if i==0:
        return Constraint.Skip
    return m.x6dot[i] == m.k5*m.x3[i]*m.x6[i]/(m.k6+m.x3[i]) - m.k9*m.x6[i]*m.x7[i]
m.x6dotcon = Constraint(m.t, rule=_x6dot)

def _x7dot(m,i):
    if i==0:
        return Constraint.Skip
    return m.x7dot[i] == m.y6*m.k5*m.x3[i]*m.x6[i]/(m.k6+m.x3[i])
m.x7dotcon = Constraint(m.t, rule=_x7dot)

def _obj(m):
    return sum((m.x1[i]-m.x1_meas[i])**2+(m.x2[i]-m.x2_meas[i])**2+(m.x3[i]-m.x3_meas[i])**2+(m.x4[i]-m.x4_meas[i])**2+(m.x5[i]-m.x5_meas[i])**2+(m.x6[i]-m.x6_meas[i])**2+(m.x7[i]-m.x7_meas[i])**2 for i in m.MEAS_t)
m.obj = Objective(rule=_obj)

m.pprint()

instance = m.create_instance('exp.dat')
instance.t.pprint()

discretizer = TransformationFactory('dae.collocation')
discretizer.apply_to(instance,nfe=30)#,ncp=3)

solver=SolverFactory('ipopt')

results = solver.solve(instance,tee=True)

但是，我正在尝试在另一个实验数据中运行相同的估计例程，这些数据在某些动态变量的一个或最多两个时间序列的末尾缺少值。

换句话说，这些完整的实验数据看起来像(在.dat 文件中):

set t := 0  6   12  18  24  30  36  42  48  54  60  66  72  84  96  120 144;
set MEAS_t := 0 6   12  18  24  30  36  42  48  54  60  66  72  84  96  120 144;
param x1_meas :=
0   51.963
6   43.884
12  24.25
18  26.098
24  11.871
30  4.607
36  1.714
42  4.821
48  5.409
54  3.701
60  3.696
66  1.544
72  4.428
84  1.086
96  2.337
120 2.837
144 3.486
;
param x2_meas :=
0   6.289
6   6.242
12  7.804
18  7.202
24  6.48
30  5.833
36  6.644
42  5.741
48  4.568
54  4.252
60  5.603
66  5.167
72  4.399
84  4.773
96  4.801
120 3.866
144 3.847
;
param x3_meas :=
0   0
6   2.97
12  9.081
18  9.62
24  6.067
30  11.211
36  16.213
42  10.215
48  20.106
54  22.492
60  5.637
66  5.636
72  13.85
84  4.782
96  9.3
120 4.267
144 7.448
;
param x4_meas :=
0   6.799
6   7.73
12  7.804
18  8.299
24  8.208
30  8.523
36  8.507
42  8.656
48  8.49
54  8.474
60  8.203
66  8.127
72  8.111
84  8.064
96  6.845
120 6.721
144 6.162
;
param x5_meas :=
0   0
6   0.267
12  0.801
18  1.256
24  1.745
30  5.944
36  3.246
42  7.787
48  7.991
54  6.943
60  8.593
66  8.296
72  6.85
84  8.021
96  7.667
120 7.209
144 8.117
;
param x6_meas :=
0   4.08
6   4.545
12  4.784
18  4.888
24  5.293
30  5.577
36  5.802
42  5.967
48  6.386
54  6.115
60  6.625
66  6.835
72  6.383
84  6.605
96  5.928
120 5.354
144 4.975
;
param x7_meas :=
0   0
6   0.152
12  1.616
18  0.979
24  4.033
30  5.121
36  2.759
42  3.541
48  4.278
54  4.141
60  6.139
66  3.219
72  5.319
84  4.328
96  3.621
120 4.208
144 5.93
;

虽然我的一个不完整的数据集可以完成所有时间序列，但像这样的数据集:

param x6_meas :=
0   4.08
6   4.545
12  4.784
18  4.888
24  5.293
30  5.577
36  5.802
42  5.967
48  6.386
54  6.115
60  6.625
66  6.835
72  6.383
84  6.605
96  5.928
120 5.354
144 .
;

我知道可以指定 Pyomo 获取某些变量相对于不同时间序列的导数。然而，尝试之后，它并没有起作用，我猜这是因为这些是耦合的 ODE。所以基本上我的问题是是否有办法在 Pyomo 中克服这个问题。

提前致谢。

Answer 1

我认为您需要做的就是稍微修改您的目标函数，如下所示:

def _obj(m):
    sum1 = sum((m.x1[i]-m.x1_meas[i])**2 for i in m.MEAS_t if i in m.x1_meas.keys())
    sum2 = sum((m.x2[i]-m.x2_meas[i])**2 for i in m.MEAS_t if i in m.x2_meas.keys())
    sum3 = sum((m.x3[i]-m.x3_meas[i])**2 for i in m.MEAS_t if i in m.x3_meas.keys())
    sum4 = sum((m.x4[i]-m.x4_meas[i])**2 for i in m.MEAS_t if i in m.x4_meas.keys())
    sum5 = sum((m.x5[i]-m.x5_meas[i])**2 for i in m.MEAS_t if i in m.x5_meas.keys())
    sum6 = sum((m.x6[i]-m.x6_meas[i])**2 for i in m.MEAS_t if i in m.x6_meas.keys())
    sum7 = sum((m.x7[i]-m.x7_meas[i])**2 for i in m.MEAS_t if i in m.x7_meas.keys())
    return sum1+sum2+sum3+sum4+sum5+sum6+sum7
m.obj = Objective(rule=_obj)

在将索引添加到总和之前，会仔细检查i是否是每组测量的有效索引。如果您事先知道哪些测量集缺少数据，那么您可以通过仅对这些测量集进行检查并像以前一样对其他测量集进行求和来简化此功能。