最小化约束的不可迭代错误来自 scipy.optimize

Question

我刚开始研究 Python 的优化，我遇到了一个问题。我有一个问题，我想使用从 scipy.optimize 最小化来最小化我的 objective 函数 (obj_fun)。我将分享一个例子：

import numpy as np

def analysis(A):
    N = []
    for i in A:
        N.append(i*3)
    return N

def cons(A):
    N = analysis(A)
    C = []
    for i in len(N):
        if N[i] < 2:
            C.append({'type': 'ineq', 'fun': lambda x: x[0]*N[i]})
        else:
            C.append({'type': 'ineq', 'fun': lambda x: x[0]-N[i]})
    return C

def obj_fun(A):
    """Objective function returns the weight of the structure"""
    w=  0.5*[1*A[0]+2*A[1]+3*A[2]]
    return w

# Initial values
A0 = np.array([0.001 for i in range(0, 3)])

N = analysis(A0)

##  Optimization

bnds = [(1e-6, None) for i in range(len(A0))]

from scipy.optimize import minimize
sol = minimize(obj_fun, x0=A0, method='trust-constr', bounds=bnds, 
constraints=cons)
print(sol)

我得到的整个错误是：运行文件（'C:/Users/Myc/Documents/Python Scripts/example stack.py'，wdir='C:/Users/Myc/Documents/Python Scripts'）回溯（最近调用最后）：

文件“C:\Users\Myc\Documents\PythonScripts\examplestack.py”，第 40 行，在溶胶=最小化（obj_fun，x0=A0，方法='trust-constr'，边界=bnds，约束=cons）

文件“C:\Users\Myc\anaconda3\lib\site-packages\scipy\optimize_minimize.py”，第 605 行，最小化 constraints = standardize_constraints(constraints, x0, meth)

文件“C:\Users\Myc\anaconda3\lib\site-packages\scipy\optimize_minimize.py”，第 825 行，在 standardize_constraints constraints = list(constraints) # 确保它是一个可变序列

类型错误：'function'对象不可迭代

我知道主要问题是我如何定义约束，如果我在优化之前定义 Cons1 = rest(A0)，我可以将 constraints=cons 替换为 constraints = Cons1。然而，这对我没有帮助，因为我需要在优化的每次迭代中执行函数 trus_analysis 以更新限制的参数 N。如何定义约束？

Answer 1

原剧本：

def obj_fun(A):
    return 7*A[0]+ 3*A[1]+ 7*A[2]

def analysis(A):
    N = []
    for i in A:
        N.append(i*3)
    return N

def cons(A):
    n = analysis(A)
    C = []
    for i in range(len(A)):
        if n[i] < 4:
            C.append({'type': 'ineq', 'fun': lambda x: x[i]**2 / n[i]})
        else:
            C.append({'type': 'ineq', 'fun': lambda x: x[i] - n[i]})
    return C


A0 = [1,2,3]
C = cons(A0)
bnds = [(1e-6, None) for i in range(len(A0))]

from scipy.optimize import minimize
sol = minimize(obj_fun, x0=A0, method='trust-constr', bounds=bnds, constraints=C)
print(sol)

运行：

/usr/local/lib/python3.8/dist-packages/scipy/optimize/_hessian_update_strategy.py:182: UserWarning: delta_grad == 0.0. Check if the approximated function is linear. If the function is linear better results can be obtained by defining the Hessian as zero instead of using quasi-Newton approximations.
  warn('delta_grad == 0.0. Check if the approximated '
 barrier_parameter: 0.00016000000000000007
 barrier_tolerance: 0.00016000000000000007
          cg_niter: 15
      cg_stop_cond: 1
            constr: [array([9.00009143]), array([4.57149698e-05]), array([4.57149698e-05]), array([2.38571416e-05, 5.43334162e-05, 9.00004571e+00])]
       constr_nfev: [40, 40, 40, 0]
       constr_nhev: [0, 0, 0, 0]
       constr_njev: [0, 0, 0, 0]
    constr_penalty: 1.0
  constr_violation: 0.0
    execution_time: 0.0873115062713623
               fun: 63.00065000502843
              grad: array([7.        , 3.        , 6.99999999])
               jac: [array([[0.        , 0.        , 2.00001017]]), array([[0., 0., 1.]]), array([[0., 0., 1.]]), array([[1., 0., 0.],
       [0., 1., 0.],
       [0., 0., 1.]])]
   lagrangian_grad: array([1.77635684e-15, 1.55431223e-14, 5.67948534e-14])
           message: '`gtol` termination condition is satisfied.'
            method: 'tr_interior_point'
              nfev: 40
              nhev: 0
               nit: 14
             niter: 14
              njev: 10
        optimality: 5.679485337974424e-14
            status: 1
           success: True
         tr_radius: 18734.614693588483
                 v: [array([-1.77775972e-05]), array([-3.49997333]), array([-3.49997333]), array([-7.00000000e+00, -3.00000000e+00, -1.77776895e-05])]
                 x: array([2.38571416e-05, 5.43334162e-05, 9.00004571e+00])

这里

In [36]: C
Out[36]: 
[{'type': 'ineq', 'fun': <function __main__.cons.<locals>.<lambda>(x)>},
 {'type': 'ineq', 'fun': <function __main__.cons.<locals>.<lambda>(x)>},
 {'type': 'ineq', 'fun': <function __main__.cons.<locals>.<lambda>(x)>}]

A0 用于创建 3 个约束函数。

analysis 函数只是将 A 乘以 3。

In [38]: analysis(A0)
Out[38]: [3, 6, 9]
In [39]: A0
Out[39]: [1, 2, 3]
In [40]: analysis(A0)
Out[40]: [3, 6, 9]
In [41]: np.array(A0)*3
Out[41]: array([3, 6, 9])

在最新的 cons 中，您删除了 range，直接使用 cons 而不是 cons(A0)。 constraints 参数应该是 dict 的 list，如 C.

所示

最小化约束的不可迭代错误来自 scipy.optimize

Not iterable Error with constraints in minimize from scipy.optimize

python

restriction

scipy