问题描述
我刚刚开始学习使用 Python 进行优化,但遇到了一个问题。 我有一个问题,我想使用 scipy.optimize 中的最小化来最小化我的目标函数 (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: runfile('C:/Users/Myc/Documents/Python Scripts/example stack.py',wdir='C:/Users/Myc/Documents/Python Scripts') 回溯(最近一次调用):
文件“C:\Users\Myc\Documents\Python Scripts\example stack.py”,第 40 行,在 sol = 最小化(obj_fun,x0=A0,method='trust-constr',bounds=bnds,constraints=cons)
文件“C:\Users\Myc\anaconda3\lib\site-packages\scipy\optimize_minimize.py”,第 605 行,最小化 约束 = 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
TypeError: '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),我可以将约束 = cons 替换为约束 = Cons1。 但是,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?
解决方法
原始脚本:
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([2.38571416e-05,5.43334162e-05,9.00004571e+00])]
constr_nfev: [40,40,0]
constr_nhev: [0,0]
constr_njev: [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.,1.]]),array([[1.,0.],[0.,1.,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([-7.00000000e+00,-3.00000000e+00,-1.77776895e-05])]
x: array([2.38571416e-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)>}]
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]: [1,3]
In [40]: analysis(A0)
Out[40]: [3,9]
In [41]: np.array(A0)*3
Out[41]: array([3,9])
在最新的 cons
中,您去掉了 range
,直接使用 cons
而不是 cons(A0)
。 constraints
参数应该是 dict
的 list,如 C
所示。