我仍然是python的初学者，所以如果这太微不足道了，我很抱歉。我想计算一个函数的最小值，该函数总共有12个变量。在这12个变量中，10个固定在给定值，其余2个可自由计算最小值。这是我的代码示例。
import numpy as np
from sympy import *
from scipy.optimize import minimize
init_printing(use_unicode=True)
X_1,X_2,Y_1,Y_2,X_c1,X_c2,Y_c1,Y_c2,a_1,a_2,b_1,b_2,t_1,t_2,psi_1,psi_2= symbols('X_1 X_2 Y_1 Y_2 X_c1 X_c2 Y_c1 Y_c2 a_1 a_2 b_1 b_2 t_1 t_2 psi_1 psi_2')
X_1=X_c1 + (a_1 * cos(t_1) * cos(psi_1)) - ((b_1) * sin(t_1)* sin(psi_1))
X_2=X_c2 + (a_2 * cos(t_2) * cos(psi_2)) - ((b_2) * sin(t_2)* sin(psi_2))
Y_1=Y_c1 + (a_1 * cos(t_1) * sin(psi_1)) + ((b_1) * sin(t_1)* cos(psi_1))
Y_2=Y_c2 + (a_2 * cos(t_2) * sin(psi_2)) + ((b_2) * sin(t_2)* sin(psi_2))
param=(t_1,t_2,X_c1,X_c2,Y_c1,Y_c2,a_1,a_2,b_1,b_2,psi_1,psi_2) #12 parameters, 10 are fixed and 2 are free.
free_param=(t_1,t_2) #These are my two free parameters
D=((X_2-X_1)**2 + (Y_2-Y_1)**2)**0.5 #Expression to be minimised
distance=lambdify(param, D, modules='numpy')
以下一段代码基于此链接： Want to do multi-variation minimize with sympy
#Build Jacobian:
jac_D=[D.diff(x) for x in param]
jac_distance=[lambdify(param, jf, modules='numpy') for jf in jac_D]
def vector_distance(zz):
""" Helper for receiving vector parameters """
return distance(zz[0], zz[1], zz[2], zz[3], zz[4], zz[5], zz[6], zz[7], zz[8], zz[9], zz[10], zz[11])
def jac_vector_distance(zz):
""" Jacobian Helper for receiving vector parameters """
return np.array([jfn(zz[0], zz[1], zz[2], zz[3], zz[4], zz[5], zz[6], zz[7], zz[8], zz[9], zz[10], zz[11]) for jfn in jac_distance])
zz0 = np.array([np.pi/2, np.p1/2]) #Guess values for t_1 and t_2
现在我想修复其他10个变量的值。我想过使用约束。 （我想要X_c1 = 150，X_c2 = 2.03等等，如下所示）
cons=({'type': 'eq',
'fun' : lambda x: np.array([X_c1-150])},
{'type': 'eq',
'fun' : lambda x:np.array([X_c2-2.03)]},
{'type': 'eq',
'fun': lambda x:np.array([Y_c1-152])},
{'type': 'eq',
'fun' : lambda x: np.array([Y_c2-2.31])},
{'type': 'eq',
'fun' : lambda x:np.array([a_1-5])},
{'type': 'eq',
'fun': lambda x:np.array([a_2-3])},
{'type': 'eq',
'fun' : lambda x: np.array([b_1-9])},
{'type': 'eq',
'fun' : lambda x:np.array([b_2-4])},
{'type': 'eq',
'fun': lambda x:np.array([psi_1-np.pi/2])},
{'type': 'eq',
'fun' : lambda x: np.array([psi_2-np.pi/4])},
)
bnds=((0,np.2pi), (0,np.2pi)) # My free parameters can take values between 0 and 2pi.
rslts = minimize(vector_distance, zz0, method='SLSQP', jac=jac_vector_distance, constraints=cons, bounds=bnds)
这会返回以下错误：
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
TypeError: can't convert expression to float
During handling of the above exception, another exception occurred:
SystemError Traceback (most recent call last)
SystemError: <built-in function hasattr> returned a result with an error set
During handling of the above exception, another exception occurred:
SystemError Traceback (most recent call last)
SystemError: <built-in function hasattr> returned a result with an error set
During handling of the above exception, another exception occurred:
SystemError Traceback (most recent call last)
SystemError: <built-in function hasattr> returned a result with an error set
During handling of the above exception, another exception occurred:
SystemError Traceback (most recent call last)
SystemError: <built-in function hasattr> returned a result with an error set
During handling of the above exception, another exception occurred:
SystemError Traceback (most recent call last)
SystemError: <built-in function hasattr> returned a result with an error set
During handling of the above exception, another exception occurred:
SystemError Traceback (most recent call last)
SystemError: <built-in function hasattr> returned a result with an error set
During handling of the above exception, another exception occurred:
SystemError Traceback (most recent call last)
SystemError: <built-in function hasattr> returned a result with an error set
During handling of the above exception, another exception occurred:
SystemError Traceback (most recent call last)
SystemError: <built-in function hasattr> returned a result with an error set
During handling of the above exception, another exception occurred:
SystemError Traceback (most recent call last)
<ipython-input-18-fc64da7d0cae> in <module>()
----> 1 rslts = minimize(vector_distance, zz0, method='SLSQP', jac=jac_vector_distance, constraints=cons)
/users/vishnu/anaconda3/lib/python3.5/site-packages/scipy/optimize/_minimize.py in minimize(fun, x0, args, method, jac, hess, hessp, bounds, constraints, tol, callback, options)
453 elif meth == 'slsqp':
454 return _minimize_slsqp(fun, x0, args, jac, bounds,
--> 455 constraints, callback=callback, **options)
456 elif meth == 'dogleg':
457 return _minimize_dogleg(fun, x0, args, jac, hess,
/users/vishnu/anaconda3/lib/python3.5/site-packages/scipy/optimize/slsqp.py in _minimize_slsqp(func, x0, args, jac, bounds, constraints, maxiter, ftol, iprint, disp, eps, callback, **unknown_options)
404
405 # Call SLSQP
--> 406 slsqp(m, meq, x, xl, xu, fx, c, g, a, acc, majiter, mode, w, jw)
407
408 # call callback if major iteration has incremented
/users/vishnu/anaconda3/lib/python3.5/site-packages/sympy/core/expr.py in __float__(self)
219 # to fail, and if it is we still need to check that it evalf'ed to
220 # a number.
--> 221 result = self.evalf()
222 if result.is_Number:
223 return float(result)
/users/vishnu/anaconda3/lib/python3.5/site-packages/sympy/core/evalf.py in evalf(self, n, subs, maxn, chop, strict, quad, verbose)
1359
1360 """
-> 1361 from sympy import Float, Number
1362 n = n if n is not None else 15
1363
/users/vishnu/anaconda3/lib/python3.5/importlib/_bootstrap.py in _handle_fromlist(module, fromlist, import_)
SystemError: <built-in function hasattr> returned a result with an error set
答案 0 :(得分：1)
您似乎正在最小化两个椭圆之间的距离。你不需要同意这样做。这是一个例子：
from math import sin, cos, hypot, pi
from scipy import optimize
import numpy as np
def ellipse(xc, yc, a, b, psi):
a_cos_p = a * cos(psi)
a_sin_p = a * sin(psi)
b_cos_p = b * cos(psi)
b_sin_p = b * sin(psi)
def f(t):
cos_t = cos(t)
sin_t = sin(t)
x = xc + cos_t * a_cos_p - sin_t * b_sin_p
y = yc + cos_t * a_sin_p + sin_t * b_cos_p
return x, y
return f
def min_dist_between_ellipses(el1, el2):
def dist(pars):
t1, t2 = pars.tolist()
x1, y1 = el1(t1)
x2, y2 = el2(t2)
return hypot(x1 - x2, y1 - y2)
r = optimize.minimize(dist, (0, 0))
return r.x.tolist(), dist(r.x)
xc1 = 150
xc2 = 2.03
yc1 = 152
yc2 = 2.31
a1 = 5
a2 = 3
b1 = 9
b2 = 4
psi1 = pi / 2
psi2 = pi / 4
elpars1 = xc1, yc1, a1, b1, psi1
elpars2 = xc2, yc2, a2, b2, psi2
el1 = ellipse(*elpars1)
el2 = ellipse(*elpars2)
print((min_dist_between_ellipses(el1, el2)))
x1, y1 = np.array([el1(t) for t in np.linspace(0, 2*np.pi, 100)]).T
x2, y2 = np.array([el2(t) for t in np.linspace(0, 2*np.pi, 100)]).T
print(np.hypot(x1[:, None] - x2[None, :], y1[:, None] - y2[None, :]).min())
输出：
([2.098535986219504, 0.03199718973020122], 200.25805791197473)
200.259630185