问题
I try to solve nonlinear programming task using scipy.optimize.minimize
max r
x1**2 + y1**2 <= (1-r)**2
(x1-x2)**2 + (y1-y2)**2 >= 4*r**2
0 <= r <= 1
So I've wrote next code:
r = np.linspace(0, 1, 100)
x1 = np.linspace(0, 1, 100)
y1 = np.linspace(0, 1, 100)
x2 = np.linspace(0, 1, 100)
y2 = np.linspace(0, 1, 100)
fun = lambda r: -r
cons = ({'type': 'ineq',
'fun': lambda x1, r: [x1[0] ** 2 + x1[1] ** 2 - (1 - r) ** 2],
'args': (r,)},
{'type': 'ineq',
'fun': lambda x2, r: [x2[0] ** 2 + x2[1] ** 2 - (1 - r) ** 2],
'args': (r,)},
{'type': 'ineq',
'fun': lambda x1, x2, r: [(x1[0] - x2[0]) ** 2 + (x1[1] - x2[1]) ** 2 - 4 * r ** 2],
'args': (x2, r,)})
bnds = ((0, 1), (-1, 1), (-1, 1), (-1, 1), (-1, 1))
x0 = [0, 0, 0, 0, 0]
minimize(fun, x0, bounds=bnds, constraints=cons)
But I've got next error
File "C:\Anaconda2\lib\site-packages\scipy\optimize\slsqp.py", line 377, in _minimize_slsqp
c = concatenate((c_eq, c_ieq))
ValueError: all the input arrays must have same number of dimensions
Please, help me to find out my mistakes and write correct code
UPD: Thx to @unutbu i've understand how to build it correctly.
fun = lambda x: -x[0]
cons = ({'type': 'ineq',
'fun': lambda x: -x[1] ** 2 - x[2] ** 2 + (1 - x[0]) ** 2},
{'type': 'ineq',
'fun': lambda x: -x[3] ** 2 - x[4] ** 2 + (1 - x[0]) ** 2},
{'type': 'ineq',
'fun': lambda x: (x[1] - x[3]) ** 2 + (x[1] - x[4]) ** 2 - 4 * x[0] ** 2})
bnds = ((0, 1), (-1, 1), (-1, 1), (-1, 1), (-1, 1))
x0 = [0.5, 0.3, 0.5, 0.3, 0.5]
answer = minimize(fun, x0, bounds=bnds, constraints=cons)
In task of minimization we have to lead constraints to such form:
g(x) >= 0
that's why constraints look like in that way.
回答1:
Your parameter space appears to be 5-dimensional. A point in your parameter
space would be z = (r, x1, y1, x2, y2). Therefore the function to be minimized
-- and also the constraint functions -- should accept a point z and
return a scalar value.
Thus instead of
fun = lambda r: -r
use
def func(z):
r, x1, y1, x2, y2 = z
return -r
and instead of
lambda x1, r: [x1[0] ** 2 + x1[1] ** 2 - (1 - r) ** 2]
use
def con1(z):
r, x1, y1, x2, y2 = z
return x1**2 + y1**2 - (1-r)**2
and so on.
Note that simple constraints such as 0 <= r <= 1 can be handled by setting the bounds parameter instead of defining a constraint. And if the bounds for x1, y1, x2, y2 are from -1 to 1, then you might also want change
x1 = np.linspace(0, 1, 100)
...
to
x1 = np.linspace(-1, 1, 100)
...
However, the arrays r, x1, y1, x2, y2 are not needed to minimize func, so you could just as well eliminate them from the script entirely.
import numpy as np
import scipy.optimize as optimize
"""
max r
x1**2 + y1**2 <= (1-r)**2
(x1-x2)**2 + (y1-y2)**2 >= 4*r**2
0 <= r <= 1
"""
def func(z):
r, x1, y1, x2, y2 = z
return -r
def con1(z):
r, x1, y1, x2, y2 = z
return x1**2 + y1**2 - (1-r)**2
def con2(z):
r, x1, y1, x2, y2 = z
return 4*r**2 - (x1-x2)**2 - (y1-y2)**2
cons = ({'type': 'ineq', 'fun': con1}, {'type': 'ineq', 'fun': con2},)
bnds = ((0, 1), (-1, 1), (-1, 1), (-1, 1), (-1, 1))
guess = [0, 0, 0, 0, 0]
result = optimize.minimize(func, guess, bounds=bnds, constraints=cons)
print(result)
yields
fun: -1.0
jac: array([-1., 0., 0., 0., 0., 0.])
message: 'Optimization terminated successfully.'
nfev: 14
nit: 2
njev: 2
status: 0
success: True
x: array([ 1., 0., 0., 0., 0.])
来源:https://stackoverflow.com/questions/37777025/scipy-optimize-minimize-function