Using scipy.optimize.minimize without lambda functions

Question

I want to use the scipy.optimize.minimize function without specifying my constraints as lambda functions. Is this possible?

i.e. for the standard example:

from scipy.optimize import minimize

def fun(x):
    return (x[0] - 1) ** 2 + (x[1] - 2.5)**2.

x   = (2, 0)

def c_0(x):
    return x[0] - 2. * x[1] + 2.

def c_1(x):
    return -x[0] - 2. * x[1] + 6.

def c_2(x):
    return -x[0] + 2. * x[1] + 2.

cons = ({'type': 'ineq', 'fun': c_0(x)},
    {'type': 'ineq', 'fun': c_1(x)},
    {'type': 'ineq', 'fun': c_2(x)})

bnds = ((0, None), (0, None))

res = minimize(fun(x), x, method='SLSQP', bounds=bnds, constraints=cons)

The reason I want to avoid using lambda functions is that my constraint number grows quite quickly for my problem (2*number of degrees of freedom), so unless there's a way of creating a "lambda" factory for my constraints, explicitly writing them will become tedious very quickly.

The above code snippet returns:

TypeError: 'float' object is not callable

Note the difference: fun is a function; fun(x) is its value at x. — user6655984
– user6655984, Commented Oct 2, 2017 at 17:06

hesham_EE · Accepted Answer · 2017-10-02 15:56:02Z

1

Call the functions without arguments. This works for me:

from scipy.optimize import minimize

def fun(x):
    return (x[0] - 1) ** 2 + (x[1] - 2.5)**2.

x   = (2, 0)

def c_0(x):
    return x[0] - 2. * x[1] + 2.

def c_1(x):
    return -x[0] - 2. * x[1] + 6.

def c_2(x):
    return -x[0] + 2. * x[1] + 2.

cons = ({'type': 'ineq', 'fun': c_0},
    {'type': 'ineq', 'fun': c_1},
    {'type': 'ineq', 'fun': c_2})

bnds = ((0, None), (0, None))

res = minimize(fun, x, method='SLSQP', bounds=bnds, constraints=cons)

answered Oct 2, 2017 at 15:56

hesham_EE

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1 Comment

user6655984 Over a year ago

Correct, but that's not calling the function, it's passing it as a parameter.

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