I don't know how to pass additional arguments through the minimize function to the constraint dictionary. I can successfully pass additional arguments to the objective function.

Documentation on minimize function is here: http://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.minimize.html#scipy.optimize.minimize

The constraints argument is a dict that has a field 'args', where args is a sequence. I'm sure this is where I need to pass in the additional arguments but I don't know the syntax. The closest I have got is below:

from scipy.optimize import minimize
def f_to_min (x, p):return (p[0]*x[0]*x[0]+p[1]*x[1]*x[1]+p[2])f_to_min([1,2],[1,1,1]) # test function to minimizep=[] # define additional args to be passed to objective function
f_to_min_cons=({'type': 'ineq', 'fun': lambda x, p : x[0]+p[0], 'args': (p,)}) # define constraintp0=np.array([1,1,1])
minimize(f_to_min, [1,2], args=(p0,), method='SLSQP', constraints=f_to_min_cons)

I get the following error

---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-19-571500063c9e> in <module>()1 p0=np.array([1,1,1])
----> 2 minimize(f_to_min, [1,2], args=(p0,), method='SLSQP', constraints=f_to_min_cons)C:\Python27\lib\site-packages\scipy\optimize\_minimize.pyc in minimize(fun, x0, args,     method, jac, hess, hessp, bounds, constraints, tol, callback, options)356     elif meth == 'slsqp':357         return _minimize_slsqp(fun, x0, args, jac, bounds,
--> 358                                constraints, **options)359     else:360         raise ValueError('Unknown solver %s' % method)C:\Python27\lib\site-packages\scipy\optimize\slsqp.pyc in _minimize_slsqp(func, x0,     args, jac, bounds, constraints, maxiter, ftol, iprint, disp, eps, **unknown_options)298     # meq, mieq: number of equality and inequality constraints299     meq = sum(map(len, [atleast_1d(c['fun'](x, *c['args'])) for c in     cons['eq']]))
--> 300     mieq = sum(map(len, [atleast_1d(c['fun'](x, *c['args'])) for c in     cons['ineq']]))301     # m = The total number of constraints302     m = meq + mieq<ipython-input-18-163ef1a4f6fb> in <lambda>(x, p)
----> 1 f_to_min_cons=({'type': 'ineq', 'fun': lambda x, p : x[0]+p[0], 'args': (p,)})IndexError: list index out of range

I'm accessing the first element of the additional parameter so I shouldn't have an out of range error.

If you remove the constraints=f_to_min_cons argument from the minimize function then the code above works.

The answer simply, is that p = [] has no elements and no length, and so p[0] is out of bounds.

The following, where we set p = [0], runs without error. What p should actually hold is of course, not something that we can answer with the information given.

import numpy as npfrom scipy.optimize import minimize
def f_to_min (x, p):return (p[0]*x[0]*x[0]+p[1]*x[1]*x[1]+p[2])f_to_min([1,2],[1,1,1]) # test function to minimizep=[0] # define additional args to be passed to the constraint
f_to_min_cons=({'type': 'ineq', 'fun': lambda x, p : x[0]+p[0], 'args': (p,)},) # define constraintp0=np.array([1,1,1]) # args to be passed to the objective function
minimize(f_to_min, [1,2], args=(p0,), method='SLSQP', constraints=f_to_min_cons)