Adding multiple constraints to scipy minimize, autogenerate constraint dictionary list?

Question

Is there a way to autogenerate a dictionary list of multiple constraints in scipy.minimize? When I use the following code (where the list constraint is a list of sage multivariate polynomials over the same ring)

cons = [{'type': 'eq', 'fun': lambda s: ((constraint[0])(*s))},
        {'type': 'eq', 'fun': lambda s: ((constraint[1])(*s))},
        {'type': 'eq', 'fun': lambda s: ((constraint[2])(*s))},
        {'type': 'eq', 'fun': lambda s: ((constraint[3])(*s))}]

y0 = [.5 for xx in x]
bnds = tuple([(0.0, 1.0) for xx in x])
ssoln = scipy.optimize.minimize(HH, y0, jac=dHH, method='SLSQP', bounds=bnds, constraints=cons)
print ssoln

My output is

status: 0
 success: True
    njev: 14
    nfev: 22
     fun: -2.2669026273652237
       x: array([ 0.034829615490635,  0.933405952554424,  0.93340765416238 ,
        0.093323548109654,  0.335713397575351,  0.413107862378296])
 message: 'Optimization terminated successfully.'
     jac: array([-3.321836605297572,  2.640225014918886,  2.640252390205999,
       -2.273713195767229, -0.682455873949375, -0.351132324172705,  0.               ])
     nit: 14

However if I try to create cons by

cons=[]
for ii in range(len(constraint)):
    cons.append({'type': 'eq', 'fun': lambda s:  ((constraint[ii])(*s))})

minimize fails with

status: 6
 success: False
    njev: 1
    nfev: 1
     fun: -4.1588830833596715
       x: array([ 0.5,  0.5,  0.5,  0.5,  0.5,  0.5])
 message: 'Singular matrix C in LSQ subproblem'
     jac: array([ 0.,  0.,  0.,  0.,  0.,  0.,  0.])
     nit: 1

My list, constraint, of sage polynomials may change in length, in number of polynomials from problem to problem and I don't want to have to hard code the cons dict list as first given above for each problem. Is there a way to automate?

The following works but I understand it is not best practices to eval strings

str1='{\'type\': \'eq\', \'fun\': lambda s: ((constraint['
str2='])(*s))},'
mystr='['
for ii in range(len(constraint)):
    mystr=mystr+str1+str(ii)+str2
mystr=mystr+']'
cons = eval(mystr)

I have recently run into this problem myself. Did you ever find a solution that didn't involve evaluating the string? — Ryan, Nov 04 '17 at 13:55

score 5 · Answer 1 · answered Nov 02 '18 at 13:00

The problem is in your loop. The lambda operator performs what is called a lazy evaluation. At the end of your loop, the lambda the cons is performing the function on the last value of ii, instead of on each index.

To perform a strict evaluation, you can use the partial object from the python functools module (in both python 2 or python 3).

To exemplify, with lambda:

constraint = (lambda x: x, lambda x: x**2, lambda x: x**3, lambda x: x**4)

cons=[]
for ii in range(len(constraint)):
    # lambda s will evaluate the last value of ii
    cons.append({'type': 'eq', 'fun': lambda s: ((constraint[ii])(s))})

print([i['fun'](2) for i in cons])
# The value of ii is 3, so it will always call lambda x: x**4
>> [16, 16, 16, 16]

from functools import partial

def f_constraint(s, index):
    return constraint[index](s)

cons=[]
for ii in range(len(constraint)):
    # the value of ii is set in each loop
    cons.append({'type': 'eq', 'fun': partial(f_constraint, index=ii)})

print([i['fun'](2) for i in cons])
>> [2, 4, 8, 16]

Replacing s by *s to match your definition:

from functools import partial

def f_constraint(s, index):
    return constraint[index](*s)

cons=[]
for ii in range(len(constraint)):
    cons.append({'type': 'eq', 'fun': partial(f_constraint, index=ii)})

Hope it helps!

score 0 · Answer 2 · answered Nov 18 '21 at 06:18

def wrapper(f, *add_args):
  def inner(*args):
    lt_args=list(args)
    lt_args.extend(add_args)#this arguments will be thus added after so see the positional argument of the parent function to 
    
    return f(*lt_args)
  return inner



def constraint(M,i):
  global bb
  return M - bb[i] 

lt_fn=[]
# const=wrapper(const,)
lt_fn=[wrapper(constraint,i) for i in range(len(bb))]


[f(100) for f in lt_fn]

This will execute like inner(100) inner(100) inner(100) inner(100) inner(100)__> constraint(100,0),constraint(100,1),constraint(100,2),constraint(100,3)constraint(100,4)

Adding multiple constraints to scipy minimize, autogenerate constraint dictionary list?

2 Answers2