List defining the list of variable of function in Python -

i want have function foo outputs function, list of variables depends on input list.

precisely:

suppose func function free variable t , 3 parameters a,gamma,x

example: func = lambda t,a,gamma,x: somefunction

i want define function foo, takes input list , outputs function. output function sum of func's, each func summand has parameters independent each other.

depending on input list variables of outputs changes in following:

if entry of list 'none' output function 'gains' variable , if entry of list float 'fixes' parameter.

example:

li=[none,none,0.1] g=foo(li) 

gives same output as

g = lambda t,a,gamma: func(t,a,gamma,0.1)

or

li=[none,none,0.1,none,0.2,0.3] g=foo(li) 

gives same output as

g = lambda t,a1,gamma1,a2:func(t,a,gamma,0.1)+func(t,a2,0.2,0.3)

note: order in list relevant , behaviour wanted.

i don't have clue on how that...

i first tried build string, depends on inout list , execute it, surely not way.

first, partition parameters li chunks. use iterator either next function parameters *args, if value in chunk none, or provided value parameters chunk.

def foo(li, func, num_params):     chunks = (li[i:i+num_params] in range(0, len(li), num_params))     def function(t, *args):         result = 0         iter_args = iter(args)         chunk in chunks:             actual_params = [next(iter_args) if x none else x x in chunk]             result += func(t, *actual_params)         return result     return function 

example:

def f(t, a, b, c):     print t, a, b, c     return + b + c  func = foo([1,2,none,4,none,6], f, 3) print func("foo", 3, 5) 

output:

foo 1 2 3  # first call f foo 4 5 6  # second call f 21         # result of func