Вы можете использовать itertools.product
:
import itertools
cont = {'perm': {'r': False, 'rw': True}, 'prig': {'sq': False, 'rot': False, 'rq': True}, 'anon': {'100': False, '500': True, '99': False, '400': False}}
keys = {a:list(b) for a, b in cont.items()}
p = list(itertools.product(*_keys.values()))
result = [[cont[[a for a, b in keys.items() if c in b][0]][c] for c in i] for i in p]
new_result = [any(i) for i in result]
Выход:
#result:
[[False, False, False], [False, False, True], [False, False, False], [False, False, False], [False, False, False], [False, False, True], [False, False, False], [False, False, False], [False, True, False], [False, True, True], [False, True, False], [False, True, False], [True, False, False], [True, False, True], [True, False, False], [True, False, False], [True, False, False], [True, False, True], [True, False, False], [True, False, False], [True, True, False], [True, True, True], [True, True, False], [True, True, False]]
#new_result
[False, True, False, False, False, True, False, False, True, True, True, True, True, True, True, True, True, True, True, True, True, True, True, True]
#list(map(','.join, prod))
['r,sq,100', 'r,sq,500', 'r,sq,99', 'r,sq,400', 'r,rot,100', 'r,rot,500', 'r,rot,99', 'r,rot,400', 'r,rq,100', 'r,rq,500', 'r,rq,99', 'r,rq,400', 'rw,sq,100', 'rw,sq,500', 'rw,sq,99', 'rw,sq,400', 'rw,rot,100', 'rw,rot,500', 'rw,rot,99', 'rw,rot,400', 'rw,rq,100', 'rw,rq,500', 'rw,rq,99', 'rw,rq,400']