django - How to create dynamic methods with python? -

for project need dynamically create custom (class) methods.

i found out not easy in python:

class userfilter(django_filters.filterset):     '''     filter used in api     '''     # legacy below, has added dynamically     #is_field_type1 = methodfilter(action='filter_field_type1')       #def filter_field_type1(self, queryset, value):     #    return queryset.filter(related_field__field_type1=value)      class meta:         model = get_user_model()         fields = [] 

but giving me errors (and headaches...). possible?

i try make code between #legacy dynamic

one option found create class dynamically

def create_filter_dict():     new_dict = {}     field in list_of_fields:          def func(queryset, value):             _filter = {'stableuser__'+field:value}             return queryset.filter(**_filter)          new_dict.update({'filter_'+field: func})          new_dict.update({'is_'+field: methodfilter(action='filter_'+field)})      return new_dict   meta_model_dict = {'model': get_user_model(), 'fields':[]} meta_type = type('meta',(), meta_model_dict)  filter_dict = create_filter_dict() filter_dict['meta'] = meta_type userfilter = type('userfilter', (django_filters.filterset,), filter_dict) 

however, giving me

typeerror @ /api/v2/users/ func() takes 2 positional arguments 3 given 

does know how solve dilemma?

exception value: 'userfilter' object has no attribute 'is_bound'

you getting error because class methods generating, not bound class. bound them class, need use setattr()

try on console:

class myclass(object):     pass  @classmethod  def unbound(cls):     print "now i'm bound ", cls   print unbound setattr(myclass, "bound", unbound)  print myclass.bound  print myclass.bound() 

traceback: userfilter = type('foo', (django_filters.filterset, ), create_filter_dict().update({'meta':type('meta',(), {'model': get_user_model(), 'fields':[]} )})) typeerror: type() argument 3 must dict, not none

now, failing because dict.update() doesn't return same instance, returns none. can fixed

class_dict = create_filter_dict() class_dict.update({'meta':type('meta',(), {'model': get_user_model(), 'fields':[]})} userfilter = type('foo', (django_filters.filterset, ), class_dict)) 

however, how messy code looks. recommend try clearer code write if requires write few lines. in long run, code easier maintain , team.

meta_model_dict = {'model': get_user_model(), 'fields':[]} meta_type = type('meta',(), meta_model_dict)  filter_dict = create_filter_dict() filter_dict['meta'] = meta_type userfilter = type('foo', (django_filters.filterset,), filter_dict) 

this code might not perfect more readable original line of code posted:

userfilter = type('foo', (django_filters.filterset, ), create_filter_dict().update({'meta':type('meta',(), {'model': get_user_model(), 'fields':[]})})) 

and removes complication on kinda difficult concept grasp.

you might want learn metaclasses. maybe can overwrite new method of class. can recommend 1 or 2 posts that.

another option maybe not adding filters correctly or in way django doesn't expect? explain why no errors none of functions gets called.