Should I handle unexpected arguments of a function?

Is there a way to use a dictionary to build up arguments for a function?

• I have a class whose constructor is something like: def __init__(self, arg1, arg2, arg3, arg4): assign_all_properties() The class has a lot of properties and hence the constructor has a lot of arguments. Then, where I need to instantiate the class, I already have all the property-value pairs in a dictionary. Is there a way to directly use that dictionary to instantiate the class? I tried to build a string and used eval just to see if that would work but not too surprisingly it didn't: args_string = "arg1 = 'value1', arg2 = 'value2'" some_instance = SomeObject(eval(args_string)) I'd hate to enumerate every single arguments. Is there a better way out? I guess, I could do something like: some_instance = SomeObject(**arguments) But I don't want my class to take just *any* parameters.

• Answer:

  Edit : Didn't read your last sentence, also find a slightly better way 1 2 3 4 5 6 7 8 91011121314151617181920212223242526272829class Base():     signature = {} # Accepted parameters     defaults = {} # Default values for instances     def __init__(self, **kwargs):         # Set the default values for instances         self.__dict__ = self.defaults         # Iterate through kwargs         for key in kwargs.keys():             if key not in self.signature:                 raise Exception("Cannot accept parameter " + key)             else:                 setattr(self, key, kwargs.get(key)) class Animal(Base):     signature = { 'species', 'age' }     defaults = {         'age' : 1     } if __name__ == "__main__":     myDog = Animal(species = 'Dog')     print(myDog.species) # Dog     print(myDog.age) # 1, use default     mySheep = Animal(species = 'Sheep', age = 10)     print(mySheep.species) # Sheep     print(mySheep.age) # 10     # This will fail     myFail = Animal(species = 'Sheep', cute = True)