Checking dict keys to ensure a required key always exists, and that the dict has no other key names beyond a defined set of names

Question

I have a dict in python that follows this general format:

{'field': ['$.name'], 'group': 'name', 'function': 'some_function'} 

I want to do some pre-check of the dict to ensure that 'field' always exists, and that no more keys exist beyond 'group' and 'function' which are both optional.

I know I can do this by using a long and untidy if statement, but I'm thinking there must be a cleaner way?

This is what I currently have:

if (('field' in dict_name and len(dict_name.keys()) == 1) or      ('group' in dict_name and len(dict_name.keys()) == 2) or      ('function' in dict_name and len(dict_name.keys()) == 2) or      ('group' in dict_name and 'function' in dict_name and len(dict_name.keys()) == 3)) 

Essentially I'm first checking if 'field' exists as this is required. I'm then checking to see if it is the only key (which is fine) or if it is a key alongside 'group' and no others, or a key alongside 'function' and no others or a key alongside both 'group' and 'function' and no others.

Is there a tidier way of checking the keys supplied are only these 3 keys where two are optional?

Answer 1

As far as I'm concerned you want to check, that

1. The set {'field'} is always contained in the set of your dict keys
2. The set of your dict keys is always contained in the set {'field', 'group', 'function'}

So just code it!

required_fields = {'field'} allowed_fields = required_fields | {'group', 'function'}  d = {'field': 123}  # Set any value here  if required_fields <= d.keys() <= allowed_fields:     print("Yes!") else:     print("No!") 

This solution is scalable for any sets of required and allowed fields unless you have some special conditions (for example, mutually exclusive keys)

(thanks to @Duncan for a very elegant code reduction)