https://stackoverflow.com/questions/43491287/elegant-way-to-check-if-a-nested-key-exists-in-a-dict

To be brief, with Python you must trust it is easier to ask for forgiveness than permission

try:
    x = s['mainsnak']['datavalue']['value']['numeric-id']
except KeyError:
    pass

---

The answer

Here is how I deal with nested dict keys:

def keys_exists(element, *keys):
    '''
    Check if *keys (nested) exists in `element` (dict).
    '''
    if not isinstance(element, dict):
        raise AttributeError('keys_exists() expects dict as first argument.')
    if len(keys) == 0:
        raise AttributeError('keys_exists() expects at least two arguments, one given.')

    _element = element
    for key in keys:
        try:
            _element = _element[key]
        except KeyError:
            return False
    return True

Example:

data = {
    "spam": {
        "egg": {
            "bacon": "Well..",
            "sausages": "Spam egg sausages and spam",
            "spam": "does not have much spam in it"
        }
    }
}

print 'spam (exists): {}'.format(keys_exists(data, "spam"))
print 'spam > bacon (do not exists): {}'.format(keys_exists(data, "spam", "bacon"))
print 'spam > egg (exists): {}'.format(keys_exists(data, "spam", "egg"))
print 'spam > egg > bacon (exists): {}'.format(keys_exists(data, "spam", "egg", "bacon"))

Output:

spam (exists): True
spam > bacon (do not exists): False
spam > egg (exists): True
spam > egg > bacon (exists): True

It loop in given element testing each key in given order.

I prefere this to all variable.get('key', {}) methods I found because it follows EAFP.

Function except to be called like: keys_exists(dict_element_to_test, 'key_level_0', 'key_level_1', 'key_level_n', ..). At least two arguments are required, the element and one key, but you can add how many keys you want.

If you need to use kind of map, you can do something like:

expected_keys = ['spam', 'egg', 'bacon'] 

keys_exists(data, *expected_keys) 

from : https://stackoverflow.com/questions/43491287/elegant-way-to-check-if-a-nested-key-exists-in-a-dict