Map and set bug hunt

Questions 1 to 5 of 8

1. 1. The author wants to count the tags for doc2 without changing the dict. What is the bug in this code? ``python from collections import defaultdict tags = defaultdict(list) tags['doc1'] = ['python'] n = len(tags['doc2']) ``

   tags['doc2'] stores an empty list under 'doc2' when the key is missing, so the dict gains a key the author never addedlen() cannot be called on a list value, so the code raises TypeErrorThe factory list must return None for a missing keyA dict does not allow new keys after the first assignment
2. 2. The author runs this code with no score for bo. What goes wrong? ``python score = {'ana': 5} bo = score['bo'] print('bo: ' + str(bo)) ``

   The program prints bo: Nonescore['bo'] raises KeyError: 'bo', because [] has no default value and the key is missingThe key 'bo' is not allowed in a dictA dict of int values cannot be created, so the code will not run
3. 3. The author wants to remove every key whose value is 0. What is the bug in this loop? ``python for item in stock: if stock[item] == 0: del stock[item] ``

   The == check compares two str objects, so it never matches an int valuedel needs the value as well as the key, so the statement always failsdel changes the dict while the loop is still walking over it, so Python raises RuntimeError: dictionary changed size during iterationA for loop cannot run over a dict, because a dict is not a sequence
4. 4. The answers this code prints are correct, but the code has a bug. What is it? ``python from collections import defaultdict friends = defaultdict(set) friends['ana'] = {'bo', 'cy'} for name, other in [('ana', 'bo'), ('dev', 'ana')]: print(other in friends[name]) ``

   in on a set is case sensitive, so it returns False for 'ana'friends['dev'] returns None for 'dev', so in raises TypeErrorA set cannot be stored as a dict value, so the dict must hold a frozensetfriends[name] stores an empty set for 'dev' when 'dev' only appears in a question, so the dict ends up with a key nobody added
5. 5. What does this code print? ``python m = {'kiwi': 2} m['pear'] = 0 m['fig'] = 0 m['lime'] = 1 m['pear'] = 4 m = {k: v for k, v in m.items() if v != 0} print(dict(sorted(m.items()))) ``

   {'fig': 0, 'kiwi': 2, 'lime': 1, 'pear': 4}{'kiwi': 2, 'lime': 1, 'pear': 4}{'kiwi': 2, 'lime': 1}{'kiwi': 2, 'pear': 0, 'lime': 1}