How to build a dict which all values are independent instances of [] in Python?

Question

>>> keys = [1, 2, 3]
>>> d = dict(zip(keys, [[]]*len(keys)))
>>> d
{1: [], 2: [], 3: []}
>>> d[1].append(100)
>>> d
{1: [100], 2: [100], 3: [100]}

even copied [] not works:

the {1: [100], 2: [], 3: []} is what I wanted exactly.

The easiest would be one of the solutions [here](https://repl.it/repls/JadedHeartfeltDeclaration) — U13-Forward, Aug 29 '18 at 01:41
Maybe a [defaultdict](https://docs.python.org/3/library/collections.html#collections.defaultdict) would be useful: `d = defaultdict(lambda: [])`. — Klaus D., Aug 29 '18 at 01:46
Your attempts are all copying the list, and then making three references to that copy, which isn’t any better than just making three references to the original list. If you move the `deepcopy` _outside_ the multiplication… but you’re still better off using a comprehension in the first place — abarnert, Aug 29 '18 at 02:21

Chillie · Answer 1 · 2018-08-29T01:44:51.200

1

keys = [1, 2, 3]
d = dict(zip(keys, [[] for _ in keys]))
d[1].append(100)

As per abarnert's comment, you could make your code even simpler by using a dictionary comprehension:

keys = [1, 2, 3]
d = {key:[] for key in keys}
d[1].append(100)

Calling d with both options gives the following output:

{1: [100], 2: [], 3: []}

edited Aug 29 '18 at 01:44

answered Aug 29 '18 at 01:20

Chillie

1

Why do this instead of just writing a dict comprehension? – abarnert Aug 29 '18 at 01:36
@abarnert good idea! I wanted to keep it as close to the OP's code as possible for some reason. Adding dict comprehension. – Chillie Aug 29 '18 at 01:38

1 Answers1