Python dict: Make a key resolve to itself

Question

Given the following dict

CV = { a: [a, b], b: [b, c], c: [c], d: [d] }

Let's say I am lazy and I want to have the dictionary return a value that references the key, something like

CV = { a: [a, b], b: [b, c], _: [_] }

Is this possible?

What's the question? If a, b, c and _ all exists, that CV line works — Koen G., May 12 '22 at 09:35
I suppose you could use something like a dict comprehension to write all the reflexive entries somewhat shorter. — Luatic, May 12 '22 at 09:36
If you want to have a dictionary that for a missing key `k` returns `[k]` by default, you can do that by subclassing `dict`. Is that what you are asking for? — khelwood, May 12 '22 at 09:37
@khelwood Yes, I am looking for a way to get a dict to return some default values implicitly. Maybe that's a bit too much for a dict, so I'm thinking about wrapping it in a function that handles the default part, instead of using `.get()` and repeat the default return value multiple times. — Hoang, May 12 '22 at 09:48
So if I understand correctly you want, for missing keys, to return a value based on a function called with the missing key? In that case look here: [python-dict-with-default-value-based-on-key](https://stackoverflow.com/questions/65901825/python-dict-with-default-value-based-on-key) — Koen G., May 12 '22 at 09:52

score 1 · Answer 1 · answered May 12 '22 at 09:41

1

Use the .get() method and set the key as the default values as well:

>>> my_dict = {"a": [1], "b": [2]}
>>> my_dict.get("a", ["a"])
[1]
>>> my_dict.get("c", ["c"])
['c']
>>>

answered May 12 '22 at 09:41

Tzane

1 Answers1