Is there any straightforward option of unpacking a dictionary?

user_15887937 0
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If I do something like this

some_obj = {"a": 1, "b": 2, "c": 3}
first, *rest = some_obj

I'll get a list, but I want it in 2 dictionaries: first = {"a": 1} and rest = {"b": 2, "c": 3}. As I understand, I can make a function, but I wonder if I can make it in one line, like in javascript with spread operator.

dictionary python
4 Answers

I don't know if there is a reliable way to achieve this in one line, But here is one method.

First unpack the keys and values(.items()). Using some_obj only iterate through the keys.

>>> some_obj = {"a":1, "b":2, "c": 3}
>>> first, *rest = some_obj.items()

But this will return a tuple,

>>> first
('a', 1)
>>> rest
[('b', 2), ('c', 3)]

But you can again convert back to dict with just a dict call.

>>> dict([first])
{'a': 1}
>>> dict(rest)
{'b': 2, 'c': 3}
user_66994470 
k = next(iter(some_obj))  # get the first key
first = {k: some_obj.pop(k)}
rest = some_obj

If you need to keep the original object intact - note this degrades from O(1) to O(n) in both time and space:

k = next(iter(some_obj))
rest = some_obj.copy()
first = {k: rest.pop(k)}
user_85441230

A oneliner inspired by Abdul Niyas P M's:

first, rest = dict([next(i := iter(some_obj.items()))]), dict(i)

Uses an assignment expression, introduced in Python 3.8 almost two years ago.

user_167591160

@AbdulNiyasPM's answer is perfectly fine, but since you asked for a one-liner, here's one way to do it (though you would have to do from operator import itemgetter first):

first, rest = map(dict, itemgetter(slice(0, 1), slice(1, None))(list(some_obj.items())))

If you prefer not to import anything, you can use a similar one-liner with a lambda function that takes one fixed argument and the rest as variable-length arguments:

first, rest = map(dict, (lambda f, *r: ((f,), r))(*some_obj.items()))

Demo: https://replit.com/@blhsing/ImpeccableEllipticalGeeklog

user_68909120
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