On 2024-03-25, Loris Bennett via Python-list <python-list@python.org> wrote:
> Grant Edwards <grant.b.edwa...@gmail.com> writes:
>
>> On 2024-03-22, Loris Bennett via Python-list <python-list@python.org> wrote:
>>
>>> Yes, I was mistakenly thinking that the popping the element would
>>> leave me with the dict minus the popped key-value pair.
>>
>> It does.
>
> Indeed, but I was thinking in the context of
>
> dict_list = [d.pop('a') for d in dict_list]
>
> and incorrectly expecting to get a list of 'd' without key 'a', instead
> of a list of the 'd['a]'.
So when you say "leave me with", you mean "return the same dictionary
with"? There's an important difference between what a function
returns and what global/local state it "leaves you with".
> Thanks for pointing out 'del'. My main problem, however, was
> failing to realise that the list comprehension is populated by the
> return value of the 'pop', not the popped dict.
OK, so perhaps you weren't execting the original dict objects to be
mutated, but rather that the pop method would return a new dict object
without the "popped" element. The whole point of the 'pop method is to
return the popped value, otherwise it wouldn't be needed. The 'del'
statement would suffice.
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