Hi Lucas.rs
You wrote:
> In its current implementation, the list type does not provide a simple and
> straightforward way to retrieve one of its elements that fits a certain
> criteria.
>
Thank you. You've asked a good question. I hope my answer will be helpful.
Here's my preferred solution, using Python builtins:
>>> users = [
... {'id': 1,'name': 'john'},
... {'id': 2, 'name': 'anna'},
... {'id': 3, 'name': 'bruce'},
... ]
>>> func = (lambda user: user['id'] == 2)
>>> next(filter(func, users))
{'id': 2, 'name': 'anna'}
For comparison, here's your solution using your subclass of list.
>> my_list.find(func)
{'id': 2, 'name': 'anna'}
I prefer using the builtin filter function because:
1. I already know what 'filter' means.
2. I already know what 'next' means.
3. I already know what will happen if the object is not found (or if there
are several solutions.
4. It's fairly easy to modify the code to given different behaviour, eg
>>> list(filter(func, users))
[{'id': 2, 'name': 'anna'}]
Finally, perhaps your code would be better if the items in the 'user' list
were members of a class, rather than being bare dicts. And once you've done
that, you could create a new class for holding a collection (or set) of
members. And now we're moving towards how Django (and other frameworks)
handle interactions with a database.
--
Jonathan
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