dict.get(key, default) evaluates default even if key exists

[jak]

Il 17/12/2020 13:33, Chris Angelico ha scritto:
> On Thu, Dec 17, 2020 at 11:16 PM jak <nospam at please.ty> wrote:
>>
>> Il 17/12/2020 12:40, Peter J. Holzer ha scritto:
>>> On 2020-12-17 12:16:29 +0100, jak wrote:
>>>> print(_ if d.get('a', None) is not None else get_default())
>>>
>>> That doesn't work:
>>>
>>>>>> print(_ if d.get('a', None) is not None else get_default())
>>> Traceback (most recent call last):
>>>     File "<stdin>", line 1, in <module>
>>> NameError: name '_' is not defined
>>>
>>> But this works:
>>>
>>>>>> print(_ if (_ := d.get('a', None)) is not None else get_default())
>>> 1
>>>
>>> (I would prefer ChrisA's solution, though.)
>>>
>>>           hp
>>>
>> this one?
>>
>> """"
>>
>> D['a'] if 'a' in D else get_default()
>>
>> ChrisA
>>
>> """"
>>
>> This solution search two times same key.
> 
> Yes, it does, but hash lookups are pretty fast. Unless your key is
> some sort of custom object with a very expensive __hash__ function,
> the double lookup isn't going to be too costly. But if that does
> bother you, you can write it as a try/except instead.
> 
> ChrisA
> 

try/except is a very expensive time. if we have a small dictionary in 
which you do little research, you definitely have reason. I would be 
curious to see the differences by taking times with 50 / 60K records by 
repeating the search 1 or 2 million times with and without try/except 
and also searching 1 or 2 times too.

Cheers