Exclude 'None' from list comprehension of dicts

[dn]

On 16/08/2022 00.56, Antoon Pardon wrote:
> Op 5/08/2022 om 07:50 schreef Loris Bennett:
>> Antoon Pardon<antoon.pardon at vub.be>  writes:
>>
>>> Op 4/08/2022 om 13:51 schreef Loris Bennett:
>>>> Hi,
>>>>
>>>> I am constructing a list of dictionaries via the following list
>>>> comprehension:
>>>>
>>>>     data = [get_job_efficiency_dict(job_id) for job_id in job_ids]
>>>>
>>>> However,
>>>>
>>>>     get_job_efficiency_dict(job_id)
>>>>
>>>> uses 'subprocess.Popen' to run an external program and this can fail.
>>>> In this case, the dict should just be omitted from 'data'.
>>>>
>>>> I can have 'get_job_efficiency_dict' return 'None' and then run
>>>>
>>>>     filtered_data = list(filter(None, data))
>>>>
>>>> but is there a more elegant way?
>>> Just wondering, why don't you return an empty dictionary in case of a
>>> failure?
>>> In that case your list will be all dictionaries and empty ones will
>>> be processed
>>> fast enough.
>> When the list of dictionaries is processed, I would have to check each
>> element to see if it is empty.  That strikes me as being less efficient
>> than filtering out the empty dictionaries in one go, although obviously
>> one would need to benchmark that.
> 
> I may be missing something but why would you have to check each element
> to see if it is empty? What would go wrong if you just treated empty
> dictionaries the same as non-empty directories?

'Truthiness':-

>>> bool( {} )
False
>>> bool( { "a":1 } )
True

-- 
Regards,
=dn