[Python-ideas] Accessing the result of comprehension's expression from the conditional
g.brandl at gmx.net
Sat Jun 20 10:13:06 CEST 2009
Lie Ryan schrieb:
> Steven D'Aprano wrote:
>> On Fri, 19 Jun 2009 05:39:32 pm Lie Ryan wrote:
>>> Aahz wrote:
>>>> On Fri, Jun 19, 2009, Lie Ryan wrote:
>>>>> In list/generator comprehension, currently we have no way to
>>>>> access the result of the expression and have to write something
>>>>> like this:
>>>>> [f(x) for x in l if f(x) > 0]
>>>>> if f() is heavy or non-pure (i.e. have side effects), calling f()
>>>>> twice might be undesirable.
>>>> Listcomps and genexps are like lambdas: run up against their limits
>>>> and you should switch to a regular for loop or generator.
>>> I think of it not as limitation but as an odd gap in functionality. I
>>> think having the semantics that the filtering is done after the
>>> expression part would be much more useful than the current behavior
>>> (filtering before expression).
>> The point of the filtering is to avoid needlessly calculating a
>> potentially expensive expression only to throw it away.
>> If you want expression first, then filter, you can get that already in a
>> filter(lambda x: x > 0, [f(x) for x in seq])
>> Don't create new syntax when there are perfectly good functions that do
>> the job already.
> That's ugly because of the same reason for using map():
> [y for y in map(lambda x: f(x), seq) if y > 0]
Especially if f is already a handy callable of one argument, no need to use
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