Returning from a multiple stacked call at once
jak
nospam at please.ty
Tue Dec 15 06:39:29 EST 2020
Il 15/12/2020 12:25, Chris Angelico ha scritto:
> On Tue, Dec 15, 2020 at 9:56 PM jak <nospam at please.ty> wrote:
>>
>> this could be a way to emulate a long_jump:
>>
>> def f(i):
>> if i < 10:
>> i += 1
>> yield from f(i)
>> else:
>> yield i
>>
>> i = 0
>> retult = 0
>> for n in f(i):
>> result = n
>> break
>> print(result)
>>
>
> Note that this requires just as much cooperation as this version does:
>
> def f(i):
> if i < 10:
> return f(i + 1)
> return i
>
> result = f(0)
>
> The only difference is that "yield from" sorta kinda gives you a
> "maybe return", but you're not actually using that in your example, so
> it doesn't showcase that. But let's say you're searching a complex
> data structure for something:
>
> def find(obj, pred):
> if pred(obj): yield obj
> if isinstance(obj, list):
> for elem in obj: yield from find(elem, pred)
> if isinstance(obj, dict):
> for elem in obj.values(): yield from find(elem, pred)
>
> Taking the first matching element could be done without worrying too
> much about whether any subsequent elements would match. I wouldn't
> really call this a "long jump", though; this is a lazy filter that can
> be efficiently used to locate the first few results without
> calculating them all.
>
> ChrisA
>
You are right. In fact I used the word 'emulate' to mean that this is
what (IMO) comes closest to a long jump.
Cheers.
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