Returning from a multiple stacked call at once

[jak]

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.