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Re: Is this PEP-able? fwhile

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  Re: Is this PEP-able? fwhile Ian Kelly <ian.g.kelly@gmail.com> - 2013-06-24 14:12 -0600
    Re: Is this PEP-able? fwhile alex23 <wuwei23@gmail.com> - 2013-06-25 09:14 +1000
      Re: Is this PEP-able? fwhile Fábio Santos <fabiosantosart@gmail.com> - 2013-06-25 00:35 +0100
        Re: Is this PEP-able? fwhile alex23 <wuwei23@gmail.com> - 2013-06-25 10:00 +1000
          Re: Is this PEP-able? fwhile Fábio Santos <fabiosantosart@gmail.com> - 2013-06-25 01:19 +0100
          Re: Is this PEP-able? fwhile wu wei <wuwei23@gmail.com> - 2013-06-25 10:41 +1000
          Re: Is this PEP-able? fwhile Ian Kelly <ian.g.kelly@gmail.com> - 2013-06-24 22:50 -0600
      Re: Is this PEP-able? fwhile rusi <rustompmody@gmail.com> - 2013-06-24 23:04 -0700

#49073 — Re: Is this PEP-able? fwhile

On Mon, Jun 24, 2013 at 1:52 PM,  <jimjhb@aol.com> wrote:
> Syntax:
>
> fwhile X in ListY and conditionZ:
>
> The following would actually exactly as:  for X in ListY:
>
> fwhile X in ListY and True:
>
> fwhile would act much like 'for', but would stop if the condition after the
> 'and' is no longer True.
>
> The motivation is to be able to make use of all the great aspects of the
> python 'for' (no indexing or explict
> end condition check, etc.) and at the same time avoiding a 'break' from the
> 'for'.

I would advocate using the break myself.  Another alternative is this:

for X in itertools.takewhile(lambda X: conditionZ, ListY):
    ...

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#49106

On 25/06/2013 6:12 AM, Ian Kelly wrote:
> On Mon, Jun 24, 2013 at 1:52 PM,  <jimjhb@aol.com> wrote:
>> Syntax:
>> fwhile X in ListY and conditionZ:
>>
>> fwhile would act much like 'for', but would stop if the condition after the
>> 'and' is no longer True.
>
> I would advocate using the break myself.  Another alternative is this:
>
> for X in itertools.takewhile(lambda X: conditionZ, ListY):
>      ...

I'd probably just go with a generator expression to feed the for loop:

     for X in (i for i in ListY if conditionZ):
         ....

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#49109

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On 25 Jun 2013 00:31, "alex23" <wuwei23@gmail.com> wrote:
>
> On 25/06/2013 6:12 AM, Ian Kelly wrote:
>>
>> On Mon, Jun 24, 2013 at 1:52 PM,  <jimjhb@aol.com> wrote:
>>>
>>> Syntax:
>>> fwhile X in ListY and conditionZ:
>>>
>>> fwhile would act much like 'for', but would stop if the condition after
the
>>> 'and' is no longer True.
>>
>>
>> I would advocate using the break myself.  Another alternative is this:
>>
>> for X in itertools.takewhile(lambda X: conditionZ, ListY):
>>      ...
>
>
> I'd probably just go with a generator expression to feed the for loop:
>
>     for X in (i for i in ListY if conditionZ):
>         ....

That is nice but it's not lazy. If the condition or the iterables took too
long to compute, it would be troublesome.

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#49112

On 25/06/2013 9:35 AM, Fábio Santos wrote:
>  > I'd probably just go with a generator expression to feed the for loop:
>  >
>  >     for X in (i for i in ListY if conditionZ):
>  >         ....
>
> That is nice but it's not lazy. If the condition or the iterables took
> too long to compute, it would be troublesome.

I'm not sure I follow. It's a generator expression, not a list 
comprehension, so the condition will be evaluated per item iterated over 
in the generator, not across all valid items in ListY at once.

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#49114

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On 25 Jun 2013 01:08, "alex23" <wuwei23@gmail.com> wrote:
>
> On 25/06/2013 9:35 AM, Fábio Santos wrote:
>>
>>  > I'd probably just go with a generator expression to feed the for loop:
>>  >
>>  >     for X in (i for i in ListY if conditionZ):
>>  >         ....
>>
>> That is nice but it's not lazy. If the condition or the iterables took
>> too long to compute, it would be troublesome.
>
>
> I'm not sure I follow. It's a generator expression, not a list
comprehension, so the condition will be evaluated per item iterated over in
the generator, not across all valid items in ListY at once.
>

for X in (i for i in open('largefile') if is_part_of_header(i)):

The above code would be wasting time on IO and processing. It would load
another line and calculate the condition for every line of the large file
and I just wanted to loop over the few header lines.

itertools.takewhile and fwhile/for..while actually stops the loop when the
condition is not meant, while your example keeps checking the condition
until the end of file, even though it is a generator expression.

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#49115

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On Tue, Jun 25, 2013 at 10:19 AM, Fábio Santos <fabiosantosart@gmail.com>
 wrote:

> for X in (i for i in open('largefile') if is_part_of_header(i)):
>
> The above code would be wasting time on IO and processing. It would load
> another line and calculate the condition for every line of the large file
> and I just wanted to loop over the few header lines.
>
> itertools.takewhile and fwhile/for..while actually stops the loop when the
> condition is not meant, while your example keeps checking the condition
> until the end of file, even though it is a generator expression.
>
Ah yes, of course, my bad.

It's still possible by raising a StopIteration within the condition
function:

    def is_part_of_header(x):
        if header_condition:
            return True
        else:
            raise StopIteration

But yes, by this point any clarity of the generator expression approach
comes with the cost of more explicit setup of the breaking condition.

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#49127

On Mon, Jun 24, 2013 at 6:41 PM, wu wei <wuwei23@gmail.com> wrote:
> It's still possible by raising a StopIteration within the condition
> function:
>
>     def is_part_of_header(x):
>         if header_condition:
>             return True
>         else:
>             raise StopIteration

Which is basically just taking the break and moving it to somewhere
else in the code, and meanwhile rendering the is_part_of_header
function non-reusable.

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#49128

On Tuesday, June 25, 2013 4:44:44 AM UTC+5:30, alex23 wrote:
> I'd probably just go with a generator expression to feed the for loop:
> 
>      for X in (i for i in ListY if conditionZ):
> 
>          ....

Nice idiom -- thanks
Yes it does not correspond to a takewhile (or break in the control structure world).  It does correspond to a filter

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