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Re: L[:]

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  Re: L[:] Terry Reedy <tjreedy@udel.edu> - 2014-01-10 17:38 -0500
    Re: L[:] Grant Edwards <invalid@invalid.invalid> - 2014-01-11 16:34 +0000

#63670 — Re: L[:]

On 1/10/2014 12:38 PM, Albert-Jan Roskam wrote:
> In Python Cookbook, one of the authors (I forgot who) consistently used the "L[:]" idiom like below. If the second line simply starts with "L =" (so no "[:]") only the name "L" would be rebound, not the underlying object. That was the authorś explanation as far as I can remember. I do not get that. Why is the "L[:]" idiom more memory-efficient here? How could the increased efficiency be demonstrated?
>
> #Python 2.7.3 (default, Sep 26 2013, 16:38:10) [GCC 4.7.2] on linux2
>>>> L = [x ** 2 for x in range(10)]
>>>> L[:] = ["foo_" + str(x) for x in L]

Unless L is aliased, this is silly code. The list comp makes a new list 
object, so if L does not have aliases, it would be best to rebind 'L' to 
the existing list object instead of copying it. To do the replacement 
'in place':
L = [x ** 2 for x in range(10)]
for i, n in enumerate(L):
   L[i] = "foo_" + str(n)
print(L)
 >>>
['foo_0', 'foo_1', 'foo_4', 'foo_9', 'foo_16', 'foo_25', 'foo_36', 
'foo_49', 'foo_64', 'foo_81']
-- 
Terry Jan Reedy

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#63715

On 2014-01-10, Terry Reedy <tjreedy@udel.edu> wrote:
> On 1/10/2014 12:38 PM, Albert-Jan Roskam wrote:
>> In Python Cookbook, one of the authors (I forgot who) consistently used the "L[:]" idiom like below. If the second line simply starts with "L =" (so no "[:]") only the name "L" would be rebound, not the underlying object. That was the author?? explanation as far as I can remember. I do not get that. Why is the "L[:]" idiom more memory-efficient here? How could the increased efficiency be demonstrated?
>>
>> #Python 2.7.3 (default, Sep 26 2013, 16:38:10) [GCC 4.7.2] on linux2
>>>>> L = [x ** 2 for x in range(10)]
>>>>> L[:] = ["foo_" + str(x) for x in L]
>
> Unless L is aliased, this is silly code.

And if L _is_ aliaised, it's probably trying to be too clever and
needs to be fixed.

-- 
Grant

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