[BangPypers] How for and list.remove() works

[Anand Balachandran Pillai]

Here is another one, this time using a dictionary ;)


>>> a=dict(zip(a, [0]*len(a))).keys()
>>> a.remove(12)
>>> a
[34, 1321, 45, 77, 23]

On Thu, Jul 10, 2008 at 2:03 PM, Anand Balachandran Pillai
<abpillai at gmail.com> wrote:
> Here is what is arguably the solution with the least code.
>
>>>> a =  [12, 12, 1321, 34, 23, 12, 34, 45, 77]
>>>> list(set(a)-set([12]))
> [1321, 34, 23, 45, 77]
>
>
> Cheers
>
> --Anand
>
> On Thu, Jul 10, 2008 at 5:14 AM, Jeff Rush <jeff at taupro.com> wrote:
>> Anand Chitipothu wrote:
>>>
>>> On Wed, Jul 9, 2008 at 8:47 PM, Kushal Das <kushaldas at gmail.com> wrote:
>>>>
>>>> Hi all,
>>>>
>>>>>>> a = [12, 12, 1321, 34, 23, 12, 34, 45, 77]
>>>>>>> for x in a:
>>>>
>>>> ...   if x == 12:
>>>> ...     a.remove(x)
>>>>>>>
>>>>>>> a
>>>>
>>>> [1321, 34, 23, 12, 34, 45, 77]
>>>>
>>>> Can any one explain me how the remove works and how it is effecting the
>>>> for
>>>> loop.
>>
>> Others have explained why it fails.  Here are two approaches for making it
>> work, in case you need ideas.
>>
>>>>> a = [12, 12, 1321, 34, 23, 12, 34, 45, 77]
>>>>> filter(lambda x: x != 12, a)
>> [1321, 34, 23, 34, 45, 77]
>>
>> The filter() build-in function is deprecated and the following is the
>> preferred way of doing it now:
>>
>>>>> [x for x in a if x != 12]
>> [1321, 34, 23, 34, 45, 77]
>>
>> Or if your list is quite large and you want to avoid making the new copy of
>> it, use the new list generator notation:
>>
>>>>> b = (x for x in a if x != 12)
>>>>> for x in b:
>> ...     print x
>> 1321
>> 34
>> 23
>> 34
>> 45
>> 77
>>>>> list(b)
>> [1321, 34, 23, 34, 45, 77]
>>
>> A list generator performs the filtering just-in-time as you need the
>> elements.
>>
>> -Jeff
>> _______________________________________________
>> BangPypers mailing list
>> BangPypers at python.org
>> http://mail.python.org/mailman/listinfo/bangpypers
>>
>
>
>
> --
> -Anand
>



-- 
-Anand