>>
>> > L = filter('a'.__ne__,L)
>>
>> And this is probably the fastest. But not all types define
>> __ne__ so a
>> more generic answer would be:
>>
>> from functools import partial
>> from operator import ne
>> L = filter(partial(ne, 'a'), L)
>>
And don't forget this "traditional" solution:
>>> L=['a', 'b', 'c', 'a']
>>> filter(lambda arg: arg != 'a', L)
['b', 'c']
-John
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