how to remove multiple occurrences of a string within a list?

[Ayaz Ahmed Khan]

"Steven Bethard" typed:
>> Or, just:
>> 
>> In [1]: l = ["0024","haha","0024"]
>> In [2]: filter(lambda x: x != "0024", l)
>> Out[2]: ['haha']
> 
> Only if you want to make your code harder to read and slower::

Slower, I can see.  But harder to read? 

> There really isn't much use for filter() anymore.  Even in the one place 
> I would have expected it to be faster, it's slower::
> 
>      $ python -m timeit -s "L = ['', 'a', '', 'b']" "filter(None, L)"
>      1000000 loops, best of 3: 0.789 usec per loop
> 
>      $ python -m timeit -s "L = ['', 'a', '', 'b']" "[i for i in L if i]"
>      1000000 loops, best of 3: 0.739 usec per loop

I am getting varying results on my system on repeated runs.  What about
itertools.ifilter()?

$ python -m timeit -s "L = ['0024', 'haha', '0024']; import itertools"
"itertools.ifilter(lambda i: i != '1024', L)" 
100000 loops, best of 3: 5.37 usec per loop

$ python -m timeit -s "L = ['0024', 'haha', '0024']" 
"[i for i in L if i != '0024']" 
100000 loops, best of 3: 5.41 usec per loop

$ python -m timeit -s "L = ['0024', 'haha', '0024']" "[i for i in L if i]"
100000 loops, best of 3: 6.71 usec per loop

$ python -m timeit -s "L = ['0024', 'haha', '0024']; import itertools" 
"itertools.ifilter(None, L)"
100000 loops, best of 3: 4.12 usec per loop

-- 
Ayaz Ahmed Khan

Do what comes naturally now.  Seethe and fume and throw a tantrum.