efficient intersection of lists with rounding

Steven Bethard steven.bethard at gmail.com
Fri Dec 3 06:43:37 CET 2004

Michael Hoffman wrote:
> Steven Bethard wrote:
>> Well, in Python 2.3, I believe sets are implemented in Python while 
>> they're implemented in C in Python 2.4.
> I think the Python 2.3 Sets implementation is likely to be quicker than 
> whatever list-manipulation answer you come up with instead. But there's 
> only one way to find out ;)

Yeah, almost certainly since he's looking at lists 3K long.  If they 
were small, you never know since the list comprehension gets the C-code 
speedup, while sets.Set is Python code:

 > python -m timeit -s "a = [(123,1.3),(123,2.4),(123,7.8),(123,10.2)]; 
b = [(123, 0.9), (123, 1.9), (123, 8.0)]" "[ (i,round(j)) for i,j in a 
for l,m in b if (i,round(j)) == (l,round(m))]"
10000 loops, best of 3: 27.5 usec per loop

 > python -m timeit -s "import sets; a = 
[(123,1.3),(123,2.4),(123,7.8),(123,10.2)]; b = [(123, 0.9), (123, 1.9
), (123, 8.0)]" "sets.Set([(i,round(j)) for i,j in 
a]).intersection(sets.Set([(i, round(j)) for i, j in b]))"
10000 loops, best of 3: 47.7 usec per loop

In the case given, the O(n**2) list comprehension is faster than the 
O(n) set intersection.  Of course, this is not likely to be true with 
any reasonable sized data.  But it's something worth keeping in mind.


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