Slow comparison between two lists
redetin at gmail.com
Thu Oct 23 14:46:05 CEST 2008
On 23 loka, 15:24, Peter Otten <__pete... at web.de> wrote:
> Jani Tiainen wrote:
> > I have rather simple 'Address' object that contains streetname,
> > number, my own status and x,y coordinates for it. I have two lists
> > both containing approximately 30000 addresses.
> > I've defined __eq__ method in my class like this:
> > def __eq__(self, other):
> > return self.xcoord == other.xcoord and \
> > self.ycoord == other.ycoord and \
> > self.streetname == other.streetname and \
> > self.streetno == other.streetno
> > But it turns out to be very, very slow.
> > Then I setup two lists:
> > list_external = getexternal()
> > list_internal = getinternal()
> > Now I need get all all addresses from 'list_external' that are not in
> > 'list_internal', and mark them as "new".
> > I did it like this:
> > for addr in list_external:
> > if addr not in list_internal:
> > addr.status = 1 # New address
> > But in my case running that loop takes about 10 minutes. What I am
> > doing wrong?
> Even if list_external and list_internal contain the same items you need
> about len(list_external)*(len(list_internal)/2), or 45 million comparisons.
> You can bring that down to 30000*some_small_factor if you make your
> addresses hashable and put the internal ones into a dict or set
> def __hash__(self):
> return hash((self.xcoord, self.yccord, self.streetname, self.streetno))
> def __eq__(self, other):
> # as above
> set_internal = set(list_internal)
> for addr in list_external:
> if addr not in set_internal:
> addr.status = 1
> Note that the attributes relevant for hash and equality must not change
> during this process.
Very complete answer, thank you very much.
I tried that hash approach and sets but it seemed to get wrong results
first time and it was all due my hash method.
Now it takes like 2-3 seconds to do all that stuff and result seem to
be correct. Apparently I have lot to learn about Python... :)
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