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
Then
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.
Peter
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