Markus added the comment:
Eleminating duplicates before processing is faster once the overhead of the set
operation is less than the time required to sort the larger dataset with
duplicates.
So we are basically comparing sort(data) to sort(set(data)).
The optimum depends on the input data.
python3 -m timeit -s "import random; import bipaddress; ips =
[bipaddress.ip_address('2001:db8::') + i for i in range(100000)];
random.shuffle(ips)" -- "bipaddress.collapse_addresses(ips)"
10 loops, best of 3: 1.49 sec per loop
vs.
10 loops, best of 3: 1.59 sec per loop
If the data is pre-sorted, possible if you retrieve from database, things are
drastically different:
python3 -m timeit -s "import random; import bipaddress; ips =
[bipaddress.ip_address('2001:db8::') + i for i in range(100000)]; " --
"bipaddress.collapse_addresses(ips)"
10 loops, best of 3: 136 msec per loop
vs
10 loops, best of 3: 1.57 sec per loop
So for my usecase, I basically have less than 0.1% duplicates (if at all),
dropping the set would be better, but ... other usecases will exist.
Still, it is easy to "emulate" the use of "sorted(set())" from a users
perspective - just call collapse_addresses(set(data)) in case you expect to
have duplicates and experience a speedup by inserting unique, possibly even
sorted, data.
On the other hand, if you have a huge load of 99.99% sorted non collapseable
addresses, it is not possible to drop the set() operation in your sorted(set())
from a users perspective, no way to speed things up, and the slowdown you get
is x10.
That said, I'd drop the set().
Optimization depends on data input, dropping the set() allows the user to
optimize base on the nature of his input data.
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