Why would set creation (and the not x in b inner loop) take O(Nlog(N)) time? I would have thought it would have just required a hash-lookup for each element (O(1)?) being added (and then, a decision to either add the element or not.
Actual times definitely more than O(N) growth.
a10k = []
for x in range(10000):
a10k.append(randint(1,20000))
%timeit b10k = set(a10k)
10k elements = 364 microseconds/loop
100k elements = 5 milliseconds/loop
1mm elements = 170 milliseconds/loop
10mm elements = 2.4 seconds/loop.
100mm elements = 34.5 seconds/loop
Presumably the jump from 100k elements to 1mm elements hit that "cache locality" boundary you were referring to.
Comments
Why would set creation (and the not x in b inner loop) take O(Nlog(N)) time? I would have thought it would have just required a hash-lookup for each element (O(1)?) being added (and then, a decision to either add the element or not.
Actual times definitely more than O(N) growth.
Presumably the jump from 100k elements to 1mm elements hit that "cache locality" boundary you were referring to.I'm assuming a set is internally a balanced tree of some sort. So lookup and insertions are O(log N). So N insertions should be O(N log N).
Edit - nevermind. It's a hash table of course. So I'm wrong.