Better to add item to a set, or convert final list to a set?

Question

I have some data that looks something like this:

ID1 ID2 ID3  
ID1 ID4 ID5  
ID3 ID5 ID7 ID6  
...  
...  

where each row is a group.

M

Answer 1

So, I timed a few different options, and after a few iterations, came up with the following strategies. I thought that sets2 would be the winner, but listToSet2 was faster for every single type of group.

All of the functions except for listFilter were in the same ballpark - listFilter was much slower.

import random
import collections

small = [[random.randint(1,25) for _ in range(5)] for i in range(100)]
medium = [[random.randint(1,250) for _ in range(5)] for i in range(1000)]
mediumLotsReps = [[random.randint(1,25) for _ in range(5)] for i in range(1000)]
bigGroups = [[random.randint(1,250) for _ in range(75)] for i in range(100)]
huge = [[random.randint(1,2500) for _ in range(5)] for i in range(10000)]

def sets(groups):
    results = collections.defaultdict(set)
    for group in groups:
        for i in group:
            for j in group:
                if i is not j:
                    results[i].add(j)
    return results

def listToSet(groups):
    results = collections.defaultdict(list)
    for group in groups:
        for i,j in enumerate(group):
            results[j] += group[:i] + group[i:]
    return {k:set(v) for k, v in results.iteritems()}

def listToSet2(groups):
    results = collections.defaultdict(list)
    for group in groups:
        for i,j in enumerate(group):
            results[j] += group
    return {k:set(v)-set([k]) for k, v in results.iteritems()}

def sets2(groups):
    results = collections.defaultdict(set)
    for group in groups:
        for i in group:
            results[i] |= set(group)
    return {k:v - set([k]) for k, v in results.iteritems()}

def listFilter(groups):
    results = collections.defaultdict(list)
    for group in groups:
        for i,j in enumerate(group):
            filteredGroup = group[:i] + group[i:]
            results[j] += ([k for k in filteredGroup if k not in results[j]])
    return results