检查集合列表是否具有任何包含关系的最快方法

Question

I hava a list of 10,000 random sets with different lengths:

import random

random.seed(99)
lst = [set(random.sample(range(1, 10000), random.randint(1, 1000))) for _ in range(10000)]

I want to know the fastest way to check if there is any set that is a subset of another set (or equivalently if there is any set that is a superset of another set). Right now I am using the following very basic code:

def any_containment(lst):
    checked_sets = []
    for st in lst:
        if any(st.issubset(s) for s in checked_sets):
            return True
        else:
            checked_sets.append(st)
    return False

%timeit any_containment(lst)
# 12.3 ms ± 230 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Clearly, my code is not utilizing previous information when checking containment in each iteration. Can anyone suggest the fastest way to do this?

Answer 1

Seems to be faster to sort by length and then try small sets as subset first (and for each, try large sets as superset first). Times in ms from ten cases, data generated like you did but without seeding:

agree yours   mine  ratio  result
True   2.24   2.98   0.75  True
True 146.25   3.10  47.19  True
True 121.66   2.90  41.91  True
True   0.21   2.73   0.08  True
True  37.01   2.82  13.10  True
True   5.86   3.13   1.87  True
True  54.61   3.14  17.40  True
True   0.86   2.81   0.30  True
True 182.51   3.06  59.60  True
True 192.93   2.73  70.65  True

Code ( Try it online! ):

import random
from timeit import default_timer as time

def original(lst):
    checked_sets = []
    for st in lst:
        if any(st.issubset(s) for s in checked_sets):
            return True
        else:
            checked_sets.append(st)
    return False

def any_containment(lst):
    remaining = sorted(lst, key=len, reverse=True)
    while remaining:
        s = remaining.pop()
        if any(s <= t for t in remaining):
            return True
    return False

for _ in range(10):
    lst = [set(random.sample(range(1, 10000), random.randint(1, 1000))) for _ in range(10000)]
    t0 = time()
    expect = original(lst)
    t1 = time()
    result = any_containment(lst)
    t2 = time()
    te = t1 - t0
    tr = t2 - t1
    print(result == expect, '%6.2f ' * 3 % (te*1e3, tr*1e3, te/tr), expect)

Improvement

The following seems further ~20% faster. Instead of first comparing the smallest set with potentially all larger sets before giving even just the second -smallest a chance, this does give other small sets an early chance.

def any_containment(lst):
    sets = sorted(lst, key=len)
    for i in range(1, len(sets)):
        for s, t in zip(sets, sets[-i:]):
            if s <= t:
                return True
    return False

Comparison with my old solution ( Try it online! ):

agree  old    new   ratio  result
True   3.13   2.46   1.27  True
True   3.36   3.31   1.02  True
True   3.10   2.49   1.24  True
True   2.72   2.43   1.12  True
True   2.86   2.35   1.21  True
True   2.65   2.47   1.07  True
True   5.24   4.29   1.22  True
True   3.01   2.35   1.28  True
True   2.72   2.28   1.19  True
True   2.80   2.45   1.14  True

Yet another idea

A shortcut could be to first collect the union of all single-element sets, and check whether that intersects with any other set (either without sorting them, or again from largest to smallest after sorting). That likely suffices. If not, then proceed as previously, but without the single-element sets.