根据列表列表检查列表中组合的存在

Question

If i have a list of list of integers S: [[1,2,3],[3,4,5],[5,6,7]] , and a single list T: [2,3,1] . I want to return true if T as a combination is contained in S. Assuming each element of S has same length as that of T .

In this case, I want to return true.

Restrictions: No sorting of any kind, and note S has all unique lists, but within a list, it can have duplicate elements.

How can I do this as efficiently as possible. I can iterate through each element of S and turn it into a set and compare it with set(T) , but that seems very slow if size of S and length of each element of S gets bigger.

Answer 1

Using itertools?

from itertools import combinations
for i in combinations(t, len(t)): if i in s: return True;

Or using sorted:

t = sorted(t)
for i in s: if sorted(i)==t: return True

Answer 2

You can use sorted :

>>> S = [[1,2,3],[3,4,5],[5,6,7]]
>>> T = [2,3,1]
>>> any(sorted(T) == sorted(x) for x in S)
True

Answer 3

Collections and counter? It'll run in O(n), whereas sorting will run in roughly O(n log n), and raw comparison will be O(n²)

import collections
T = collections.Counter([2,3,1])
any(T == collections.Counter(x) for x in S)

Answer 4

If a stipulation is without sorting, you might want to look into an order-independent hash function for the combinations. Two starting points I liked were (1) the sum of the hashes of the individual elements of the combination, and (2) a tuple (sum(combination), product(combination)) .