《Python数据挖掘入门与实践》中Apriori算法代码如何避免频繁项集重复计数？

from collections import defaultdict

def find_frequent_itemsets(favorable_reviews_by_users, k_1_itemsets, min_support):
    counts = defaultdict(int)
    for user, reviews in favorable_reviews_by_users.items():
        for itemset in k_1_itemsets:
            if itemset.issubset(reviews):
                for other_reviewed_movie in reviews - itemset:
                    current_superset = itemset | frozenset((other_reviewed_movie,))
                    counts[current_superset] += 1
    return dict([(itemset, frequency) for itemset, frequency in counts.items() if frequency >= min_support])

from collections import defaultdict

def find_frequent_itemsets_improved1(favorable_reviews_by_users, k_1_itemsets, min_support):
    counts = defaultdict(int)
    for user, reviews in favorable_reviews_by_users.items():
        for itemset in k_1_itemsets:
            if itemset.issubset(reviews):
                for other_reviewed_movie in reviews - itemset:
                    current_superset = itemset | frozenset((other_reviewed_movie,))
                    # Sort the frozenset for consistent key
                    sorted_superset = frozenset(sorted(list(current_superset))) 
                    counts[sorted_superset] += 1
    return dict([(itemset, frequency) for itemset, frequency in counts.items() if frequency >= min_support])

from collections import defaultdict
from itertools import combinations

def find_frequent_itemsets_improved2(favorable_reviews_by_users, k_1_itemsets, min_support):
    counts = defaultdict(int)
    for user, reviews in favorable_reviews_by_users.items():
        for itemset in combinations(reviews, 2): #Generate all 2-item combinations
            counts[frozenset(itemset)] += 1
    return dict([(itemset, frequency) for itemset, frequency in counts.items() if frequency >= min_support])