亲和性分析——推荐电影

# 选择训练集200
ratings = all_ratings[all_ratings[‘UserID‘].isin(range(200))]
# 选出喜欢的数据
favorable_ratings = ratings[ratings[‘Favorable‘]]
# 将数记录成 一个用户对应多部 喜欢的电影
favorable_reviews_by_users = dict((k, frozenset(v.values)) for k, v in favorable_ratings.groupby("UserID")["MovieID"])
# print(favorable_reviews_by_users)
len(favorable_reviews_by_users)
# 得到最后训练集的人数 199



# 生成一个电影有多少人喜欢
num_favorable_by_movie = ratings[["MovieID", "Favorable"]].groupby("MovieID").sum()
# print(num_favorable_by_movie)
num_favorable_by_movie.sort_values("Favorable", ascending=False)[:5]

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.keys():
            if itemset.issubset(reviews):
                for other_reviewed_movie in reviews - itemset:
#                     用到了集合的特性 | 表示取与 解决了重复的问题（集合有去重的功能）
                    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])

import sys
frequent_itemsets = {}
min_support = 50
# print(type(frequent_itemsets))   dict
# k=1 
frequent_itemsets[1] = dict((frozenset((movie_id,)), row["Favorable"])
                                for movie_id, row in num_favorable_by_movie.iterrows()
                                if row["Favorable"] > min_support)
# print(frequent_itemsets)
# exit()
print("There are {} movies with more than {} favorable reviews".format(len(frequent_itemsets[1]), min_support))
sys.stdout.flush()
# 通过循环，求出了所有情况
for k in range(2, 20):
    # k代表每个数据集中 值得推荐电影总共得个数，看看结果就懂了
    # 得到结果
    cur_frequent_itemsets = find_frequent_itemsets(favorable_reviews_by_users, frequent_itemsets[k-1],
                                                   min_support)
    if len(cur_frequent_itemsets) == 0:
        print("Did not find any frequent itemsets of length {}".format(k))
        sys.stdout.flush()
        break
    else:
        print("I found {} frequent itemsets of length {}".format(len(cur_frequent_itemsets), k))
        #print(cur_frequent_itemsets)
        sys.stdout.flush()
        frequent_itemsets[k] = cur_frequent_itemsets
# 去除个数为1 的，这样的数据没办法推荐
del frequent_itemsets[1]
print(frequent_itemsets[9])

# 现在创建一个列表，
candidate_rules = []
for itemset_length, itemset_counts in frequent_itemsets.items():
    for itemset in itemset_counts.keys():
        for conclusion in itemset:
            premise = itemset - set((conclusion,))       # 减号为集合itemset除去conclusion之后得
            candidate_rules.append((premise, conclusion))
print("There are {} candidate rules".format(len(candidate_rules)))