基于Pandas实现热门商品推荐系统：现有代码优化方案咨询

现有两个DataFrame：

df1 #

cust_id           product_list
1                 ['phone', 'tv']
2                 ['ball', 'bat']
3                 ['bat']
4                 ['ball', 'bat', 'phone', 'tv']
5                 ['tv']
6                 ['bat']
7                 ['ball', 'bat', 'phone', 'tv']       
8                 ['phone', 'tv']

df2 #

product          support
ball             0.18
bat              0.29
phone            0.24
tv               0.29   

需要生成如下目标DataFrame：

预期输出 #

cust_id      product_list             recommended_dictionary
0   1            [phone, tv]              {'bat': 0.29, 'ball': 0.18}
1   2            [ball, bat]              {'tv': 0.29, 'phone': 0.24}
2   3            [bat]                    {'tv': 0.29, 'phone': 0.24, 'ball': 0.18}
3   4            [ball, bat, phone, tv]   {}
4   5            [tv]                     {'bat': 0.29, 'phone': 0.24, 'ball': 0.18}
5   6            [bat]                    {'tv': 0.29, 'phone': 0.24, 'ball': 0.18}
6   7            [ball, bat, phone, tv]   {}
7   8            [phone, tv]              {'bat': 0.29, 'ball': 0.18}

已实现的可正常运行代码如下：

创建热门度字典 #

popularity_dict = dict(df2.values)

定义推荐函数：按热门度推荐用户未购买的商品 #

def f(x):
    out = {}
    dif = [i for i in popularity_dict.keys() if i not in x]
    for i in dif:
        out[i] = popularity_dict[i]
        
    return out

计算推荐字典 #

df1['popularity_based_recommended_dictionary'] = df1['product_list'].apply(f)

可以从效率和代码简洁性两方面优化：

优化后的代码 #

# 1. 构建热门度字典（这部分和原代码一致）
popularity_dict = dict(df2.values)

# 2. 简化推荐逻辑：用集合查找+字典推导式
def recommend_unowned(x):
    owned = set(x)
    return {prod: score for prod, score in popularity_dict.items() if prod not in owned}

# 3. 应用到DataFrame
df1['recommended_dictionary'] = df1['product_list'].apply(recommend_unowned)

优化点说明 #

1. 成员检查效率提升：将用户已购商品列表转为集合set(x)，集合的成员查找操作是O(1)，远快于列表的O(n)，当商品数量或用户规模较大时，性能提升明显。
2. 代码简洁性提升：用字典推导式直接构建结果字典，替代原有的列表推导+循环赋值，代码更紧凑且可读性更强。
3. 可选：如果追求极致简洁，也可以用lambda表达式替代函数定义：

df1['recommended_dictionary'] = df1['product_list'].apply(
    lambda x: {p: s for p, s in popularity_dict.items() if p not in set(x)}
)

内容的提问来源于stack exchange，提问作者Danish