Question

我想根据存储在字典列表中的标签来匹配django数据库中的类似文章：

myarticle = {'pk': 17, 'tags': [0, 1, 0, 1, 0]}
allarticles = [{'pk': 1, 'tags': [0, 0, 0, 1, 0]}, 
               {'pk': 2, 'tags': [0, 1, 0, 1, 0]},
               {'pk': 3, 'tags': [1, 1, 0, 0, 0]},
               {'pk': 4, 'tags': [1, 0, 1, 0, 1]},
               {'pk': 5, 'tags': [0, 0, 0, 0, 1]}]

最方便的方法是获取列表，该列表根据输入的Myarticle对匹配标签的数量进行排名。预期结果：

result = [2, 1, 3, 5, 4]

Answer 1

您可以使用sorted：

myarticle = {'pk': 17, 'tags': [0, 1, 0, 1, 0]}
allarticles = [{'pk': 1, 'tags': [0, 0, 0, 1, 0]}, 
           {'pk': 2, 'tags': [0, 1, 0, 1, 0]},
           {'pk': 3, 'tags': [1, 1, 0, 0, 0]},
           {'pk': 4, 'tags': [1, 0, 1, 0, 1]},
           {'pk': 5, 'tags': [0, 0, 0, 0, 1]}]
new_articles = sorted(allarticles, key=lambda x:sum(a == b for a, b in zip(myarticle['tags'], x['tags'])), reverse=True)
final_results = [i['pk'] for i in new_articles]

输出：

[2, 1, 3, 5, 4]

Answer 2

您可以通过numpy.argsort将第三方NumPy用于矢量化解决方案。

对于更大的输入列表，这应该更有效：

allarticles = allarticles*10000

import numpy as np

def jp(myarticle, allarticles):
    arr = np.argsort((np.array([d['tags'] for d in allarticles]) == myarticle['tags']).sum(1))[::-1]
    return [allarticles[i]['pk'] for i in arr]

def ajax(myarticle, allarticles):
    new_articles = sorted(allarticles, key=lambda x:sum(a == b for a, b in zip(myarticle['tags'], x['tags'])), reverse=True)
    return [i['pk'] for i in new_articles]

%timeit jp(myarticle, allarticles)    # 49.3 ms per loop
%timeit ajax(myarticle, allarticles)  # 112 ms per loop

根据值在词典列表中查找相似的项目

2 个答案: