基于字典列表的文本-分类最佳匹配最优方案咨询

问题说明 #

需要根据输入文本，从给定的字典列表中匹配最相关的分类，而非严格的完全文本相等。现有代码仅判断文本完全匹配，无法处理类似"computer update"、"update operating system"这类部分匹配的场景。

示例数据 #

data=[{
"categories":"application update",
"options":["application","update computer application","update computer application","update application computer"],    
},
{
"categories":"computer information",
"options":["computer information","computer information","computer properties"],    
},
{
"categories":"operating system",
"options":["operating system","update computer operating system","computer operating system"],    
},
{
"categories":"adobe software",
"options":["adobe software issue","application adobe issue","fix my adobe application"],    
},
]

目标文本示例 #

• text="computer update"
• text="update operating system"（应匹配operating system分类）
• text="fix adobe software"（应匹配adobe software分类）

现有无效代码 #

for i in data:
  for j in i['options']:
    if text== j:
      print(i['categories'])

解决方案：关键词匹配得分法 #

通过计算输入文本与分类选项的单词匹配度，得分最高的分类即为最佳匹配。核心思路是：将文本拆分为单词集合，统计交集数量作为匹配得分，最终取最高分对应的分类。

基础实现代码 #

def get_best_category(text, data):
    max_score = -1
    best_category = None
    # 统一转小写并拆分单词，避免大小写干扰
    text_words = set(text.lower().split())
    
    for item in data:
        current_score = 0
        # 遍历当前分类的所有选项，累加匹配得分
        for option in item['options']:
            option_words = set(option.lower().split())
            # 计算两个单词集合的交集大小，作为单条选项的匹配度
            current_score += len(text_words & option_words)
        
        # 更新最高得分与对应分类
        if current_score > max_score:
            max_score = current_score
            best_category = item['categories']
    
    return best_category

# 测试示例
print(get_best_category("computer update", data))          # 输出: application update
print(get_best_category("update operating system", data))  # 输出: operating system
print(get_best_category("fix adobe software", data))       # 输出: adobe software

优化版（去重选项提升效率） #

原数据中部分选项存在重复，可先去重减少冗余计算：

def get_best_category(text, data):
    max_score = -1
    best_category = None
    text_words = set(text.lower().split())
    
    for item in data:
        current_score = 0
        # 对选项去重，降低计算量
        unique_options = list(set(item['options']))
        for option in unique_options:
            option_words = set(option.lower().split())
            current_score += len(text_words & option_words)
        
        if current_score > max_score:
            max_score = current_score
            best_category = item['categories']
    
    return best_category

内容的提问来源于stack exchange，提问作者new_dev