Pandas DataFrame数组列如何精确匹配3个列表元素生成对应新列

核心需要满足精确全词匹配规则，禁止子串误匹配，通过正则单词边界实现匹配逻辑即可，完整流程如下： #

1. 导入依赖、初始化样例数据 #

import pandas as pd
import re
import numpy as np

# 构造样例数据集
df = pd.DataFrame({
    'id': [1,2,3,5,6,7,8],
    'desxription': [
        ['this is bad', 'summerfull'],
        ['city tehran, country iran'],
        ['uA is a country', 'winternice'],
        ['this, is, summer'],
        ['this is winter','uAsal'],
        ['this is canada' ,'great'],
        ['this is toronto']
    ]
})

# 预定义关键词列表
L1 = ['summer', 'winter', 'fall']
L2 = ['iran', 'uA']
L3 = ['tehran', 'canada', 'toronto']

2. 编写通用匹配函数 #

函数接收单行的描述数组、对应关键词列表，返回第一个匹配到的关键词，无匹配则返回空值：

def match_exact_word(desc_list, keyword_list):
    # 拼接当前行所有描述文本，空格分隔避免跨字符串连字误判
    full_content = ' '.join(desc_list)
    for kw in keyword_list:
        # 用单词边界\b包裹关键词，re.escape转义关键词内可能存在的正则特殊字符
        match_pattern = re.compile(rf'\b{re.escape(kw)}\b')
        if match_pattern.search(full_content):
            return kw
    return np.nan

3. 批量生成三列结果 #

对每一列关键词分别调用匹配函数赋值即可：

df['L1'] = df['desxription'].apply(lambda x: match_exact_word(x, L1))
df['L2'] = df['desxription'].apply(lambda x: match_exact_word(x, L2))
df['L3'] = df['desxription'].apply(lambda x: match_exact_word(x, L3))

匹配规则说明 #

• 正则\b代表单词边界，只会匹配独立完整的词，既不会把summerfull里的summer误判为命中，也能正确识别tehran,这类带标点的独立词
• 关键词匹配顺序和列表内顺序一致，如果一行存在多个同列表匹配关键词，会返回排在列表前面的结果
• 拼接文本时加空格分隔，避免两个相邻字符串的首尾字符拼接后生成非预期的词造成误判

最终输出结果 #

执行代码后得到的DataFrame和预期完全一致：

id                 desxription      L1   L2      L3
0   1   [this is bad, summerfull]     NaN  NaN     NaN
1   2  [city tehran, country iran]     NaN  iran  tehran
2   3  [uA is a country, winternice]  NaN  uA      NaN
3   5          [this, is, summer]  summer  NaN     NaN
4   6       [this is winter, uAsal]  winter  NaN     NaN
5   7     [this is canada, great]     NaN  NaN  canada
6   8            [this is toronto]     NaN  NaN  toronto

内容的提问来源于stack exchange，提问作者user15649753