如何在pandas中通过字符串搜索为字符串列分配多个额外标签

问题根因 #

你当前使用的str.extract()方法仅会返回每条文本中第一个匹配到的正则分组，因此只会输出首个命中的标签，无法捕获同一条文本中多个匹配的分类。

解决方案1：按分类逐次判断命中（易读易维护） #

遍历所有分类，直接判断每条文本是否包含该分类下的任意关键词，最后聚合所有命中的分类：

import pandas as pd

entertainment_dict = {
  "Food": ["McDonald", "Five Guys", "KFC"],
  "Music": ["Taylor Swift", "Jay Z", "One Direction"],
  "TV": ["Big Bang Theory", "Queen of South", "Ted Lasso"]
}

data = {'text':["Kevin Lee has bought a Taylor Swift's CD and eaten at McDonald.", 
                "The best burger in McDonald is cheeze buger.",
                "Kevin Lee is planning to watch the Big Bang Theory and eat at KFC."]}

df = pd.DataFrame(data)

# 逐个分类判断是否命中
for label, keywords in entertainment_dict.items():
    keyword_regex = '|'.join(keywords)
    df[label] = df['text'].str.contains(keyword_regex, regex=True)

# 聚合所有命中的标签
df['labels'] = df.apply(lambda row: ','.join([k for k in entertainment_dict.keys() if row[k]]), axis=1)

# 可选：删除中间生成的分类判断列
df = df.drop(columns=entertainment_dict.keys())

输出结果：

text       labels
0  Kevin Lee has bought a Taylor Swift's CD and e...  Music,Food
1       The best burger in McDonald is cheeze buger.         Food
2  Kevin Lee is planning to watch the Big Bang Th...     TV,Food

解决方案2：基于str.extractall实现 #

如果你希望继续沿用原有正则分组匹配的逻辑，可以改用str.extractall()捕获所有匹配项，再聚合标签，和原有代码兼容性更高：

import pandas as pd

entertainment_dict = {
  "Food": ["McDonald", "Five Guys", "KFC"],
  "Music": ["Taylor Swift", "Jay Z", "One Direction"],
  "TV": ["Big Bang Theory", "Queen of South", "Ted Lasso"]
}

data = {'text':["Kevin Lee has bought a Taylor Swift's CD and eaten at McDonald.", 
                "The best burger in McDonald is cheeze buger.",
                "Kevin Lee is planning to watch the Big Bang Theory and eat at KFC."]}

df = pd.DataFrame(data)
regex = '|'.join(f'(?P<{k}>{"|".join(v)})' for k,v in entertainment_dict.items())

# 提取所有匹配项，按行聚合标签
matches = df['text'].str.extractall(regex).notnull().groupby(level=0).any()
df['labels'] = matches.apply(lambda r: ','.join([k for k in entertainment_dict.keys() if r[k]]), axis=1)

两种方案均支持任意数量的多标签匹配：方案1代码更易懂，适合需要频繁调整分类规则的场景；方案2执行效率更高，适合分类数量多、数据量大的场景。

内容的提问来源于stack exchange，提问作者codedancer