如何用Pandas关联两个DataFrame按文本规则过滤生成标志列

推荐用Pandas向量化操作实现，比遍历效率高得多，代码如下：

import pandas as pd

# 1. 先构造id到text的映射字典，关联查询更高效
id_text_map = meta_df.set_index('id')['text'].to_dict()

# 2. 给df匹配id1和id2对应的text值
df['text1'] = df['id1'].map(id_text_map)
df['text2'] = df['id2'].map(id_text_map)

# 3. 按规则生成flag字段
def check_flag(row):
    has_lower1 = 'lower' in row['text1']
    has_upper1 = 'upper' in row['text1']
    has_lower2 = 'lower' in row['text2']
    has_upper2 = 'upper' in row['text2']
    # 满足恰好一个含lower，一个含upper
    if (has_lower1 and has_upper2) or (has_upper1 and has_lower2):
        return 'yes'
    return 'no'

df['flag'] = df.apply(check_flag, axis=1)

# 4. 提取需要的三列作为结果
result = df[['id1', 'id2', 'flag']]

如果要修正你原来的遍历写法，错误点在于之前的过滤条件写错了，修正后代码如下（仅适合小数据量场景，大数据量不推荐）：

flags = []
for row in df.itertuples():
    # 取id1对应的text
    text1 = meta_df[meta_df['id'] == row.id1].iloc[0]['text']
    # 取id2对应的text
    text2 = meta_df[meta_df['id'] == row.id2].iloc[0]['text']
    if ('lower' in text1 and 'upper' in text2) or ('upper' in text1 and 'lower' in text2):
        flags.append('yes')
    else:
        flags.append('no')
df['flag'] = flags
result = df[['id1', 'id2', 'flag']]

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