如何在Python中使用正则替换句子列表中的多个子字符串

问题描述 #

我有如下句子列表：

sentences = ["I am learning to code", "coding seems to be intresting in python", "how to code in python", "practicing how to code is the key"]

现在我想通过存储了待替换字符串和对应替换值的字典，替换该句子列表中的若干子字符串：

word_list = {'intresting': 'interesting', 'how to code': 'learning how to code', 'am learning':'love learning', 'in python': 'using python'}

我尝试了如下代码：

replaced_sentences = [' '.join([word_list.get(w, w) for w in sentence.split()]) for sentence in sentences]

但运行后只有单个单词的字符串被替换，多单词的键没有生效。原因是我使用了sentence.split()按单词拆分句子，导致长度超过1个单词的子字符串没有被匹配到。

请问我要如何通过正则或其他方案实现子字符串的精确匹配替换？

预期输出 #

sentences = ["I love learning to code", "coding seems to be interesting using python", "learning how to code using python", "practicing learning how to code is the key"]

解决方案 #

你可以用正则匹配实现多词子串的替换，核心要注意优先匹配更长的键，避免短键先匹配破坏长键的完整结构，具体实现如下：

import re

# 原始数据
sentences = ["I am learning to code", "coding seems to be intresting in python", "how to code in python", "practicing how to code is the key"]
word_list = {'intresting': 'interesting', 'how to code': 'learning how to code', 'am learning':'love learning', 'in python': 'using python'}

# 1. 按键的长度降序排序，保证长键优先匹配
sorted_keys = sorted(word_list.keys(), key=lambda x: len(x), reverse=True)
# 2. 生成正则匹配模式，对每个键做转义处理避免特殊字符干扰
pattern = re.compile('|'.join(re.escape(key) for key in sorted_keys))
# 3. 批量替换
replaced_sentences = [pattern.sub(lambda x: word_list[x.group()], sentence) for sentence in sentences]

print(replaced_sentences)

运行后输出和预期完全一致：

["I love learning to code", "coding seems to be interesting using python", "learning how to code using python", "practicing learning how to code is the key"]

原理解释 #

• 按键长度降序排序：如果存在a和ab两个键，排序后会先匹配ab，避免a先被替换后ab无法匹配的问题
• re.escape处理：如果替换键里包含.、*这类正则特殊字符，也可以正常匹配
• re.sub回调函数：每匹配到一个键，就直接从替换字典中取出对应值做替换

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