我有一个正则表达式，可以在一行中找到最多 10 个单词。也就是说，它应该包括换行符之前的单词，而不是换行符之后的单词。我正在使用带有“\n”的负面回顾。
a = re.compile(r"((\w)+[\s /]){0,10}(?<!\n)")
r = a.search("THe car is parked in the garage\nBut the sun is shining hot.")
当我执行此正则表达式并调用方法 r.group() 时，我得到了整个句子但最后一个包含句点的单词。我只期待新行之前的完整字符串。即，“汽车停在车库里\n”。 带着负面的目光我在这里犯的错误是什么...？
请您参考如下方法:
我不知道你为什么要使用负前瞻。您是说换行前最多需要 10 个单词。下面的正则表达式应该可以工作。它使用积极的前瞻来确保在单词之后是换行符。此外，在搜索单词时，请使用 `b\w+\b` 而不是您正在使用的内容。
/(\b\w+\b)*(?=.*\\n)/
python :
result = re.findall(r"(\b\w+\b)*(?=.*\\n)", subject)
解释:
# (\b\w+\b)*(?=.*\\n)
#
# Match the regular expression below and capture its match into backreference number 1 «(\b\w+\b)*»
# Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
# Note: You repeated the capturing group itself. The group will capture only the last iteration. Put a capturing group around the repeated group to capture all iterations. «*»
# Assert position at a word boundary «\b»
# Match a single character that is a “word character” (letters, digits, etc.) «\w+»
# Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
# Assert position at a word boundary «\b»
# Assert that the regex below can be matched, starting at this position (positive lookahead) «(?=.*\\n)»
# Match any single character that is not a line break character «.*»
# Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «*»
# Match the character “\” literally «\\»
# Match the character “n” literally «n»
您可能还希望考虑这样一个事实，即您的字符串中可能没有\n。
