是否存在正则表达式使re.findall结果等价于指定split循环加搜索的代码？

Absolutely! Let's break down what your original code does first, then craft the regex that replicates its behavior perfectly.

Original Code Logic #

Your code works in two key steps:

• First, it trims leading/trailing whitespace from the input, then splits the string into words using one or more whitespace characters as separators.
• Second, it filters out any word that meets either of these criteria:
  • Has three or more consecutive identical characters (e.g., iii, whaaaat)
  • Contains underscores, digits, or non-word characters (e.g., th_is, and2, kn-ow)

The remaining valid words are those made up solely of letters, with no runs of three identical characters.

The Regex You Need #

Here's the regex that will return the exact same result as your code when used with re.findall():

r'(?<!\S)(?!.*(.)\1{2,})[a-zA-Z]+(?!\S)'

Breakdown of the Regex #

Let's unpack each part to understand how it matches your requirements:

• (?<!\S): A negative lookbehind that ensures the word starts at either the beginning of the string or right after whitespace (mimicking the split behavior for word boundaries).
• (?!.*(.)\1{2,}): A negative lookahead that checks the entire word to ensure there are no sequences of three or more identical consecutive characters.
• [a-zA-Z]+: Matches one or more letters only—this automatically excludes underscores, digits, and non-word characters (since none of those fall into the a-z/A-Z range).
• (?!\S): A negative lookahead that ensures the word ends at either the end of the string or right before whitespace (again, aligning with the split behavior).

Testing It Out #

If you run this regex against your sample doc:

doc = ' th_is is stuff. and2 3things if y4ou kn-ow ___ whaaaat iii mean)'
result = re.findall(r'(?<!\S)(?!.*(.)\1{2,})[a-zA-Z]+(?!\S)', doc)

You'll get ['is', 'if']—exactly the same output as your original code.

内容的提问来源于stack exchange，提问作者aweeeezy