提高清洁功能性能

Question

我对 python 相当陌生，我试图了解什么是紧固清洁功能的最佳方法。 （我可以使用多处理吗？异步函数？将此函数拆分为较小的函数并使用多线程？）

def clean(s):
"""
:param s: string to be processed
:return: processed string: see comments in the source code for more info
"""
   s = re.sub(r'([a-z])([A-Z])', r'\1\. \2', s)  # before lower case
   s = s.lower()
   s = re.sub(r'etc\.?','',s)
   s = re.sub(r'\(.*?\)', ' ', s)
   s = re.sub(r'(on?\s?\d+\s?\.?\:?\d*\s?am)|(on?\s?\d+\s?\.?\:?\d*\s?pm)','',s)
   s = re.sub(r'(at?\s?\d+\s?\.?\:?\d*\s?am)|(at?\s?\d+\s?\.?\:?\d*\s?pm)','',s)
   s = re.sub(r'[^\D+\s?]', ' ',s)
   s = re.sub(r'^\-\s?', '',s)
   s = re.sub(r'[$€£]+', '',s)
   s = re.sub(r'(\s+\-\s+)', ' ',s)
   s = re.sub(r'(\,+\s?\,?)', ',',s)
   s = re.sub(r'\#+', ' ',s)
   s = re.sub(r'[.!]+', '.',s)
   s = re.sub(r'\s?\&+\s?', ' and ', s)
   s = re.sub(r'([a-z])\1{2,}', r'\1', s)
   s = re.sub(r'([\W+])\1{1,}', r'\1', s)
   s = re.sub(r'\*|\W\*|\*\W', '', s)
   s = re.sub(r'\W+?\.', '.', s)
   s = re.sub(r'(\.|\?|!)(\w)', r'\1 \2', s)
   s = re.sub(r'(.{2,}?)\1{1,}', r'\1', s)
   s = re.sub(r'\s{2,}', ' ', s)
   s = re.sub(r'\([^)]*\)','',s)
return s.strip()

df['Sentences'].apply(clean)

感谢您的帮助