最后尝试使用以下模式，它使用以下代码涵盖了上述所有场景，

import re
  def contains_acronym(text):
  pattern = r"\([A-Za-z0-9]{2,}\)"
  result = re.search(pattern, text)
  return result != None

print(contains_acronym("Instant messaging (IM) is a set of communication technologies used for text-based communication")) # True
print(contains_acronym("American Standard Code for Information Interchange (ASCII) is a character encoding standard for electronic communication")) # True
print(contains_acronym("Please do NOT enter without permission!")) # False
print(contains_acronym("PostScript is a fourth-generation programming language (4GL)")) # True
print(contains_acronym("Have fun using a self-contained underwater breathing apparatus (Scuba)!")) # True

将此用作模式

`pattern = r"\(\w.*\w\)" `

“w”表示字母和数字

import re
def contains_acronym(text):
  pattern = r'\([A-Za-z0-9]{2,}\)' 
  result = re.search(pattern, text)
  return result != None

print(contains_acronym("Instant messaging (IM) is a set of communication technologies used for text-based communication")) # True
print(contains_acronym("American Standard Code for Information Interchange (ASCII) is a character encoding standard for electronic communication")) # True
print(contains_acronym("Please do NOT enter without permission!")) # False
print(contains_acronym("PostScript is a fourth-generation programming language (4GL)")) # True
print(contains_acronym("Have fun using a self-contained underwater breathing apparatus (Scuba)!")) # True

import re
def contains_acronym(text):
  pattern = r"\([A-Z0-9][A-Z0-9a-z]+\)"
  result = re.search(pattern, text)
  return result != None

你的正则表达式几乎是正确的，你只是忘记了它必须有至少2个**，所以只要把你的第一个范围作为字符串的常量部分，然后用小写的+（一个或多个）重复同样的匹配：

pattern = r"\(([A-Z0-9_][A-Za-z0-9_]+)\)"

∮这应该行得通∮

pattern = r"\([A-Z0-9].*\)"

请注意
1."+"是匹配一次或多次的元字符。
1.应使用"*"。
1.也不需要双括号。
1.\w还包括小写字符、大写字符和整数。

pattern = r"\(\w{2,}\)"

这个图案很好用，看起来也更好。

模式= r”（[A-Z 0 -9]）*”
这个很适合我，很简单

这个怎么样？

pattern = '\(\w+\)'