试试这个

original_list = ['A. Person', 'B. House', 'H. Normal', 'My House', 'Hello World']

pattern = r'^[A-Z]\.\s*'
filtered_list = [re.sub(pattern, '', item) for item in original_list]

print(filtered_list)

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假设只有一个时期，你可以用“”来分割。“并取最后一个元素。

l = ['A. Person', 'B. House', 'H. Normal', 'My House', 'Hello World']
l = [i.split(". ")[-1] for i in l]
print(l)
>>> ['Person', 'House', 'Normal', 'My House', 'Hello World']

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如果格式总是类似于A. Person，那么您可以简单地像下面这样迭代列表

letters = ["A.", "B.", ...]

for entry in list:
  for letter in letters:
    if letter in entry:
      entry = entry.replace(letter, "")

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更改数据以考虑其他可能性：

import re

p = re.compile(r'[A-Z]\.\s*')

_list = ['Mr A. P. Person', 'B.House', 'H. Normal ', 'My House', 'Hello World']

print(*(p.sub('', e) for e in _list), sep='\n')

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输出：

Mr Person
House
Normal 
My House
Hello World

型