因为你用regex标记了这个问题，这里有一个替代方法，使用相同的方法来分隔三个组：

import re
OGList = ['A10', 'BMW320i', 'Nissan NSX200', 'Benz 220c']
types = {"chars": "[a-zA-Z]", "nums": "[0-9]", "other": "[^a-zA-Z0-9]"}
res = {k: [''.join(re.findall(types[k], s)) for s in OGList] for k in types}

之后，res是一个包含不同列表的dict：

{
  'chars': ['A', 'BMWi', 'NissanNSX', 'Benzc'],
  'nums': ['10', '320', '200', '220'],
  'other': ['', '', ' ', ' ']
}

请注意，这里没有elif和else，您必须确保模式是互斥的，并涵盖所有情况，否定^，其他组应该做得很好。

OGList = ['A10', 'BMW320i', 'Nissan NSX200', 'Benz 220c']
    
    dig_lst = []
    letter_lst = []
    
 # with tobias_k suggestion of using negation classes the code becomes very simple.
    for i in OGList:
      
       letter_lst.append(re.sub(r"[^a-zA-Z]", "", i))
       dig_lst.append(re.sub(r'[^0-9]', '', i))
      
    
 print(dig_lst)
 print(letter_lst)
        
 ['10', '320', '200', '220']
 ['A', 'BMWi', 'NissanNSX', 'Benzc']

您需要为每个输入字符串创建一个单独的输出字符串，而不仅仅是将每个字符追加到输出列表中。

OGList = ['A10', 'BMW320i', 'Nissan NSX200', 'Benz 220c']

numlist = []
strlist = []
otherlist = []

for i in OGList:
    numstring = strstring = otherstring = ""
    for x in i:
        if x.isalpha():
            strstring += x
        elif x.isdigit():
            numstring += x
        else:
            otherstring += x
    if strstring:
        strlist.append(strstring)
    if numstring:
        numlist.append(numstring)
    if otherstring:
        otherlist.append(otherstring)

print(numlist)
print(strlist)

答案是，在每次插入列表之前，都需要添加一个临时str

OGList = ['A10', 'BMW320i', 'Nissan NSX200', 'Benz 220c']

numlist = []
strlist = []
otherlist = []

for i in OGList:
    mystr = ''
    myint = ''
    myother = ''
    for x in i:
        if x.isalpha():
            mystr += x
        elif x.isdigit():
            myint += x
        else:
            myother += x
   if mystr:
       strlist.append(mystr)
   if myint:
       numlist.append(myint)
   if myother:
       otherlist.append(myother)

print(numlist)
print(strlist)

一个奇怪的基于itertools.groupby()的方法怎么样？

from itertools import groupby

OGList = ['A10', 'BMW320i', 'Nissan NSX200', 'Benz 220c']

def get_char_class(char):
    if char.isalpha():
        return 'a'
    if char.isdigit():
        return 'n'
    return 'o'

grouped = [groupby(word, key=get_char_class) for word in OGList]
group_dict = [{k: ''.join(g) for k, g in group} for group in grouped]
result = {}
for d in group_dict:
    for k, v in d.items():
        result.setdefault(k, []).append(v)
# or you can go bonkers and cook up this monstrosity instead of proper loops
# [result.setdefault(k, []).append(v) for d in group_dict for k, v in d.items()]

{'a': ['A', 'i', 'NSX', 'c'], 'n': ['10', '320', '200', '220'], 'o': [' ', ' ']}