是的，你可以通过使用正则表达式和过滤器函数这样做.

import re
def filter_str(s):
    res_num = [int(match) for match in re.findall(r"\d+", s)]
    res_str = "".join(filter(lambda x: not x.isdigit(), s))
    
    return res_num, res_str

s = "U(4) Leiknir"
a, b = filter_str(s)
print("Numbers: ", a , "\nStr: ", b)

输出：

Numbers:  [4] 
Str:  U() Leiknir

是的，你可以用正则表达式来实现。
Import re

textelement ="U(4) Leiknir"
number=re.findall(r'\d+', textelement)[0]
print(number)
chars = " ".join(re.findall("[a-zA-Z]+", textelement))
print(chars)

输出：

4
U Leiknir

更新日期：

textelement ="U(3.45) Leiknir"
number=re.findall(r'(\d+(?:\.\d+)?)', textelement)[0]
print(number)
chars = " ".join(re.findall("[a-zA-Z]+", textelement))
print(chars)