处理字母和数字组合的简单方法

Question

Let's say I have two strings:

string_ex1 = 'AbC024'
string_ex2 = 'aBc24'

string_ex3 = 'AbC24'
string_ex4 = 'aBc24'

And I want a result that the two strings are equal if I compare each other. For example 'AbC' == 'aBc', '024' == '24'

I already know if I distinguish them with \w+ and \d+ and convert to lowercase and to int respectively, I can get a result saying two strings are identical. But I want to know if there's some simpler function to do it.

string1_str = lower(re.findall('\w+', string_ex1))
string1_int = int(re.findall('\d+', string_ex1))
string2_str = lower(re.findall('\w+', string_ex2))
string2_int = int(re.findall('\d+', string_ex2))

if string1_str == string2_str and string1_int == string2_int:
    print('identical')

*Edit The comparison should work both for string_ex1, string_ex2 and string_ex3, string_ex4

Answer 1

You can use a regex that removes leading zeros, then use casefold comparison:

import re

string_ex1 = 'AbC024'
string_ex2 = 'aBc24'

string_ex1 = re.sub(r'(?<=\D)0+(?=\d)', '', string_ex1)
string_ex2 = re.sub(r'(?<=\D)0+(?=\d)', '', string_ex2)

print(string_ex1.casefold() == string_ex2.casefold())
# True

Alternatively, you can call lower on both strings when calling re.sub :

import re

string_ex1 = 'AbC024'
string_ex2 = 'aBc24'

string_ex1 = re.sub(r'(?<=\D)0+(?=\d)', '', string_ex1.lower())
string_ex2 = re.sub(r'(?<=\D)0+(?=\d)', '', string_ex2.lower())

print(string_ex1 == string_ex2)

Answer 2

There is no built-in way to do it. I'd suggest that for both strings you find the groups: only letters, or only digits and compare them in lower case and without leading zeros

def test(str1, str2):
    values1 = re.findall("([a-z]+|[0-9]+)", str1, flags=re.I)
    values2 = re.findall("([a-z]+|[0-9]+)", str2, flags=re.I)
    clean = lambda x: x.lower().lstrip("0")
    return all(a == b for a, b in zip(map(clean, values1), map(clean, values2)))

print(test('AbC024', 'aBc24'))  # True