如何根据其版本对这些字符串列表进行排序

import re

def extract_version(string):
    # 使用正则表达式提取版本号
    version = re.findall(r'\d+(\.\d+)*', string)
    if version:
        return version[0]
    else:
        return ""

def compare_versions(version1, version2):
    # 将版本号拆分为数字列表
    version1_nums = list(map(int, version1.split('.')))
    version2_nums = list(map(int, version2.split('.')))

    # 逐个比较对应位置的数字
    for i in range(max(len(version1_nums), len(version2_nums))):
        v1 = version1_nums[i] if i < len(version1_nums) else 0
        v2 = version2_nums[i] if i < len(version2_nums) else 0
        if v1 < v2:
            return -1
        elif v1 > v2:
            return 1

    return 0

def sort_strings_by_version(strings):
    # 提取版本号并与字符串关联
    versions = [(extract_version(s), s) for s in strings]

    # 根据版本号进行排序
    sorted_versions = sorted(versions, key=lambda x: x[0], cmp=compare_versions)

    # 返回排序后的字符串列表
    return [v[1] for v in sorted_versions]

# 示例用法
strings = ["v2.0.1", "v1.5", "v2.1.0", "v1.0"]
sorted_strings = sort_strings_by_version(strings)
print(sorted_strings)