有没有一种Python方法可以从URL链接的数据集中选择那些只有200个状态代码的链接？

import requests

def filter_links(url_list):
    filtered_links = []
    for url in url_list:
        response = requests.get(url)
        if response.status_code == 200:
            filtered_links.append(url)
    return filtered_links

# 示例数据集
url_list = [
    'https://www.example.com',
    'https://www.example.com/nonexistent',
    'https://www.example.com/another',
    'https://www.example.com/valid'
]

filtered_links = filter_links(url_list)
print(filtered_links)