通过删除不需要的查询参数同时保留其他参数，返回干净的URL的最好方法是什么？

from urllib.parse import urlparse, parse_qs, urlencode, urlunparse

def clean_url(url, params_to_remove):
    parsed_url = urlparse(url)
    query_params = parse_qs(parsed_url.query)

    # 删除不需要的查询参数
    for param in params_to_remove:
        query_params.pop(param, None)

    # 重新构建查询参数
    encoded_params = urlencode(query_params, doseq=True)

    # 重新组合URL
    clean_url_parts = list(parsed_url)
    clean_url_parts[4] = encoded_params
    clean_url = urlunparse(clean_url_parts)

    return clean_url

# 示例用法
url = "https://www.example.com/path?param1=value1&param2=value2&param3=value3"
params_to_remove = ["param2", "param3"]
cleaned_url = clean_url(url, params_to_remove)
print(cleaned_url)