如何在抓取时避免IP封禁

import random
import time
import requests

# s4m datacenter proxies (authenticated). Placeholder host only.
USER, PASS = "USER", "PASS"
HTTP  = f"http://{USER}:{PASS}@proxy.s4m.online:3128"
SOCKS = f"socks5://{USER}:{PASS}@proxy.s4m.online:1080"  # pip install requests[socks]

PROXIES = {"http": HTTP, "https": HTTP}

HEADERS = {
    "User-Agent": (
        "Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 "
        "(KHTML, like Gecko) Chrome/125.0 Safari/537.36"
    ),
    "Accept": "text/html,application/xhtml+xml;q=0.9,*/*;q=0.8",
    "Accept-Language": "en-US,en;q=0.9",
}


def polite_get(url, session, max_retries=5):
    for attempt in range(max_retries):
        r = session.get(url, headers=HEADERS, proxies=PROXIES, timeout=20)
        if r.status_code == 429 or r.status_code >= 500:
            wait = float(r.headers.get("Retry-After", 2 ** attempt))
            time.sleep(wait + random.uniform(0, 1))  # backoff + jitter
            continue
        r.raise_for_status()
        return r
    raise RuntimeError(f"gave up after {max_retries} tries: {url}")


def crawl(urls):
    # One Session = one sticky exit for this run; open a new one to rotate.
    with requests.Session() as s:
        for url in urls:
            resp = polite_get(url, s)
            yield url, resp.text
            time.sleep(random.uniform(1.5, 4.0))  # rate-limit + jitter


if __name__ == "__main__":
    for url, html in crawl(["https://example.com/page/1"]):
        print(url, len(html))