在Python中,使用requests库进行网页抓取时,可以通过以下方法来提高抓取速度:
import requests url = "https://example.com" response = requests.get(url, timeout=5) import requests from concurrent.futures import ThreadPoolExecutor urls = ["https://example.com"] * 10 def fetch(url): response = requests.get(url) return response.text with ThreadPoolExecutor(max_workers=5) as executor: results = list(executor.map(fetch, urls)) import aiohttp import asyncio async def fetch(url): async with aiohttp.ClientSession() as session: async with session.get(url) as response: return await response.text() async def main(): urls = ["https://example.com"] * 10 tasks = [fetch(url) for url in urls] results = await asyncio.gather(*tasks) asyncio.run(main()) import requests url = "https://example.com" headers = { "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.110 Safari/537.3" } response = requests.get(url, headers=headers) import requests url = "https://example.com" proxies = { "http": "http://proxy.example.com:8080", "https": "https://proxy.example.com:8080", } response = requests.get(url, proxies=proxies) import requests import time url = "https://example.com" for _ in range(10): response = requests.get(url) time.sleep(1) # 等待1秒 import requests from functools import lru_cache @lru_cache(maxsize=128) def fetch(url): response = requests.get(url) return response.text url = "https://example.com" result = fetch(url) 通过以上方法,可以在一定程度上提高Python爬虫的抓取速度。但请注意,过于频繁的请求可能会导致目标网站的服务器负载过大,甚至被封禁。在进行爬虫开发时,请遵守相关法律法规和网站的使用条款。