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。