使用 Selenium、BeautifulSoup 抓取网页耗时过长

Question

I want to scrape a website and its sub-pages, but it is taking too long. How can I optimize the request or use an alternative solution?

Below is the code I am using. It takes 10s for just loading the Google home page. So it's clearly not scalable if I were to give it 280 links

from selenium import webdriver
import time
# prepare the option for the chrome driver
options = webdriver.ChromeOptions()
options.add_argument('headless')

# start chrome browser
browser = webdriver.Chrome("/usr/lib/chromium-browser/chromedriver" ,chrome_options=options)
start=time.time()
browser.get('http://www.google.com/xhtml')
print(time.time()-start)
browser.quit()

Answer 1

Use python requests and Beautiful soup module.

import requests
from bs4 import BeautifulSoup
url="https://tajinequiparle.com/dictionnaire-francais-arabe-marocain/"
url1="https://tajinequiparle.com/dictionnaire-francais-arabe-marocain/{}/"
req = requests.get(url,verify=False)
soup = BeautifulSoup(req.text, 'html.parser')
print("Letters : A")
print([item['href'] for item in soup.select('.columns-list a[href]')])

letters=['B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z']

for letter in letters:

    req = requests.get(url1.format(letter), verify=False)
    soup = BeautifulSoup(req.text, 'html.parser')
    print('Letters : ' + letter)
    print([item['href'] for item in soup.select('.columns-list a[href]')])

Answer 2

you can use that script for the speed. multithread crawler better than all:

https://edmundmartin.com/multi-threaded-crawler-in-python/

After that you must change that code:

def run_scraper(self):
    with open("francais-arabe-marocain.csv", 'a') as file:
        file.write("url")
        file.writelines("\n")
        for i in range(50000):
            try:
                target_url = self.to_crawl.get(timeout=600)
                if target_url not in self.scraped_pages and "francais-arabe-marocain" in target_url:
                    self.scraped_pages.add(target_url)
                    job = self.pool.submit(self.scrape_page, target_url)
                    job.add_done_callback(self.post_scrape_callback)
                    df = pd.DataFrame([{'url': target_url}])
                    df.to_csv(file, index=False, header=False)
                    print(target_url)
            except Empty:
                return
            except Exception as e:
                print(e)
                continue

If url include "francais-arabe-marocain" save urls in a csv file.

After that you can scrape that urls in one for loop reading csv line by line with same way

Answer 3

try to use urllib just like this

import urllib.request
start=time.time()
page = urllib.request.urlopen("https://google.com/xhtml")
print(time.time()-start)

it took only 2s. However, it depends also on the quality of connection you have