python - 如何从 Instagram 网络浏览器中抓取关注者？

Question

谁能告诉我如何访问基础 URL 以查看给定用户的 Instagram 关注者？我可以使用 Instagram API 做到这一点，但考虑到审批流程的待定更改，我决定改用抓取。

Instagram 网络浏览器允许您查看任何给定公共(public)用户的关注者列表 - 例如，要查看 Instagram 的关注者，请访问“https://www.instagram.com/instagram”，然后单击关注者 URL 以打开一个按查看者分页的窗口(注意:您必须登录到您的帐户才能查看)。

我注意到弹出此窗口时 URL 更改为“https://www.instagram.com/instagram/followers”，但我似乎无法查看此 URL 的基础页面源。

因为它出现在我的浏览器窗口中，所以我认为我可以抓取它。但是我必须使用像 Selenium 这样的包吗？有谁知道底层 URL 是什么，所以我不必使用 Selenium？

例如，我可以通过访问“instagram.com/instagram/media/”直接访问底层提要数据，我可以从中抓取所有迭代并对其进行分页。我想对关注者列表做一些类似的事情，并直接访问这些数据(而不是使用 Selenium)。

Answer 1

编辑:2018 年 12 月更新:

自从发布以来，Insta 土地上的情况发生了变化。这是一个更新后的脚本，它更像 pythonic 并且更好地利用了 XPATH/CSS 路径。

注意，要使用这个更新的脚本，你必须安装explicit包(pip install explicit)，或者用waiter转换每一行到纯 Selenium 显式等待。

import itertools

from explicit import waiter, XPATH
from selenium import webdriver


def login(driver):
    username = ""  # <username here>
    password = ""  # <password here>

    # Load page
    driver.get("https://www.instagram.com/accounts/login/")

    # Login
    waiter.find_write(driver, "//div/input[@name='username']", username, by=XPATH)
    waiter.find_write(driver, "//div/input[@name='password']", password, by=XPATH)
    waiter.find_element(driver, "//div/button[@type='submit']", by=XPATH).click()

    # Wait for the user dashboard page to load
    waiter.find_element(driver, "//a/span[@aria-label='Find People']", by=XPATH)


def scrape_followers(driver, account):
    # Load account page
    driver.get("https://www.instagram.com/{0}/".format(account))

    # Click the 'Follower(s)' link
    # driver.find_element_by_partial_link_text("follower").click()
    waiter.find_element(driver, "//a[@href='/instagram/followers/']", by=XPATH).click()

    # Wait for the followers modal to load
    waiter.find_element(driver, "//div[@role='dialog']", by=XPATH)

    # At this point a Followers modal pops open. If you immediately scroll to the bottom,
    # you hit a stopping point and a "See All Suggestions" link. If you fiddle with the
    # model by scrolling up and down, you can force it to load additional followers for
    # that person.

    # Now the modal will begin loading followers every time you scroll to the bottom.
    # Keep scrolling in a loop until you've hit the desired number of followers.
    # In this instance, I'm using a generator to return followers one-by-one
    follower_css = "ul div li:nth-child({}) a.notranslate"  # Taking advange of CSS's nth-child functionality
    for group in itertools.count(start=1, step=12):
        for follower_index in range(group, group + 12):
            yield waiter.find_element(driver, follower_css.format(follower_index)).text

        # Instagram loads followers 12 at a time. Find the last follower element
        # and scroll it into view, forcing instagram to load another 12
        # Even though we just found this elem in the previous for loop, there can
        # potentially be large amount of time between that call and this one,
        # and the element might have gone stale. Lets just re-acquire it to avoid
        # that
        last_follower = waiter.find_element(driver, follower_css.format(follower_index))
        driver.execute_script("arguments[0].scrollIntoView();", last_follower)


if __name__ == "__main__":
    account = 'instagram'
    driver = webdriver.Chrome()
    try:
        login(driver)
        # Print the first 75 followers for the "instagram" account
        print('Followers of the "{}" account'.format(account))
        for count, follower in enumerate(scrape_followers(driver, account=account), 1):
            print("\t{:>3}: {}".format(count, follower))
            if count >= 75:
                break
    finally:
        driver.quit()

我做了一个快速基准测试来展示性能如何随着您尝试以这种方式抓取的关注者越多而呈指数下降:

$ python example.py
Followers of the "instagram" account
Found    100 followers in 11 seconds
Found    200 followers in 19 seconds
Found    300 followers in 29 seconds
Found    400 followers in 47 seconds
Found    500 followers in 71 seconds
Found    600 followers in 106 seconds
Found    700 followers in 157 seconds
Found    800 followers in 213 seconds
Found    900 followers in 284 seconds
Found   1000 followers in 375 seconds

原帖:你的问题有点困惑。例如，我不太确定“我可以通过所有迭代从中抓取和分页”到底是什么意思。您目前使用什么来抓取和分页？

无论如何，instagram.com/instagram/media/ 与 instagram.com/instagram/followers 不是同一类型的端点。 media 端点似乎是一个 REST API，配置为返回一个易于解析的 JSON 对象。

据我所知，followers 端点并不是真正的 RESTful 端点。相反，在您单击“关注者”按钮后，Instagram AJAX 会将信息发送到页面源(使用 React？)。我认为如果不使用 Selenium 之类的东西，您将无法获得该信息，Selenium 可以加载/呈现向用户显示关注者的 javascript。

此示例代码将起作用:

from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC


def login(driver):
    username = ""  # <username here>
    password = ""  # <password here>

    # Load page
    driver.get("https://www.instagram.com/accounts/login/")

    # Login
    driver.find_element_by_xpath("//div/input[@name='username']").send_keys(username)
    driver.find_element_by_xpath("//div/input[@name='password']").send_keys(password)
    driver.find_element_by_xpath("//span/button").click()

    # Wait for the login page to load
    WebDriverWait(driver, 10).until(
        EC.presence_of_element_located((By.LINK_TEXT, "See All")))


def scrape_followers(driver, account):
    # Load account page
    driver.get("https://www.instagram.com/{0}/".format(account))

    # Click the 'Follower(s)' link
    driver.find_element_by_partial_link_text("follower").click()

    # Wait for the followers modal to load
    xpath = "//div[@style='position: relative; z-index: 1;']/div/div[2]/div/div[1]"
    WebDriverWait(driver, 10).until(
        EC.presence_of_element_located((By.XPATH, xpath)))

    # You'll need to figure out some scrolling magic here. Something that can
    # scroll to the bottom of the followers modal, and know when its reached
    # the bottom. This is pretty impractical for people with a lot of followers

    # Finally, scrape the followers
    xpath = "//div[@style='position: relative; z-index: 1;']//ul/li/div/div/div/div/a"
    followers_elems = driver.find_elements_by_xpath(xpath)

    return [e.text for e in followers_elems]


if __name__ == "__main__":
    driver = webdriver.Chrome()
    try:
        login(driver)
        followers = scrape_followers(driver, "instagram")
        print(followers)
    finally:
        driver.quit()

由于多种原因，这种方法存在问题，其中最主要的原因是它相对于 API 来说有多慢。