How to web scrape followers from Instagram web browser?

Question

Can anyone tell me how to access the underlying URL to view a given user's Instagram followers? I am able to do this with Instagram API, but given the pending changes to the approval process, I have decided to switch to scraping.

The Instagram web browser allows you to view the follower list for any given public user - for example, to view Instagram's followers, visit "https://www.instagram.com/instagram", and then click on the followers URL to open a window that paginates through viewers (note: you must be logged in to your account to view this).

I note that the URL changes to "https://www.instagram.com/instagram/followers" when this window pops up, but I can't seem to view the underlying page source for this URL.

Since it appears on my browser window, I assume that I will be able to scrape. But do I have to use a package like Selenium? Does anyone know what the underlying URL is, so I don't have to use Selenium?

As an example, I am able to directly access the underlying feed data by visiting "instagram.com/instagram/media/", from which I can scrape and paginate through all iterations. I would like to do something similar with the list of followers, and access this data directly (rather than using Selenium).

Answer 1

EDIT: Dec 2018 Update:

Things have changed in Insta land since this was posted. Here is an updated script that is a bit more pythonic and better utilizes XPATH/CSS paths.

Note that to use this updated script, you must install the explicit package (pip install explicit), or convert each line with waiter to a pure selenium explicit wait.

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()

I did a quick benchmark to show how performance decreases exponentially the more followers you attempt to scrape this way:

$ 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

Original post: Your question is a little confusing. For instance, I'm not really sure what "from which I can scrape and paginate through all iterations" actually means. What are you currently using to scrape and paginate?

Regardless, instagram.com/instagram/media/ is not the same type of endpoint as instagram.com/instagram/followers. The media endpoint appears to be a REST API, configured to return an easily parseable JSON object.

The followers endpoint isn't really a RESTful endpoint from what I can tell. Rather, Instagram AJAXs in the information to the page source (using React?) after you click the Followers button. I don't think you will be able to get that information without using something like Selenium, which can load/render the javascript that displays the followers to the user.

This example code will work:

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()

This approach is problematic for a number of reasons, chief among them being how slow it is, relative to the the API.

Answer 2

Update: March 2020

This is just the Levi answer with a small updates in some parts because as it is now, it didn't quit the driver successfully. This also gets by default all the followers, as everyone else have said, it's not intended for a lot of followers.

import itertools

from explicit import waiter, XPATH
from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By
from time import sleep

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

    # Load page
    driver.get("https://www.instagram.com/accounts/login/")
    sleep(3)
    # Login
    driver.find_element_by_name("username").send_keys(username)
    driver.find_element_by_name("password").send_keys(password)
    submit = driver.find_element_by_tag_name('form')
    submit.submit()

    # Wait for the user dashboard page to load
    WebDriverWait(driver, 15).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
    sleep(2)
    driver.find_element_by_partial_link_text("follower").click()

    # Wait for the followers modal to load
    waiter.find_element(driver, "//div[@role='dialog']", by=XPATH)
    allfoll = int(driver.find_element_by_xpath("//li[2]/a/span").text)
    # 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):
            if follower_index > allfoll:
                raise StopIteration
            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
        # tha
        last_follower = waiter.find_element(driver, follower_css.format(group+11))
        driver.execute_script("arguments[0].scrollIntoView();", last_follower)


if __name__ == "__main__":
    account = ""  # <account to check>
    driver = webdriver.Firefox(executable_path="./geckodriver")
    try:
        login(driver)
        print('Followers of the "{}" account'.format(account))
        for count, follower in enumerate(scrape_followers(driver, account=account), 1):
            print("\t{:>3}: {}".format(count, follower))
    finally:
        driver.quit()