私有 InstagramApi

Question

我正在尝试获取某个用户的关注者并根据特定条件过滤他们，因为我想缩小值得与之互动的用户列表。如果我不包括 time.sleep()，我会得到一个 KeyError。这样我得到 429、500 等...我知道这意味着请求太多，但是没有办法避免这种情况还是我做错了？有更多 effective/pythonic 的方法吗？提前致谢。

import imageio
imageio.plugins.ffmpeg.download()
from InstagramAPI import InstagramAPI
import re
import time

def get_user_followers(api, user_id):
    followers = []
    next_max_id = True
    while next_max_id:
        # first iteration hack
        if next_max_id is True:
            next_max_id = ''
        _ = api.getUserFollowers(user_id, maxid=next_max_id)
        followers.extend(api.LastJson.get('users', []))
        next_max_id = api.LastJson.get('next_max_id', '')
    return followers

#returns user id of the user you want to get followers from
def get_user_id(self, user):
    self.SendRequest('users/' + str(user) + '/usernameinfo/')
    userid = self.LastJson['user']['pk']
    return userid


follower_list = []
def scrape_followers(self, user):
    # gets the id from a user
    self.SendRequest('users/' + str(user) + '/usernameinfo/')
    userid = self.LastJson['user']['pk']
    self.SendRequest('users/' + str(userid) + '/info/')

    # filter users based on criteria
    pp = self.LastJson['user']['has_anonymous_profile_picture']             #has profile pic
    fer = self.LastJson['user']['follower_count']                           #number of followers
    fing = self.LastJson['user']['following_count']                         #number of followings
    bio = self.LastJson['user']['biography']                                #certain words in bio
    b = re.compile(r'free|shop|download', re.IGNORECASE).search(bio)
    business = self.LastJson['user']['is_business']                         #isn't a business profile
    story = self.LastJson['user']['has_highlight_reels']                    #is active/has a story 
    if not pp and 1500 > fer > 50 and 1000 > fing > 100 and not business and story and b == None:
        follower_list.append(userid)
    # return follower_list


# ACTUAL CALL
# Your login details
api = InstagramAPI("xxx", "xxx")
api.login()
#Call the function
user = 'GlitteryHell'
userid = get_user_id(api, user)
followers = get_user_followers(api, userid)
for i in range(10):
    user = list(followers[i].values())[1]
    try:
        scrape_followers(api, user)
        time.sleep(1)
    except KeyError:
        time.sleep(10)
print(follower_list)

Answer 1

最近 Instagram 弃用了几个 API 并更改了几个 API 的响应所以现在您无法在 API 响应中获取用户的个人资料照片和其他信息。

关注和关注相关 API 也已弃用。

您可以在此处阅读更多相关信息。

https://www.instagram.com/developer/changelog/

此外，请求数量也发生了巨大变化。它从每个用户每小时 4000 个减少到 200 个。

您可以添加更多授权令牌以增加每小时可以发送的请求数。

Answer 2

Instagram 具有内置限制以防止滥用其 API。 "fix" 您对此无能为力，因为问题不在于您的代码。

您可以在请求之间设置延迟以防止出现 429。万一你得到 429，你应该暂时不要尝试做任何事情（除了每隔几分钟检查一次块是否消失）。

私有 InstagramApi

Private InstagramApi

python

for-loop

instagram-api