解析用户名以提取用户位置 Twitter

Question

我正在尝试从 twitter 中获取与用户名相关的用户位置。

输入：用户列表有超过50K个用户名

AkkiPritam,6.77E+17,12/15/2015,#chennaifloods
AkkiPritam,6.77E+17,12/15/2015,#bhoomikatrust
AkkiPritam,6.77E+17,12/15/2015,#akshaykumar
gischethans,6.77E+17,12/15/2015,#chennaifloods
mid_day,6.77E+17,12/15/2015,#bollywood
mid_day,6.77E+17,12/15/2015,#chennaifloods
Nanthivarman16,6.77E+17,12/15/2015,#admkfails
Nanthivarman16,6.77E+17,12/15/2015,#jayafails
Nanthivarman16,6.77E+17,12/15/2015,#stickergovt
Nanthivarman16,6.77E+17,12/15/2015,#chennaifloods
AdilaMatra,6.77E+17,12/15/2015,#chennaifloods
AdilaMatra,6.77E+17,12/15/2015,#climatechange
AdilaMatra,6.77E+17,12/15/2015,#delhichokes
AdilaMatra,6.77E+17,12/15/2015,#smog
HDFCERGOGIC,6.77E+17,12/15/2015,#chennaifloods
HDFCERGOGIC,6.77E+17,12/15/2015,#tnfloods
ImSoorej,6.77E+17,12/15/2015,#chennaifloods
ImSoorej,6.77E+17,12/15/2015,#chennaimicr

代码：我想查找地理位置，可能是地理坐标。

from __future__ import print_function
import tweepy
from tweepy import OAuthHandler
from tweepy import Stream
from tweepy.streaming import StreamListener
import pandas as pd
import csv

consumer_key = 'xyz'
consumer_secret = 'xyz'
access_token = 'xyz'
access_token_secret = 'xyz'

data = pd.read_csv('user_keyword.csv')
df = ['user_name', 'user_id', 'date', 'keyword']

def get_user_details(username):
        userobj = api.get_user(username)
        return userobj

if __name__ == '__main__':
    #authenticating the app (https://apps.twitter.com/)
    auth = tweepy.auth.OAuthHandler(consumer_key, consumer_secret)
    auth.set_access_token(access_token, access_token_secret)
    api = tweepy.API(auth)

    username = df['user_name']
    userOBJ = get_user_details(username)
    print(userOBJ.location)

错误：无法将用户名解析为程序。

Traceback (most recent call last):
  File "user_profile_location.py", line 38, in <module>
    username = df['user_name']
TypeError: list indices must be integers, not str

Answer 1

您正在使用“数据”来定义您的 DataFrame 和“df”，我认为应该是 DataFrame 的列

data = pd.read_csv('user_keyword.csv')
df = ['user_name', 'user_id', 'date', 'keyword']

我假设 user_keyword.csv 文件没有标题，尝试添加：

data.columns = df

它会将列名更改为存储在 df 中的值。 然后稍后代替：

username = df['user_name']

试试：

username = data['user_name']

请记住，现在用户名是一整列，因此 get_user_details(username) 不应期待单个字符串。