Question

好的，请注意这是我的第一篇文章

所以，我正在尝试使用Python来获取Twitter趋势，我正在使用python 2.7和Tweepy。

我想要这样的东西（有效）：

#!/usr/bin/python
# -*- coding: utf-8 -*-
import tweepy
consumer_key = 'secret'
consumer_secret = 'secret'
access_token = 'secret'
access_token_secret = 'secret'
# OAuth process, using the keys and tokens
auth = tweepy.OAuthHandler(consumer_key, consumer_secret)
auth.set_access_token(access_token, access_token_secret)
api = tweepy.API(auth)
trends1 = api.trends_place(1)

print trends1

它提供了一个巨大的JSON字符串，

我想将每个趋势名称提取为变量，理想情况下为字符串格式str(trendsname)。

理想情况下，哪个趋势的名称会如此： trendsname = str(trend1) + " " +str(trend2) + " " 等等，对于每个趋势名称。

请注意我只学习Python。

Answer 1

看起来Tweepy会为您反序列化JSON。因此，trends1只是一个普通的Python列表。在这种情况下，您可以简单地执行以下操作：

trends1 = api.trends_place(1) # from the end of your code
# trends1 is a list with only one element in it, which is a 
# dict which we'll put in data.
data = trends1[0] 
# grab the trends
trends = data['trends']
# grab the name from each trend
names = [trend['name'] for trend in trends]
# put all the names together with a ' ' separating them
trendsName = ' '.join(names)
print(trendsName)

结果：

#PolandNeedsWWATour #DownloadElyarFox #DünMürteciBugünHaşhaşi #GalatasaraylılıkNedir #KnowTheTruth Tameni Video Anisa Rahma Mikaeel Manado JDT

Answer 2

我认为以下代码也可以正常运行：

trends1 = api.trends_place(1)
trends = set([trend['name'] for trend in trends1[0]['trends']])
print trends

Python：以tweepy获取Twitter趋势，并解析JSON

2 个答案: