python - 推文的日期和时间？

Question

我想获取特定推文的标准格式的日期和时间。这是我正在使用的以下代码:-

import tweepy
import time
import datetime
from tweepy import OAuthHandler
from tweepy import Stream
from tweepy.streaming import StreamListener
import string
from nltk.corpus import stopwords
from nltk.tokenize import TweetTokenizer

consumer_key = '' #your consumer key
consumer_secret = '' # your consumer secret
access_token = '' # your access token
access_secret = '' # your access secret
auth = OAuthHandler(consumer_key, consumer_secret)
auth.set_access_token(access_token, access_secret)

api = tweepy.API(auth)


class MyListener(StreamListener):


    def on_data(self, data):
        try:
            tweet=data.split('","text":"')[1].split('","source":"')[0]
            saveThis=str(time.time())+'::'+tweet
            with open('twitDB.csv', 'a') as f:
                f.write(saveThis)
                f.write('\n')
                f.close()
                return True
        except BaseException as e:
            print(str(e))
        return True

    def on_error(self, status):
        print(status)
        return True


twitter_stream = Stream(auth, MyListener())
twitter_stream.filter(track=['cars'], languages=['en'])

这就是我从中得到的:-

1495596971.6034188::automotive auto ebc greenstuff 6000 series supreme truck 
and suv brake pads dp61603 https:\/\/t.co\/jpylzjyd5o cars\u2026 
https:\/\/t.co\/gfsbz6pkj7""display_text_range:[0140]source:""\u003ca 
href=\""https:\/\/dlvrit.com\/\"" 
rel=\""nofollow\""\u003edlvr.it\u003c\/a\u003e"""
"1495596972.330948::new free stock photo of city cars road 
https:\/\/t.co\/qbkgvkfgpp""display_text_range:[0"
"1495596972.775966::ebay: 1974 volkswagen beetle - classic 1952 custom 
 conversion extremely rare 1974 vw beetle\u2026\u2026 
 https:\/\/t.co\/wdsnf2pmo7""display_text_range:[0140]source:""\u003ca 
 href=\""https:\/\/dlvrit.com\/\"" 
rel=\""nofollow\""\u003edlvr.it\u003c\/a\u003e"""

我想要标准格式的日期和时间，而不是“1495596971.6034188”。你知道我该怎么做吗？

Answer 1

以下是将 unix 时间戳转换为可读日期的基本示例:

import datetime
print(
    datetime.datetime.fromtimestamp(
        int("1495596971")
    ).strftime('%Y-%m-%d %H:%M:%S')
)
>>2017-05-24 03:36:11

您需要将“”1495596971.6034188”拆分为“.”并使用第一个索引[0]