如何告诉python不要打印列表中的项目?

如何告诉python不要打印列表中的项目?,python,python-2.7,feedparser,goose,Python,Python 2.7,Feedparser,Goose,我的python脚本解析来自多个RSS提要的标题和链接。我将这些标题存储在一个列表中,我想确保我从不打印重复的标题。我该怎么做 #!/usr/bin/python from twitter import * from goose import Goose import feedparser import time from pyshorteners import Shortener import pause import newspaper dr = feedparse

我的python脚本解析来自多个RSS提要的标题和链接。我将这些标题存储在一个列表中,我想确保我从不打印重复的标题。我该怎么做

    #!/usr/bin/python
 from twitter import *
 from goose import Goose
 import feedparser
 import time
 from pyshorteners import Shortener
 import pause
 import newspaper

 dr = feedparser.parse("http://www.darkreading.com/rss_simple.asp") 
 sm =feedparser.parse("http://www.securitymagazine.com/rss/topic/2654-cyber-tactics.rss")



dr_posts =["CISO Playbook: Games of War & Cyber Defenses",
         "SWIFT Confirms Cyber Heist At Second Bank; Researchers Tie Malware Code to Sony Hack","The 10 Worst Vulnerabilities of The Last 10 Years",
         "GhostShell Leaks Data From 32 Sites In 'Light Hacktivism' Campaign",
          "OPM Breach: 'Cyber Sprint' Response More Like A Marathon",
        "Survey: Customers Lose Trust In Brands After A Data Breach",
       "Domain Abuse Sinks 'Anchors Of Trust'",
       "The 10 Worst Vulnerabilities of The Last 10 Years",
]

sm_posts = ["10 Steps to Building a Better Cybersecurity Plan"]

x = 1

while True:

    try:

        drtitle = dr.entries[x]["title"]
        drlink = dr.entries[x]["link"]
        if drtitle in dr_posts:
            x += 1
            drtitle = dr.entries[x]["title"]
            drtitle = dr.entries[x]["link"]
            print drtitle + "\n" + drlink
            dr_posts.append(drtitle)
            x -= 1
            pause.seconds(10)
        else:
            print drtitle + "\n" + drlink
            dr_posts.append(drtitle)
            pause.seconds(10)

        smtitle = sm.entries[x]["title"]
        smlink = sm.entries[x]["link"]
        if smtitle in sm_posts:
            x +=1
            smtitle = sm.entries[x]["title"]
            smtitle = sm.entries[x]["title"]
            print smtitle + "\n" + smlink
            sm_posts.append(smtitle)
            pause.seconds(10)
    else:
        print smtitle + "\n" + smlink
        sm_posts.append(smtitle)
        x+=1
        pause.seconds(10)



except IndexError:
    print "FAILURE"
    break

目前我只允许它跳过条目。这将是一个问题,因为如果RSS提要中还有另一个副本,那么我将有更多的副本。

您可以利用数据结构,因为它的“唯一性”属性将为您提供帮助。基本上,我们可以将您的列表设置为一个集合,然后再将集合设置为一个列表,这样可以确保您的列表现在填充了严格唯一的值

如果你有一个列表l,那么你可以通过

l = list(set(l))

如果不想打印重复链接,可以使用或

好的是,您还可以通过执行以下操作获得链接的重复次数

sm_posts[sm_links]
>>> link_counts

试试看。

谢谢!这真的帮助了我!
sm_posts[sm_links]
>>> link_counts