Python:报纸模块 - 任何直接从URL获取文章的方法?

时间:2016-05-24 02:47:01

标签: python multithreading parsing nlp python-newspaper

我使用报纸模块为python找到了here

在教程中,它描述了如何汇集不同报纸的建设。它会同时生成它们。 (参见"多线程文章下载"在上面的链接中)

有没有办法直接从网址列表中提取文章?也就是说,有什么方法可以将多个URL添加到以下设置中并让它同时下载和解析它们?

from newspaper import Article
url = 'http://www.bbc.co.uk/zhongwen/simp/chinese_news/2012/12/121210_hongkong_politics.shtml'
a = Article(url, language='zh') # Chinese
a.download()
a.parse()
print(a.text[:150])

4 个答案:

答案 0 :(得分:4)

我可以通过为每篇文章网址创建Source来完成此操作。 (免责声明:不是python开发人员)

import newspaper

urls = [
  'http://www.baltimorenews.net/index.php/sid/234363921',
  'http://www.baltimorenews.net/index.php/sid/234323971',
  'http://www.atlantanews.net/index.php/sid/234323891',
  'http://www.wpbf.com/news/funeral-held-for-gabby-desouza/33874572',  
]

class SingleSource(newspaper.Source):
    def __init__(self, articleURL):
        super(StubSource, self).__init__("http://localhost")
        self.articles = [newspaper.Article(url=url)]

sources = [SingleSource(articleURL=u) for u in urls]

newspaper.news_pool.set(sources)
newspaper.news_pool.join()

for s in sources:
  print s.articles[0].html

答案 1 :(得分:2)

我知道这个问题确实很老,但这是我用Google搜索如何获取多线程报纸时显示的第一个链接之一。虽然Kyles的回答非常有帮助,但它并不完整,我认为它有一些错别字...

import newspaper

urls = [
'http://www.baltimorenews.net/index.php/sid/234363921',
'http://www.baltimorenews.net/index.php/sid/234323971',
'http://www.atlantanews.net/index.php/sid/234323891',
'http://www.wpbf.com/news/funeral-held-for-gabby-desouza/33874572',  
]

class SingleSource(newspaper.Source):
def __init__(self, articleURL):
    super(SingleSource, self).__init__("http://localhost")
    self.articles = [newspaper.Article(url=articleURL)]

sources = [SingleSource(articleURL=u) for u in urls]

newspaper.news_pool.set(sources)
newspaper.news_pool.join()

我将 Stubsource 更改为 Singlesource ,并将 urls 之一更改为 articleURL 。 当然,这只是下载网页,您仍然需要解析它们才能获取文本。

multi=[]
i=0
for s in sources:
    i+=1
    try:
        (s.articles[0]).parse()
        txt = (s.articles[0]).text
        multi.append(txt)
    except:
        pass

在我的100个URL样本中,与仅按顺序处理每个URL相比,这花费了一半的时间。 (编辑:将样本量增加到2000后,减少了大约四分之一。)

(编辑:整个过程都可以使用多线程!)我为实现使用了this很好的解释。对于100个URL的样本大小,使用4个线程花费的时间与上面的代码相当,但是将线程数增加到10可进一步减少大约一半。较大的样本量需要更多线程才能产生可​​比的差异。

import newspaper
from multiprocessing.dummy import Pool as ThreadPool

def getTxt(url):
    article = Article(url)
    article.download()
    try:
        article.parse()
        txt=article.text
        return txt
    except:
        return ""

pool = ThreadPool(10)

# open the urls in their own threads
# and return the results
results = pool.map(getTxt, urls)

# close the pool and wait for the work to finish 
pool.close() 
pool.join()

答案 2 :(得分:1)

以约瑟夫·瓦尔斯的答案为基础。我假设原始发布者想要使用多线程来提取一堆数据并将其正确存储在某处。经过大量的尝试,我认为我找到了一个解决方案,它可能不是最有效的,但是它可以起作用,但是我试图使其变得更好,但是,我认为报纸3k插件可能有点bug。但是,这可以将所需的元素提取到DataFrame中。

import newspaper
from newspaper import Article
from newspaper import Source
import pandas as pd

gamespot_paper = newspaper.build('https://www.gamespot.com/news/', memoize_articles=False)
bbc_paper = newspaper.build("https://www.bbc.com/news", memoize_articles=False)
papers = [gamespot_paper, bbc_paper]
news_pool.set(papers, threads_per_source=4) 
news_pool.join()

#Create our final dataframe
df_articles = pd.DataFrame()

#Create a download limit per sources
limit = 100

for source in papers:
    #tempoary lists to store each element we want to extract
    list_title = []
    list_text = []
    list_source =[]

    count = 0

    for article_extract in source.articles:
        article_extract.parse()

        if count > limit:
            break

        #appending the elements we want to extract
        list_title.append(article_extract.title)
        list_text.append(article_extract.text)
        list_source.append(article_extract.source_url)

        #Update count
        count +=1


    df_temp = pd.DataFrame({'Title': list_title, 'Text': list_text, 'Source': list_source})
    #Append to the final DataFrame
    df_articles = df_articles.append(df_temp, ignore_index = True)
    print('source extracted')

请提出任何改进建议!

答案 3 :(得分:-1)

我不熟悉报纸模块,但以下代码使用URL列表,应该与链接页面中提供的相同:

crossfilter