所以我试图为搜索引擎制作一个抓取工具。我的代码大部分都是完整的。它的工作方式是:
我遇到的问题是,有时它会多次索引网址。
我做的一个示例运行是使用初始pages = ["http://www.ece.udel.edu"]
。在输出文件中,每行包含docID和url。格式为docID url\n
。这会在对结果进行排名时引起问题,因为虽然它应该对这些结果进行排名相同,但我会将同一页面的多个(每个都有不同的docID)排名。
这是我到目前为止的代码。我试图对其进行彻底评论,以便更容易理解。
def cleanText(soup):
#remove all unnecessary tags & punctuation leaving text of all lowercase letters & words
texts = soup.findAll(text=True)
visible_elements=[]
for elem in texts:
if elem.parent.name in ['style', 'script', '[document]', 'head', 'title']:
visible_elements.append('')
elem = elem.encode('utf-8','replace')
result = re.sub('<!--.*-->|\r|\n', '', str(elem), flags=re.DOTALL)
result = re.sub('\s{2,}| ', ' ', result)
visible_elements.append(result)
visible_text = ' '.join(visible_elements)
visible_text = re.sub('[^0-9a-zA-Z]+', ' ', visible_text)
visible_text = visible_text.lower()
return visible_text
def naming(pagename, pathToOutput):
#used to create filename to save output text file
cantBeInFilename = ("/", "\\",":","*","?","<",">","|")
for k in range(len(cantBeInFilename)):
pagename = pagename.replace(cantBeInFilename[k], "-")
filename = "\\".join((pathToOutput, pagename))
filename = ".".join((filename, "txt"))
return filename
def crawl(pages, pathToOutput, pathToLinks):
depth = len(pages)
docID = 1
for i in range(depth): #for each link on the page
newpages=set() #create an empty set for gathered pages
for page in pages: #each page in pages, initially atleast 1
linksText = open(pathToLinks, 'a') #open a file
linksText.write(str(docID) + " "); #write a docID to identify page
docID = docID + 1 #increment docID
linksText.write(str(page) + '\n') #append link into end of file
linksText.close() #close file to update
try:
c = urllib2.urlopen(page) #open the page
print ("Opening " + page)
except:
print ("Could not open " + page)
continue
soup = BeautifulSoup(c.read())
clean_text = cleanText(soup)
#get all links off page
links = soup.findAll('a')
#write web page to text file in pathToOutput directory
filename = naming(page, pathToOutput)
f = open(filename, 'w')
f.write(clean_text)
f.close()
depth = 0
for link in links:
if('href' in dict(link.attrs)):
#set depth equal to the # of links in the pages
depth = depth+1
url = urljoin(page, link['href'])
#remove unnecessary portions of url
if url.find("'") != -1:
continue
url = url.split('#')[0]
#get each line (each link) from linksText
linksText = open(pathToLinks, 'r')
lines = linksText.readlines()
linksText.close()
#remove docIDs just to have
for j in range(len(lines)):
lines[j-1] = lines[j-1].split()[1]
#FOR ALL LINKS
if url[0:4] == "http":
#check to see if the url ends with a /, if so remove it
if url[len(url)-1]=="/":
url = url[0:len(url)-1]
#check to see if the url is already in links
present = 0
for line in lines:
if (url == line):
present = 1
else:
present = 0
#if the url isn't present, add to pages to be cycled through
if (present == 0):
print ("Indexing " + url)
newpages.add(url)
#add newpages to pages
for newpage in newpages:
pages.append(newpage)
newpages = set()
#remove already read page from pages
pages.remove(page)
我会附上我已经从这里得到的输出,但我刚开始在网站上,不能发布一个带有两个以上链接的问题:(对不起,如果这很长,令人困惑,只是简单不做感
答案 0 :(得分:0)
此代码段没有按您的想法执行
present = 0
for line in lines:
if (url == line):
present = 1
else:
present = 0
#if the url isn't present, add to pages to be cycled through
if (present == 0):
print ("Indexing " + url)
newpages.add(url)
如果匹配项位于字符串的最后一项,则只会离开present = 1
。一种更简单的测试方法是:
#if the url isn't present, add to pages to be cycled through
if url not in lines:
print ("Indexing " + url)
newpages.add(url)
您可能还需要更改
#get each line (each link) from linksText
linksText = open(pathToLinks, 'r')
lines = linksText.readlines()
linksText.close()
进入
#get each line (each link) from linksText
linksText = open(pathToLinks, 'r')
lines = linksText.read().split('\n')
linksText.close()
以便line
中的每个lines
都不会以\n
结尾