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【Python】 seo市场竞争度查询工具

作者:otine / 发布于2012/12/4/ 677

# -*- coding: utf-8 -*-
from bs4 import BeautifulSoup
import urllib2,re,time
keys=["玩具","儿童玩具","少儿玩具","智力玩具","益智玩具","安全玩具"]

urls={}#放排名总合#小数部分存放找到次数整数部分存放排名总数
urls1={}#放排名次数
urls2={}#放平均排名
urls3={}#计算关键词覆盖率
for key in keys:
    key=urllib2.quote(key)#汉字转码
    html=urllib2.urlopen("http://www.baidu.com/s?word=%s" %key)
    soup=BeautifulSoup(html)
    n=1

    for tables in soup.find_all("table",class_=re.compile("result"),id=re.compile("\d{1,2}")):
        if tables.get("mu") and not tables.find("span","g"):#百度产品
    #mu="baidu.com/link?url=为百度知道baidu.com/s?tn=baidurt&rtt=1是新闻
            print n,
            if tables.get("mu").find("link?url=")+1:
                print "zhidao.baidu.com"
                url="zhidao.baidu.com"
            elif tables.get("mu").find("s?tn=baidurt")+1:
                print "news.baidu.com"
                url="news.baidu.com"
            else:
                print tables.get("mu").split("http://")[1].split("/")[0]
                url=tables.get("mu").split("http://")[1].split("/")[0]

        else:
            print n,
            if tables.find("span","g").get_text().find("wenku.baidu")+1:
                print "wenku.baidu.com"
                url="wenku.baidu.com"
            else:
                print tables.find("span","g").get_text().split(" ")[2].split("/")[0]
                url=tables.find("span","g").get_text().split(" ")[2].split("/")[0]
        try:
            urls[url]+=n
            urls1[url]+=1
        except:
            urls[url]=n
            urls1[url]=1

        n+=1

print urls
print "\n"*3
for each in urls:
    urls2[each]=urls[each]/urls1[each]
    print each
    print "total sum position is "+str(urls[each])
    print "Ranked "+str(urls1[each])+" Keywords"
    print "Keywords coverage is "+str(float(urls1[each])/len(keys)*100).split(".")[0]+"%"
    print "average position is "+str(urls2[each])
    print "\n"*4

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.seo市场竞争度查询工具