python -- 题目不看别人的自己写然后比较

题目一

'''
编写Python脚本，分析xx.log文件，按域名统计访问次数倒序输出

xx.log文件内容如下：
https://www.sogo.com/ale.html
https://www.qq.com/3asd.html
https://www.sogo.com/teoans.html
https://www.bilibili.com/2
https://www.sogo.com/asd_sa.html
https://y.qq.com/
https://www.bilibili.com/1
https://dig.chouti.com/
https://www.bilibili.com/imd.html
https://www.bilibili.com/

输出：
www.bilibili.com
www.sogo.com
www.qq.com
y.qq.com
dig.chouti.com

'''
import re

domain_dict = {}
with open('./visit.log','r') as fr:
    for line in fr.readlines():
        pattern = re.compile(r'(http.*?com).*')
        domain = pattern.match(line).group(1)
        if domain in domain_dict:
            domain_dict[domain] = domain_dict[domain]+1
        else:
            domain_dict[domain] = 1
print(domain_dict)
sorted(domain_dict.items(),key=lambda domain_dict:domain_dict[1],reverse=True)

改进版，优化内存

import re
def buffered_read(file_opened,block_size=4096):
    while True:
        data = file_opened.read(block_size)
        if not data:
            break
        yield data
        
domain_dict = {}        
with open('./visit.log') as f:
    for block in buffered_read(f):
        pattern = re.compile(r'https:.*?com')
        domain_list = pattern.findall(block)
        #domain_dict = [{domain:1} for domain in domain_list]
        for key in domain_list:
            if key in domain_dict:
                domain_dict[key] = domain_dict[key]+1
            else:
                domain_dict[key] = 1
                
sorted(domain_dict.items(),key=lambda d:d[1],reverse=True)

# 别人家的方法
#第一种方式
import re
from collections import Counter
with open("xx.log","r",encoding="utf-8") as f:
    data=f.read()
    res=re.findall(r"https://(.*?)/.*?",data)
    dic=Counter(res)
      
ret=sorted(dic.items(),key=lambda x:x[1],reverse=True)

for k,v in ret:
    print(v,k)

#第二种方式
dic={}
with open("xx.log","r",encoding="utf-8") as f:
    for line in f:
        line=line.split("/")[2]
        if line not in dic:
            dic[line]=1
        else:
            dic[line]+=1
ret=sorted(dic.items(),key=lambda x:x[1],reverse=True)
for k,v in ret:
    print( v,k)