使用正则表达式从4个列表中创建多个词典 [英] Create multiple dictionaries from 4 lists using regex

问题描述

我有以下txt文件:

197.109.77.178 - kertzmann3129 [21/Jun/2019:15:45:25 -0700] "DELETE /virtual/solutions/target/web+services HTTP/2.0" 203 26554
156.127.178.177 - okuneva5222 [21/Jun/2019:15:45:27 -0700] "DELETE /interactive/transparent/niches/revolutionize HTTP/1.1" 416 14701
100.32.205.59 - ortiz8891 [21/Jun/2019:15:45:28 -0700] "PATCH /architectures HTTP/1.0" 204 6048
168.95.156.240 - stark2413 [21/Jun/2019:15:45:31 -0700] "GET /engage HTTP/2.0" 201 9645
71.172.239.195 - dooley1853 [21/Jun/2019:15:45:32 -0700] "PUT /cutting-edge HTTP/2.0" 406 24498
180.95.121.94 - mohr6893 [21/Jun/2019:15:45:34 -0700] "PATCH /extensible/reinvent HTTP/1.1" 201 27330

我想创建一个函数，将它们转换成多个字典，其中每一行都是一个字典:

I want to create a function that converts these to multiple dictionaries where each line is a dictionary :

example_dict = {主机":"146.204.224.152"，用户名":"feest6811"，时间":"21/Jun/2019:15:45:24-0700&"，"request":"POST/incentivize HTTP/1.1"}

到目前为止，我能够做到这一点，为所有项目创建4个列表，但我不知道如何为每行创建多个dic:

I was able to do this so far, create 4 lists for all the items but I didn't know how to create multiple dics for each line:

import re
def logs():
    with open("assets/logdata.txt", "r") as file:
        logdata = file.read()
        host = (re.findall('(.*?)\-',logdata))
        username = re.findall('\-(.*?)\[',logdata)
        time = re.findall('\[(.*?)\]', logdata)
        request = re.findall('\"(.*?)\"',logdata)
        #for line in range(len(logdata)):
            #dc = {'host':host[line], 'user_name':user_name[line], 'time':time[line], 'request':request[line]}
       

推荐答案

我现在正在做这门课程，得到的答案是

I am doing this course rightnow and the answer i got is

import re
def logs():
with open("assets/logdata.txt", "r") as file:
    logdata = file.read()

# YOUR CODE HERE

pattern='''
(?P<host>[\w.]*)
(\ -\ )
(?P<user_name>([a-z\-]*[\d]*))
(\ \[)
(?P<time>\w.*?)
(\]\ \")
(?P<request>\w.*)
(\")
'''

lst=[]

for item in re.finditer(pattern,logdata,re.VERBOSE):
    lst.append(item.groupdict())
print(lst)
return lst

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