用Python排序apache日志

Question

例如，我有一个简单的apache日志：

192.168.1.1 GET /index.php
192.168.1.1 GET /pilt.png
192.168.1.1 GET /index.php
192.168.1.5 GET /index.php
192.168.1.5 GET /pilt.png
192.168.1.7 GET /index.php
192.168.1.7 GET /index.php
192.168.1.7 GET /index.php
192.168.1.7 GET /kaust/index.php
192.168.1.7 GET /index.php

我该如何编写一个Python代码来整理所有相似的IP地址，并计算出有多少IP地址

w = open("C:\\Users\\xxx\\Desktop\\test.txt","r")

for i in w:
  log=i.split(' ')
  print log[0]
w.close()

已经尝试了这么多，但无法进一步编写代码。你知道吗

谢谢你！你知道吗

Answer 1

您可以将defaultdict(int)用于您的目的：

from collections import defaultdict
my_dict = defaultdict(int)
w = open("C:\\Users\\xxx\\Desktop\\test.txt", "r")
for line in w:
    ip = line.split(' ')[0]
    my_dict[ip]+=1

my_dict  # defaultdict(<class 'int'>, {'192.168.1.7': 5, '192.168.1.1': 3, '192.168.1.5': 2})

Answer 2

具体操作如下：

x = open('PATH_TO_FILE').read()

from itertools import groupby
from operator import itemgetter
x = x.split('\n')
for i in range(len(x)):
    x[i] = x[i].split(' ')

j = 0

for elt, items in groupby(x, itemgetter(0)):
    j += 1
    k = 0
    print elt, items
    for i in items:
        k += 1
        print i
    print 'Total count for IP ',i[0],' is :',k

print 'Total unique IP address are : ',j