Implement a log analysis program with 70 lines of code​

高洛峰
Release: 2016-10-18 14:38:44
Original
1028 people have browsed it

Another masterpiece of python, feel the power of python. Use 70 lines of code to implement the log analysis program

Function introduction: It can directly group and sort the text date. After the output results are pasted into excel, charts can be directly generated, which is very useful for troubleshooting some production environment problems.


Code:

#encoding=utf-8
from optparse import OptionParser
import re
  
def get_args():
    def get_parser():
        usage = u"""%prog -f filename -r rule [-d] [-c]
用途:对文本文件按照指定模式进行分组并排序,主要分析文本日志用
注意:如果正则表达式里有分组,则提取第一分组,
      如果不希望这样,请使用正则的无捕获分组(?:)
示例:统计日志里每分钟的日志量,默认按时间正序排列
      python group.py -f log.txt -r "\d\d\d\d\-\d\d\-\d\d \d\d:\d\d"
      统计日志里每个ip出现的次数,并按出现次数倒序排列
      python group.py -f input.txt -r "\d+\.\d+.\d+.\d+" -c -d"""
        return OptionParser(usage)
  
    def add_option(parser):
        parser.add_option("-f", "--file", dest="filename",
           help=u"需要分组的文本文件", metavar="FILE")
        parser.add_option("-r", "--rule", dest="rule",
           help=u"分组的正则表达式", metavar="REGEX")
        parser.add_option("-d", action="store_true",
            dest="reverse", default=False, help=u"反序排列")
        parser.add_option("-c", action="store_true",
            dest="orderbycount", default=False,
            help=u"按数量排序,默认按匹配字符串排序")
  
    def get_options(parser):
        options, args = parser.parse_args()
        if not options.filename:
            parser.error('没有指定文件名')
        if not options.rule:
            parser.error('没有指定分组规则')
        return options
  
    parser = get_parser()
    add_option(parser)
    return get_options(parser)
  
options      = get_args()
filename     = options.filename
rule         = options.rule
reverse      = options.reverse
orderbycount = options.orderbycount
regex        = re.compile(rule, re.IGNORECASE)
keys         = {}
  
def counter_key(key):
    keys.setdefault(key, 0)
    keys[key] += 1
  
def print_keys():
    sort_key = (lambda d:d[1]) if orderbycount else (lambda d:d[0])
    temp_items = sorted(keys.items(), key=sort_key, reverse=reverse)
    for item in temp_items:
        key = item[0]
        print key, keys[key]
  
def get_key(line):
    m = regex.search(line)
    if m:
        return m.group() if regex.groups == 0 else m.group(1)
    return '!NotMatch!'
  
with open(filename) as f:
    for line in f:
        key = get_key(line)
        counter_key(key)
  
print_keys()
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