排序和组织字典

Question

我有很多很多的键/值对的字典。

的关键是日期和值是全球顶级域名。

我要输出的词典到一个文本文件，以便它计数和alpha排序相似的价值观，但只能在同一关键

for example: 
*key:  value1:count value2:count* 
date1: au:4 be:12 com:44 
date2: az:4 com:14 net:5 

代码：

with open('access_logshort.txt','rU') as f: 
for line in f: 
    list1 = re.search(r'(?P<Date>[0-9]{2}/[a-zA-Z]{3}/[0-9]{4})(.+)(GET|POST)\s(http://|https://)([a-zA-Z.]+)(\.)(?P<tld>[a-zA-Z]+)(/).+?"\s200',line) 

    if list1 != None: 
     print list1.groupdict() 
     one_tuple = list1.group(1,7) 
     my_dict[one_tuple[0]]=one_tuple[1] 

output: 
print my_dict 
{'09/Mar/2004': 'hu'} 
{'09/Mar/2004': 'hu'} 
{'09/Mar/2004': 'com'} 
{'09/Mar/2004': 'ru'} 
{'09/Mar/2004': 'ru'} 
{'09/Mar/2004': 'com'} 

牛逼

Answer 1

这应该适合你的情况。

from collections import defaultdict 
from dateutil.parser import parse 
import csv 
import re 

data = defaultdict(lambda: defaultdict(int)) 

with open('access_logshort.txt','rU') as f: 
    for line in f: 
     list1 = re.search(r'(?P<Date>[0-9]{2}/[a-zA-Z]{3}/[0-9]{4})(.+)(GET|POST)\s(http://|https://)([a-zA-Z.]+)(\.)(?P<tld>[a-zA-Z]+)(/).+?"\s200',line) 

     if list1 is not None: 
      date, domain = list1.group(1,7) 
      data[date.lower()][domain.lower()] += 1 

with open('my_data.csv', 'wb') as ofile: 
    # add delimiter='\t' to the argument list of csv.writer if you want 
    # tsv rather than csv 
    writer = csv.writer(ofile) 

    for key, value in sorted(data.iteritems(), key=lambda x: parse(x[0])): 
     domains = sorted(value.iteritems()) 
     writer.writerow([key] + ['{}:{}'.format(*d) for d in domains]) 

输出：

10/Mar/2004,com:2,hu:2,ru:2 
09/Mar/2004,com:2,hu:2,ru:2