在python字典中为每个唯一键计算唯一值 [英] Count unique values per unique keys in python dictionary

问题描述

我有这样的字典：

yahoo.com|98.136.48.100
yahoo.com|98.136.48.105
 yahoo.com|98.136.48.110
 yahoo.com|98.136.48.114
 yahoo.com|98.136.48.66
 yahoo.com|98.136.48.71
 yahoo.com|98.136.48.73
 yahoo.com|98.136.48.75
 yahoo.net|98.136.48.100
g03.msg.vcs0|98.136.48.105

其中我有重复的键和值。而我想要的是一个唯一键（ips）和唯一值计数（域）的最终字典。我有laready以下代码：

in which I have repetitive keys and values. And what I want is a final dictionary with unique keys (ips) and count of unique values (domains). I have laready below code:

for dirpath, dirs, files in os.walk(path):
    for filename in fnmatch.filter(files, '*.txt'):
        with open(os.path.join(dirpath, filename)) as f:
            for line in f:
                if line.startswith('.'):
                    ip = line.split('|',1)[1].strip('\n')
                    semi_domain = (line.rsplit('|',1)[0]).split('.',1)[1]
                    d[ip]= semi_domains
                    if ip not in d:
                        key = ip
                        val = [semi_domain]
                        domains_per_ip[key]= val

但这不能正常工作。有人可以帮助我吗？

but this is not working properly. Can somebody help me out with this?

推荐答案

使用defaultdict：

Use a defaultdict:

from collections import defaultdict

d = defaultdict(set)

with open('somefile.txt') as thefile:
   for line in the_file:
      if line.strip():
          value, key = line.split('|')
          d[key].add(value)

for k,v in d.iteritems():  # use d.items() in Python3
    print('{} - {}'.format(k, len(v)))

这篇关于在python字典中为每个唯一键计算唯一值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！