python:读取和拆分文件到词典列表

Question

我将文件内容转换为字典列表时遇到了麻烦,你能建议吗？

File content:
host1.example.com#192.168.0.1#web server 
host2.example.com#192.168.0.5#dns server
host3.example.com#192.168.0.7#web server 
host4.example.com#192.168.0.9#application server 
host5.example.com#192.168.0.10#database server

文件夹旁边有多个文件具有相同的格式.最后,我希望收到以下格式的词典列表:

[ {'dns': 'host1.example.com', 'ip': '192.168.0.1', 'description': 'web_server'},
{'dns': 'host2.example.com', 'ip': '192.168.0.5', 'description': 'dns server'}, 
{'dns': 'host3.example.com', 'ip': '192.168.0.7', 'description': 'web server'}, 
{'dns': 'host4.example.com', 'ip': '192.168.0.9', 'description': 'application server'},
{'dns': 'host5.example.com', 'ip': '192.168.0.10', 'description': 'database server'} ]

先感谢您!

Answer 1

首先,您要分开每一行#.然后,您可以使用zip它们与标签一起压缩它们,然后将其转换为字典.

out = []
labels = ['dns', 'ip', 'description']
for line in data:
    out.append(dict(zip(labels, line.split('#'))))

那个追加线有点复杂,所以要分解它:

# makes the list ['host2.example.com', '192.168.0.7', 'web server']
line.split('#')  

# takes the labels list and matches them up:
# [('dns', 'host2.example.com'),
#  ('ip', '192.168.0.7'),
#  ('description', 'web server')]
zip(labels, line.split('#'))  

# takes each tuple and makes the first item the key,
#  and the second item the value
dict(...)