Question

我在将文件内容转换为词典列表方面遇到了麻烦，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格式相同。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'} ]

提前谢谢！

Answer 1

首先，您要分割#上的每一行。然后，您可以使用zip将它们与标签一起压缩，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('#'))))

这一个追加线有点复杂，所以要将其分解：

# 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(...)

Answer 2

rows = []
for line in input_file:
    r = line.split('#')
    rows.append({'dns':r[0],'ip':r[1],'description':r[2]})

Answer 3

假设您的文件为infile.txt

>>> entries = (line.strip().split("#") for line in open("infile.txt", "r"))
>>> output = [dict(zip(("dns", "ip", "description"), e)) for e in entries]
>>> print output
[{'ip': '192.168.0.1', 'description': 'web server', 'dns': 'host1.example.com'}, {'ip': '192.168.0.5', 'description': 'dns server', 'dns': 'host2.example.com'}, {'ip': '192.168.0.7', 'description': 'web server', 'dns': 'host3.example.com'}, {'ip': '192.168.0.9', 'description': 'application server', 'dns': 'host4.example.com'}, {'ip': '192.168.0.10', 'description': 'database server', 'dns': 'host5.example.com'}]

Answer 4

>>> map(lambda x : dict(zip(("dns", "ip", "description"), tuple(x.strip().split('#')))), open('input_file'))

python：读取和拆分文件到词典列表

4 个答案: