更新字典而不覆盖现有的键值对？

Question

我有一个如下的输入文件，我可以从中构建我的字典

一般格式

<IP_1>
KEY_1=VALUE_1
KEY_2=VALUE_2

<IP_2>
KEY_1=VALUE_1
KEY_2=VALUE_2

例子

192.168.1.1
USER_NAME=admin
PASSWORD=admin123

192.168.1.2
USER_NAME=user
PASSWORD=user123

预期的字典应该是这样的：

>>print dictionary_of_ip
{'192.168.1.1':{'USER_NAME'='admin','PASSWORD'='admin123'},
 '192.168.1.2':{'USER_NAME'='user','PASSWORD'='user123'}}

本质上是字典中的字典

下面是我的代码：

def generate_key_value_pair(filePath, sep='='):
    dict_of_ip = {}
    slave_properties = {}
    with open(filePath, "rt") as f:
        for line in f:
            stripped_line = line.strip()
            if stripped_line and stripped_line[0].isdigit():
                #print 'Found Ip'
                ip = stripped_line
                dict_of_ip[ip] = ''
            elif stripped_line and stripped_line[0].isupper():
                #print "Found attributes")
                key_value = stripped_line.split(sep)
                key = key_value[0].strip()
                value = key_value[1].strip()
                slave_properties[key] = value
                dict_of_ip[ip] = slave_properties

    return dict_of_ip

我能够按预期获取第一个 IP 及其属性，但是来自第二个 IP 的第二组值覆盖了第一个。

>>print dict_of_ip
{'192.168.1.1': {'USER_NAME': 'user', 'PASSWORD': 'user123'},
 '192.168.1.2': {'USER_NAME': 'user', 'PASSWORD': 'user123'}}

dict_of_ip[ip] = slave_properties 导致覆盖。如何防止 '192.168.1.2' 键中的值覆盖第一个？

Answer 1

您可以选择正则表达式方式与 dict 理解相结合：

import re

string = """
192.168.1.1
USER_NAME=admin
PASSWORD=admin123

192.168.1.2
USER_NAME=user
PASSWORD=user123
"""

regex = re.compile(r"""
    ^
    (?P<ip>\d+\.\d+\.\d+\.\d+)[\n\r]
    USER_NAME=(?P<user>.+)[\r\n]
    PASSWORD=(?P<password>.+)
    """, re.MULTILINE | re.VERBOSE)

users = {match.group('ip'):
            {'USER_NAME': match.group('user'), 
            'PASSWORD': match.group('password')}
            for match in regex.finditer(string)}

print(users)
# {'192.168.1.2': {'USER_NAME': 'user', 'PASSWORD': 'user123'}, '192.168.1.1': {'USER_NAME': 'admin', 'PASSWORD': 'admin123'}}

看到它working on ideone.com。这是对应的demo on regex101.com。

Answer 2

使用切片和 dictcomp 的紧凑解决方案：

with open("data.txt", "r") as f:
    f = [i.strip() for i in f.readlines()]
    ipdict = {ip: {'USER_NAME': user[10:], 'PASSWORD': password[9:]} 
              for ip, user, password in zip(f[0::4], f[1::4], f[2::4])}

Answer 3

只需在识别 IP 地址后将 slave_properties 的初始化移动到右侧即可。这样一来，遇到的每个问题都会开始为空（我还删除了您在dict_of_ip[ip] 那里进行的不必要的初始化）。

from pprint import pprint

def generate_key_value_pair(filePath, sep='='):
    dict_of_ip = {}

    with open(filePath, "rt") as f:
        for line in f:
            line = line.strip()
            if line and line[0].isdigit():  # ip?
                slave_properties = {}  # initialize
                ip = line
            elif line and line[0].isupper():
                key_value = line.split(sep)
                key = key_value[0].strip()
                value = key_value[1].strip()
                slave_properties[key] = value
                dict_of_ip[ip] = slave_properties

    return dict_of_ip

result = generate_key_value_pair('ips.txt')
pprint(result)

输出：

{'192.168.1.1': {'PASSWORD': 'admin123', 'USER_NAME': 'admin'},
 '192.168.1.2': {'PASSWORD': 'user123', 'USER_NAME': 'user'}}

Answer 4

如果您使用 Python 的高性能数据类型，这将变得更容易和更高效。例如，这里是与defaultdict 相同的代码，而不是创建您自己的集合类型。

from collections import defaultdict

dict_of_ip = defaultdict(dict)   # This creates a dictionary of dictionaries for you.

ip = None  
for line in f:
    stripped_line = line.strip() 
    if stripped_line and stripped_line[0].isdigit():
        ip = stripped_line  # The value for the ip only changes if a new 
                            # IP is detected. Otherwise the algorithm proceeds
                            # with the older IP address to the nested dict. 

    elif stripped_line and stripped_line[0].isupper():
        key_value = stripped_line.split(sep)  
        key = key_value[0].strip()
        value = key_value[1].strip()
        dict_of_ip[ip][key] = value # IP set in the earlier if-loop. 

更具体地说，您收到错误的原因是您正在为每个子词典编辑相同的 slave_properties 词典。因此，一个人的变化会传播到另一个人。

Answer 5

你没有在循环中做slave_properties = {}，所以当你认为你正在创建一个新的dict对象时，你指的是同一个dict对象

slave_properties[key] = value
dict_of_ip[ip] = slave_properties

Answer 6

试试这个：

def generate_key_value_pair(filePath, sep='='):
    dict_of_ip = {}
    with open(filePath, "rt") as f:
        for line in f:
            stripped_line = line.strip()
            if stripped_line and stripped_line[0].isdigit():
                #print 'Found Ip'
                slave_properties = {}
                ip = stripped_line
                dict_of_ip[ip] = ''
            elif stripped_line and stripped_line[0].isupper():
                #print "Found attributes")
                key_value = stripped_line.split(sep)
                key = key_value[0].strip()
                value = key_value[1].strip()
                slave_properties[key] = value
                dict_of_ip[ip] = slave_properties

    return dict_of_ip

您使用相同的（修改后的）dict。我没有故意更改您的代码逻辑，只是将slave_properties = {} 更改为应该在的位置

您甚至可以删除您的 slave_properties 并单独使用 dict

def generate_key_value_pair(filePath, sep='='):
    dict_of_ip = {}
    with open(filePath, "rt") as f:
        for line in f:
            stripped_line = line.strip()
            if stripped_line and stripped_line[0].isdigit():
                #print 'Found Ip'
                ip = stripped_line
                dict_of_ip[ip] = {}
            elif stripped_line and stripped_line[0].isupper():
                #print "Found attributes")
                key_value = stripped_line.split(sep)
                key = key_value[0].strip()
                value = key_value[1].strip()
                dict_of_ip[ip][key] = value

    return dict_of_ip