如何对存储在 Python 字典中的 IP 地址进行排序? [英] How to sort IP addresses stored in dictionary in Python?

问题描述

我有一段如下所示的代码:

I have a piece of code that looks like this:

ipCount = defaultdict(int)

for logLine in logLines:
    date, serverIp, clientIp = logLine.split(" ")
    ipCount[clientIp] += 1

for clientIp, hitCount in sorted(ipCount.items), key=operator.itemgetter(0)):
    print(clientIp)

它有点像 IP，但就像这样:

and it kind of sorts IP's, but like this:

192.168.102.105
192.168.204.111
192.168.99.11

这还不够好，因为它不能识别 99 是小于 102 或 204 的数字.我希望输出是这样的:

which is not good enough since it does not recognize that 99 is a smaller number than 102 or 204. I would like the output to be like this:

192.168.99.11
192.168.102.105
192.168.204.111

我找到了这个，但我不知道如何实现它在我的代码中，或者甚至可能因为我使用字典.我在这里有哪些选择?谢谢..

I found this, but I am not sure how to implement it in my code, or if it is even possible since I use dictionary. What are my options here? Thank you..

推荐答案

您可以使用自定义 key 函数返回字符串的可排序表示:

You can use a custom key function to return a sortable representation of your strings:

def split_ip(ip):
    """Split a IP address given as string into a 4-tuple of integers."""
    return tuple(int(part) for part in ip.split('.'))

def my_key(item):
    return split_ip(item[0])

items = sorted(ipCount.items(), key=my_key)

split_ip() 函数接受一个像 '192.168.102.105' 这样的 IP 地址字符串，并将它变成一个整数元组 (192, 168, 102, 105) .Python 内置支持按字典顺序对元组进行排序.

The split_ip() function takes an IP address string like '192.168.102.105' and turns it into a tuple of integers (192, 168, 102, 105). Python has built-in support to sort tuples lexicographically.

更新:使用 实际上可以更轻松地完成此操作socket 模块中的 inet_aton() 函数:

UPDATE: This can actually be done even easier using the inet_aton() function in the socket module:

import socket
items = sorted(ipCount.items(), key=lambda item: socket.inet_aton(item[0]))

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