时间戳

首页 > 解决方案 > 将 IP 列表转换为相应 IP 范围的列表（python）

python - 将 IP 列表转换为相应 IP 范围的列表（python）

问题描述

我想将 IP 列表转换为相应 IP 范围的列表。例如：

iplist = ['137.226.161.121', '134.130.4.1', '137.226.161.149', '137.226.161.221', '137.226.161.240', '137.226.161.237', '8.8.8.8', '8.8.4.4', '137.226.161.189', '137.226.161.245', '137.226.161.172', '137.226.161.241', '137.226.161.234', '137.226.161.236', '134.130.5.1']

到

ipranges = ['137.226.161.0/24', '134.130.4.0/24', '8.8.8.0/24', '8.8.4.0/24', '134.130.5.0/24']

这样做最有效的方法是什么？我还没有找到提供类似功能的模块。使用此功能的原因是应将一长串 IP（超过 1000 个 ips）转换为子网列表以提高可读性。

谢谢

标签： pythonipsubnet

解决方案

如果我理解正确，您只想根据前 24 位 (/24) 相同的情况进行匹配。对于这些任务，我推荐一个set：

iplist = ['137.226.161.121', '134.130.4.1', '137.226.161.149', '137.226.161.221', '137.226.161.240', '137.226.161.237', '8.8.8.8', '8.8.4.4', '137.226.161.189', '137.226.161.245', '137.226.161.172', '137.226.161.241', '137.226.161.234', '137.226.161.236', '134.130.5.1']

ipset = set()
for i in iplist:
    ipset.add(".".join(i.split(".")[:-1]))

ipranges = [p + ".0/24" for p in ipset]
print(ipranges)

这打印： ['134.130.5.0/24', '8.8.4.0/24', '8.8.8.0/24', '134.130.4.0/24', '137.226.161.0/24']

那么这段代码有什么作用呢？

首先，我们遍历列表，并切断每个 IP 的最后一段：

segments = "8.8.8.8".split(".")  # segments == ["8", "8", "8", "8"]
segments_cut = segments[:-1]     # segments_cut == ["8", "8", "8"]
prefix = ".".join(segments_cut)  # prefix == "8.8.8"

现在我们将这些前缀添加到set. Pythonset只允许唯一元素。这导致：ìpset == {'134.130.5', '8.8.4', '8.8.8', '134.130.4', '137.226.161'}

最后，我们遍历集合并附加后缀“.0/24”来表示子网。

编辑：关于“效率”

我喜欢darkless 的答案，但只知道我的解决方案要快得多（1.2 s vs 0.09 s）：

>>> import timeit
>>> # darkless' ipaddress solution
>>> timeit.timeit("[str(ipaddress.ip_network('{}/24'.format(ip), strict=False)) for ip in iplist]", setup="import ipaddress;iplist = ['137.226.161.121', '134.130.4.1', '137.226.161.149', '137.226.161.221', '137.226.161.240', '137.226.161.237', '8.8.8.8', '8.8.4.4', '137.226.161.189', '137.226.161.245', '137.226.161.172', '137.226.161.241', '137.226.161.234', '137.226.161.236', '134.130.5.1']", number=10000)
1.186...
>>> # My solution
>>> timeit.timeit("[p + '.0/24' for p in {'.'.join(i.split('.')[:-1]) for i in iplist}]", setup="import ipaddress;iplist = ['137.226.161.121', '134.130.4.1', '137.226.161.149', '137.226.161.221', '137.226.161.240', '137.226.161.237', '8.8.8.8', '8.8.4.4', '137.226.161.189', '137.226.161.245', '137.226.161.172', '137.226.161.241', '137.226.161.234', '137.226.161.236', '134.130.5.1']", number=10000)
0.096...

推荐阅读