【问题标题】：Convert list of IPs to list of corresponding IP Ranges (python)将 IP 列表转换为相应 IP 范围的列表（python）
【发布时间】：2020-01-21 11:49:06
【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']

最有效的方法是什么？我还没有找到提供类似功能的模块。这个函数的原因是为了提高可读性，应该将一长串 IP（超过 1000 个 ip）转换为子网列表。

谢谢

【问题讨论】：

是所有 IP 范围 /24，还是还有子网掩码？
不管怎样，我会看看Built-in IP-Address 模块。

标签： python ip subnet

【解决方案1】：

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，截断每个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。 Python set 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answer by darkless，但只知道我的解决方案明显更快（1.2 秒对 0.09 秒）：

>>> 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...

【讨论】：

这就是我想要的！谢谢
我很高兴听到这个消息！
关于效率的好消息；）对于 /24 范围（或 /8 /16 /32）确实可以更好地解决您的问题。

【解决方案2】：

正如 Hampus Larsson 所说，您可以使用 python ipaddress 模块：

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']

ipranges = [str(ipaddress.ip_network('{}/24'.format(ip), strict=False)) for ip in iplist]

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

【讨论】：