将IP列表转换为相应IP范围的列表（python）

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

最有效的方法是什么？我还没有找到一个模块提供这样的功能。此功能的原因是，应将一长串IP（超过1000个IP）转换为子网列表，以提高可读性

---

谢谢

如果我理解正确，您只希望根据前24位（/24）相同进行匹配。对于这些任务，我建议设置

：

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)

这张照片是：
['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']

那么这段代码是做什么的呢？
首先，我们遍历该列表并切断每个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
集合
。Python
set
只允许唯一的元素。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”以表示子网

编辑：关于“效率”
我喜欢，但只知道我的解决方案要快得多（1.2秒vs 0.09秒）：

正如Hampus Larsson提到的，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']


是所有的IP作用域都是/24，还是也有子网掩码？不管怎样，我会查看模块。这就是我要找的！谢谢，听到这个消息我很高兴！关于效率的好发现；）对于/24范围（或/8/16/32），您的解决方案确实更好。
>>> 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...

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']