將 IP 列表轉換為相應的 IP 范圍列表（python）

Question

我想將 IP 列表轉換為相應 IP 范圍的列表。 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']

至

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']

這樣做最有效的方法是什么？ 我還沒有找到提供這樣的 function 的模塊。 這個 function 的原因是應該將一長串 IP（超過 1000 個 ip）轉換為子網列表以提高可讀性。

謝謝

Answer 1

如果我理解正確，您只想根據前 24 位 (/24) 相同的情況進行匹配。 對於這些任務，我推薦一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)

這打印： ['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']

那么這段代碼有什么作用呢？

首先，我們遍歷列表並截斷每個 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"

現在我們將這些前綴添加到set中。 Python set僅允許唯一元素。 這導致： ìpset == {'134.130.5', '8.8.4', '8.8.8', '134.130.4', '137.226.161'}

最后我們遍歷集合和 append 后綴“.0/24”來表示子網。

編輯：關於“效率”

我喜歡darkless 的答案，但只知道我的解決方案要快得多（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...

Answer 2

正如 Hampus Larsson 提到的，您可以使用 python ipaddress 模塊：

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']