我编造了以下内容：([0-9a-fA-F]:?){12}以匹配文本中的MAC地址。
下面是它的工作原理：

• [0-9a-fA-F]匹配用于表示十六进制数字的字符
• :?与可选冒号匹配
• (...){12}-然后将所有这些内容分组并重复12次。12因为MAC地址由6对十六进制数字组成，中间用冒号分隔

您可以在here的操作中看到它。
Python代码将变为：

import re
p = re.compile(r'(?:[0-9a-fA-F]:?){12}')
test_str = u"TEXT WITH SOME MAC ADDRESSES 00:24:17:b1:cc:cc TEXT CONTINUES WITH SOME MORE TEXT 20:89:86:9a:86:24"

re.findall(p, test_str)

结果产生：

[u'00:24:17:b1:cc:cc', u'20:89:86:9a:86:24']

([0-9a-f]{2}(?::[0-9a-f]{2}){5})

试试这个。看演示。
http://regex101.com/r/kP8uF5/5

import re
p = re.compile(ur'([0-9a-f]{2}(?::[0-9a-f]{2}){5})', re.IGNORECASE)
test_str = u"TEXT WITH SOME MAC ADDRESSES 00:24:17:b1:cc:cc TEXT CONTINUES WITH SOME MORE TEXT 20:89:86:9a:86:24"

re.findall(p, test_str)

我还必须匹配MAC地址，这是有效的：((?:[\da-fA-F]{2}[:\-]){5}[\da-fA-F]{2})
我用这个regex测试器测试了它：https://regex101.com/#python它对每个正则表达式做了很好的分类。

text = "this is aa:bb:cc:dd:01:02 test for aa-bb-cc-dd-ee-ff and AABBCCDDEEFF is a mac address without separator"

让我们提取MAC地址

def extract_mac_address(text):
    pattern = '(([0-9a-fA-F]{2}[:]){5}([0-9a-fA-F]{2})|([0-9a-fA-F]{2}[-]){5}([0-9a-fA-F]{2})|[0-9a-fA-F]{12})'
    mac_addr_list = re.findall(pattern, text)
    return list(map(lambda x: x[0], mac_addr_list))

print(extract_mac_address(text))

输出为['aa:bb:cc:dd:01:02', 'aa-bb-cc-dd-ee-ff', 'AABBCCDDEEFF']

单个MAC地址的最佳正则表达式匹配，且结尾不溢出：

import re

regex = r"^((([a-f0-9]{2}:){5})|(([a-f0-9]{2}-){5}))[a-f0-9]{2}$"

test_str = "89:89:89:89:89:89"

subst = ""

# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, subst, test_str, 1, re.IGNORECASE)

if result:
    print (result)

参考：https://regexpattern.com/mac-address/

import re
print(re.search("([a-f0-9A-F]{4}[.]){2}[a-f0-9A-F]{4}", "0000.aaaa.bbbb").group())

[a-f0-9A-F]{4} -> matches for four occurrences of a-f or 0-9, A-F 

here it will search only for one mac in a string.if you want to search for more than one occurance we need to use re.findall.

从任何输入中查找有效MAC地址的最佳方法是：
r'([0-9A-Fa-f]{2}[:]){5}([0-9A-Fa-f]{2})[\s]'