拆分字符串然后根据结果再次拆分的 Pythonic 方法？

Question

有没有更pythonic的方法来做到这一点

def parse_address(hostname, addresses):
    netmask=''
    for address in addresses:
        if hostname in address:
            _hostname, _netmask = address.strip().split('/')
            hostname = _hostname.split()[-1]
            netmask = '/' + _netmask.split()[0]
            break

    return netmask

测试用例

如果你做 TDD

def test_parse_netmask(self):
        hostname = '127.0.0.1'

        stdout = [
            "1: lo    inet 127.0.0.1/8 scope host lo\       valid_lft forever preferred_lft forever",
            "3: wlp4s0    inet 192.168.2.133/24 brd 192.168.2.255 scope global dynamic wlp4s0\       valid_lft 58984sec preferred_lft 58984sec",
            "4: docker0    inet 172.17.0.1/16 scope global docker0\       valid_lft forever preferred_lft forever",
            "5: br-a49026d1a341    inet 172.18.0.1/16 scope global br-a49026d1a341\       valid_lft forever preferred_lft forever",
            "6: br-d26f2005f732    inet 172.19.0.1/16 scope global br-d26f2005f732\       valid_lft forever preferred_lft forever",
        ]

        netmask = scanner.parse_address(hostname, stdout)

        self.assertEqual(netmask, '/8')

Answer 1

import re
hostname = '127.0.0.1'
stdout = [
            "1: lo    inet 127.0.0.1/8 scope host lo\       valid_lft forever preferred_lft forever",
            "3: wlp4s0    inet 192.168.2.133/24 brd 192.168.2.255 scope global dynamic wlp4s0\       valid_lft 58984sec preferred_lft 58984sec",
            "4: docker0    inet 172.17.0.1/16 scope global docker0\       valid_lft forever preferred_lft forever",
            "5: br-a49026d1a341    inet 172.18.0.1/16 scope global br-a49026d1a341\       valid_lft forever preferred_lft forever",
            "6: br-d26f2005f732    inet 172.19.0.1/16 scope global br-d26f2005f732\       valid_lft forever preferred_lft forever",
        ]

print [item.split('/')[-1] for item in re.findall(r'(?:\d+\.){3}\d+\/\d+',''.join(stdout)) if hostname  in item]

['8']

Answer 2

def x(hostname,addresses):
    import re 
    for address in addresses:
        result = re.search(hostname+r"/\d", address)
        if result:
            return result.group(0).split(hostname)[1]

不知道是否更多 'Pythonic' 但我会这样做。它对其他人来说是可读的，但它足够短，不会拖累功能。

Answer 3

你能以这样的代码为基础吗？

from urllib.parse import urlparse

parseResult = urlparse('http://www.fake.ca/185')
print ( parseResult )

parseResult 是一个结构，其元素显示在 print 语句的输出中。

ParseResult(scheme='http', netloc='www.fake.ca', path='/185', params='', query='', fragment='')

拆分字符串然后根据结果再次拆分的 Pythonic 方法？

Pythonic way to split string then split again on result?

python

split

idioms

测试用例