在可能出现多次的关键字上拆分列表

Question

我读过那些看似相似的例子，但我并不是在那个层次去理解答案。我希望获得列表输出，并将每个接口写成一个单独的行(aka list I write to a csv)。我需要分割关键字‘接口Vlan*’上的初始返回列表

我希望将关键字接口vlan*上返回的列表vlanlist拆分为单独的列表。

from ciscoconfparse import CiscoConfParse
import os

for filename in os.listdir():
    if filename.endswith(".cfg"):
        p = CiscoConfParse(filename)
        vlanlist=(p.find_all_children('^interface Vlan'))
        vlanlist.insert(0,filename)

        print(vlanlist) 

这是一个输出线。我需要将关键字"interface vlanxxx"上的列表拆分成不同的行

[ 'interface Vlan1', ' no ip address', ' shutdown', 'interface Vlan2003', ' description XXXXXX', ' ip address 10.224.6.130 255.255.255.224', ' no ip redirects', ' no ip unreachables', ' no ip proxy-arp', ' load-interval 60', ' arp timeout 420']

所需的输出(这可能有2-20个不同的接口，我想根据配置文件来分割)

['interface Vlan1' ' no ip address', ' shutdown']
['interface Vlan2003', ' description XXXXXX', ' ip address 10.224.6.130 255.255.255.224', ' no ip redirects', ' no ip unreachables', ' no ip proxy-arp', ' load-interval 60', ' arp timeout 420']

Answer 1

在追加文件名之前，可以进一步分隔返回的vlanlist：

# First, find the index in the list where "interface Vlan" exists:
# Also, append None at the end to signify index for end of list
indices = [i for i, v in enumerate(l) if v.startswith('interface Vlan')] + [None]

# [0, 3, None]

# Then, create the list of lists based on the extracted indices and prepend with filename
newlist = [[filename] + vlanlist[indices[i]:indices[i+1]] for i in range(len(indices)-1)]

for l in newlist: print(l)

# ['test.cfg', 'interface Vlan1', ' no ip address', ' shutdown']
# ['test.cfg', 'interface Vlan2003', ' description XXXXXX', ' ip address 10.224.6.130 255.255.255.224', ' no ip redirects', ' no ip unreachables', ' no ip proxy-arp', ' load-interval 60', ' arp timeout 420']

对第二个清单理解的解释：

newlist = [
    [filename] +                   # prepend single-item list of filename
    vlanlist[                      # slice vlanlist
        indices[i]:                # starting at the current index
        indices[i+1]               # up to the next index
    ] 
    for i in range(len(indices)-1) # iterate up to the second last index so i+1 doesn't become IndexError
]

如果您不喜欢索引方法，可以尝试zip：

lists = [[filename] + vlanlist[start:end] for start, end in zip(indices[:-1], indices[1:])]

Answer 2

这里有一个与您的单个测试用例高度耦合的解决方案。如果完整的数据集不能代表您的单个测试用例，那么您将不得不通过更多的测试来改进它。

def extract(items):
  result, filename, idx = [], items[0], -1

  for x in items[1:]:
    if x.startswith('interface Vlan'):
      idx += 1
      result.append([filename])
    result[idx].append(x)

  return result

# given & expected are your example and output 
assert expected == extract(given)

编辑：

..。你已经改变了输入和输出。

def extract(items):
  result, idx = [], -1

  for x in items:
    if x.startswith('interface Vlan'):
      idx += 1
      result.append([])

    if not result: continue  # assuming possible unwanted items before 'interface Vlan'
    result[idx].append(x)

  return result

assert expected == extract(given)

Answer 3

一种快速而直接的解决方案。检查列表中的interface Vlan项，如果是的话，它会创建一个新列表，否则会追加到旧列表和一些.strip()上，以获得良好的度量。

output = ['interface Vlan1', ' no ip address', ' shutdown', 'interface Vlan2003', ' description XXXXXX', ' ip address 10.224.6.130 255.255.255.224', ' no ip redirects', ' no ip unreachables', ' no ip proxy-arp', ' load-interval 60', ' arp timeout 420']

results = []

for i in output:
    if 'interface Vlan' in i:
        results.append([i.strip()])
    else:
        results[-1].append(i.strip())

>> results
 [['interface Vlan1', 'no ip address', 'shutdown'],
 ['interface Vlan2003',
  'description XXXXXX',
  'ip address 10.224.6.130 255.255.255.224',
  'no ip redirects',
  'no ip unreachables',
  'no ip proxy-arp',
  'load-interval 60',
  'arp timeout 420']]