Python - Compare two lists for matching items where item contains not equals?

Question

Is it possible to search for matches in two lists where a list item contains a matching string, does not equal?

for example:

list_a = [
    'ip prefix-list PL_ABBA seq 5 permit 10.10.10.0/24',
    'ip prefix-list PL_ABBA seq 10 permit 10.20.10.0/24',
    ]

list_b = [
    '10.10.10.0/24',
    '10.20.10.0/24',
    '10.30.10.0/24',
    '10.40.10.0/24',
    ]

10.30.10.0/24 and 10.40.10.0/24 are missing from list_a so I want to return these two as missing?

I could regex out the subnets from list_a to make new_list_a then compare using set? but was wondering if there was an easier method?

Thanks

Already answered in https://stackoverflow.com/questions/3462143/get-difference-between-two-lists — Serge, Nov 01 '18 at 17:02
`filter(lambda x: x not in [x.split(' ')[-1] for x in list_a], list_b)` — Nir Alfasi, Nov 01 '18 at 17:05

score 5 · Accepted Answer · answered Nov 01 '18 at 17:01

5

You can use

>>> IPs = {s.rsplit(' ')[-1] for s in list_a}
>>> IPs
>>> {'10.10.10.0/24', '10.20.10.0/24'}
>>> [ip for ip in list_b if ip not in IPs]
>>> ['10.30.10.0/24', '10.40.10.0/24']

IPs is a set (for the O(1) membership test) of the ip addresses, where s.rsplit(' ')[-1] is used to get the IP part of the elements in list_a.

answered Nov 01 '18 at 17:01

timgeb

76,762
20
123
145

@AlexW we only have to iterate once over `list_a` for splitting off the IPs. After that, all `in` checks on `IPs` are O(1). In other words, this code has complexity O(n + m) where n is the length of `list_a` and `m` is the length of `list_b` - thus linear. – timgeb Nov 01 '18 at 17:10

score 0 · Answer 2 · answered Nov 01 '18 at 17:10

0

Try below example.

result = []

for item in list_b:
    for st in list_a:
        if item in st:
            break
    else:
        result.append(item)
print(result)

Output:

['10.30.10.0/24', '10.40.10.0/24']

answered Nov 01 '18 at 17:10

Chandella07

2,089
14
22

Python - Compare two lists for matching items where item contains not equals?

2 Answers2