CODEWITHSUNDEEP
pythonsequencemembership
user2870222
user2870222

Reputation: 269

python list element sequence membership

Let's say we have a reference sequence which is a list or tuple:

reference = ['a', 'b', 'c', 'd']

and I want to check if some sequence does exist inside the reference sequence:

target1 = ['a', 'b', 'c']  # true
target2 = ['a', 'c', 'b']  # false
target3 = ['a', 'c', 'd']  # false
target4 = ['c', 'd', 'e']  # false
target5 = ['c', 'd']       # true

Is there any built-in function to check this kind of sequence membership? Thank you!

Upvotes: 0

Views: 408

Answers (2)

user2870222
user2870222

Reputation: 269

I found a relevant module difflib.SequenceMatcher.

def sequence_membership(target, reference):
    get_eq = lambda x: [c for c in x if c[0]=='equal']
    get_length = lambda x: x[2]-x[1]
    s = SequenceMatcher(None, target, reference)
    match_result = s.get_opcodes()
    overlapped = get_eq(match_result)
    if len(overlapped) ==1 and get_length(overlapped[0]) == len(target):
        return True
    else: return False

>>> sequence_membership(target1,reference)
True
>>> sequence_membership(target2,reference)
False
>>> sequence_membership(target3,reference)
False
>>> sequence_membership(target4,reference)
False
>>> sequence_membership(target5,reference)
True

Upvotes: 1

ichbinblau
ichbinblau

Reputation: 4809

On the first thought, you can convert the list or tuple to string and find if the substring exists in the reference. A simple code snippet to explain my thought:

reference = ['a', 'b', 'c', 'd']
target1 = ['a', 'b', 'c']  # true

list2str = lambda ref: "".join(ref)

base = list2str(reference)
t1 = list2str(target1)

if t1 in base:
    print 'found'
else:
    print 'unfound'

Upvotes: 0

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