Finding subsequence (nonconsecutive)

Question

If I have string needle and I want to check if it exists contiguously as a substring in haystack, I can use:

if needle in haystack:
    ...

What can I use in the case of a non-continuous subsequence? Example:

>>> haystack = "abcde12345"
>>> needle1 = "ace13"
>>> needle2 = "123abc"
>>> is_subsequence(needle1, haystack)
True
>>> is_subsequence(needle2, haystack)  # order is important!
False

Asked By: user4847061

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Source

Answer 1

I don’t know if there’s builtin function, but it is rather simple to do manually

def exists(a, b):
    """checks if b exists in a as a subsequence"""
    pos = 0
    for ch in a:
        if pos < len(b) and ch == b[pos]:
            pos += 1
    return pos == len(b)

>>> exists("moo", "mo")
True
>>> exists("moo", "oo")
True
>>> exists("moo", "ooo")
False
>>> exists("haystack", "hack")
True
>>> exists("haystack", "hach")
False
>>>

Answered By: Ishamael

Answer 2

Another possibility: You can create iterators for both, needle and haystack, and then pop elements from the haystack-iterator until either all the characters in the needle are found, or the iterator is exhausted.

def is_in(needle, haystack):
    try:
        iterator = iter(haystack)
        for char in needle:
            while next(iterator) != char:
                pass
        return True
    except StopIteration:
        return False

Answered By: tobias_k

Answer 3

Using an iterator trick:

it = iter(haystack)
all(x in it for x in needle)

This is only a concise version of the same idea presented in tobias_k’s answer.

Answered By: wim

Answer 4

We can try simple for loop and break method and pass on substring once the match is found

def substr(lstr,sstr):
lenl = len(lstr)
for i in sstr:
    for j in range(lenl):
        if i not in lstr:
            return False
        elif i == lstr[j]:
            lstr = lstr[j+1:]
            break
        else:
            pass
return True

Answered By: Shri

Finding subsequence (nonconsecutive)

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