How to test if one string is a subsequence of another?

Question

This is a weaker condition than being a substring. For example ‘iran’ is not a substring of ‘ireland’, but it is a subsequence IRelANd. The difference is a subsequence doesn’t have to be contiguous.

More examples:

‘indonesia’ contains ‘india’. INDonesIA
‘romania’ contains ‘oman’. rOMANia
‘malawi’ contains ‘mali’. MALawI

Movitation: My friends like word games. Yesterday we played ‘countries within countries’. I am curious if there are any pairs we missed.

Edit: If you aren’t familiar with the mathematical definition of subsequence

A subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements

Asked By: Colonel Panic

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Source

Answer 1

I did

def is_subsequence(x, y):
    """Test whether x is a subsequence of y"""
    x = list(x)
    for letter in y:
        if x and x[0] == letter:
            x.pop(0)

    return not x

Answered By: Colonel Panic

Answer 2

Just keep on looking for the next character of your potential subsequence, starting behind the last found one. As soon as one of the characters can’t be found in the remainder of the string, it’s no subsequence. If all characters can be found this way, it is:

def is_subsequence(needle, haystack):
    current_pos = 0
    for c in needle:
        current_pos = haystack.find(c, current_pos) + 1
        if current_pos == 0:
            return False
    return True

An advantage over the top answer is that not every character needs to be iterated in Python here, since haystack.find(c, current_pos) is looping in C code. So this approach can perform significantly better in the case that the needle is small and the haystack is large:

>>> needle = "needle"
>>> haystack = "haystack" * 1000
>>> %timeit is_subseq(needle, haystack)
296 µs ± 2.09 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %timeit is_subsequence(needle, haystack)
334 ns ± 1.51 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Answered By: KillianDS

Answer 3

def is_subseq(x, y):
    it = iter(y)
    return all(any(c == ch for c in it) for ch in x)

assert is_subseq('india', 'indonesia')
assert is_subseq('oman', 'romania')
assert is_subseq('mali', 'malawi')
assert not is_subseq('mali', 'banana')
assert not is_subseq('ais', 'indonesia')
assert not is_subseq('ca', 'abc')

Also works for any iterables:

assert is_subseq(['i', 'n', 'd', 'i', 'a'],
                 ['i', 'n', 'd', 'o', 'n', 'e', 's', 'i', 'a'])

UPDATE

Stefan Pochmann suggested this.

def is_subseq(x, y):
    it = iter(y)
    return all(c in it for c in x)

Both versions uses iterators; Iterator yields items that was not yielded in previous iteration.

For example:

>>> it = iter([1,2,3,4])
>>> for x in it:
...     print(x)
...     break
...
1
>>> for x in it:  # `1` is yielded in previous iteration. It's not yielded here.
...     print(x)
...
2
3
4

Answered By: falsetru

Answer 4

def subsequence(seq, subseq):
    seq = seq.lower()
    subseq = subseq.lower()
    for char in subseq:
        try:      
            seq = seq[seq.index(char)+1:]            
        except:
            return False
    return True

Answered By: Josip Grggurica

How to test if one string is a subsequence of another?

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