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

Answer 1

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

Answer 2

I was trying to understand falsetru's solution, and essentially using iter simplifies what the following would be doing:

def is_subseq(sub, string):
    i = 0
    for c in string:
        if i == len(sub):
            break
        if sub[i] == c:
            i += 1
    if i != len(sub):
        return False
    return True


def is_subseq2(sub, string):
    j = 0
    count = 0
    for c in sub:
        if j == len(string):
            return False
        while j < len(string) and string[j] != c:
            j += 1
        if string[j] == c:
            count += 1
    return count == len(sub)

Answer 3

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

Answer 4

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

Answer 5

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