seq_in_seq.py

def seq_in_seq(subseq, seq):
    """Return an index of `subseq`uence in the `seq`uence.
    Or `-1` if `subseq` is not a subsequence of the `seq`.

    The time complexity of the algorithm is O(n*m), where
        n, m = len(seq), len(subseq)

    >>> index([1,2], range(5))
    1
    >>> index(range(1, 6), range(5))
    -1
    >>> index(range(5), range(5))
    0
    >>> index([1,2], [0, 1, 0, 1, 2])
    3

    Args:
        subseq (list):
        seq (list):

    Returns:
        int: position of subseq on seq.
    """
    i, n, m = -1, len(seq), len(subseq)
    if m == 0:
        return -1
    try:
        while True:
            i = seq.index(subseq[0], i + 1, n - m + 1)
            if subseq == seq[i:i + m]:
                return i
    except ValueError:
        return -1