Best way to determine if a sequence is in another sequence?

I second the Knuth-Morris-Pratt algorithm. By the way, your problem (and the KMP solution) is exactly recipe 5.13 in Python Cookbook 2nd edition. You can find the related code at http://code.activestate.com/recipes/117214/

It finds all the correct subsequences in a given sequence, and should be used as an iterator:

>>> for s in KnuthMorrisPratt([4,'a',3,5,6], [5,6]): print s
3
>>> for s in KnuthMorrisPratt([4,'a',3,5,6], [5,7]): print s
(nothing)

Here's a brute-force approach O(n*m) (similar to @mcella's answer). It might be faster than the Knuth-Morris-Pratt algorithm implementation in pure Python O(n+m) (see @Gregg Lind answer) for small input sequences.

#!/usr/bin/env python
def index(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
    """
    i, n, m = -1, len(seq), len(subseq)
    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

if __name__ == '__main__':
    import doctest; doctest.testmod()

I wonder how large is the small in this case?

A simple approach: Convert to strings and rely on string matching.

Example using lists of strings:

 >>> f = ["foo", "bar", "baz"]
 >>> g = ["foo", "bar"]
 >>> ff = str(f).strip("[]")
 >>> gg = str(g).strip("[]")
 >>> gg in ff
 True

Example using tuples of strings:

>>> x = ("foo", "bar", "baz")
>>> y = ("bar", "baz")
>>> xx = str(x).strip("()")
>>> yy = str(y).strip("()")
>>> yy in xx
True

Example using lists of numbers:

>>> f = [1 , 2, 3, 4, 5, 6, 7]
>>> g = [4, 5, 6]
>>> ff = str(f).strip("[]")
>>> gg = str(g).strip("[]")
>>> gg in ff
True