v = [1,2,3,4,3,1,2]
any([2,3] == v[i:i+2] for i in xrange(len(v) - 1))

虽然@PaoloCapriotti的版本可以解决问题，但此方法更快，因为一旦找到匹配项，它将停止解析v.

While @PaoloCapriotti's version does the trick, this one is faster, because it stops parsing the v as soon as a match is found.