How to test if a list contains another list as a contiguous subsequence?

Question

How can I test if a list contains another list (ie. it's a contiguous subsequence). Say there was a function called contains:

contains([1,2], [-1, 0, 1, 2]) # Returns [2, 3] (contains returns [start, end]) contains([1,3], [-1, 0, 1, 2]) # Returns False contains([1, 2], [[1, 2], 3]) # Returns False contains([[1, 2]], [[1, 2], 3]) # Returns [0, 0] 

Edit:

contains([2, 1], [-1, 0, 1, 2]) # Returns False contains([-1, 1, 2], [-1, 0, 1, 2]) # Returns False contains([0, 1, 2], [-1, 0, 1, 2]) # Returns [1, 3] 

Answer 1

If all items are unique, you can use sets.

>>> items = set([-1, 0, 1, 2]) >>> set([1, 2]).issubset(items) True >>> set([1, 3]).issubset(items) False 

Answer 2

There's an all() and any() function to do this. To check if big contains ALL elements in small

result = all(elem in big for elem in small) 

To check if small contains ANY elements in big

result = any(elem in big for elem in small) 

the variable result would be boolean (TRUE/FALSE).