list match in python: get indices of a sub-list in a larger list

Question

For two lists,

a = [1, 2, 9, 3, 8, ...]   (no duplicate values in a, but a is very big)
b = [1, 9, 1,...]          (set(b) is a subset of set(a), 1<

        

Answer 1

Presuming we are working with smaller lists, this is as easy as:

>>> a = [1, 2, 9, 3, 8] 
>>> b = [1, 9, 1] 
>>> [a.index(item) for item in b]
[0, 2, 0]

On larger lists, this will become quite expensive.

(If there are duplicates, the first occurrence will always be the one referenced in the resulting list, if not set(b) <= set(a), you will get a ValueError).

Answer 2

A fast method (when a is a large list) would be using a dict to map values in a to indices:

>>> index_dict = dict((value, idx) for idx,value in enumerate(a))
>>> [index_dict[x] for x in b]
[0, 2, 0]

This will take linear time in the average case, compared to using a.index which would take quadratic time.