python中的列表匹配：获取更大列表中的子列表的索引

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<<len(b)<<len(a)) 

indices = get_indices_of_a(a, b)

how to let get_indices_of_a return indices = [0, 2, 0,...] with array(a)[indices] = b ? Is there a faster method than using a.index , which is taking too long?

Making b a set is a fast method of matching lists and returning indices (see compare two lists in python and return indices of matched values ), but it will lose the index of the second 1 as well as the sequence of the indices in this case.

Answer 1

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.

Answer 2

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).