问题
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.
回答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.
回答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).
来源：https://stackoverflow.com/questions/10385647/list-match-in-python-get-indices-of-a-sub-list-in-a-larger-list