查找列表中最后一次出现唯一值的索引

Question

Consider the list a = [1,2,3,5,6,1,2,4,5]

output is the index of last occurrence of unique value. idx = [5,6,2,7,8,4]

I have actually tried this out

if len(a) != 0:
    #just keep unique ind_prof(col5) (last occurance=default)
    unique_val = np.unique(a)
    idx = [(len(a) - 1 - a[::-1].index(i)) for i in unique_val]

This seems to be working but just wanted to know if there's a better way to do this

Answer 1

Because you wany index of last occurrence you can reverse list and find index and minus this index from len(a) and find you want like below:

a = [1,2,3,5,6,1,2,4,5]
b = a[::-1]
print(b)

[(len(a) - b.index(i) -1) for i in set(a)] 

Output:

# b
[5, 4, 2, 1, 6, 5, 3, 2, 1]
# last occurrence index
[5, 6, 2, 7, 8, 4]