How to get the sums of unique values with respective occurrence count from a list in Python?

Question

With a list like this:

[1,1,1,2,2,3,3,3]

I would like to get the sums of each unique number [1,2,3] which is [3, 4, 9]. Using an approach from the related post How to get unique values with respective occurrence count from a list in Python? I'm able to get the count of the occurence of each uniqe number using:

L = [1,1,1,2,2,3,3,3]
uniq, counts = np.unique(L, return_counts=True)
counts

Which gives the output [3, 2, 3]. With this, I'm able to get what I'm looking for using a cumbersome approach with an enumerated For Loop and some rather cryptic conditions:

L = [1,1,1,2,2,3,3,3]
elements = [3,2,3]
    
sums = []
index = 0
for i, e in enumerate(elements):
    if i == 0:
        sums.append(sum(L[0:e]))
        index = index + e
    else:
        sums.append(sum(L[index:index + e]))
        index = index + e
print(sums)

Which gives the desired output [3, 4, 9]. Does anyone know if it's possible to do the same thing a bit more elegantly?

Answer 1

Since you are using Numpy, you can just multiply the results you already have from np.unique:

import numpy as np

L = [1,1,1,2,2,3,3,3]    
uniq, counts = np.unique(L, return_counts=True)

uniq * counts
# array([3, 4, 9])