Вы можете использовать itertools.groupby
с zip
:
from itertools import groupby
sells = [25, 30, 1, 5, 15, 12]
date = ['July 18', 'July 18', 'August 18', 'September 18', 'September 18', 'September 18']
new_results = groupby(sorted(zip(date, sells), key=lambda x:x[0]), key=lambda x:x[0])
final_data = {a:sum(b for _, b in c) for a, c in new_results}
Выход:
{'August 18': 1, 'July 18': 55, 'September 18': 32}