I have a list of lists (string,integer)
eg:
my_list=[["apple",5],["banana",6],["orange",6],["banana",9],["orange",3],["apple",111]]
I'd like to sum the same items and finally get this:
my2_list=[["apple",116],["banana",15],["orange",9]]
You can use itertools.groupby on the sorted list:
from itertools import groupby
my_list=[["apple",5],["banana",6],["orange",6],["banana",9],["orange",3],["apple",111]]
my_list2 = []
for i, g in groupby(sorted(my_list), key=lambda x: x[0]):
my_list2.append([i, sum(v[1] for v in g)])
print(my_list2)
# [['apple', 116], ['banana', 15], ['orange', 9]]
Speaking of SQL Group By and pre-sorting:
The operation of
groupby()is similar to the uniq filter in Unix. It generates a break or new group every time the value of the key function changes (which is why it is usually necessary to have sorted the data using the same key function). That behavior differs from SQL’s GROUP BY which aggregates common elements regardless of their input order.
Emphasis Mine
If you dont want the order to preserved, then plz use the below code.
my_list=[["apple",5],["banana",6],["orange",6],["banana",9],["orange",3],["apple",111]]
my_dict1 = {}
for d in my_list:
if d[0] in my_dict1.keys():
my_dict1[d[0]] += d[1]
else:
my_dict1[d[0]] = d[1]
my_list2 = [[k,v] for (k,v) in my_dict1.items()]