Sum list of list elements in python like sql group by [duplicate]

Answer 1

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()]

Answer 2

Using Pandas and groupby:

import pandas as pd

>>> pd.DataFrame(my_list, columns=['fruit', 'count']).groupby('fruit').sum()
        count
fruit        
apple     116
banana     15
orange      9

Answer 3

from itertools import groupby
[[k, sum(v for _, v in g)] for k, g in groupby(sorted(my_list), key = lambda x: x[0])]

# [['apple', 116], ['banana', 15], ['orange', 9]]

Answer 4

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

Answer 5

from collections import defaultdict

my_list= [["apple",5],["banana",6],["orange",6],["banana",9],["orange",3],["apple",111]]

result = defaultdict(int)
for fruit, value in my_list:
    result[fruit] += value
result = result.items()

print result

Or you can keep result as dictionary