group data with python

Question

given an 2dList
[[oma,50],[HonFu, 20],[oma,20]]

I need a short hack for beginner to make it

[[oma,70],[HonFu, 20]]

maybe also with numpy or pandas

Answer 1

Based on this answer:

https://stackoverflow.com/a/38648218

https://code.sololearn.com/cCHGrz92f99M/?ref=app

Answer 2

Is this a 'hack'?

https://code.sololearn.com/cRm1573u2GNy/?ref=app

Answer 3

items = [["oma", 50], ["HonFu", 20], ["oma", 20]]
d = {item[0] : 0 for item in items}
for item in items:
d[item[0]] += item[1]
items = [[k, v] for k,v in d.items()]
print(items)

Answer 4

If using dictionary;

lst = [['Oma', 50], ['Honfu', 50], ['Oma', 20]]

dct = {}

for i in lst:
dct[i[0]] = i[1] + dct.get(i[0], 0)

print(*dct.items())

Answer 5

no hack, done with only python module:

from collections import Counter

lst = [['oma',50], ['honfu',20], ['oma',20], ['honfu',17]]
print("input list : ", lst)
res = list(Counter(key for key, num in lst for _ in range(num)).items())
print("aggregated list : ", res)
# output: [('oma', 70), ('honfu', 37)]

Answer 6

Here an other version, just with dict.update

lst = [['oma',50], ['honfu',20], ['oma',20], ['honfu',7]]
dict = {}
for i in lst:
dict.update({i[0]:i[1]+dict.get(i[0])}) if dict.get(i[0]) else dict.update({i[0]:i[1]})
print(dict)

Answer 7

Not sure if that meets the requirements "for beginner" but I had to make a oneliner challenge out of it 😂
https://code.sololearn.com/cjfFvOPoraBh/?ref=app

Answer 8

https://www.sololearn.com/post/1429641/?ref=app

Answer 9

Just another variant with defaultdict

https://code.sololearn.com/cx3dnPKTG7FN/?ref=app

Answer 10

Haw many possible