Python sum elements in 2d list with the same first

Question

I'm trying to find an efficient way to do the following:

I have this sample:

sample = [['no',2, 6], ['ja',5,7], ['no',4,9], ['ja',10,11], ['ap',7,12]]

and would need

res = [['no', 6, 15], ['ja', 15, 18], ['ap',7,12]]

i.e. sum the corresponding values of the sublists where the first element is the same.

Thanks a lot

My code is:

codes = list(set([element[0] for element in sample]))
res=[]
for code in codes:
    aux=[code]
    res01 = 0
    res02 = 0
    for element in sample:
        if element[0] == code:
            res01 += element[1]
            res02 += element[2]
    aux += [res01, res02]
    res.append(aux) 

Answer 1

Using defaultdict:

>>> from collections import defaultdict

>>> d = defaultdict(lambda: [0,0], list())
>>> for a,b,c in sample: 
        d[a][0]+=b 
        d[a][1]+=c 

#driver values :

IN : sample = [['no',2, 6], ['ja',5,7], ['no',4,9], ['ja',10,11], ['ap',7,12]]

OUT : d = defaultdict(<function <lambda> at 0x7f4349f17620>, 
           {'no': [6, 15], 'ja': [15, 18], 'ap': [7, 12]})

Since the output is structured as such, I would suggest you utilise the dict type for storing your output as future processing with it will be easier.

In case you still want the output as a list, just map the dict, as follows:

>>> [ [key]+ele for key,ele in d.items()]

=> [['no', 6, 15], ['ja', 15, 18], ['ap', 7, 12]]

Answer 2

import pandas as pd
x=pd.DataFrame(sample).groupby(0).agg({1:"sum", 2:"sum"})
d=x.to_dict(orient="split")
#{'columns': [1, 2], 'data': [[7, 12, 'ap'], [15, 18, 'ja'], [6, 15, 'no']],'index': ['ap', 'ja', 'no']}

[d["data"][i]+[d["index"][i]] for i  in range(0, len(d["data"]))]
-----OUTPUT-----------
[[7, 12, 'ap'], [15, 18, 'ja'], [6, 15, 'no']]