Addition 2 lists in Python

Question

I have a list :

A = [[33, 0], [34, 0], [35, 5], [36, 0], [38, 0], [39, 0], [40, 0], [41, 0], [42, 0], [43, 0], [44, 0], [45, 1], [46, 0], [47, 0], [48, 0], [49, 0], [50, 0], [51, 1], [52, 0], [56, 0], [57, 0], [58, 0], [59, 0], [62, 1]]

And a second list :

B = [[35, 1], [35, 2], [35, 1], [35, 2], [35, 0], [45, 7], [51, 0], [62, 0]]

My goal is to create the list C like :

if A[0] = B[0] :
  do A[1] + B[1]

The result should be :

C = [[33, 0], [34, 0], [35, 11], [36, 0], [38, 0], [39, 0], [40, 0], [41, 0], [42, 0], [43, 0], [44, 0], [45, 8], [46, 0], [47, 0], [48, 0], [49, 0], [50, 0], [51, 1], [52, 0], [56, 0], [57, 0], [58, 0], [59, 0], [62, 1]]

Hope it's understanding. I don't know how to write it properly in Python.

Could you explain your algorithm for adding the two lists together better? — user
– user, Commented May 15, 2020 at 15:33
The two lists have not equal length. And what should happen if A[0]!=B[0]? — Code Pope
– Code Pope, Commented May 15, 2020 at 15:40
@OmarAflak, it seems he's adding all the B's that have B[0]==35. Some kind of a bucket sum. — Jean-Didier
– Jean-Didier, Commented May 15, 2020 at 15:42

wilkben · Accepted Answer · 2020-05-15 15:40:04Z

1

C=[]
for a in A:
    c = a.copy()
    for b in B:
        if b[0] == a[0]:
            c[1] += b[1]
    C.append(c)

Or, using list comprehensions:

C = [[a[0], a[1] + sum(b[1] for b in B if b[0] == a[0])] for a in A]

answered May 15, 2020 at 15:40

wilkben

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wjandrea · Accepted Answer · 2020-05-15 16:14:32Z

0

Use a Counter to sum B, then add that to A.

from collections import Counter

d = Counter()
for x, count in B:
    d[x] += count

C = [[y, count+d[y]] for y, count in A]

Afterwards:

>>> C
[[33, 0], [34, 0], [35, 11], [36, 0], [38, 0], [39, 0], [40, 0], [41, 0], [42, 0], [43, 0], [44, 0], [45, 8], [46, 0], [47, 0], [48, 0], [49, 0], [50, 0], [51, 1], [52, 0], [56, 0], [57, 0], [58, 0], [59, 0], [62, 1]]
>>> [c for c in C if c[1] != 0]  # The interesting bits
[[35, 11], [45, 8], [51, 1], [62, 1]]
>>> d
Counter({35: 6, 45: 7, 51: 0, 62: 0})

You could also use defaultdict(int) instead, but I think Counter is the more natural fit here. As well, a defaultdict would be mutated just by looking up a non-existent element (e.g. d[33]), which IMO is not ideal.

edited May 15, 2020 at 16:14

answered May 15, 2020 at 15:56

wjandrea

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