Summing elements of list of lists in Python

Question

I have the following list of lists:

myList = [[8100, 3], [8200, 5], [8400, 8]]

I would like to sum the second element of every sub-list to the next second element of the next list and so on.

Here is how it should look:

myList = [[8100, 3], [8200, 5], [8400, 8]]
myNewList = [[8100, 3], [8200, 8], [8400, 16]]

Because 3+5=8, 8+8=16 and so on. Is there any method to do this? Or am i supposed to write my own function for that, using NumPy for example?

Answer 1

Use np.cumsum given your array is large or you wanna do further manipulations. For smaller ones custom python function would be better.

a = np.array(myList)
a[:, 1] = a[:, 1].cumsum()

Output

array([[8100,    3],
       [8200,    8],
       [8400,   16]])

Answer 2

You can achieve the results in one pass O(n) and without allocating any additional space O(1) , unless you want to, by looping from the first element, progressively updating the items to be summed as follows:

In [1]: A = [[8100, 3], [8200, 5], [8400, 8]]

In [2]: for i in range(1, len(A)):
            A[i][-1] += A[i-1][-1]

In [3]: A
Out[3]: [[8100, 3], [8200, 8], [8400, 16]]

Answer 3

this method should work:

myNewList = []
value = 0
for sublist in myList:
    value = sublist[1]+value
    newSublist = [sublist[0],value]
    myNewList.append(newSublist) 

so simply store the previous value in a temprary variable adn add that to the next sublists second element.

Answer 4

col1, col2 = zip(*myList)
# col1 == (8100, 8200, 8400), col2 == (3, 5, 8)
sums = [ sum(col2[:i]) for i, _ in enumerate(col2, 1) ]
# sums == [3, 8, 16]
myNewList = [ list(x) for x in zip(col1, sums) ]
# myNewList == [[8100, 3], [8200, 8], [8400, 16]]

Answer 5

A very simple way to solve this, is to maintain a total variable and replace the second value of each sub-list with this sum.

myList = [[8100, 3], [8200, 5], [8400, 8]]
total, new_list = 0, []

for sublist in myList:
    total += sublist[1]
    new_list.append([sublist[0],total])

print(new_list)

Output

[[8100, 3], [8200, 8], [8400, 16]]

Answer 6

You can use the new assignment operator (or walrus operator ) to create an accumulator inside a comprehension:

myList = [[8100, 3], [8200, 5], [8400, 8]]

acc=0
new_list=[[sl1, acc := acc + sl2] for sl1, sl2 in myList]
>>> new_list
[[8100, 3], [8200, 8], [8400, 16]]

This is only for Python 3.8+

Answer 7

I made an example for you as clear as possible: you can make the function drier and more efficient, if you think about merging the loops.

L = [[8100, 3], [8200, 5], [8400, 8]]

def newList(L):
    gen = []
    for subL in L:
        if len(gen) > 0:
            gen.append(gen[-1] + subL[1])
        else:
            gen.append(subL[1])

    for idx, subL in enumerate(L):
        subL[1] = gen[idx]

    print(L)

newList(L)

Output:

[[8100, 3], [8200, 8], [8400, 16]]