Running Average using nested arrays and for-loops in Python without numpy

Question

I'm trying to find a way to get a final running average of elements in a nested list using a for loop and I'm getting stuck on how to go about it. I have some sample data below, where the element[0] of each sublist is an id marker to be grouped, element[1] is the value to be summed, and element[2] is acting as a counter

listA = [[u001,4,1],[u002,6,1],[u003,3,1],[u001,12,1],[u002,1,1]
listB = []

for data in listA:
    if data[0] not in listB:
        listB.append(data)
    else: # essentailly if the value is in listB then do:
        listB[1] += data[1] #but this won't work because listB[1] is a list, and data[1] is a value
                  # i need to find a way to append the value data in listB WHERE listB[0] == data[0]
        AND
        listB[2] += 1 #at that specific location

eg pf each for loop pass:

listA:[[u001,4,1],[u002,6,1],[u003,3,1],[u001,12,1],[u002,1,1]
listB:[]
listB:[[u001,4,1]]
listB:[[u001,4,1],[u002,6,1]]
listB:[[u001,4,1],[u002,6,1],[u003,3,1]]
listB:[[u001,16,2],[u002,6,1],[u003,3,1]]
listB:[[u001,16,2],[u002,7,2],[u003,3,1]] #final desired result

python_ged · Accepted Answer · 2021-05-10 08:31:00Z

2

If you are willing to work with a dictionary, you can do it this way:

listA = [['u001',4,1],['u002',6,1],['u003',3,1],['u001',12,1],['u002',1,1]]
listB = {}

for a in listA:
    listB[a[0]] = [listB.get(a[0],[0,0])[0] + a[1],listB.get(a[0],[0,0])[1] + a[2]]

answered May 10, 2021 at 8:31

python_ged

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Aeglas · Accepted Answer · 2021-05-10 08:49:47Z

1

The problem of the solution I'm posting is that has O(n²) time complexity. If you know that your data is not so big it could be good to use.

listA = [['u001',4,1],['u002',6,1],['u003',3,1],['u001',12,1],['u002',1,1]]
listB = []

for data in listA:
    found = False

    for elem in listB:
        if elem[0] == data[0]:
            elem[2] += 1
            elem[1] += data[1]
            found = True
            break

    if not found:
        listB.append(data)

edited May 10, 2021 at 8:49

answered May 10, 2021 at 8:39

Aeglas

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MaPy · Accepted Answer · 2021-05-10 10:13:00Z

1

Another solution using the itertools.groupby:

from operator import itemgetter
from itertools import groupby
listA = [['u001', 4, 1], ['u002', 6, 1],
         ['u003', 3, 1], ['u001', 12, 1],
         ['u002', 1, 1]]
listA.sort(key=itemgetter(0))
listB = []
for k, g in groupby(listA, key=itemgetter(0)):
    suma = 0
    counta = 0
    for x, y, z in g:
        suma += y
        counta += 1
    listB.append([x, suma, counta])

answered May 10, 2021 at 10:13

MaPy

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