Python find indexwise maximum average minimum for list of 100 sublists

Question

I have a list containing 100+ sublists. Need to get index wise maximum for this scenario using python.

L1= [3,2,7]
L2= [1,5,7]
L3= [10,3,6]

1. Get max value from index 0 of all lists

2. Get min value from index 1 of all lists

3. Get avg value from index 2 of all lists

Result list is [10,2,6.666]

I don't have further code, there are at least 100 such lists for which this calculation is required

Answer 1

I can share with you one approach

let us say there are only 2 lists.. you should try to extend this logic to 100 lists

L1: [2,8, 1]
L2: [3, 1, 6]

so the answer will be

[3,8,6]

to solve this make a list which will store the answer

ans = [0,0,0]

now iterate over every list and just replace every index with the largest value you see at that index

after first list ans will look like

[2,8,1]

coz all values are greater than zero

when you iterate over 2nd list, 3 is greater than 2 and 6 is greater than 1 so you just update them

and your answer will become

[3,8,6]

Answer 2

Pretty simple: https://numpy.org/

Transform your list of lists "LoL" into an array...

import numpy as np
LoL = np.asarray([L1, L2, L3])

...and then apply those aggregations...

Check the docs for details. Eg, "max": https://numpy.org/doc/stable/reference/generated/numpy.ndarray.max.html#numpy.ndarray.max

Answer 3

using zip():

Code:

L1= [3,2,7]
L2= [1,5,7]
L3= [10,3,6]
counter=0
res=[]
for lis in zip(L1,L2,L3):  #It brings same index element together in the form of tuples  [i.e (3, 1, 10),(2, 5, 3),(7, 7, 6)]
    if counter==0:
        res.append(max(lis))
        counter+=1
    elif counter==1:
        res.append(min(lis))
        counter+=1
    else:
        res.append(sum(lis)/len(lis)) #mean=sum_of_all_element/total_number_of_element
print(res) #Output [10, 2, 6.666666666666667]