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Python - 2D list - find duplicates in one column and sum values in another column

I have a 2D list that contains soccer player names, the number of times they scored a goal, and the number of times they attempted a shot on goal, respectively.

player_stats = [['Adam', 5, 10], ['Kyle', 12, 18], ['Jo', 20, 35], ['Adam', 15, 20], ['Charlie', 31, 58], ['Jo', 6, 14], ['Adam', 10, 15]]

From this list, I'm trying to return another list that shows only one instance of each player with their respective total goals and total attempts on goal, like so:

player_stats_totals = [['Adam', 30, 45], ['Kyle', 12, 18], ['Jo', 26, 49], ['Charlie', 31, 58]]

After searching on Stack Overflow I was able to learn (from this thread) how to return the indexes of the duplicate players

x = [player_stats[i][0] for i in range (len(player_stats))]

for i in range (len(x)):
    if (x[i] in x[:i]) or (x[i] in x[i+1:]): print (x[i], i)

but got stuck on how to proceed thereafter and if indeed this method is strictly relevant for what I need(?)

What's the most efficient way to return the desired list of totals?

like image 870
Mikey Avatar asked Nov 19 '25 19:11

Mikey


1 Answers

Use a dictionary to accumulate the values for a given player:

player_stats = [['Adam', 5, 10], ['Kyle', 12, 18], ['Jo', 20, 35], ['Adam', 15, 20], ['Charlie', 31, 58], ['Jo', 6, 14], ['Adam', 10, 15]]

lookup = {}
for player, first, second in player_stats:
    
    # if the player has not been seen add a new list with 0, 0 
    if player not in lookup:
        lookup[player] = [0, 0]
    
    # get the accumulated total so far 
    first_total, second_total = lookup[player]
    
    # add the current values to the accumulated total, and update the values 
    lookup[player] = [first_total + first, second_total + second]

# create the output in the expected format
res = [[player, first, second] for player, (first, second) in lookup.items()]
print(res)

Output

[['Adam', 30, 45], ['Kyle', 12, 18], ['Jo', 26, 49], ['Charlie', 31, 58]]

A more advanced, and pythonic, version is to use a collections.defaultdict:

from collections import defaultdict

player_stats = [['Adam', 5, 10], ['Kyle', 12, 18], ['Jo', 20, 35],
                ['Adam', 15, 20], ['Charlie', 31, 58], ['Jo', 6, 14], ['Adam', 10, 15]]

lookup = defaultdict(lambda: [0, 0])
for player, first, second in player_stats:
    # get the accumulated total so far
    first_total, second_total = lookup[player]

    # add the current values to the accumulated total, and update the values
    lookup[player] = [first_total + first, second_total + second]

# create the output in the expected format
res = [[player, first, second] for player, (first, second) in lookup.items()]

print(res)

This approach has the advantage of skipping the initialisation. Both has approaches are O(n).

Notes

The expression:

res = [[player, first, second] for player, (first, second) in lookup.items()]

is a list comprehension, equivalent to the following for loop:

res = []
for player, (first, second) in lookup.items():
    res.append([player, first, second])

Additionally, read this for understanding unpacking.

like image 165
Dani Mesejo Avatar answered Nov 21 '25 09:11

Dani Mesejo



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