Sum categories by unique values in list in python

Question

I have this list:

[('2023-03-15', 'paris', 'flight', 4),
('2023-03-21', 'berlin', 'flight', 2),
('2023-03-01', 'madrid', 'drive', 10),
('2023-03-04', 'madrid', 'cycling', 3),
('2023-03-08', 'rome', 'train', 9),
('2023-03-11', 'amsterdam', 'flight', 5),
('2023-03-14', 'london', 'boat', 1)]

How do you reproduce the same list syntax summing similar activities like "flight", taking the latest date as the date for the totals for each activity in the new list? Summing the integers associated.

score 1 · Accepted Answer · answered Mar 20 '23 at 11:26

Code:

lis=[('2023-03-15', 'paris', 'flight', 4),
     ('2023-03-21', 'berlin', 'flight', 2),
     ('2023-03-01', 'madrid', 'drive', 10),
     ('2023-03-04', 'madrid', 'cycling', 3),
     ('2023-03-08', 'rome', 'train', 9),
     ('2023-03-11', 'amsterdam', 'flight', 5),
     ('2023-03-14', 'london', 'boat', 1)]

totals={}
for dt,loc,activ,num in lis:
    if activ in totals:
        totals[activ]['total']+=num
        totals[activ]['latest_date']=max(totals[activ]['latest_date'],dt)
    else:
        totals[activ]={'total': num, 'latest_date': dt}

res=[(totals[activ]['latest_date'], loc, activ, totals[activ]['total']) for dt,loc,activ,num in lis if activ in totals]

print(res)

Output:

[('2023-03-21', 'paris', 'flight', 11), 
 ('2023-03-21', 'berlin', 'flight', 11), 
 ('2023-03-01', 'madrid', 'drive', 10), 
 ('2023-03-04', 'madrid', 'cycling', 3), 
 ('2023-03-08', 'rome', 'train', 9), 
 ('2023-03-21', 'amsterdam', 'flight', 11), 
 ('2023-03-14', 'london', 'boat', 1)]

Thanks guys, Yash Mehta got it right. I should have provided more information on the final output. — Zen4ttitude, Mar 20 '23 at 11:53

score 0 · Answer 2 · answered Mar 20 '23 at 11:22



original_list = [('2023-03-15', 'paris', 'flight', 4),
                 ('2023-03-21', 'berlin', 'flight', 2),
                 ('2023-03-01', 'madrid', 'drive', 10),
                 ('2023-03-04', 'madrid', 'cycling', 3),
                 ('2023-03-08', 'rome', 'train', 9),
                 ('2023-03-11', 'amsterdam', 'flight', 5),
                 ('2023-03-14', 'london', 'boat', 1)]

activity_totals = {}
for date, city, activity, num in original_list:
    if activity not in activity_totals:
        activity_totals[activity] = (date, num)
    else:
        latest_date, total_num = activity_totals[activity]
        if date > latest_date:
            activity_totals[activity] = (date, total_num + num)
        else:
            activity_totals[activity] = (latest_date, total_num + num)

new_list = [(latest_date, city, activity, total_num) for activity, (latest_date, total_num) in activity_totals.items()]

print(new_list)

In all sublists.. the `city` is outputting `london` – Yash Mehta Mar 20 '23 at 11:32 — Yash Mehta, Mar 20 '23 at 11:32

score 0 · Answer 3 · answered Mar 20 '23 at 11:35

You didn't really specify what you were looking for in the output, but I took it as you are trying to condense all the entries of a given key into the entry with the latest date and the sum of the integers.

import itertools

input_list = [('2023-03-15', 'paris', 'flight', 4),
('2023-03-21', 'berlin', 'flight', 2),
('2023-03-01', 'madrid', 'drive', 10),
('2023-03-04', 'madrid', 'cycling', 3),
('2023-03-08', 'rome', 'train', 9),
('2023-03-11', 'amsterdam', 'flight', 5),
('2023-03-14', 'london', 'boat', 1)

key = 'flight'
sorted_list = sorted(input_list, key=lambda x: x[2])
groups = itertools.groupby(sorted_list, key=lambda x: x[2])
for k, g in groups:
    if k == key:
        group_list = list(g)
        sum_value = sum(x[3] for x in group_list)
        selected_value = list(max(group_list, key=lambda x: x[0]))
        selected_value[-1] = sum_value
print(selected_value)

Output

['2023-03-21', 'berlin', 'flight', 11]

Sum categories by unique values in list in python

3 Answers3