How to sum a column grouped by other columns in a list?

Question

I have a list as follows.

[['Andrew', '1', '9'], ['Peter', '1', '10'], ['Andrew', '1', '8'], ['Peter', '1', '11'], ['Sam', '4', '9'], ['Andrew', '2', '2']]

I would like sum up the last column grouped by the other columns.The result is like this

[['Andrew', '1', '17'], ['Peter', '1', '21'], ['Sam', '4', '9'], ['Andrew', '2', '2']]

which is still a list.

In real practice, I would always like to sum up the last column grouped by many other columns. Is there a way I can do this in Python? Much appreciated.

Answer 1

dynamically grouping by all columns except the last one:

In [24]: df = pd.DataFrame(data)

In [25]: df.groupby(df.columns[:-1].tolist(), as_index=False).agg(lambda x: x.astype(int).sum()).values.tolist()
Out[25]: [['Andrew', '1', 17], ['Andrew', '2', 2], ['Peter', '1', 21], ['Sam', '4', 9]]

Answer 2

This is an O(n) solution via collections.defaultdict, adaptable to any number of keys.

If your desired output is a list, then this may be preferable to a solution via Pandas, which requires conversion to and from a non-standard type.

from collections import defaultdict

lst = [['Andrew', '1', '9'], ['Peter', '1', '10'], ['Andrew', '1', '8'],
       ['Peter', '1', '11'], ['Sam', '4', '9'], ['Andrew', '2', '2']]

d = defaultdict(int)

for *keys, val in lst:
    d[tuple(keys)] += int(val)

res = [[*k, v] for k, v in sorted(d.items())]

Result

[['Andrew', '1', 17], ['Andrew', '2', 2], ['Peter', '1', 21], ['Sam', '4', 9]]

Explanation

• Cycle through your list of lists, define keys / value and add to your defaultdict of lists.
• Use a list comprehension to convert dictionary to desired output.

Answer 3

Op1

You can pass a index sum and add tolist convert back to list

pd.DataFrame(L).\
   set_index([0,1])[2].astype(int).sum(level=[0,1]).\
        reset_index().values.tolist()
Out[78]: [['Andrew', '1', 17], ['Peter', '1', 21], ['Sam', '4', 9], ['Andrew', '2', 2]]

---

Op2

For list of list you can using groupby from itertools

from itertools import groupby
[k+[sum(int(v) for _,_, v in g)] for k, g in groupby(sorted(l), key = lambda x: [x[0],x[1]])]
Out[98]: [['Andrew', '1', 17], ['Andrew', '2', 2], ['Peter', '1', 21], ['Sam', '4', 9]]

Answer 4

Create to DataFrame and aggregate third column converted to integers by first and second columns, last convert back to lists:

df = pd.DataFrame(L)
L = df[2].astype(int).groupby([df[0], df[1]]).sum().reset_index().values.tolist()
print (L)
[['Andrew', '1', 17], ['Andrew', '2', 2], ['Peter', '1', 21], ['Sam', '4', 9]]

And solution with defaultdict, python 3.x only:

from collections import defaultdict

d = defaultdict(int)
#https://stackoverflow.com/a/10532492
for *head, tail in L:
    d[tuple(head)] += int(tail)

d = [[*i, j] for i, j in sorted(d.items())]
print (d)
[['Andrew', '1', 17], ['Andrew', '2', 2], ['Peter', '1', 21], ['Sam', '4', 9]]