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Home/ Questions/Q 8129213
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T08:13:15+00:00 2026-06-06T08:13:15+00:00

Given the following list: [ (‘A’, ”, Decimal(‘4.0000000000’), 1330, datetime.datetime(2012, 6, 8, 0, 0)),

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Given the following list:

[
    ('A', '', Decimal('4.0000000000'), 1330, datetime.datetime(2012, 6, 8, 0, 0)),
    ('B', '', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 6, 4, 0, 0)),
    ('AA', 'C', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 5, 31, 0, 0)),
    ('B', '', Decimal('7.0000000000'), 1330, datetime.datetime(2012, 5, 24, 0, 0)),
    ('A', '', Decimal('21.0000000000'), 1330, datetime.datetime(2012, 5, 14, 0, 0))
]

I would like to group these by the first, second, fourth and fifth columns in the tuple and sum the 3rd.
For this example I’ll name the columns as col1, col2, col3, col4, col5.

In SQL I would do something like this:

select col1, col2, sum(col3), col4, col5 from my table
group by col1, col2, col4, col5

Is there a “cool” way to do this or is it all a manual loop?

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  1. Editorial Team
    Editorial Team
    2026-06-06T08:13:22+00:00Added an answer on June 6, 2026 at 8:13 am
    >>> [(x[0:2] + (sum(z[2] for z in y),) + x[2:5]) for (x, y) in
          itertools.groupby(sorted(L, key=operator.itemgetter(0, 1, 3, 4)),
          key=operator.itemgetter(0, 1, 3, 4))]
    [
      ('A', '', Decimal('21.0000000000'), 1330, datetime.datetime(2012, 5, 14, 0, 0)),
      ('A', '', Decimal('4.0000000000'), 1330, datetime.datetime(2012, 6, 8, 0, 0)),
      ('AA', 'C', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 5, 31, 0, 0)),
      ('B', '', Decimal('7.0000000000'), 1330, datetime.datetime(2012, 5, 24, 0, 0)),
      ('B', '', Decimal('31.0000000000'), 1330, datetime.datetime(2012, 6, 4, 0, 0))
    ]
    

    (NOTE: output reformatted)

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