Django - How to group multiple records based on multiple columns

Question

I am new to python/django so please bear with me.

I have QuerySet qrySet.values_list('group_id', 'date', 'time', 'mobile')

with that I get list like this

[1, datetime.date(2018, 9, 8), datetime.time(18, 39), 'phone1']
[2, datetime.date(2018, 9, 12), datetime.time(4, 0), 'phone1']
[2, datetime.date(2018, 9, 19), datetime.time(4, 0), 'phone2']
[2, datetime.date(2018, 9, 19), datetime.time(4, 0), 'phone4']
[9, datetime.date(2018, 9, 10), datetime.time(4, 35), 'phone3']
[9, datetime.date(2018, 9, 17), datetime.time(4, 35), 'phone3']
[9, datetime.date(2018, 9, 18), datetime.time(4, 35), 'phone3']

I am trying to put all phone numbers together that have same group_id, date and time, something like this [[group_id, date, time, [phone1, phone2, phone9]]

Is there anything built-in in Django or python for this kind of thing?

I tried some SO solutions using lambda but they group based on one value while I have three.

Thank you

Is this a sorting question or are you trying to have just one record per phone? — PhoebeB
– PhoebeB, Commented Sep 5, 2018 at 10:46
What's your database? Postgres supports an ArrayAgg: stackoverflow.com/questions/43203014/… Otherwise you'll have to end up doing some post processing to get a phone number list. — wholevinski
– wholevinski, Commented Sep 5, 2018 at 10:57

Will Keeling · Accepted Answer · 2018-09-05 11:06:22Z

0

You could try itertools.groupby:

from itertools import groupby

[list(key) + [[g[3] for g in group]] 
    for key, group in groupby(
        qrySet.values_list('group_id', 'date', 'time', 'mobile'), 
        key=lambda x: (x[0], x[1], x[2]))]

With your example data, that gives:

[[1, datetime.date(2018, 9, 8), datetime.time(18, 39), ['phone1']],
 [2, datetime.date(2018, 9, 12), datetime.time(4, 0), ['phone1']],
 [2, datetime.date(2018, 9, 19), datetime.time(4, 0), ['phone2', 'phone4']],
 [9, datetime.date(2018, 9, 10), datetime.time(4, 35), ['phone3']],
 [9, datetime.date(2018, 9, 17), datetime.time(4, 35), ['phone3']],
 [9, datetime.date(2018, 9, 18), datetime.time(4, 35), ['phone3']]]

edited Sep 5, 2018 at 11:06

answered Sep 5, 2018 at 10:46

Will Keeling

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