如果某个字典中的某个键具有相同的月份，该如何计算字典中的值？

Question

I have Django queryset:

summary = Contracts.objects.all() \
                   .filter(datestart__gte=time_threshold) \
                   .values('datestart', 'id_deposits__id_valuta__name') \
                   .annotate(sum=Sum('suma')) \
                   .order_by('-id_deposits__id_valuta__name')

And I catch qs by:

for s in summary:
    for k, v in CURRENCY.items():
        if s['id_deposits__id_valuta__name'] == k:
            s['sum'] *= CURRENCY[k]

It updates currency sum, as needed. It returns dicts

{'sum': Decimal('400000.00'), 'datestart': datetime.date(2016, 3, 1), 'id_deposits__id_valuta__name': 'Гривня'}
{'sum': Decimal('2500000.00'), 'datestart': datetime.date(2016, 4, 25), 'id_deposits__id_valuta__name': 'Долар'}
{'sum': Decimal('2500000.00'), 'datestart': datetime.date(2016, 4, 25), 'id_deposits__id_valuta__name': 'Долар'}

Now i need to sum that value of "sum" key in dicts, when months in "datestart" value are equal.

Ex. some dicts has sum, datestart keys. I need to have sum of 'sum' values when dicts has same month: 2016-04-03, 2016-04-14

So how to do this in minimalistic way?

Answer 1

You can convert every dictionary to Counter where month is key and value is s['sum'] and then add them together:

from collections import Counter

sum((Counter({x['datestart'].month: x['sum']}) for x in summary), Counter()) 
# Counter({4: Decimal('5000000.00'), 3: Decimal('400000.00')})

Note that if you need to differentiate the month and year use {(x['datestart'].month, x['datestart'].year): x['sum']} instead.

Answer 2

Use a dictionary to track the sum per month; you could use a defaultdict here to manage initial values:

from collections import defaultdict
from decimal import Decimal

per_month = defaultdict(Decimal)

for s in summary:
    s['sum'] *= CURRENCY.get(s['id_deposits__id_valuta__name'], 1)
    month = s['datestart'].year, s['datestart'].month
    per_month[month] += s['sum']

I've removed your loop over CURRENCY.items() ; you can just use dict.get() to get the value instead; if the key isn't in the dictionary, 1 is used instead.

Note that I am grouping your sums on (year, month) tuples, assuming that you want months in different years to be treated as separate sums.