Sum values of each key in list of dictionaries python - django

Question

I have a list of dictionaries which is coming from Django query set.
Like this:
email_sent_count = [
{
'second_follow_count': 1,
'first_follow_count': 1,
'initial_count': 1,
'third_follow_count': 0
},
{
'second_follow_count': 1,
'first_follow_count': 0,
'initial_count': 1,
'third_follow_count': 1
},
{
'second_follow_count': 1,
'first_follow_count': 1,
'initial_count': 1,
'third_follow_count': 1
}
]
Now, I want the sum of each key separately.
like this:
inital_contact = 3
first_followup = 2
second_followup = 3
third_followup = 2
I am trying the following solution:
initial_contact = sum(map(lambda x: x['initial_count'], email_sent_count))
first_followup = sum(map(lambda x: x['first_follow_count'], email_sent_count))
second_followup = sum(map(lambda x: x['second_follow_count'], email_sent_count))
third_followup = sum(map(lambda x: x['third_follow_count'], email_sent_count))
But now, I'm getting 11 keys in all dictionaries and I'm implementing 11 lambda function so is there any good way to solve this issue rather than calling 11 times lambda function
By following ORM I getting above email_sent_count
i = Q(inital_contact__range=(from_date, to_date))
f1 = Q(first_followup__range=(from_date, to_date))
f2 = Q(second_followup__range=(from_date, to_date))
f3 = Q(third_followup__range=(from_date, to_date))
email_count = campaign_contact.filter(i | f1 | f2 | f3).annotate(initial_count=Count('inital_contact'),
first_follow_count=Count('first_followup'),
second_follow_count=Count('second_followup'),
third_follow_count=Count('third_followup'),
).values('initial_count', 'first_follow_count',
'second_follow_count', 'third_follow_count'
So, is there is a solution which is directly working with ORM ?

Answer 1

if you don't mind getting a dictionary as result you could use collections.defaultdict like this:
from collections import defaultdict
sums = defaultdict(int)
for item in email_sent_count:
for key, value in item.items():
sums[key] += value
which results in
defaultdict(<class 'int'>,
{'second_follow_count': 3, 'initial_count': 3,
'first_follow_count': 2, 'third_follow_count': 2})
and you can access the individual sums just like a dictionary: sums['second_follow_count'].
...or maybe even better with collections.Counter:
from collections import Counter
sums = Counter()
for item in email_sent_count:
for key, value in item.items():
sums[key] += value
# Counter({'second_follow_count': 3, 'initial_count': 3,
# 'first_follow_count': 2, 'third_follow_count': 2})

Answer 2

Or if you prefer do do it yourself without using Counter or DefaultDict:
from pprint import pprint
email_sent_count = [
{
'second_follow_count': 1,
'first_follow_count': 1,
'initial_count': 1,
'third_follow_count': 0
},
{
'second_follow_count': 1,
'first_follow_count': 0,
'initial_count': 1,
'third_follow_count': 1
},
{
'second_follow_count': 1,
'first_follow_count': 1,
'initial_count': 1,
'third_follow_count': 1
}
]
# create empty dict with all keys from any inner dict and initialized to 0
alls = dict( (x,0) for y in email_sent_count for x in y)
pprint(alls)
for d in email_sent_count:
for k in d:
alls[k] += d[k]
pprint(alls)
Output:
{'first_follow_count': 0,
'initial_count': 0,
'second_follow_count': 0,
'third_follow_count': 0}
{'first_follow_count': 2,
'initial_count': 3,
'second_follow_count': 3,
'third_follow_count': 2}

Answer 3

Finally, I changed the ORM query that gets me the exact result:
ORM looks like this:
email_sent_count = campaign_contact.filter(i | f1 | f2 | f3
).annotate(initial_count=Count('inital_contact'),
first_follow_count=Count('first_followup'),
second_follow_count=Count('second_followup'),
third_follow_count=Count('third_followup')
).aggregate(initial_sum=Sum('initial_count'),
first_follow_sum=Sum('first_follow_count'),
second_follow_sum=Sum('second_follow_count'),
third_follow_sum=Sum('third_follow_count'))
output...
{'third_follow_sum': 2, 'second_follow_sum': 3, 'first_follow_sum': 2, 'initial_sum': 3}
So, there is not for loops, no lambda... I think performance wise it will work.
And Thanks to all to provide me the solutions, by looking other solution I'm able to solve this. :)