Sum of key values in list of dictionary grouped by particular key

Since you don’t provide anything about your previous attempts I’m assuming this question is about where to start.

First thing I’d look for is a data structure that makes solving your problem simple. In this case I’d create a dictionary of sums:

sums = {
  'internal': {'invoiced': …, 'initial_boq': …},
  # …
}

Especially suited for this would be a defaultdict:

from collections import defaultdict

sums = defaultdict(lamdba: defaultdict(lambda:0))

With this definition you can add your values like this:

sums['internal']['invoiced'] += one_value

You can use itertools.groupby, like so. With an extra function to sum up the grouped items.

from itertools import groupby
from operator import itemgetter


d = [
    {'scope': u'internal', 'invoiced': 1000, 'initial_boq': 2800},
    {'scope': u'internal', 'invoiced': 2000, 'initial_boq': 1000}, 
    {'scope': u'internal', 'invoiced': 2000, 'initial_boq': 500},
    {'scope': u'external', 'invoiced': 500,  'initial_boq': 1800},
    {'scope': u'external', 'invoiced': 150,  'initial_boq': 200},
    {'scope': u'both',     'invoiced': 5000, 'initial_boq': 7000},
]


def getsummed(scope, elems):

    d = {'scope': scope, 'invoiced': 0, 'initial_boq': 0}

    for e in elems:
        d['invoiced'] += e['invoiced']
        d['initial_boq'] += e['initial_boq']
    return d


def sortedgroupby(iterable, key):

    return groupby(sorted(iterable, key=key), key=key)


print([getsummed(gpr, groups) for gpr, groups in sortedgroupby(d, key=itemgetter('scope'))])

result is

[{'scope': 'internal', 'invoiced': 5000, 'initial_boq': 4300}, {'scope': 'external', 'invoiced': 650, 'initial_boq': 2000}, {'scope': 'both', 'invoiced': 5000, 'initial_boq': 7000}]

Here is a one-liner 🙂

from collections import Counter
from itertools import groupby

key = lambda d: d['scope']
res = [dict(sum((Counter({k:v for k,v in grp.items() if k!='scope'}) for grp in grps), Counter()), scope=scope) for scope,grps in groupby(sorted(lst, key=key), key=key)]
print (res)

[{'invoiced': 5000, 'initial_boq': 7000, 'scope': 'both'}, {'invoiced': 650, 'initial_boq': 2000, 'scope': 'external'}, {'invoiced': 5000, 'initial_boq': 4300, 'scope': 'internal'}]

And here is the equivalent code for the that one-liner

key = lambda d: d['scope']
res = []
for scope,grps in groupby(sorted(lst, key=key), key=key):
    c = Counter()
    for grp in grps:
         grp.pop('scope')
         c += Counter(grp)

    res.append(dict(c, scope=scope))

pprint(res)

Much less spectacular than many solutions already posted here but very clear

def removeDuplicatedScopesFrom(startingData): 
    differentScopes = [] 
    for x in startingData:
        scope = x["scope"]
        if scope not in differentScopes: 
            differentScopes.append(scope) 
    return differentScopes
    
def composeDictionaryElement(scope, invoiced, initial_boq):
    return("{'scope': u'" + scope + "', 'invoiced': " + str(invoiced) + ", 'initial_boq': " + str(initial_boq) + "}")

def main():
    var = [
        {'scope': u'internal', 'invoiced': 1000, 'initial_boq': 2800},
        {'scope': u'internal', 'invoiced': 2000, 'initial_boq': 1000},   
        {'scope': u'internal', 'invoiced': 2000, 'initial_boq': 500},
        {'scope': u'external', 'invoiced': 500,  'initial_boq': 1800},
        {'scope': u'external', 'invoiced': 150,  'initial_boq': 200},
        {'scope': u'both',     'invoiced': 5000, 'initial_boq': 7000}
    ]
    
    # empty list for the final result
    finalList = []
    # identifying the different scopes involved
    scopes = removeDuplicatedScopesFrom(var)
    
    # scanning the input and joining data from the same scope
    for scope in scopes:
        
        # resetting values for each different scope
        invoiced = 0;
        initial_boq = 0;
        
        # checking all the elements in the list
        for y in var:
            if y["scope"] == scope:
                invoiced = invoiced + y["invoiced"]
                initial_boq = initial_boq + y["initial_boq"]
                
        # when list is over we ask to compose the related dictionary element
        finalDictionaryElement = composeDictionaryElement(scope, invoiced, initial_boq)
        # adding it to the final list
        finalList.append(finalDictionaryElement)
    
    # results out without surrounding quotes
    print("[%s]" % (', '.join(finalList)))
    
if __name__== "__main__":
    main()

Output

[{'scope': u'internal', 'invoiced': 5000, 'initial_boq': 4300}, {'scope': u'external', 'invoiced': 650, 'initial_boq': 2000}, {'scope': u'both', 'invoiced': 5000, 'initial_boq': 7000}]