如何使用Counter在defaultdict中汇总列表中的元组值？

Question

I have a defaultdict which groups values as a list of tuples. I'd like to add values of each key using Counter .

So, the question is how to tranform this:

dict_items([
    ('Key_One', [('1', 7), ('1', 2), ('1', 2), ('1', 12), ('5', 1)]),
    ('Key_Two', [('1', 13), ('1', 9), ('1', 7), ('1', 12), ('1', 2), ('1', 10)])
])

into this:

[
    'Key_One': Counter({'1': 23, '5': 1})
    'Key_Two': Counter({'1': 53})
]

Answer 1

Using collections.defaultdict and a simple iteration

Ex:

from collections import defaultdict

d = dict([
    ('Key_One', [('1', 7), ('1', 2), ('1', 2), ('1', 12), ('5', 1)]),
    ('Key_Two', [('1', 13), ('1', 9), ('1', 7), ('1', 12), ('1', 2), ('1', 10)])
])

result = {}
for k, v in d.items():
    temp = defaultdict(int)
    for m, n in v:
        temp[m] += n
    result[k] = temp
print(result)

Output:

{'Key_One': defaultdict(<class 'int'>, {'1': 23, '5': 1}), 'Key_Two': defaultdict(<class 'int'>, {'1': 53})}

---

Using collections.Counter

Ex:

from collections import Counter

d = dict([
    ('Key_One', [('1', 7), ('1', 2), ('1', 2), ('1', 12), ('5', 1)]),
    ('Key_Two', [('1', 13), ('1', 9), ('1', 7), ('1', 12), ('1', 2), ('1', 10)])
])

result = {}
for k, v in d.items():
    temp = Counter()
    for m, n in v:
        temp.update(Counter({m:n}))
    result[k] = temp
print(result)

Output:

{'Key_One': Counter({'1': 23, '5': 1}), 'Key_Two': Counter({'1': 53})}

Answer 2

Using Counter and defaultdict. Python 3.6.7

from collections import Counter
from collections import defaultdict

a = dict([
    ('Key_One', [('1', 7), ('1', 2), ('1', 2), ('1', 12), ('5', 1)]),
    ('Key_Two', [('1', 13), ('1', 9), ('1', 7), ('1', 12), ('1', 2), ('1', 10)])
          ])

l = {}
for i in a:
    output = defaultdict(int)
    for k, v in a[i]:
        output[k] += v
    l.update({i: Counter(dict(output.items()))})

print(l)

Output:

{'Key_One': Counter({'1': 23, '5': 1}), 'Key_Two': Counter({'1': 53})}