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Counter + sum -> one-liner solution in Clear category for Aggregate and Count by DenisKnyazev
from collections import Counter
def aggregate_and_count(items: list) -> dict:
return dict(Counter(sum([[x[0]]*x[1] for x in items],[])))
print("Example:")
print(aggregate_and_count([["a", 1], ["b", 2], ["c", 3], ["a", 5]]))
assert aggregate_and_count([["a", 1], ["b", 2]]) == {"a": 1, "b": 2}
assert aggregate_and_count([["a", 1], ["a", 2]]) == {"a": 3}
assert aggregate_and_count([["a", 1], ["b", 2], ["c", 3], ["a", 5]]) == {
"a": 6,
"b": 2,
"c": 3,
}
assert aggregate_and_count([["a", 1]]) == {"a": 1}
print("The aggregation is done! Click 'Check' to earn cool rewards!")
July 23, 2021
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