Second clear solution for Aggregate and Count by viktor.chyrkin - python coding challenges

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Second solution in Clear category for Aggregate and Count by viktor.chyrkin

def aggregate_and_count(items: list) -> dict:
    md = dict()
    for i, e in items:
        md[i] = md.setdefault(i, 0) + e
    return md

July 11, 2022

Comments:

mikaeldovbnia on Jan. 22, 2023, 11:11 p.m.
<p>Very short. Cool.</p>

lisovsky on Jan. 27, 2023, 9:53 p.m.
<p>int() is zero Therefore, you can write like this:</p> <pre class='brush: python'>d = defaultdict(int) d[key] += value </pre>

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