python - 通过总结 Python 中的现有列表来创建一个新列表

Question

我在 Python 中有一个这样的列表:

my_list = [
            {"id":1,"symbol":"ABC","quantity":100},
            {"id":2,"symbol":"PQR","quantity":500},
            {"id":3,"symbol":"ABC","quantity":300},
            {"id":1,"symbol":"XYZ","quantity":320},
            {"id":3,"symbol":"PQR","quantity":800},
         ]

我想总结一个具有相似符号的数量并创建一个新列表。 new_list 将如下所示:

new_list = [
             { "symbol":"ABC","total_quantity":400,"ids":"1 3"}
             { "symbol":"PQR","total_quantity":1300,"ids":"2 3"}
             { "symbol":"XYZ","total_quantity":320,"ids":"1"}
           ]

我对 Python 相当陌生，尝试了一些方法但没有奏效。我可以尝试什么来实现这一目标？

Answer 1

一个不错的解决方案是使用 defaultdict这有助于按 symbol 分组，您只需要最后一步来构建 ids

from collections import defaultdict

my_list = [{"id": 1, "symbol": "ABC", "quantity": 100},
           {"id": 2, "symbol": "PQR", "quantity": 500},
           {"id": 3, "symbol": "ABC", "quantity": 300},
           {"id": 1, "symbol": "XYZ", "quantity": 320},
           {"id": 3, "symbol": "PQR", "quantity": 800}, ]

expected = defaultdict(lambda: {'symbol': '', 'ids': [], "total_quantity": 0})
for row in my_list:
    expected[row['symbol']]['symbol'] = row['symbol']
    expected[row['symbol']]['ids'].append(str(row['id']))
    expected[row['symbol']]['total_quantity'] += row['quantity']

result = [{**x, 'ids': " ".join(x['ids'])} for x in expected.values()]