Python 匹配 2 个字典列表中的值 - 最快的解决方案

Question

我有 2 个带有产品的字典列表。 第一个列表包含带有产品信息的字典（一些基本信息+股票）第二个列表包含带有产品价格的字典。 我需要根据“product_code”字段合并字典。 我的列表包含 200k+ 产品。 有没有办法让它更快？

merged_list_of_dicts = []
for price in prices:
    for stock in stocks:
        if price['product_code'] == stock['product_code']:
            dict_ = {}
            dict_['product_code'] = price['product_code']
            dict_['price'] = price['price']
            dict_['stock'] = stock['stock']
            all_together_list.append(dict_)
            break

return all_together_list

Answer 1

您可以提前对列表进行排序，如果它们匹配 1:1，您可以将它们一起迭代 -

prices = sorted(prices, key = lambda x: x["product_code"])
stocks = sorted(stocks, key = lambda x: x["product_code"])
merged_list_of_dics = []
for p, s in zip(prices, stocks):
    dic = {}
    dic['product_code'] = price['product_code']
    dic['price'] = price['price']
    dic['stock'] = stock['stock']
    all_together_list.append(dic)

return all_together_list