找到相同的键并合并到字典列表中？ Python

Question

此代码生成字典列表。

 watchlist = r.get_open_option_positions()
    for x in watchlist:
        print('Symbol: {}, Average Price: {}, Quantity: {}'.format(x['chain_symbol'], 
    x['average_price'], x['quantity']))

Output：

Symbol: PG, Average Price: -46.5714, Quantity: 35.0000
Symbol: PG, Average Price: 33.7142, Quantity: 35.0000
Symbol: MSFT, Average Price: -80.0000, Quantity: 6.0000
Symbol: MSFT, Average Price: 53.0000, Quantity: 6.0000

如何编码以下标准：

if symbol is the same and quantity of both symbols is the same, then subtract average prices and multiply by quantity

因此，例如，结果应如下所示：

Symbol: PG, Average Price: (-12.8572 * 35), Quantity: 35.000
Symbol: MSFT, Average Price: (-27 * 6), Quantity: 6.000

Answer 1

• 设置一个字典（为方便起见，一个默认字典）来跟踪每个组：

   groups = collections.defaultdict(list)

• 遍历watchlist将每个x添加到一个组中：

   for x in watchlist: groups[(x["chain_symbol"], x["quantity"])].append(x)

• 遍历每个组并对价格求和（实际上与在这里减去它们是一样的）：

   for group_key, group in groups.items(): final_price = sum(x["average_price"] for x in group) print(group_key, final_price)

Answer 2

您可以将符号和数量的每个组合的所有价格值存储在字典中，如下所示：

product = {}

for x in watchlist:
    if not x['chain_symbol'], x['quantity'] in product.keys():
        product[x['chain_symbol'], x['quantity']] = []
    product[x['chain_symbol'], x['quantity']].append(x['average_price'])

然后您遍历所有产品（符号和数量的组合），您可以对所有现有价格实施您想要的操作。 以下代码确实是一个意思，但您可以将其更改为您需要的。

for k in product.keys():
    symbol = k[0]
    quantity = k[1]
    all_the_prices = product[k]
    price = sum(all_the_prices)/len(all_the_prices) # Change here to your operation
    print('Symbol: {}, Average Price: {}, Quantity: {}'.format(symbol, price, quantity)