如何使用列表推导式获取值内键对值的值？
How can I reach values of a key pair value within a value with list comprehensions?
我想遍历“价格”字典（我的意思是 t1 或 t2，而不是 e1 或 e2）并通过使用列表推导来获取它们的值
def combine2(ticks: Mapping[str, Generator[Tick, None, None]]) -> None:
[????]
t1 = Tick(time=datetime.now(), prices={'i1': 11.11, 'i2': 12.12, })
t2 = Tick(time=datetime.now(), prices={'i1': 13.13, 'i2': 14.14, })
t3 = Tick(time=datetime.now(), prices={'i1': 15.15, 'i2': 16.16, })
combine2({'e1': t1, 'e2': t2, 'e3': t3})
感谢您的提问，试试这个：
def combine2(ticks: Mapping[str, Tick]) -> None:
prices = [price for tick in ticks.values() for price in tick.prices.values()]
print(prices)
我想遍历“价格”字典（我的意思是 t1 或 t2，而不是 e1 或 e2）并通过使用列表推导来获取它们的值
def combine2(ticks: Mapping[str, Generator[Tick, None, None]]) -> None:
[????]
t1 = Tick(time=datetime.now(), prices={'i1': 11.11, 'i2': 12.12, })
t2 = Tick(time=datetime.now(), prices={'i1': 13.13, 'i2': 14.14, })
t3 = Tick(time=datetime.now(), prices={'i1': 15.15, 'i2': 16.16, })
combine2({'e1': t1, 'e2': t2, 'e3': t3})
感谢您的提问，试试这个：
def combine2(ticks: Mapping[str, Tick]) -> None:
prices = [price for tick in ticks.values() for price in tick.prices.values()]
print(prices)