Python每隔一段定义的时间清除lru_cache

Question

我对一个应用程序接口进行了很多多余的调用，所以我决定使用lru_cache。但是，我也希望经常清除缓存，以防给定输入的响应值发生变化。因此，作为一种解决方案，我创建了自己的装饰器，以便每隔一段时间清除缓存，但现在我的函数api_call()似乎总是返回None。下面是我的代码：

def cache_maintainer(clear_time: int):
    """
    A function decorator that clears lru_cache clear_time seconds
    :param clear_time: In seconds, how often to clear cache (only checks when called)
    """
    def inner(func):
        def wrapper(*args, **kwargs):
            if hasattr(func, 'next_clear'):
                if time.time() > func.next_clear:
                    func.cache_clear()
                    func.next_clear = time.time() + clear_time
            else:
                func.next_clear = time.time() + clear_time

            func(*args, **kwargs)
        return wrapper
    return inner


@cache_maintainer(86400)  # clear cache every 1 day
@lru_cache(maxsize=1000)
def api_call(param):
    return param

print(api_call("test"))

我无论如何也找不出我做错了什么

Answer 1

这个答案是使用python包- cachetools

安装cachetools pip install cachetools

from cachetools import TTLCache

cache = TTLCache(maxsize=5, ttl=100)
cache['dog'] = 'dog'
a = cache['dog'] # dog
... after 100 seconds...
cache['dog']
KeyError exception thrown

Answer 2

非常简单的错误，花了太长时间才弄清楚。正如Tim Peters提到的，func()被调用，但结果被丢弃(face palm)。我所需要做的就是在调用func()的地方添加一个返回值，它就像我所期望的那样工作。以下是修改后的正常工作代码：

def cache_maintainer(clear_time: int):
    """
    A function decorator that clears lru_cache clear_time seconds
    :param clear_time: In seconds, how often to clear cache (only checks when called)
    """
    def inner(func):
        def wrapper(*args, **kwargs):
            if hasattr(func, 'next_clear'):
                if time.time() > func.next_clear:
                    func.cache_clear()
                    func.next_clear = time.time() + clear_time
            else:
                func.next_clear = time.time() + clear_time

            return func(*args, **kwargs)
        return wrapper
    return inner


@cache_maintainer(86400)  # clear cache every 1 day
@lru_cache(maxsize=1000)
def api_call(param):
    return param

print(api_call("test"))