函数缓存简单解决方案

TTL(时间到生命)和max_entries

当修饰函数接受不可哈希类型作为输入(例如dicts)时，不工作 可选参数:TTL(每个条目的生存时间) 可选参数:max_entries(如果缓存参数组合太多，不会使存储混乱) 确保该函数没有重要的副作用

示例使用

import time

@cache(ttl=timedelta(minutes=3), max_entries=300)
def add(a, b):
    time.sleep(2)
    return a + b

@cache()
def substract(a, b):
    time.sleep(2)
    return a - b

a = 5
# function is called with argument combinations the first time -> it takes some time
for i in range(5):
    print(add(a, i))

# function is called with same arguments again? -> will answer from cache
for i in range(5):
    print(add(a, i))

复制装饰器代码

from datetime import datetime, timedelta

def cache(**kwargs):
  def decorator(function):
    # static function variable for cache, lazy initialization
    try: function.cache
    except: function.cache = {}
    def wrapper(*args):
        # if nothing valid in cache, insert something
        if not args in function.cache or datetime.now() > function.cache[args]['expiry']:
            if 'max_entries' in kwargs:
                max_entries = kwargs['max_entries']
                if max_entries != None and len(function.cache) >= max_entries:
                    now = datetime.now()
                    # delete the the first expired entry that can be found (lazy deletion)
                    for key in function.cache:
                        if function.cache[key]['expiry'] < now:
                            del function.cache[key]
                            break
                    # if nothing is expired that is deletable, delete the first
                    if len(function.cache) >= max_entries:
                        del function.cache[next(iter(function.cache))]
            function.cache[args] = {'result': function(*args), 'expiry': datetime.max if 'ttl' not in kwargs else datetime.now() + kwargs['ttl']}

        # answer from cache
        return function.cache[args]['result']
    return wrapper
  return decorator

functools。缓存已经在Python 3.9 (docs)中发布:

from functools import cache

@cache
def factorial(n):
    return n * factorial(n-1) if n else 1

在以前的Python版本中，早期的答案之一仍然是有效的解决方案:使用lru_cache作为普通缓存，没有限制和lru特性。(文档)

如果maxsize设置为None，将禁用LRU特性，并将缓存 可以不受束缚地成长。

这里有一个更漂亮的版本:

cache = lru_cache(maxsize=None)

@cache
def func(param1):
   pass

class memorize(dict):
    def __init__(self, func):
        self.func = func

    def __call__(self, *args):
        return self[args]

    def __missing__(self, key):
        result = self[key] = self.func(*key)
        return result

示例使用:

>>> @memorize
... def foo(a, b):
...     return a * b
>>> foo(2, 4)
8
>>> foo
{(2, 4): 8}
>>> foo('hi', 3)
'hihihi'
>>> foo
{(2, 4): 8, ('hi', 3): 'hihihi'}

创建自己的装饰器并使用它

from django.core.cache import cache
import functools

def cache_returned_values(func):
    @functools.wraps(func)
    def wrapper(*args, **kwargs):
        key = "choose a unique key here"
        results = cache.get(key)
        if not results:
            results = func(*args, **kwargs)
            cache.set(key, results)
        return results

    return wrapper

现在看函数

@cache_returned_values
def get_some_values(args):
  return x

函数缓存简单解决方案

TTL(时间到生命)和max_entries

当修饰函数接受不可哈希类型作为输入(例如dicts)时，不工作 可选参数:TTL(每个条目的生存时间) 可选参数:max_entries(如果缓存参数组合太多，不会使存储混乱) 确保该函数没有重要的副作用

示例使用

import time

@cache(ttl=timedelta(minutes=3), max_entries=300)
def add(a, b):
    time.sleep(2)
    return a + b

@cache()
def substract(a, b):
    time.sleep(2)
    return a - b

a = 5
# function is called with argument combinations the first time -> it takes some time
for i in range(5):
    print(add(a, i))

# function is called with same arguments again? -> will answer from cache
for i in range(5):
    print(add(a, i))

复制装饰器代码

from datetime import datetime, timedelta

def cache(**kwargs):
  def decorator(function):
    # static function variable for cache, lazy initialization
    try: function.cache
    except: function.cache = {}
    def wrapper(*args):
        # if nothing valid in cache, insert something
        if not args in function.cache or datetime.now() > function.cache[args]['expiry']:
            if 'max_entries' in kwargs:
                max_entries = kwargs['max_entries']
                if max_entries != None and len(function.cache) >= max_entries:
                    now = datetime.now()
                    # delete the the first expired entry that can be found (lazy deletion)
                    for key in function.cache:
                        if function.cache[key]['expiry'] < now:
                            del function.cache[key]
                            break
                    # if nothing is expired that is deletable, delete the first
                    if len(function.cache) >= max_entries:
                        del function.cache[next(iter(function.cache))]
            function.cache[args] = {'result': function(*args), 'expiry': datetime.max if 'ttl' not in kwargs else datetime.now() + kwargs['ttl']}

        # answer from cache
        return function.cache[args]['result']
    return wrapper
  return decorator

从Python 3.2开始，有一个内置的装饰器:

@functools。lru_cache(最大容量= 100,输入= False)

装饰器使用一个可记忆可调用对象来包装函数，该可调用对象最多保存maxsize最近的调用。当使用相同的参数周期性地调用昂贵的或I/O绑定的函数时，它可以节省时间。

用于计算斐波那契数的LRU缓存示例:

from functools import lru_cache

@lru_cache(maxsize=None)
def fib(n):
    if n < 2:
        return n
    return fib(n-1) + fib(n-2)

>>> print([fib(n) for n in range(16)])
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610]

>>> print(fib.cache_info())
CacheInfo(hits=28, misses=16, maxsize=None, currsize=16)

如果你被Python 2困住了。X，这里是其他兼容的内存库列表:

functools32 | PyPI |源代码 repoze。lru | PyPI |源代码 pylru | PyPI |源代码 补丁。functools_lru_cache | PyPI |源代码