字典值的惰性评估？

Question

假设我有以下字典 d={'a': heavy_expression1, 'b': heavy_expression2}.

如何包装表达式，以便在访问它们时对其进行求值，之后不执行求值？

d['a'] # only here heavy_expression1 is executed
d['a'] # no execution here, already calculated

我需要使用 lambda 还是生成器？

Answer 1

带有 lambda 的版本：

class LazyDict(dict):
    def __init__(self, lazies):
        self.lazies = lazies
    def __missing__(self, key):
        value = self[key] = self.lazies[key]()
        return value

d = LazyDict({'a': lambda: print('heavy_expression1') or 1,
              'b': lambda: print('heavy_expression2') or 2})
print(d['a'])
print(d['a'])
print(d['b'])
print(d['b'])

输出：

heavy_expression1
1
1
heavy_expression2
2
2

Answer 2

另一种方法是中断子类中的 __getitem__ 方法并在那里进行缓存：

class LazyDict(dict):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        self.cache = {}

    def __getitem__(self, item):
        if item in self.cache:
            return self.cache[item]
        result = super().__getitem__(item)()

        self.cache[item] = result
        return result

这是带有函数的测试用例：（lambda 如果您不想定义函数，这里非常适合）

def heavy_expresion1():
    print('heavy expression1 is calculated')
    return 10

def heavy_expresion2():
    print('heavy expression2 is calculated')
    return 20

d = LazyDict({'a': heavy_expresion1, 'b': heavy_expresion2})
print(d)

print(d['a'])
print(d['a'])

print(d['b'])
print(d['b'])

输出：

{'a': <function heavy_expresion1 at 0x000001AFE783ED30>, 'b': <function heavy_expresion2 at 0x000001AFE8036940>}
heavy expression1 is calculated
10
10
heavy expression2 is calculated
20
20

字典值的惰性评估？

Lazy evaluation of dict values?

python

lazy-evaluation

python-3.x