python - 字典值的惰性评估？

Question

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

我如何包装表达式，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