Python惰性评估器

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是否有Python方式封装惰性函数调用，从而在第一次使用函数f()时调用了先前绑定的函数g(Z)，而在后续调用f()中返回了缓存的值?请注意，记忆可能并不完美.

我有:

f = g(Z) if x: return 5 elif y: return f elif z: return h(f)

该代码有效，但我想对其进行重组，以便仅在使用该值的情况下调用g(Z).我不想更改g(...)的定义，并且Z有点大.

我认为f必须是一个函数，但事实并非如此.

解决方案

无论您是寻求缓存还是懒惰的评估，我都有些困惑.对于后者，请查看Alberto Bertogli的模块 lazy.py .

Is there a Pythonic way to encapsulate a lazy function call, whereby on first use of the function f(), it calls a previously bound function g(Z) and on the successive calls f() returns a cached value?

Please note that memoization might not be a perfect fit.

I have:

f = g(Z) if x: return 5 elif y: return f elif z: return h(f)

The code works, but I want to restructure it so that g(Z) is only called if the value is used. I don't want to change the definition of g(...), and Z is a bit big to cache.

EDIT: I assumed that f would have to be a function, but that may not be the case.

解决方案

I'm a bit confused whether you seek caching or lazy evaluation. For the latter, check out the module lazy.py by Alberto Bertogli.