在 Python 中处理时,如何确保任意函数调用列表不会急切地评估超过短路点?

How do I ensure a list of arbitrary function calls is not eagerly evaluated past the short circuit point while processing in Python?

例如,给定

def expensive_call(x):
    print(x)
    if x == "d":
        return x
def expensive_call_2(x, y):
    print(x)
    print(y)
    return x + y

a = [expensive_call("a"), expensive_call_2("b", "c"), expensive_call("d")]
next((e for e in a if e is not None), 'All are Nones')

输出是

a
b
c
d
Out[22]: 'bc'

由于 expensive_call("d") 被急切求值,请注意即使 next 调用在第二次调用时短路并输出 "bc",也会打印 "d"。

我正在对列表 a 中的调用进行硬编码,并且 a 不必是列表数据结构。

一种可能的解决方案如下:

a = ['expensive_call("a")', 'expensive_call_2("b", "c")', 'expensive_call("d")']
def generator():
    for e in a:
        r = eval(e)
        if r is not None:
            yield r
next(generator(), 'All are Nones')

输出是

a
b
c
Out[23]: 'bc'

随心所欲。但是,我真的不喜欢必须使用 eval。我也不想使用任何最初将函数指针和参数分开的解决方案,如 (expensive_call, ("a"))。理想情况下我会有类似

的东西
a = lazy_magic([expensive_call("a"), expensive_call_2("b", "c"), expensive_call("d")])
next((e for e in a if e is not None), 'All are Nones')

请注意, 是一个类似的问题,但仅适用于函数具有相同方法签名的情况。

您可以使用以下装饰器:

def lazy_fn(fn):
    return lambda *args: lambda: fn(*args)

(如果您喜欢 lambda,也可以表示为 lazy_fn = lambda fn: lambda *args: lambda: fn(*args) 。)

这样使用:

@lazy_fn
def expensive_call(x):
    print(x)
    if x == "d":
        return x

@lazy_fn
def expensive_call_2(x, y):
    print(x)
    print(y)
    return x + y

a = [expensive_call("a"), expensive_call_2("b", "c"), expensive_call("d")]
print(next((e for e in map(lambda i: i(), a) if e is not None), 'All are Nones'))

输出:

a
b
c
bc

请注意,您需要使用 for e in map(lambda i: i(), a),而不是 for e in a

您可以将它们全部放在一个函数中并产生结果:

def gen():
    yield expensive_call("a")
    yield expensive_call_2("b", "c")
    yield expensive_call("d")


result = next(
    (value for value in gen() if value is not None),
    'All are Nones')

另一种解决方案是使用 partial 应用程序:

from functools import partial

calls = [partial(expensive_call, 'a'),
         partial(expensive_call_2, 'b', 'c'),
         partial(expensive_call, 'd')]

然后评估:

next((result for call in calls
      for result in [call()]
      if result is not None),
     'All results None')