在Python中进行处理时,如何确保不急于通过短路点评估任意函数调用的列表?

问题描述 投票:2回答:2

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')

输出为

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

由于急切地评估了expensive_call("d"),请注意,即使next调用在第二次调用时短路且输出为“ bc”,也会打印“ d”。

我正在对列表a中的呼叫进行硬编码,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')

输出为

a
b
c
Out[23]: 'bc'

根据需要。但是,我真的不喜欢必须使用eval。我也不想使用任何最初使函数指针和参数分开的解决方案,例如(expensive_call, ("a"))。理想情况下,我会喜欢

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,https://stackoverflow.com/a/3405828/2750819是类似的问题,但仅在函数具有相同的方法签名时才适用。

python python-3.x list yield short-circuiting
2个回答
2
投票

您可以将它们全部放入函数中并产生结果:

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')

另一个解决方案是使用partial应用程序:

partial

然后评估:

from functools import partial

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

2
投票

您可以使用以下装饰器:

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

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

像这样使用它:

lazy_fn = lambda fn: lambda *args: lambda: fn(*args)

输出:

@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'))

请注意,您无需使用a b c bc ,而需要使用for e in a

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