Question

我正在尝试在python 2.6运行时从http://www.ibm.com/developerworks/linux/library/l-prog3.html运行此代码段。

from functional import *

taxcalc = lambda income,rate,deduct: (income-(deduct))*rate

taxCurry = curry(taxcalc)
taxCurry = taxCurry(50000)
taxCurry = taxCurry(0.30)
taxCurry = taxCurry(10000)
print "Curried taxes due =",taxCurry

print "Curried expression taxes due =", \
      curry(taxcalc)(50000)(0.30)(10000)

好的，所以我从http://www.python.org/dev/peps/pep-0309/了解到，功能被重命名为functools而curry为partial，但只是进行重命名并没有帮助。我收到错误：

taxCurry = taxCurry(50000)
TypeError: <lambda>() takes exactly 3 arguments (1 given)

以下确实有效，但我真的需要改变它吗？

from functools import partial

taxcalc = lambda income,rate,deduct: (income-(deduct))*rate

taxCurry = partial(taxcalc)
taxCurry = partial(taxCurry, 50000)
taxCurry = partial(taxCurry, 0.30)
taxCurry = partial(taxCurry, 10000)
print "Curried taxes due =", taxCurry()

print "Curried expression taxes due =", \
      taxcalc(50000, 0.30, 10000)

有没有更好的方法来保留原始示例的机制？最后原始的例子是真正的currying还是只是部分应用？（根据http://www.uncarved.com/blog/not_currying.mrk）

感谢您的时间

Answer 1

我猜他们改变它的原因是因为Python是动态类型的。这意味着如果出现任何问题，调试原始curry代码将非常困难 - 比像Haskell这样的语言更难以直接获得一个好的类型错误。所以我认为用更明确的partial版本替换它是一个合理的决定（对我来说看起来更加pythonic）。

您的示例也有点奇怪，因为您只是将部分应用的函数重新分配给相同的名称。通常，部分应用的函数将被赋予另一个函数。至少这是我能想到的Python中唯一合理的用例。

Answer 2

toolz project中curry的实施应该是替代品。

$ pip install toolz
>>> from toolz import curry

Answer 3

我写了an implementation of a curry decorator，效果很好：

def curry(func):
    """
    Decorator to curry a function, typical usage:

    >>> @curry
    ... def foo(a, b, c):
    ...    return a + b + c

    The function still work normally:
    >>> foo(1, 2, 3)
    6

    And in various curried forms:
    >>> foo(1)(2, 3)
    6
    >>> foo(1)(2)(3)
    6

    This also work with named arguments:
    >>> foo(a=1)(b=2)(c=3)
    6
    >>> foo(b=1)(c=2)(a=3)
    6
    >>> foo(a=1, b=2)(c=3)
    6
    >>> foo(a=1)(b=2, c=3)
    6

    And you may also change your mind on named arguments,
    But I don't know why you may want to do that:
    >>> foo(a=1, b=0)(b=2, c=3)
    6

    Finally, if you give more parameters than expected, the exception
    is the expected one, not some garbage produced by the currying
    mechanism:

    >>> foo(1, 2)(3, 4)
    Traceback (most recent call last):
       ...
    TypeError: foo() takes exactly 3 arguments (4 given)
    """
    def curried(*args, **kwargs):
        if len(args) + len(kwargs) >= func.__code__.co_argcount:
            return func(*args, **kwargs)
        return (lambda *args2, **kwargs2:
                curried(*(args + args2), **dict(kwargs, **kwargs2)))
    return curried


if __name__ == "__main__":
    import doctest
    doctest.testmod()

是否有旧功能的确切替代品？

3 个答案: