Currying decorator in python

不羁岁月 提交于 2019-11-27 04:04:46

问题


I am trying to write a currying decorator in python, and I think I've got the general idea down, but still got some cases that aren't working right...

def curry(fun):

    cache = []
    numargs = fun.func_code.co_argcount

    def new_fun(*args, **kwargs):
        print args
        print kwargs
        cache.extend(list(args))

        if len(cache) >= numargs:   # easier to do it explicitly than with exceptions

            temp = []
            for _ in xrange(numargs):
                temp.append(cache.pop())
            fun(*temp)

    return new_fun


@curry
def myfun(a,b):
    print a,b

While for the following case this works fine:

myfun(5)
myfun(5)

For the following case it fails:

myfun(6)(7)

Any pointers on how to correctly do this would be greatly appreciated!

Thanks!


回答1:


The below implementation is naive, google for "currying python" for more accurate examples.

def curry(x, argc=None):
    if argc is None:
        argc = x.func_code.co_argcount
    def p(*a):
        if len(a) == argc:
            return x(*a)
        def q(*b):
            return x(*(a + b))
        return curry(q, argc - len(a))
    return p

@curry
def myfun(a,b,c):
    print '%d-%d-%d' % (a,b,c)



myfun(11,22,33)
myfun(44,55)(66)
myfun(77)(88)(99)



回答2:


The source code for curry in the toolz library is available at the following link.

https://github.com/pytoolz/toolz/blob/master/toolz/functoolz.py

It handles args, kwargs, builtin functions, and error handling. It even wraps the docstrings back onto the curried object.




回答3:


Many of the answers here fail to address the fact that a curried function should only take one argument.

A quote from Wikipedia:

In mathematics and computer science, currying is the technique of translating the evaluation of a function that takes multiple arguments (or a tuple of arguments) into evaluating a sequence of functions, each with a single argument (partial application).

Choosing to decorate it with recursion and without co_argcount makes for a decently elegant solution.

from functools import partial, wraps, reduce

def curry(f):
    @wraps(f)
    def _(arg):
        try:
            return f(arg)
        except TypeError:
            return curry(wraps(f)(partial(f, arg)))
    return _

def uncurry(f):
    @wraps(f)
    def _(*args):
        return reduce(lambda x, y: x(y), args, f)
    return _

As shown above, it is also fairly trivial to write an uncurry decorator. :) Unfortunately, the resulting uncurried function will allow any number of arguments instead of requiring a specific number of arguments, as may not be true of the original function, so it is not a true inverse of curry. The true inverse in this case would actually be something like unwrap, but it would require curry to use functools.wraps or something similar that sets a __wrapped__ attribute for each newly created function:

def unwrap(f):
    try:
        return unwrap(f.__wrapped__)
    except AttributeError:
        return f



回答4:


As it's cool to write currying decorators in python, I tried mine: 5 lines of code, readable, and tested curry function.

def curry(func):
    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



回答5:


Here is my version of curry that doesn't use partial, and makes all the functions accept exactly one parameter:

def curry(func):
"""Truly curry a function of any number of parameters
returns a function with exactly one parameter
When this new function is called, it will usually create
and return another function that accepts an additional parameter,
unless the original function actually obtained all it needed
at which point it just calls the function and returns its result
""" 
def curried(*args):
    """
    either calls a function with all its arguments,
    or returns another functiont that obtains another argument
    """
    if len(args) == func.__code__.co_argcount:
        ans = func(*args)
        return ans
    else:
        return lambda x: curried(*(args+(x,)))

return curried



回答6:


This one is fairly simple and doesn't use inspect or examine the given function's args

import functools


def curried(func):
    """A decorator that curries the given function.

    @curried
    def a(b, c):
        return (b, c)

    a(c=1)(2)  # returns (2, 1)
    """
    @functools.wraps(func)
    def _curried(*args, **kwargs):
        return functools.partial(func, *args, **kwargs)
    return _curried



回答7:


I think I've got a better one:

def curried (function):
    argc = function.__code__.co_argcount

    # Pointless to curry a function that can take no arguments
    if argc == 0:
        return function

    from functools import partial
    def func (*args):
        if len(args) >= argc:
            return function(*args)
        else:
            return partial(func, *args)
    return func

This solution uses Python's own functools.partial function instead of effectively recreating that functionality. It also allows you to pass in more arguments than the minimum, -allows keyword arguments,- and just passes through functions that don't have to take arguments, since those are pointless to curry. (Granted, the programmer should know better than to curry zero-arity or multi-arity functions, but it's better than creating a new function in that case.)

UPDATE: Whoops, the keyword argument part doesn't actually work right. Also, optional arguments are counted in the arity but *args are not. Weird.




回答8:


Simplest way to curry a function in python is like this:

from functools import partial
curry = lambda f, g: partial(
    lambda F, G, *args, **kwargs: F(G(*args,**kwargs)),
    f, g
)

https://gist.github.com/hkupty/0ba733c0374964d41dec

One can use it as follows:

_list = []
mask = "Test {}"
append_masked = curry(_list.append, mask.format)
for i in range(10):
    append_masked(i)

which will produce:

['Test 1', 'Test 2', 'Test 3' ... 'Test 10']



回答9:


The solution from Roger Christman will not work with every constellation. I applied a small fix to also handle this situation:

curried_func(1)(2,3)

The small fix that makes it work with every constellation lies in the returned lambda:

def curried(func):
    def curry(*args):
        if len(args) == func.__code__.co_argcount:
            ans = func(*args)
            return ans
        else:
            return lambda *x: curry(*(args+x))
    return curry


来源:https://stackoverflow.com/questions/9458271/currying-decorator-in-python

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