Can one partially apply the second argument of a function that takes no keyword arguments?

徘徊边缘 提交于 2019-12-17 07:14:44

问题


Take for example the python built in pow() function.

xs = [1,2,3,4,5,6,7,8]

from functools import partial

list(map(partial(pow,2),xs))

>>> [2, 4, 8, 16, 32, 128, 256]

but how would I raise the xs to the power of 2?

to get [1, 4, 9, 16, 25, 49, 64]

list(map(partial(pow,y=2),xs))

TypeError: pow() takes no keyword arguments

I know list comprehensions would be easier.


回答1:


No

According to the documentation, partial cannot do this (emphasis my own):

partial.args

The leftmost positional arguments that will be prepended to the positional arguments


You could always just "fix" pow to have keyword args:

_pow = pow
pow = lambda x, y: _pow(x, y)



回答2:


I think I'd just use this simple one-liner:

import itertools
print list(itertools.imap(pow, [1, 2, 3], itertools.repeat(2)))

Update:

I also came up with a funnier than useful solution. It's a beautiful syntactic sugar, profiting from the fact that the ... literal means Ellipsis in Python3. It's a modified version of partial, allowing to omit some positional arguments between the leftmost and rightmost ones. The only drawback is that you can't pass anymore Ellipsis as argument.

import itertools
def partial(func, *args, **keywords):
    def newfunc(*fargs, **fkeywords):
        newkeywords = keywords.copy()
        newkeywords.update(fkeywords)
        return func(*(newfunc.leftmost_args + fargs + newfunc.rightmost_args), **newkeywords)
    newfunc.func = func
    args = iter(args)
    newfunc.leftmost_args = tuple(itertools.takewhile(lambda v: v != Ellipsis, args))
    newfunc.rightmost_args = tuple(args)
    newfunc.keywords = keywords
    return newfunc

>>> print partial(pow, ..., 2, 3)(5) # (5^2)%3
1
>>> print partial(pow, 2, ..., 3)(5) # (2^5)%3
2
>>> print partial(pow, 2, 3, ...)(5) # (2^3)%5
3
>>> print partial(pow, 2, 3)(5) # (2^3)%5
3

So the the solution for the original question would be with this version of partial list(map(partial(pow, ..., 2),xs))




回答3:


Why not just create a quick lambda function which reorders the args and partial that

partial(lambda p, x: pow(x, p), 2)



回答4:


You could create a helper function for this:

from functools import wraps
def foo(a, b, c, d, e):
    print('foo(a={}, b={}, c={}, d={}, e={})'.format(a, b, c, d, e))

def partial_at(func, index, value):
    @wraps(func)
    def result(*rest, **kwargs):
        args = []
        args.extend(rest[:index])
        args.append(value)
        args.extend(rest[index:])
        return func(*args, **kwargs)
    return result

if __name__ == '__main__':
    bar = partial_at(foo, 2, 'C')
    bar('A', 'B', 'D', 'E') 
    # Prints: foo(a=A, b=B, c=C, d=D, e=E)

Disclaimer: I haven't tested this with keyword arguments so it might blow up because of them somehow. Also I'm not sure if this is what @wraps should be used for but it seemed right -ish.




回答5:


you could use a closure

xs = [1,2,3,4,5,6,7,8]

def closure(method, param):
  def t(x):
    return method(x, param)
  return t

f = closure(pow, 2)
f(10)
f = closure(pow, 3)
f(10)



回答6:


One way of doing it would be:

def testfunc1(xs):
    from functools import partial
    def mypow(x,y): return x ** y
    return list(map(partial(mypow,y=2),xs))

but this involves re-defining the pow function.

if the use of partial was not 'needed' then a simple lambda would do the trick

def testfunc2(xs):
    return list(map(lambda x: pow(x,2), xs))

And a specific way to map the pow of 2 would be

def testfunc5(xs):
    from operator import mul
    return list(map(mul,xs,xs))

but none of these fully address the problem directly of partial applicaton in relation to keyword arguments




回答7:


You can do this with lambda, which is more flexible than functools.partial():

pow_two = lambda base: pow(base, 2)
print(pow_two(3))  # 9

More generally:

def bind_skip_first(func, *args, **kwargs):
  return lambda first: func(first, *args, **kwargs)

pow_two = bind_skip_first(pow, 2)
print(pow_two(3))  # 9

One down-side of lambda is that some libraries are not able to serialize it.




回答8:


The very versatile funcy includes an rpartial function that exactly addresses this problem.

xs = [1,2,3,4,5,6,7,8]
from funcy import rpartial
list(map(rpartial(pow, 2), xs))
# [1, 4, 9, 16, 25, 36, 49, 64]

It's just a lambda under the hood:

def rpartial(func, *args):
    """Partially applies last arguments."""
    return lambda *a: func(*(a + args))



回答9:


As already said that's a limitation of functools.partial if the function you want to partial doesn't accept keyword arguments.

If you don't mind using an external library 1 you could use iteration_utilities.partial which has a partial that supports placeholders:

>>> from iteration_utilities import partial
>>> square = partial(pow, partial._, 2)  # the partial._ attribute represents a placeholder
>>> list(map(square, xs))
[1, 4, 9, 16, 25, 36, 49, 64]

1 Disclaimer: I'm the author of the iteration_utilities library (installation instructions can be found in the documentation in case you're interested).




回答10:


If you can't use lambda functions, you can also write a simple wrapper function that reorders the arguments.

def _pow(y, x):
    return pow(x, y)

and then call

list(map(partial(_pow,2),xs))

>>> [1, 4, 9, 16, 25, 36, 49, 64]


来源:https://stackoverflow.com/questions/11173660/can-one-partially-apply-the-second-argument-of-a-function-that-takes-no-keyword

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!