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

Question

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))

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Answer 1

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

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

Answer 2

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]

Answer 3

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)

Answer 4

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