Can one partially apply the second argument of a f

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))

>>> [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.

Answer 1

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).

Answer 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))

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

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.