Using functools partial with multiple arguments

Question

Let's assume, I have a function

def func(u,v,w,x, alpha = 4, beta = 5):
    print('u ',u)
    print('v ',v)
    print('x ',w)
    print('u ',x)
    print('** kwarqs: alpha ',alpha)
    print('** kwarqs: beta ',beta)
    return u*4 + v*3 + w*2 + x

I now want to make it a partial-function by using functools.partial. I set the u,v,w variable to a constant and create a function $$part=f(x)|_{u,v,w = \text{const}}$$

u = 2
v = 4
w = 5

part = partial(func,u,v,w, alpha = 'Hello 1', beta = 'Hello 2')
print(part(4))

The result is

u  2
v  4
w  5
x  4   # This is the remaining free variable
** kwarqs: alpha  Hello 1
** kwarqs: beta  Hello 2
34

How can I create functions

f(u) #or f(v),f(w)
f(v,x) #or all other variable combinations
f(v,alpha) # a combination of args and kwargs

?

Best regards

Work arounds are also welcome.

Does this help: https://stackoverflow.com/questions/11173660/can-one-partially-apply-the-second-argument-of-a-function-that-takes-no-keyword — user2390182, Mar 29 '20 at 14:34
@schwobaseggl - Thank you, I wonder wether something has changed in the last 8 years — Uwe.Schneider, Mar 29 '20 at 14:38
@albusSimba The third wanted function seems to be overloading. Are partial functions also considered to be overloaded functions? — Uwe.Schneider, Mar 29 '20 at 14:42
Nope, it hasn't. Whatever arguments your partial function takes must be kwargs or the last postional arguments of the function passed to partial. — user2390182, Mar 29 '20 at 14:42

score 2 · Accepted Answer · answered Mar 29 '20 at 14:42

2

Can I recommend the use of lambda for your case.

f1 = lambda u : func(u,v,w,x, alpha = 4, beta = 5) 
f2 = lambda v, x: func(u,v,w,x, alpha = 4, beta = 5) 
f3 = lambda v, alpha: func(u,v,w,x, alpha = alpha, beta = 5)

The usage is the same as what you wanted:

f1(u) #or f(v),f(w)
f2(v,x) #or all other variable combinations
f3(v,alpha) # a combination of args and kwargs

answered Mar 29 '20 at 14:42

albusSimba

441
4
14

1

I got no idea but is this what u wanted? – albusSimba Mar 29 '20 at 14:45
1

How is this a useful abstraction when you have to write a specific lambda function (or convential function) with reordered arguments for each individual case? – user2390182 Mar 29 '20 at 14:46
you can always fix the values before hand similar to partial – albusSimba Mar 29 '20 at 14:48
Yes, partial has its charme, but if it is not possible to use it in those cases, I would regard this the smoothest solution. – Uwe.Schneider Mar 29 '20 at 14:49

user2390182 · Answer 2 · 2020-03-29T15:27:35.187

Interesting problem. You could go and turn your function into a kwarg-only function:

from functools import wraps, partial

def kwarged(func):
    @wraps(func)
    def wrapper(**kwargs):
        return func(**kwargs)
    return wrapper

Which you can then use to build any partial version:

>>> func = kwarged(func)

>>> f_u = partial(func, v=5, w=2, x=4)
>>> f_u(u=7)
u  7
v  5
w  2
x  4
** kwarqs: alpha  4
** kwarqs: beta  5
51

>>> f_vx = partial(func, u=5, w=2) 
>>> f_vx(v=3, x=6)
u  5
v  3
w  2
x  6
** kwarqs: alpha  4
** kwarqs: beta  5
39
>>> f_vx(v=3, x=6, alpha=9)
u  5
v  3
w  2
x  6
** kwarqs: alpha  9
** kwarqs: beta  5
39

And even though all arguments are kwargs with defaults, you will still have to pass all original positional args not set in partial:

>>> f_vx(v=3, alpha=9)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 10, in wrapper
TypeError: func() missing 1 required positional argument: 'x'

Using functools partial with multiple arguments

2 Answers2