Passing default parameters into nested functions python

Question

I have a function

foo(x, setting1=def1, settings2=def2)

(actually this is a third party library function with a bunch of default parameters of which I need to set only 1 externally, but for the sake of example..)

I am calling this from

bar(y, settings2_in=def2)
   x = get_x(y)
   foo(x, settings2=settings2_in)

This is ok, but stylistically I'd rather call name the parameter settings2_in settings2. When I keep passing another layer down, I'll have to keep renaming the parameter, and it gets ugly.

bletch(z, settings2_in_2=def2)
   y = get_y(z)
   bar(y, settings2_in=settings_in_2)

Is there a "nice"/Pythonic way to pass a subset of default parameters down many layers of functions in this way?

Is Python clever enough to work out that if I do:

bar(y, settings2=def2)
   x = get_x(y)
   foo(x, settings2=settings2)

That the 2 uses of settings2 are different from the context?

just bar(y, **kwargs) then foo(x, **kwargs) I guess

Jean-François Fabre
– Jean-François Fabre ♦

2017-09-16 11:54:48 +00:00
Commented Sep 16, 2017 at 11:54 — Jean-François Fabre
– Jean-François Fabre ♦, Commented Sep 16, 2017 at 11:54
**kwargs looks like the way to do it nicely. Thanks!

Tom Walker
– Tom Walker

2017-09-16 11:59:41 +00:00
Commented Sep 16, 2017 at 11:59 — Tom Walker
– Tom Walker, Commented Sep 16, 2017 at 11:59

Jean-François Fabre · Accepted Answer · 2017-09-16 11:58:37Z

3

you could avoid redefining all the parameters in your functions, just use **kwargs to capture all keyword arguments.

Then pass the keyword arguments to your foo function

def foo(x, setting1="def1", settings2="def2"):
    print(setting1,settings2)


def bar(y,**kwargs):
    foo(y,**kwargs)

bar(12)
bar(y=12)
bar(12,setting1="other")

result:

def1 def2
def1 def2
other def2

note that since y is outside kwargs, passing y as keyword also works.

the only drawback is that you don't know what can be actually passed to bar, but it's logical as it depends on what foo accepts. Make that up by adding a docstring to your function.

answered Sep 16, 2017 at 11:58

Jean-François Fabre♦

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Comments

Chris Johnson · Accepted Answer · 2017-09-16 12:16:25Z

1

To answer your question directly: yes, Python is clever enough to not confuse a local variable and an outbound function parameter with the same name. Example:

def x(a):
    ...

def y(a, b):
    return x(a=a)

In the return line, the right-hand a is the local variable from the parameter passed into function y and the left-hand a is the parameter to function x.

Names of parameters to called functions are in the namespace of the called function's parameter list, which is unrelated to the local and global variables of the calling context.

answered Sep 16, 2017 at 12:16

Chris Johnson

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