How to tell a function to use the default argument values?

Question

I have a function foo that calls math.isclose:

import math def foo(..., rtol=None, atol=None):     ...     if math.isclose(x, y, rel_tol=rtol, abs_tol=atol):         ...     ... 

The above fails in math.isclose if I do not pass rtol and atol to foo:

TypeError: must be real number, not NoneType 

I do not want to put the system default argument values in my code (what if they change down the road?)

Here is what I came up with so far:

import math def foo(..., rtol=None, atol=None):     ...     tols = {}     if rtol is not None:         tols["rel_tol"] = rtol     if atol is not None:         tols["abs_tol"] = atol     if math.isclose(x, y, **tols):         ...     ... 

This looks long and silly and allocates a dict on each invocation of foo (which calls itself recursively so this is a big deal).

So, what is the best way to tell math.isclose to use the default tolerances?

PS. There are several related questions. Note that I do not want to know the actual default arguments of math.isclose - all I want it to tell it to use the defaults whatever they are.

Answer 1

One way to do it would be with variadic argument unpacking:

def foo(..., **kwargs):     ...     if math.isclose(x, y, **kwargs):         ... 

This would allow you to specify atol and rtol as keyword arguments to the main function foo, which it would then pass on unchanged to math.isclose.

However, I would also say that it is idiomatic that arguments passed to kwargs modify the behaviour of a function in some way other than to be merely passed to sub-functions being called. Therefore, I would suggest that instead, a parameter is named such that it is clear that it will be unpacked and passed unchanged to a sub-function:

def foo(..., isclose_kwargs={}):     ...     if math.isclose(x, y, **isclose_kwargs):         ... 

You can see an equivalent pattern in matplotlib (example: plt.subplots - subplot_kw and gridspec_kw, with all other keyword arguments being passed to the Figure constructor as **fig_kw) and seaborn (example: FacetGrid - subplot_kws, gridspec_kws).

This is particularly apparent when there are mutiple sub-functions you might want to pass keyword arguments, but retain the default behaviour otherwise:

def foo(..., f1_kwargs={}, f2_kwargs={}, f3_kwargs={}):     ...     f1(**f1_kwargs)     ...     f2(**f2_kwargs)     ...     f3(**f3_kwargs)     ... 

Caveat:

Note that default arguments are only instantiated once, so you should not modify the empty dicts in your function. If there is a need to, you should instead use None as the default argument and instantiate a new empty dict each time the function is run:

def foo(..., isclose_kwargs=None):     if isclose_kwargs is None:         isclose_kwargs = {}     ...     if math.isclose(x, y, **isclose_kwargs):         ... 

My preference is to avoid this where you know what you're doing since it is more brief, and in general I don't like rebinding variables. However, it is definitely a valid idiom, and it can be safer.

Answer 2

The correct solution would be to use the same defaults as math.isclose(). There is no need to hard-code them, as you can get the current defaults with the inspect.signature() function:

import inspect import math  _isclose_params = inspect.signature(math.isclose).parameters  def foo(..., rtol=_isclose_params['rel_tol'].default, atol=_isclose_params['abs_tol'].default):     # ... 

Quick demo:

>>> import inspect >>> import math >>> params = inspect.signature(math.isclose).parameters >>> params['rel_tol'].default 1e-09 >>> params['abs_tol'].default 0.0 

This works because math.isclose() defines its arguments using the Argument Clinic tool:

[T]he original motivation for Argument Clinic was to provide introspection “signatures” for CPython builtins. It used to be, the introspection query functions would throw an exception if you passed in a builtin. With Argument Clinic, that’s a thing of the past!

Under the hood, the math.isclose() signature is actually stored as a string:

>>> math.isclose.__text_signature__ '($module, /, a, b, *, rel_tol=1e-09, abs_tol=0.0)' 

This is parsed out by the inspect signature support to give you the actual values.

Not all C-defined functions use Argument Clinic yet, the codebase is being converted on a case-by-case basis. math.isclose() was converted for Python 3.7.0.

You could use the __doc__ string as a fallback, as in earlier versions this too contains the signature:

>>> import math >>> import sys >>> sys.version_info sys.version_info(major=3, minor=6, micro=8, releaselevel='final', serial=0) >>> math.isclose.__doc__.splitlines()[0] 'isclose(a, b, *, rel_tol=1e-09, abs_tol=0.0) -> bool' 

so a slightly more generic fallback could be:

import inspect  def func_defaults(f):     try:         params = inspect.signature(f).parameters     except ValueError:         # parse out the signature from the docstring         doc = f.__doc__         first = doc and doc.splitlines()[0]         if first is None or f.__name__ not in first or '(' not in first:             return {}         sig = inspect._signature_fromstr(inspect.Signature, math.isclose, first)         params = sig.parameters     return {         name: p.default for name, p in params.items()         if p.default is not inspect.Parameter.empty     } 

I'd see this as a stop-gap measure only needed to support older Python 3.x releases. The function produces a dictionary keyed on parameter name:

>>> import sys >>> import math >>> sys.version_info sys.version_info(major=3, minor=6, micro=8, releaselevel='final', serial=0) >>> func_defaults(math.isclose) {'rel_tol': 1e-09, 'abs_tol': 0.0} 

Note that copying the Python defaults is very low risk; unless there is a bug, the values are not prone to change. So another option could be to hardcode the 3.5 / 3.6 known defaults as a fallback, and use the signature provided in 3.7 and newer:

try:     # Get defaults through introspection in newer releases     _isclose_params = inspect.signature(math.isclose).parameters     _isclose_rel_tol = _isclose_params['rel_tol'].default     _isclose_abs_tol = _isclose_params['abs_tol'].default except ValueError:     # Python 3.5 / 3.6 known defaults     _isclose_rel_tol = 1e-09     _isclose_abs_tol = 0.0 

Note however that you are at greater risk of not supporting future, additional parameters and defaults. At least the inspect.signature() approach would let you add an assertion about the number of parameters that your code expects there to be.