How to tell a function to use the default argument values?
There really aren't many ways to make a function use its default arguments... You only have two options:
- Pass the real default values
- Don't pass the arguments at all
Since none of the options are great, I'm going to make an exhaustive list so you can compare them all.
Use
**kwargs
to pass through argumentsDefine your method using
**kwargs
and pass those tomath.isclose
:def foo(..., **kwargs): ... if math.isclose(x, y, **kwargs):
Cons:
- the parameter names of both functions have to match (e.g.
foo(1, 2, rtol=3)
won't work)
- the parameter names of both functions have to match (e.g.
Manually construct a
**kwargs
dictdef foo(..., rtol=None, atol=None): ... kwargs = {} if rtol is not None: kwargs["rel_tol"] = rtol if atol is not None: kwargs["abs_tol"] = atol if math.isclose(x, y, **kwargs):
Cons:
- ugly, a pain to code, and not fast
Hard-code the default values
def foo(..., rtol=1e-09, atol=0.0): ... if math.isclose(x, y, rel_tol=rtol, abs_tol=atol):
Cons:
- hard-coded values
Use introspection to find the default values
You can use the
inspect
module to determine the default values at run time:import inspect, math signature = inspect.signature(math.isclose) DEFAULT_RTOL = signature.parameters['rel_tol'].default DEFAULT_ATOL = signature.parameters['abs_tol'].default def foo(..., rtol=DEFAULT_RTOL, atol=DEFAULT_ATOL): ... if math.isclose(x, y, rel_tol=rtol, abs_tol=atol):
Cons:
inspect.signature
may fail on builtin functions or other functions defined in C
Delegate the recursion to a helper function so that you only need to build the dict
once.
import math
def foo_helper(..., **kwargs):
...
if math.isclose(x, y, **kwargs):
...
...
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
return foo_helper(x, y, **tols)
Or, instead of passing tolerances to foo
, pass a function which already incorporates the desired tolerances.
from functools import partial
# f can take a default value if there is a common set of tolerances
# to use.
def foo(x, y, f):
...
if f(x,y):
...
...
# Use the defaults
foo(x, y, f=math.isclose)
# Use some other tolerances
foo(x, y, f=partial(math.isclose, rtol=my_rtel))
foo(x, y, f=partial(math.isclose, atol=my_atol))
foo(x, y, f=partial(math.isclose, rtol=my_rtel, atol=my_atol))
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.
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.