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












16















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?










share|improve this question


















  • 1





    you can grab the values of the default args: Get a function argument's default value? and come up with a way to define your function using those values. There's probably a way using functools.partial

    – pault
    12 hours ago











  • possible duplicate of stackoverflow.com/questions/4670665/… but you will not find better answer in it.

    – naivepredictor
    12 hours ago






  • 2





    @pault I tried inspect.signature(math.isclose) and it fails with ValueError: no signature found for builtin <built-in function isclose>.

    – Sanyash
    12 hours ago











  • @Sanyash It works for me in python 3.7.2.

    – Aran-Fey
    12 hours ago











  • @Aran-Fey indeed, in 3.7.2 it works, but not in 3.6.6

    – Sanyash
    12 hours ago
















16















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?










share|improve this question


















  • 1





    you can grab the values of the default args: Get a function argument's default value? and come up with a way to define your function using those values. There's probably a way using functools.partial

    – pault
    12 hours ago











  • possible duplicate of stackoverflow.com/questions/4670665/… but you will not find better answer in it.

    – naivepredictor
    12 hours ago






  • 2





    @pault I tried inspect.signature(math.isclose) and it fails with ValueError: no signature found for builtin <built-in function isclose>.

    – Sanyash
    12 hours ago











  • @Sanyash It works for me in python 3.7.2.

    – Aran-Fey
    12 hours ago











  • @Aran-Fey indeed, in 3.7.2 it works, but not in 3.6.6

    – Sanyash
    12 hours ago














16












16








16


3






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?










share|improve this 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?







python python-3.x






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 12 hours ago









sdssds

40.3k1498176




40.3k1498176








  • 1





    you can grab the values of the default args: Get a function argument's default value? and come up with a way to define your function using those values. There's probably a way using functools.partial

    – pault
    12 hours ago











  • possible duplicate of stackoverflow.com/questions/4670665/… but you will not find better answer in it.

    – naivepredictor
    12 hours ago






  • 2





    @pault I tried inspect.signature(math.isclose) and it fails with ValueError: no signature found for builtin <built-in function isclose>.

    – Sanyash
    12 hours ago











  • @Sanyash It works for me in python 3.7.2.

    – Aran-Fey
    12 hours ago











  • @Aran-Fey indeed, in 3.7.2 it works, but not in 3.6.6

    – Sanyash
    12 hours ago














  • 1





    you can grab the values of the default args: Get a function argument's default value? and come up with a way to define your function using those values. There's probably a way using functools.partial

    – pault
    12 hours ago











  • possible duplicate of stackoverflow.com/questions/4670665/… but you will not find better answer in it.

    – naivepredictor
    12 hours ago






  • 2





    @pault I tried inspect.signature(math.isclose) and it fails with ValueError: no signature found for builtin <built-in function isclose>.

    – Sanyash
    12 hours ago











  • @Sanyash It works for me in python 3.7.2.

    – Aran-Fey
    12 hours ago











  • @Aran-Fey indeed, in 3.7.2 it works, but not in 3.6.6

    – Sanyash
    12 hours ago








1




1





you can grab the values of the default args: Get a function argument's default value? and come up with a way to define your function using those values. There's probably a way using functools.partial

– pault
12 hours ago





you can grab the values of the default args: Get a function argument's default value? and come up with a way to define your function using those values. There's probably a way using functools.partial

– pault
12 hours ago













possible duplicate of stackoverflow.com/questions/4670665/… but you will not find better answer in it.

– naivepredictor
12 hours ago





possible duplicate of stackoverflow.com/questions/4670665/… but you will not find better answer in it.

– naivepredictor
12 hours ago




2




2





@pault I tried inspect.signature(math.isclose) and it fails with ValueError: no signature found for builtin <built-in function isclose>.

– Sanyash
12 hours ago





@pault I tried inspect.signature(math.isclose) and it fails with ValueError: no signature found for builtin <built-in function isclose>.

– Sanyash
12 hours ago













@Sanyash It works for me in python 3.7.2.

– Aran-Fey
12 hours ago





@Sanyash It works for me in python 3.7.2.

– Aran-Fey
12 hours ago













@Aran-Fey indeed, in 3.7.2 it works, but not in 3.6.6

– Sanyash
12 hours ago





@Aran-Fey indeed, in 3.7.2 it works, but not in 3.6.6

– Sanyash
12 hours ago












4 Answers
4






active

oldest

votes


















7














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.






share|improve this answer


























  • Further, document that any additional keyword arguments to foo are passed directly to math.isclose.

    – chepner
    11 hours ago











  • @chepner Yup, that's how the FacetGrid example does it. In addition, I think that in a case where keyword arguments are being passed on, an explicit isclose_kwargs or similar argument would be preferred to naked **kwargs.

    – gmds
    11 hours ago













  • Don't use mutable default arguments, however. f1_kwargs=None is better.

    – Martijn Pieters
    11 hours ago











  • @MartijnPieters if you don't mutate it in your function itself, it makes your code less verbose to use {} as a default argument.

    – gmds
    11 hours ago






  • 1





    @gmds: I can still use foo.__defaults__[0]['bar'] = 42 to mess with the defaults. And future refactorings of the function could easily miss that you defined these as defaults. Don't make it that easy to introduce errors.

    – Martijn Pieters
    11 hours ago



















5














There really aren't many ways to make a function use its default arguments... You only have two options:




  1. Pass the real default values

  2. 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 arguments



    Define your method using **kwargs and pass those to math.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)




  • Manually construct a **kwargs dict



    def 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








share|improve this answer
























  • inspect.signature(math.isclose) does not work in Python 3.6

    – sds
    11 hours ago



















5














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.






share|improve this answer





















  • 1





    inspect.signature(math.isclose) does not work in Python 3.6

    – sds
    11 hours ago











  • @sds: right, the Argument Clinic conversion was done for 3.7. Tough luck for you!

    – Martijn Pieters
    11 hours ago








  • 1





    Also math.isclose.__text_signature__ contains None for me on Python3.6.6.

    – Sanyash
    11 hours ago











  • @Sanyash: yes, that's what my answer is saying. That it requires 3.7 or newer for this to work.

    – Martijn Pieters
    11 hours ago











  • And I personally am surprised that someone felt that the 3.7+ compatibility requirement for inspect.signature(math.isclose) was worthy of downvoting. Or is there some other aspect that I missed? In which case, please do leave me some feedback in a comment so I can improve my answer.

    – Martijn Pieters
    11 hours ago



















0














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))





share|improve this answer


























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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7














    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.






    share|improve this answer


























    • Further, document that any additional keyword arguments to foo are passed directly to math.isclose.

      – chepner
      11 hours ago











    • @chepner Yup, that's how the FacetGrid example does it. In addition, I think that in a case where keyword arguments are being passed on, an explicit isclose_kwargs or similar argument would be preferred to naked **kwargs.

      – gmds
      11 hours ago













    • Don't use mutable default arguments, however. f1_kwargs=None is better.

      – Martijn Pieters
      11 hours ago











    • @MartijnPieters if you don't mutate it in your function itself, it makes your code less verbose to use {} as a default argument.

      – gmds
      11 hours ago






    • 1





      @gmds: I can still use foo.__defaults__[0]['bar'] = 42 to mess with the defaults. And future refactorings of the function could easily miss that you defined these as defaults. Don't make it that easy to introduce errors.

      – Martijn Pieters
      11 hours ago
















    7














    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.






    share|improve this answer


























    • Further, document that any additional keyword arguments to foo are passed directly to math.isclose.

      – chepner
      11 hours ago











    • @chepner Yup, that's how the FacetGrid example does it. In addition, I think that in a case where keyword arguments are being passed on, an explicit isclose_kwargs or similar argument would be preferred to naked **kwargs.

      – gmds
      11 hours ago













    • Don't use mutable default arguments, however. f1_kwargs=None is better.

      – Martijn Pieters
      11 hours ago











    • @MartijnPieters if you don't mutate it in your function itself, it makes your code less verbose to use {} as a default argument.

      – gmds
      11 hours ago






    • 1





      @gmds: I can still use foo.__defaults__[0]['bar'] = 42 to mess with the defaults. And future refactorings of the function could easily miss that you defined these as defaults. Don't make it that easy to introduce errors.

      – Martijn Pieters
      11 hours ago














    7












    7








    7







    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.






    share|improve this answer















    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.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 4 hours ago

























    answered 12 hours ago









    gmdsgmds

    3,923323




    3,923323













    • Further, document that any additional keyword arguments to foo are passed directly to math.isclose.

      – chepner
      11 hours ago











    • @chepner Yup, that's how the FacetGrid example does it. In addition, I think that in a case where keyword arguments are being passed on, an explicit isclose_kwargs or similar argument would be preferred to naked **kwargs.

      – gmds
      11 hours ago













    • Don't use mutable default arguments, however. f1_kwargs=None is better.

      – Martijn Pieters
      11 hours ago











    • @MartijnPieters if you don't mutate it in your function itself, it makes your code less verbose to use {} as a default argument.

      – gmds
      11 hours ago






    • 1





      @gmds: I can still use foo.__defaults__[0]['bar'] = 42 to mess with the defaults. And future refactorings of the function could easily miss that you defined these as defaults. Don't make it that easy to introduce errors.

      – Martijn Pieters
      11 hours ago



















    • Further, document that any additional keyword arguments to foo are passed directly to math.isclose.

      – chepner
      11 hours ago











    • @chepner Yup, that's how the FacetGrid example does it. In addition, I think that in a case where keyword arguments are being passed on, an explicit isclose_kwargs or similar argument would be preferred to naked **kwargs.

      – gmds
      11 hours ago













    • Don't use mutable default arguments, however. f1_kwargs=None is better.

      – Martijn Pieters
      11 hours ago











    • @MartijnPieters if you don't mutate it in your function itself, it makes your code less verbose to use {} as a default argument.

      – gmds
      11 hours ago






    • 1





      @gmds: I can still use foo.__defaults__[0]['bar'] = 42 to mess with the defaults. And future refactorings of the function could easily miss that you defined these as defaults. Don't make it that easy to introduce errors.

      – Martijn Pieters
      11 hours ago

















    Further, document that any additional keyword arguments to foo are passed directly to math.isclose.

    – chepner
    11 hours ago





    Further, document that any additional keyword arguments to foo are passed directly to math.isclose.

    – chepner
    11 hours ago













    @chepner Yup, that's how the FacetGrid example does it. In addition, I think that in a case where keyword arguments are being passed on, an explicit isclose_kwargs or similar argument would be preferred to naked **kwargs.

    – gmds
    11 hours ago







    @chepner Yup, that's how the FacetGrid example does it. In addition, I think that in a case where keyword arguments are being passed on, an explicit isclose_kwargs or similar argument would be preferred to naked **kwargs.

    – gmds
    11 hours ago















    Don't use mutable default arguments, however. f1_kwargs=None is better.

    – Martijn Pieters
    11 hours ago





    Don't use mutable default arguments, however. f1_kwargs=None is better.

    – Martijn Pieters
    11 hours ago













    @MartijnPieters if you don't mutate it in your function itself, it makes your code less verbose to use {} as a default argument.

    – gmds
    11 hours ago





    @MartijnPieters if you don't mutate it in your function itself, it makes your code less verbose to use {} as a default argument.

    – gmds
    11 hours ago




    1




    1





    @gmds: I can still use foo.__defaults__[0]['bar'] = 42 to mess with the defaults. And future refactorings of the function could easily miss that you defined these as defaults. Don't make it that easy to introduce errors.

    – Martijn Pieters
    11 hours ago





    @gmds: I can still use foo.__defaults__[0]['bar'] = 42 to mess with the defaults. And future refactorings of the function could easily miss that you defined these as defaults. Don't make it that easy to introduce errors.

    – Martijn Pieters
    11 hours ago













    5














    There really aren't many ways to make a function use its default arguments... You only have two options:




    1. Pass the real default values

    2. 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 arguments



      Define your method using **kwargs and pass those to math.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)




    • Manually construct a **kwargs dict



      def 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








    share|improve this answer
























    • inspect.signature(math.isclose) does not work in Python 3.6

      – sds
      11 hours ago
















    5














    There really aren't many ways to make a function use its default arguments... You only have two options:




    1. Pass the real default values

    2. 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 arguments



      Define your method using **kwargs and pass those to math.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)




    • Manually construct a **kwargs dict



      def 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








    share|improve this answer
























    • inspect.signature(math.isclose) does not work in Python 3.6

      – sds
      11 hours ago














    5












    5








    5







    There really aren't many ways to make a function use its default arguments... You only have two options:




    1. Pass the real default values

    2. 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 arguments



      Define your method using **kwargs and pass those to math.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)




    • Manually construct a **kwargs dict



      def 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








    share|improve this answer













    There really aren't many ways to make a function use its default arguments... You only have two options:




    1. Pass the real default values

    2. 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 arguments



      Define your method using **kwargs and pass those to math.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)




    • Manually construct a **kwargs dict



      def 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









    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 11 hours ago









    Aran-FeyAran-Fey

    20.1k53971




    20.1k53971













    • inspect.signature(math.isclose) does not work in Python 3.6

      – sds
      11 hours ago



















    • inspect.signature(math.isclose) does not work in Python 3.6

      – sds
      11 hours ago

















    inspect.signature(math.isclose) does not work in Python 3.6

    – sds
    11 hours ago





    inspect.signature(math.isclose) does not work in Python 3.6

    – sds
    11 hours ago











    5














    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.






    share|improve this answer





















    • 1





      inspect.signature(math.isclose) does not work in Python 3.6

      – sds
      11 hours ago











    • @sds: right, the Argument Clinic conversion was done for 3.7. Tough luck for you!

      – Martijn Pieters
      11 hours ago








    • 1





      Also math.isclose.__text_signature__ contains None for me on Python3.6.6.

      – Sanyash
      11 hours ago











    • @Sanyash: yes, that's what my answer is saying. That it requires 3.7 or newer for this to work.

      – Martijn Pieters
      11 hours ago











    • And I personally am surprised that someone felt that the 3.7+ compatibility requirement for inspect.signature(math.isclose) was worthy of downvoting. Or is there some other aspect that I missed? In which case, please do leave me some feedback in a comment so I can improve my answer.

      – Martijn Pieters
      11 hours ago
















    5














    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.






    share|improve this answer





















    • 1





      inspect.signature(math.isclose) does not work in Python 3.6

      – sds
      11 hours ago











    • @sds: right, the Argument Clinic conversion was done for 3.7. Tough luck for you!

      – Martijn Pieters
      11 hours ago








    • 1





      Also math.isclose.__text_signature__ contains None for me on Python3.6.6.

      – Sanyash
      11 hours ago











    • @Sanyash: yes, that's what my answer is saying. That it requires 3.7 or newer for this to work.

      – Martijn Pieters
      11 hours ago











    • And I personally am surprised that someone felt that the 3.7+ compatibility requirement for inspect.signature(math.isclose) was worthy of downvoting. Or is there some other aspect that I missed? In which case, please do leave me some feedback in a comment so I can improve my answer.

      – Martijn Pieters
      11 hours ago














    5












    5








    5







    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.






    share|improve this answer















    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.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 11 hours ago

























    answered 11 hours ago









    Martijn PietersMartijn Pieters

    724k14325452348




    724k14325452348








    • 1





      inspect.signature(math.isclose) does not work in Python 3.6

      – sds
      11 hours ago











    • @sds: right, the Argument Clinic conversion was done for 3.7. Tough luck for you!

      – Martijn Pieters
      11 hours ago








    • 1





      Also math.isclose.__text_signature__ contains None for me on Python3.6.6.

      – Sanyash
      11 hours ago











    • @Sanyash: yes, that's what my answer is saying. That it requires 3.7 or newer for this to work.

      – Martijn Pieters
      11 hours ago











    • And I personally am surprised that someone felt that the 3.7+ compatibility requirement for inspect.signature(math.isclose) was worthy of downvoting. Or is there some other aspect that I missed? In which case, please do leave me some feedback in a comment so I can improve my answer.

      – Martijn Pieters
      11 hours ago














    • 1





      inspect.signature(math.isclose) does not work in Python 3.6

      – sds
      11 hours ago











    • @sds: right, the Argument Clinic conversion was done for 3.7. Tough luck for you!

      – Martijn Pieters
      11 hours ago








    • 1





      Also math.isclose.__text_signature__ contains None for me on Python3.6.6.

      – Sanyash
      11 hours ago











    • @Sanyash: yes, that's what my answer is saying. That it requires 3.7 or newer for this to work.

      – Martijn Pieters
      11 hours ago











    • And I personally am surprised that someone felt that the 3.7+ compatibility requirement for inspect.signature(math.isclose) was worthy of downvoting. Or is there some other aspect that I missed? In which case, please do leave me some feedback in a comment so I can improve my answer.

      – Martijn Pieters
      11 hours ago








    1




    1





    inspect.signature(math.isclose) does not work in Python 3.6

    – sds
    11 hours ago





    inspect.signature(math.isclose) does not work in Python 3.6

    – sds
    11 hours ago













    @sds: right, the Argument Clinic conversion was done for 3.7. Tough luck for you!

    – Martijn Pieters
    11 hours ago







    @sds: right, the Argument Clinic conversion was done for 3.7. Tough luck for you!

    – Martijn Pieters
    11 hours ago






    1




    1





    Also math.isclose.__text_signature__ contains None for me on Python3.6.6.

    – Sanyash
    11 hours ago





    Also math.isclose.__text_signature__ contains None for me on Python3.6.6.

    – Sanyash
    11 hours ago













    @Sanyash: yes, that's what my answer is saying. That it requires 3.7 or newer for this to work.

    – Martijn Pieters
    11 hours ago





    @Sanyash: yes, that's what my answer is saying. That it requires 3.7 or newer for this to work.

    – Martijn Pieters
    11 hours ago













    And I personally am surprised that someone felt that the 3.7+ compatibility requirement for inspect.signature(math.isclose) was worthy of downvoting. Or is there some other aspect that I missed? In which case, please do leave me some feedback in a comment so I can improve my answer.

    – Martijn Pieters
    11 hours ago





    And I personally am surprised that someone felt that the 3.7+ compatibility requirement for inspect.signature(math.isclose) was worthy of downvoting. Or is there some other aspect that I missed? In which case, please do leave me some feedback in a comment so I can improve my answer.

    – Martijn Pieters
    11 hours ago











    0














    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))





    share|improve this answer






























      0














      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))





      share|improve this answer




























        0












        0








        0







        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))





        share|improve this answer















        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))






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 11 hours ago

























        answered 11 hours ago









        chepnerchepner

        260k34250343




        260k34250343






























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