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Python docstrings templated

Why doesn't dynamically formatting docstrings work? Is there an acceptable workaround for doing this at function definition time?

>>> DEFAULT_BAR = "moe's tavern"
>>> def foo(bar=DEFAULT_BAR):
...     """
...     hello this is the docstring
...     
...     Args:
...       bar (str)    the bar argument (default: {})
...     """.format(DEFAULT_BAR)
... 
>>> foo.__doc__
>>> foo.__doc__ is None
True

I tried with old-skool style %s formatting and that didn't work either.

like image 568
wim Avatar asked Dec 15 '22 06:12

wim


1 Answers

Your string requires the function to be called, but function bodies are not executed when a function is created.

A proper docstring is not executed, it is simply taken from the parsed sourcecode and attached to the function object, no code is ever executed for this. Python stores the docstring as the first constant value in a code object:

>>> def f():
...     """docstring"""
...     pass
...
>>> f.__code__.co_consts
('docstring', None)

where the code object was passed to the function type when constructing a new function (see the PyFunction_New() function).

See the Function definitions reference documentation:

The function definition does not execute the function body; this gets executed only when the function is called. [3]

[...]

[3] A string literal appearing as the first statement in the function body is transformed into the function’s __doc__ attribute and therefore the function’s docstring.

Your definition is otherwise valid; there is just no stand-alone string literal at the top of the function body. Your string literal is instead part of the function itself, and is only executed when the function is called (and the result discarded as you don't store that).

Note that the __doc__ attribute on a function object is writable; you can always apply variables after creating the function:

>>> DEFAULT_BAR = "moe's tavern"
>>> def foo(bar=DEFAULT_BAR):
...     """
...     hello this is the docstring
...
...     Args:
...       bar (str)    the bar argument (default: {})
...     """
...
>>> foo.__doc__ = foo.__doc__.format(DEFAULT_BAR)
>>> print(foo.__doc__)

    hello this is the docstring

    Args:
      bar (str)    the bar argument (default: moe's tavern)

You could do that in a decorator with the help of functionobject.__globals__ and inspect.getargspec() perhaps, but then do use named slots in the template so you can apply everything as a dictionary and have the docstring choose what to interpolate:

from inspect import getargspec

def docstringtemplate(f):
    """Treat the docstring as a template, with access to globals and defaults"""
    spec = getargspec(f)
    defaults = {} if not spec.defaults else dict(zip(spec.args[-len(spec.defaults):], spec.defaults))
    f.__doc__ = f.__doc__ and f.__doc__.format(**dict(f.__globals__, **defaults))
    return f

Demo:

>>> @docstringtemplate
... def foo(bar=DEFAULT_BAR):
...     """
...     hello this is the docstring
...
...     Args:
...       bar (str)    the bar argument (default: {bar!r}, or {DEFAULT_BAR!r})
...
...     """
...
>>> print(foo.__doc__)

    hello this is the docstring

    Args:
      bar (str)    the bar argument (default: "moe's tavern", or "moe's tavern")

Function keyword arguments override globals, as they would in the function.

like image 72
Martijn Pieters Avatar answered Jan 05 '23 19:01

Martijn Pieters