python - _functools module

Question

How does this import work, what file does it use?

import _functools

In python 2.5:

import _functools
print _functools.__file__

Gives:

Traceback (most recent call last):
  File "D:\zjm_code\mysite\zjmbooks\a.py", line 5, in <module>
    print _functools.__file__
AttributeError: 'module' object has no attribute '__file__'

How can I get the meaning of partial (from _functools import partial) if I can't read C code?

Answer 1

C-coded modules can be built-in (lacking __file__) or live in a .so or .pyd dynamic library (which their __file__ will indicate) -- that's an implementation detail that you should not care about.

If you want to understand how a Python-callable, C-coded function works by studying code, learning to read C is generally best (far less hard than actually productively coding in C;-). However, often you'll find (suggestive, non-authoritative) "sample Python implementations" of C-coded functionality, and you can study those.

A particularly fruitful repository of Python-coded equivalents to Python standard library functionality that's normally coded in C is the pypy project (which does Python implementations coded in Python) -- its sources are browseable here and of course you can download and peruse them on your machine.

In particular, this is pypy's _functools.py implementation:

""" Supplies the internal functions for functools.py in the standard library """

class partial:
    """
    partial(func, *args, **keywords) - new function with partial application
    of the given arguments and keywords.
    """
    __slots__ = ['func', 'args', 'keywords']

    def __init__(self, func, *args, **keywords):
        if not callable(func):
            raise TypeError("the first argument must be callable")
        self.func = func
        self.args = args
        self.keywords = keywords

    def __call__(self, *fargs, **fkeywords):
        newkeywords = self.keywords.copy()
        newkeywords.update(fkeywords)
        return self.func(*(self.args + fargs), **newkeywords)

Pretty trivial to read and understand, I hope!

Answer 2

please be more clear when asking questions next time. I assume you want this

>>> import _functools
>>> _functools.__file__
'/usr/lib/python2.6/lib-dynload/_functools.so'

Answer 3

It is a shared object _functools.so, written in C. I guess, it is imported by the actual functools module. Why are you trying to import from it? Sure, it is possible, however, you can just as well from functools import partial.

Answer 4

For Python2.6, The _functools module is a builtin. You can see that if you simply type import _functools ; repr(_functools) and hit enter at your interpreter prompt.

If you want to see the C source of the module, check out http://hg.python.org/cpython/file/d7e85ddb1336/Modules/_functoolsmodule.c

This _functools module doesn't have a __file__ attribute (see next paragraph) since it's compiled into the interpreter.

For Python2.5, The _functools module is a standard library module implemented in C and is available at http://hg.python.org/cpython/file/a78381ead4cf/Modules/_functoolsmodule.c if you want to see it. You can see the location from where the module is loaded up by typing import _functools ; print _functools.__file__ at your interpreter prompt

Answer 5

The behaviour of the functools.partial() function is described in PEP 309 in which it was defined. The although the actual implementations of built-ins are often in C PEP's including this one usually contain example implementations in Python.

This example code should be sufficient for you to understand the behaviour of functools.partial(). If there is a specific issue about the C implementation that is concerning you are going to have to read the C code. Or perhaps describe it in your question and someone might know the answer.