Subclass `pathlib.Path` fails

问题

I would like to enhance the class pathlib.Path but the simple example above dose not work.

from pathlib import Path

class PPath(Path):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)

test = PPath("dir", "test.txt")

Here is the error message I have.

Traceback (most recent call last):
  File "/Users/projetmbc/test.py", line 14, in <module>
    test = PPath("dir", "test.txt")
  File "/anaconda/lib/python3.4/pathlib.py", line 907, in __new__
    self = cls._from_parts(args, init=False)
  File "/anaconda/lib/python3.4/pathlib.py", line 589, in _from_parts
    drv, root, parts = self._parse_args(args)
  File "/anaconda/lib/python3.4/pathlib.py", line 582, in _parse_args
    return cls._flavour.parse_parts(parts)
AttributeError: type object 'PPath' has no attribute '_flavour'

What I am doing wrong ?

回答1:

You can subclass the concrete implementation, so this works:

class Path(type(pathlib.Path())):

Here's what I did with this:

import pathlib

class Path(type(pathlib.Path())):
    def open(self, mode='r', buffering=-1, encoding=None, errors=None, newline=None):
        if encoding is None and 'b' not in mode:
            encoding = 'utf-8'
        return super().open(mode, buffering, encoding, errors, newline)

Path('/tmp/a.txt').write_text("я")

回答2:

Here is the definition of the Path class. It does something rather clever. Rather than directly returning an instance of Path from its __new__(), it returns an instance of a subclass, but only if it's been invoked directly as Path() (and not as a subclass).

Otherwise, it expects to have been invoked via either WindowsPath() or PosixPath(), which both provide a _flavour class attribute via multiple inheritance. You must also provide this attribute when subclassing. You'll probably need to instantiate and/or subclass the _Flavour class to do this. This is not a supported part of the API, so your code might break in a future version of Python.

TL;DR: This idea is fraught with peril, and I fear that my answers to your questions will be interpreted as approval rather than reluctant assistance.

回答3:

Note

I have opened a bug track here after a little discussion on the Python dev. list.

A temporary solution

Sorry for this double answer but here is a way to achieve what I want. Thanks to Kevin that points me to the source of pathlib and the fact we have here constructors.

import pathlib
import os

def _extramethod(cls, n):
    print("=== "*n)

class PathPlus(pathlib.Path):
    def __new__(cls, *args):
        if cls is PathPlus:
            cls = pathlib.WindowsPath if os.name == 'nt' else pathlib.PosixPath

        setattr(cls, "extramethod", _extramethod)

        return cls._from_parts(args)

test = PathPlus("C:", "Users", "projetmbc", "onefile.ext")

print("File ?", test.is_file())
print("Dir  ?", test.is_dir())
print("New name:", test.with_name("new.name"))
print("Drive ?", test.drive)

test.extramethod(4)

This prints the following lines.

File ? False
Dir  ? False
New name: C:/Users/projetmbc/new.name
Drive ? 
=== === === ===

回答4:

I have been struggling with this too.

Here is what i did, studying from the pathlib module. Seems to me that is the cleaner way to do it, but if the pathlib module changes its implementation, it probably won't hold.

from pathlib import Path
import os
import pathlib

class PPath(Path):

    _flavour = pathlib._windows_flavour if os.name == 'nt' else pathlib._posix_flavour

    def __new__(cls, *args):
        return super(PPath, cls).__new__(cls, *args)

    def __init__(self, *args):
        super().__init__() #Path.__init__ does not take any arg (all is done in new)
        self._some_instance_ppath_value = self.exists() #Path method

    def some_ppath_method(self, *args):
        pass

test = PPath("dir", "test.txt")

回答5:

Here is a simple way to do things regarding to the observation made by Kevin.

class PPath():
    def __init__(self, *args, **kwargs):
        self.path = Path(*args, **kwargs)

Then I will need to use a trick so as to automatically bind all the Path's methods to my PPpath class. I think that will be funny to do.

回答6:

It's work too.

from pathlib import Path

class SystemConfigPath(type(Path())):
    def __new__(cls, **kwargs):
        path = cls._std_etc()
        return super().__new__(cls, path, **kwargs)

    @staticmethod
    def _std_etc():
        return '/etc'

name = SystemConfigPath()
name = name / 'apt'
print(name)

Printed:

/etc/apt

@staticmethod can be replaced by @classmethod

来源：https://stackoverflow.com/questions/29850801/subclass-pathlib-path-fails

标签

oop

python-3.4