python 3中execfile的替代方法是什么？

It seems they canceled in Python 3 all the easy way to quickly load a script by removing execfile()

Is there an obvious alternative I’m missing?

According to the documentation, instead of

execfile("./filename") 

Use

exec(open("./filename").read())

See:

• What’s New In Python 3.0

You are just supposed to read the file and exec the code yourself. 2to3 current replaces

execfile("somefile.py", global_vars, local_vars)

as

with open("somefile.py") as f:
    code = compile(f.read(), "somefile.py", 'exec')
    exec(code, global_vars, local_vars)

(The compile call isn’t strictly needed, but it associates the filename with the code object making debugging a little easier.)

See:

• http://docs.python.org/release/2.7.3/library/functions.html#execfile
• http://docs.python.org/release/3.2.3/library/functions.html#compile
• http://docs.python.org/release/3.2.3/library/functions.html#exec

While exec(open("filename").read()) is often given as an alternative to execfile("filename"), it misses important details that execfile supported.

The following function for Python3.x is as close as I could get to having the same behavior as executing a file directly. That matches running python /path/to/somefile.py.

def execfile(filepath, globals=None, locals=None):
    if globals is None:
        globals = {}
    globals.update({
        "__file__": filepath,
        "__name__": "__main__",
    })
    with open(filepath, 'rb') as file:
        exec(compile(file.read(), filepath, 'exec'), globals, locals)

# execute the file
execfile("/path/to/somefile.py")

Notes:

• Uses binary reading to avoid encoding issues
• Guaranteed to close the file (Python3.x warns about this)
• Defines __main__, some scripts depend on this to check if they are loading as a module or not for eg. if __name__ == "__main__"
• Setting __file__ is nicer for exception messages and some scripts use __file__ to get the paths of other files relative to them.

• Takes optional globals & locals arguments, modifying them in-place as execfile does – so you can access any variables defined by reading back the variables after running.

• Unlike Python2’s execfile this does not modify the current namespace by default. For that you have to explicitly pass in globals() & locals().

As suggested on the python-dev mailinglist recently, the runpy module might be a viable alternative. Quoting from that message:

https://docs.python.org/3/library/runpy.html#runpy.run_path

import runpy
file_globals = runpy.run_path("file.py")

There are subtle differences to execfile:

• run_path always creates a new namespace. It executes the code as a module, so there is no difference between globals and locals (which is why there is only a init_globals argument). The globals are returned.

  execfile executed in the current namespace or the given namespace. The semantics of locals and globals, if given, were similar to locals and globals inside a class definition.

• run_path can not only execute files, but also eggs and directories (refer to its documentation for details).

This one is better, since it takes the globals and locals from the caller:

import sys
def execfile(filename, globals=None, locals=None):
    if globals is None:
        globals = sys._getframe(1).f_globals
    if locals is None:
        locals = sys._getframe(1).f_locals
    with open(filename, "r") as fh:
        exec(fh.read()+"\n", globals, locals)

You could write your own function:

def xfile(afile, globalz=None, localz=None):
    with open(afile, "r") as fh:
        exec(fh.read(), globalz, localz)

If you really needed to…

If the script you want to load is in the same directory than the one you run, maybe “import” will do the job ?

If you need to dynamically import code the built-in function __ import__ and the module imp are worth looking at.

>>> import sys
>>> sys.path = ['/path/to/script'] + sys.path
>>> __import__('test')
<module 'test' from '/path/to/script/test.pyc'>
>>> __import__('test').run()
'Hello world!'

test.py:

def run():
        return "Hello world!"

If you’re using Python 3.1 or later, you should also take a look at importlib.

Here’s what I had (file is already assigned to the path to the file with the source code in both examples):

execfile(file)

Here’s what I replaced it with:

exec(compile(open(file).read(), file, 'exec'))

My favorite part: the second version works just fine in both Python 2 and 3, meaning it’s not necessary to add in version dependent logic.

Note that the above pattern will fail if you’re using PEP-263 encoding declarations that aren’t ascii or utf-8. You need to find the encoding of the data, and encode it correctly before handing it to exec().

class python3Execfile(object):
    def _get_file_encoding(self, filename):
        with open(filename, 'rb') as fp:
            try:
                return tokenize.detect_encoding(fp.readline)[0]
            except SyntaxError:
                return "utf-8"

    def my_execfile(filename):
        globals['__file__'] = filename
        with open(filename, 'r', encoding=self._get_file_encoding(filename)) as fp:
            contents = fp.read()
        if not contents.endswith("\n"):
            # http://bugs.python.org/issue10204
            contents += "\n"
        exec(contents, globals, globals)

Also, while not a pure Python solution, if you’re using IPython (as you probably should anyway), you can do:

%run /path/to/filename.py

Which is equally easy.

I’m just a newbie here so maybe it’s pure luck if I found this :

After trying to run a script from the interpreter prompt >>> with the command

    execfile('filename.py')

for which I got a “NameError: name ‘execfile’ is not defined” I tried a very basic

    import filename

it worked well :-)

I hope this can be helpful and thank you all for the great hints, examples and all those masterly commented pieces of code that are a great inspiration for newcomers !

I use Ubuntu 16.014 LTS x64. Python 3.5.2 (default, Nov 17 2016, 17:05:23) [GCC 5.4.0 20160609] on linux

To me, the cleanest approach is to use importlib and import the file as a module by path, like so:

from importlib import util

def load_file(name, path):
    spec = util.spec_from_file_location(name, path)
    module = util.module_from_spec(spec)
    spec.loader.exec_module(module)
    return module

Usage example

Let’s have a file foo.py:

print('i got imported')
def hello():
    print('hello from foo')

Now just import and use it like a normal module:

>>> foo = load_file('foo', './foo.py')
i got imported
>>> foo.hello()
hello from foo

I prefer this technique over direct approaches like exec(open(...)) because it does not clutter your namespaces or unnecessarily messes with $PATH.