# What is an alternative to execfile in Python 3?

## The Question :

375 people think this question is useful

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?

• reload is back, as imp.reload, since 3.2.
• If you are using Python interactively consider using IPython: %run script_name works with all version of Python.
• Since 3.4 imp is importlib (which must be imported): importlib.reload(mod_name) imports and executes mod_name.
• what’s wrong with runfile(“filename.py”) ?
• Thanks @mousomer!! I was precisely looking for the functionality of runfile() since I needed to run a Python script that executes in its own namespace (as opposed to executing on the calling namespace). My application: add the directory of the called script to the system path (sys.path) using the __file__ attribute: if we use execfile() or its equivalent in Python 3 (exec(open('file.py').read())) the included script is run in the calling namespace and thus __file__ resolves to the calling file name.

415 people think this answer is useful

According to the documentation, instead of

execfile("./filename")



Use

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



See:

222 people think this answer is useful

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:
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:

75 people think this answer is useful

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:

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



Notes:

• Uses binary reading to avoid encoding issues
• 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().

69 people think this answer is useful

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).

21 people think this answer is useful

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:



17 people think this answer is useful

You could write your own function:

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



If you really needed to…

13 people think this answer is useful

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.

9 people think this answer is useful

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.

5 people think this answer is useful

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:
except SyntaxError:
return "utf-8"

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



4 people think this answer is useful

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.

1 people think this answer is useful

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

1 people think this answer is useful

Avoid exec() if you can. For most applications, it’s cleaner to make use of Python’s import system.

This function uses built-in importlib to execute a file as an actual module:

from importlib import util

spec = util.spec_from_file_location(name, location)
module = util.module_from_spec(spec)
return module



## Usage example

Let’s have a file foo.py:

def hello():
return 'hi from module!'
print('imported from', __file__, 'as', __name__)



And import it as a regular module:

>>> mod = load_file_as_module('mymodule', './foo.py')
imported from /tmp/foo.py as mymodule
>>> mod.hello()
hi from module!
>>> type(mod)
<class 'module'>



This approach doesn’t pollute namespaces or messes with your \$PATH whereas exec() runs code directly in the context of the current function, potentially causing name collisions. Also, module attributes like __file__ and __name__ will be set correctly, and code locations are preserved. So, if you’ve attached a debugger or if the module raises an exception, you will get usable tracebacks.
Note that one minor difference from static imports is that the module gets imported (executed) every time you run load_file_as_module(), and not just once as with the import keyword.