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?
12 Answers
According to the documentation, instead of
execfile("./filename")
Use
exec(open("./filename").read())
See:
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:
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
execfiledoes - so you can access any variables defined by reading back the variables after running.Unlike Python2's
execfilethis does not modify the current namespace by default. For that you have to explicitly pass inglobals()&locals().
As suggested on the python-dev mailinglist recently, the runpy module might be a viable alternative. Quoting from that message:
import runpy file_globals = runpy.run_path("file.py")
There are subtle differences to execfile:
run_pathalways 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 ainit_globalsargument). The globals are returned.execfileexecuted in the current namespace or the given namespace. The semantics oflocalsandglobals, if given, were similar to locals and globals inside a class definition.run_pathcan 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.
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
def load_file_as_module(name, location):
spec = util.spec_from_file_location(name, location)
module = util.module_from_spec(spec)
spec.loader.exec_module(module)
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'>