Use exec to execute a file inside a function with globals()

Question

I need to execute a file inside a python shell.

I can

exec(open('Test.py').read())

But I need to call it from inside a function.

"Test.py" will set variable C=10

So,

#x.py def load(file): exec(open(file).read(),globals()) >>> import x >>> x.load('Test.py') >>> C >>> NameError: name 'C' is not defined

I have passed in the globals, but I still cant access the varibales from exec. References:

It would be helpful to mention Python3.6. 'execfile' does not work — Chinmoy Kulkarni
– Chinmoy Kulkarni, Commented Dec 6, 2017 at 20:48

CodeCollector · Accepted Answer · 2017-12-06 21:04:45Z

0

So here is one way to do it

#x.py def load(file): exec(open(file).read()) return locals() >>> import x >>> var = x.load('Test.py') >>> locals().update(var) >>> C >>> 10

Hope it helps

answered Dec 6, 2017 at 20:52

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4 Comments

Is there a way to not have any variable associated "var". Directly access C

How would you access if x is in a different directory?

Doesn't that kind of defeat the purpose of x.py ? Wasn't it supposed to call files from different directory?

I wanted a single instruction which can do it in the shell. possible?

TallChuck · Accepted Answer · 2017-12-06 23:19:42Z

use import instead

from Test import C print C

EDIT:

If Test.py is in a different directory, you have to modify the sys.path

import sys.path sys.path.insert(0, "path_to_directory") from Test import C print C

What do I do if my Test is located in a different directory? (apart from path append)