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, Dec 06 '17 at 20:48

TallChuck · Answer 1 · 2017-12-06T23:19:42.857

0

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

edited Dec 06 '17 at 23:19

answered Dec 06 '17 at 20:36

TallChuck

What do I do if my Test is located in a different directory? (apart from path append) – Chinmoy Kulkarni Dec 06 '17 at 20:47

CodeCollector · Answer 2 · 2017-12-06T21:04:45.263

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

edited Dec 06 '17 at 21:04

answered Dec 06 '17 at 20:52

CodeCollector

Is there a way to not have any variable associated "var". Directly access C – Chinmoy Kulkarni Dec 06 '17 at 20:59
How would you access if x is in a different directory? – Chinmoy Kulkarni Dec 06 '17 at 21:11
Doesn't that kind of defeat the purpose of x.py ? Wasn't it supposed to call files from different directory? – CodeCollector Dec 06 '17 at 21:36
I wanted a single instruction which can do it in the shell. possible? – Chinmoy Kulkarni Dec 08 '17 at 22:04

2 Answers2