hierarchical namespaces in custom python modules

Question

Tried searching the site, but cannot find an answer to my problem:

Lets say I have a module, named mymodule.py that contains:

def a():
    return 3

def b():
    return 4 + a()

Then the following works:

import mymodule
print(mymodule.b())

However, when I try defining the module contents dynamically:

import imp

my_code = '''
def a():
    return 3

def b():
    return 4 + a()
'''

mymodule = imp.new_module('mymodule')
exec(my_code, globals(), mymodule.__dict__)
print(mymodule.b())

Then it fails in function b():

Traceback (most recent call last):
File "", line 13, in <module>
File "", line 6, in b
NameError: global name 'a' is not defined

I need a way to preserve the hierarchical namespace searching in modules, which seems to fail unless the module resides on disk.

Any clues as to whats the difference?

Thanks, Rob.

Adam Wagner · Accepted Answer · 2012-06-27 04:15:47Z

3

You're close. You need tell exec to work in a different namespace like so (see note at bottom for python 3.x):

exec my_code in mymodule.__dict__

Full example:

import imp

my_code = '''
def a():
    return 3

def b():
    return 4 + a()
'''

mymodule = imp.new_module('mymodule')

exec my_code in mymodule.__dict__
print(mymodule.b())

This said, I've not used this before, so I'm not sure if there are any weird side-effects to this, but it looks like it works to me.

Also, there's a small blurb about 'exec in ...' in the python docs here: http://docs.python.org/reference/simple_stmts.html#the-exec-statement

Update

The reason your original attempt didn't work correctly is you were passing your current module's globals() dictionary, which is different than the globals() your new module should use.

This exec line also works (but isn't as pretty as the 'exec in ...' style):

exec(my_code, mymodule.__dict__, mymodule.__dict__)

Update 2: Since exec is now a function in python 3.x, it has no 'exec in ...' style, so the line above must be used.

edited Jun 27, 2012 at 4:15

answered Jun 27, 2012 at 3:59

Adam Wagner

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4 Comments

rbairos Over a year ago

is there a python 3.x version of "exec ... in " ? Can't seem to find it in the exec() function. Thanks

Adam Wagner Over a year ago

it doesn't look like it... see my update, the 'less pretty' version should work for you in 3.x (I've not tried it yet, so let me know if it works)

Adam Wagner Over a year ago

Alright, I've tested exec(my_code, mymodule.__dict__, mymodule.__dict__) in python3 and it seems to work fine. I've updated my answer to reflect this as well.

rbairos Over a year ago

Okay Ill give that a shot it seems very promising (I have to modify some things for my original project). Thanks very much.

Amber · Accepted Answer · 2012-06-27 03:45:25Z

0

That is not how exec() is designed to be used. exec() doesn't actually put the code it runs "into" any module - it runs it in the module-space that exec() was called from.

answered Jun 27, 2012 at 3:45

Amber

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1 Comment

rbairos Over a year ago

so are there any alternatives to exec() that reproduce the module behaviour I need? thanks

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