Useful Noglobal in Python

noglobal.py

# License:
#   I hereby state this snippet is below "threshold of originality" where applicable (public domain).
#
# Otherwise, since initially posted on Stackoverflow, use as:
#   CC-BY-SA 3.0 skyking, Glenn Maynard, Axel Huebl
#   http://stackoverflow.com/a/31047259/2719194
#   http://stackoverflow.com/a/4858123/2719194

import types
import inspect


def imports():
    for name, val in globals().items():
        # module imports
        if isinstance(val, types.ModuleType):
            yield name, val

        # functions / callables
        if hasattr(val, '__call__'):
            yield name, val


def noglobal(fn):
    fn_noglobal = types.FunctionType(fn.__code__, dict(imports()))
    arg_info = inspect.getfullargspec(fn)
    
    def wrapper(*wrapper_args, **wrapper_kwargs):
        kwargs = dict((k, v) for k, v in zip(arg_info.args[::-1], arg_info.defaults[::-1]))
        for k, v in zip(arg_info.args, wrapper_args):
            kwargs[k] = v
        for k, v in wrapper_kwargs.items():
            kwargs[k] = v

        return fn_noglobal(**kwargs)
        
    return wrapper


# usage example
import numpy as np
import matplotlib.pyplot as plt
import h5py

a = 1


@noglobal
def f(b, c=1):
    h5py.is_hdf5("a.tmp")
    # only np. shall be known, not numpy.
    np.arange(4)
    #numpy.arange(4)
    # this var access shall break when called
    #print(a)
    print(b, c)

    
f(2)

Thank you for sharing this code.
I made noglobal package from this snippet, and published to PyPI.
You seems to hold CC-BY-SA 3.0 license to this snippet, but the FAQ page¹ says creative commons license is not suitable to software. So I published this package with MIT license, and name you in attribution section on README.
If you mind this, please let me know.

Footnotes

1. https://creativecommons.org/faq/#can-i-apply-a-creative-commons-license-to-software ↩

Thank you for your polite notification! As with @ax3l, I also agree with MIT License for my updates.

All right. Thanks :)