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Can I call a global function from a function that has the same name?

For example:

def sorted(services):
    return {sorted}(services, key=lambda s: s.sortkey())

By {sorted} I mean the global sorted function. Is there a way to do this? I then want to call my function with the module name: service.sorted(services)

I want to use the same name, because it does the same thing as the global function, except that it adds a default argument.

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Python's name-resolution scheme which sometimes is referred to as LEGB rule, implies that when you use an unqualified name inside a function, Python searches up to four scopes— First the local (L) scope, then the local scopes of any enclosing (E) defs and lambdas, then the global (G) scope, and finally the built-in (B) scope. (Note that it will stops the search as soon as it finds a match)

So when you use sorted inside the functions interpreter considers it as a Global name (your function name) so you will have a recursion function. if you want to access to built-in sorted you need to specify that for Python . by __builtin__ module (in Python-2.x ) and builtins in Python-3.x (This module provides direct access to all ‘built-in’ identifiers of Python)


python 2 :

import __builtin__
def sorted(services):
    return __builtin__.sorted(services, key=lambda s: s.sortkey())

python 3 :

import builtins
def sorted(services):
    return builtins.sorted(services, key=lambda s: s.sortkey())

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