CODEWITHSUNDEEP
pythonpython-2.7decoratorpython-decorators
con-f-use
con-f-use

Reputation: 4028

How can one attach a decorator to a function "after the fact" in python?

The way I understand decorators of function in python (and I might be wrong), is that they are supposed to add side effects and modify the return value of a function. Now decorators are added above the function definition of the function to be decorated or by an assignment. Here is a small example:

def print_args_decor(function):
    def wrapper(*args, **kwargs):
        print 'Arguments:', args, kwargs         # Added side-effect
        return function(*args, **kwargs)*5       # Modified return value
    return wrapper

@print_args_decor
def do_stuff(strg, n=10):
    """Repeats strg a few times."""
    return strg * n

new_decorated_func = print_args_decor(do_stuff)  # Decoration by assignment

print do_stuff('a', 2) # Output: aaaaaaaaaa

Now, how does one attach a decorator to a function defined elsewhere, ideally retaining the original function's name and docstring (like functools.wraps does)? For example, I'm importing the sqrt() function from Python's math module, and want to decorate it, how do I go about that?

from functools import wraps
from math import sqrt

def print_args_decor(function):
    @wraps(function)
    def wrapper(*args, **kwargs):
        print 'Arguments:', args, kwargs         # Added side-effect
        return function(*args, **kwargs)*5       # Modified return value
    return wrapper

# Decorate the sqrt() function from math module somehow
@print_args_decor #???
sqrt #???

print sqrt(9)   
# Output: 
# Arguments: ([9],) {}
# 15                   # <--- sqrt(9)*5

How about decorate methods within classes after the fact? How about decorating classes themselves?

Upvotes: 37

Views: 13348

Answers (1)

Martijn Pieters
Martijn Pieters

Reputation: 1123710

You imported sqrt into your module, just apply the decorator there in your own global namespace:

sqrt = print_args_decor(sqrt)

This sets the name sqrt in your module namespace to the result of the decorator. There is no requirement that sqrt was originally defined in this module.

It is up to the decorator to uses the functools.wraps() decorator to preserve function metadata such as the name and docstring.

Decorating a class is no different in this respect:

ClassName = decorator(ClassName)

On Python 2, for methods you need to be careful to grab the original unbound function; easiest is to use the method.__func__ attribute:

try:
    # Python 2
    ClassName.function_name = decorator(ClassName.function_name.__func__)
except AttributeError:
    # Python 3
    ClassName.function_name = decorator(ClassName.function_name)

I've wrapped the above in a try...except to make the pattern work across Python versions. The alternative is to grab the function object out of the class __dict__ to avoid the descriptor protocol from kicking in:

ClassName.function_name = decorator(ClassName.__dict__['function_name'])

Upvotes: 40

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