how to do a conditional decorator in python
Is it possible to decorator a function conditionally. For example, I want to decorate the function foo()
with a timer function (timeit
) only doing_performance_analysis is True
(see the psuedo-code below).
if doing_performance_analysis:
@timeit
def foo():
"""
do something, timeit function will return the time it takes
"""
time.sleep(2)
else:
def foo():
time.sleep(2)
Solution 1:
Decorators are simply callables that return a replacement, optionally the same function, a wrapper, or something completely different. As such, you could create a conditional decorator:
def conditional_decorator(dec, condition):
def decorator(func):
if not condition:
# Return the function unchanged, not decorated.
return func
return dec(func)
return decorator
Now you can use it like this:
@conditional_decorator(timeit, doing_performance_analysis)
def foo():
time.sleep(2)
The decorator could also be a class:
class conditional_decorator(object):
def __init__(self, dec, condition):
self.decorator = dec
self.condition = condition
def __call__(self, func):
if not self.condition:
# Return the function unchanged, not decorated.
return func
return self.decorator(func)
Here the __call__
method plays the same role as the returned decorator()
nested function in the first example, and the closed-over dec
and condition
parameters here are stored as arguments on the instance until the decorator is applied.
Solution 2:
A decorator is simply a function applied to another function. You can apply it manually:
def foo():
# whatever
time.sleep(2)
if doing_performance_analysis:
foo = timeit(foo)
Solution 3:
How about:
def foo():
...
if doing_performance_analysis:
foo = timeit(foo)
I imagine you could even wrap this into a decorator that would take a boolean flag and another decorator, and would only apply the latter if the flag is set to True
:
def cond_decorator(flag, dec):
def decorate(fn):
return dec(fn) if flag else fn
return decorate
@cond_decorator(doing_performance_analysis, timeit)
def foo():
...
Solution 4:
use_decorator = False
class myDecorator(object):
def __init__(self, f):
self.f = f
def __call__(self):
print "Decorated running..."
print "Entering", self.f.__name__
self.f()
print "Exited", self.f.__name__
def null(a):
return a
if use_decorator == False :
myDecorator = null
@myDecorator
def CoreFunction():
print "Core Function running"
CoreFunction()
Solution 5:
Blckknght's answer is great if you want to do the check every time you call the function, but if you have a setting that you can read once and never changes you may not want to check the setting every time the decorated function is called. In some of our high performance daemons at work I have written a decorator that checks a setting file once when the python file is first loaded and decides if it should wrap it or not.
Here is a sample
def timed(f):
def wrapper(*args, **kwargs):
start = datetime.datetime.utcnow()
return_value = f(*args, **kwargs)
end = datetime.datetime.utcnow()
duration = end - start
log_function_call(module=f.__module__, function=f.__name__, start=__start__, end=__end__, duration=duration.total_seconds())
if config.get('RUN_TIMED_FUNCTIONS'):
return wrapper
return f
Assuming that log_function_call logs your call to a database, logfile, or whatever and that config.get('RUN_TIMED_FUNCTIONS') checks your global configuration, then adding the @timed decorator to a function will check once on load to see if you are timing on this server, environment, etc. and if not then it won't change the execution of the function on production or the other environments where you care about performance.