I would like to do this:
def foo():
if <a magical condition>:
return x
else:
poof()
# or...
def foo():
x = <a magical object>
return x
def poof():
print 'poof!'
bar = foo() # bar points to <a magical object> but poof() is not called
foo() # prints 'poof!'
I guess it comes down to what the circumstanses are when the returned object's __del__ method is called. But maybe there is a better way. Like if the function itself knew it's returned value was being assigned. I guess I'm worried about relying on the timing of the garbage collection. Also I don't like that global at_end_of_program flag.
My solution:
class Magic:
def __del__(s):
poof()
def foo():
x = Magic()
return x
def poof():
if not at_end_of_program:
print 'poof!'
bar = foo() # No poof.
foo() # prints 'poof!'
I'm pretty confused by your question, but I think what you are trying to do is run a function when a value is reassigned.
Instead of doing tricky things with a __del__() method function, I suggest you just put your value into a class instance, and then overload __setattr__(). You could also overload __delattr__() to make sure you catch del(object.x) for your value x.
The very purpose of __setattr__() is to give you a hook to catch when something assigns to a member of your class. And you won't need any strange end_of_program flag. At the end of your program, just get rid of your overloaded function for __delattr__() so it doesn't get called for end-of-program cleanup.
A function can't tell what its return value is used for. Your solution will print poof if you re-assign to bar for example.
What's the real problem you are trying to solve?
Related
To start with, my question here is about the semantics and the logic behind why the Python language was designed like this in the case of chained decorators. Please notice the nuance how this is different from the question
How decorators chaining work?
Link: How decorators chaining work? It seems quite a number of other users had the same doubts, about the call order of chained Python decorators. It is not like I can't add a __call__ and see the order for myself. I get this, my point is, why was it designed to start from the bottom, when it comes to chained Python decorators?
E.g.
def first_func(func):
def inner():
x = func()
return x * x
return inner
def second_func(func):
def inner():
x = func()
return 2 * x
return inner
#first_func
#second_func
def num():
return 10
print(num())
Quoting the documentation on decorators:
The decorator syntax is merely syntactic sugar, the following two function definitions are semantically equivalent:
def f(arg):
...
f = staticmethod(f)
#staticmethod
def f(arg):
...
From this it follows that the decoration in
#a
#b
#c
def fun():
...
is equivalent to
fun = a(b(c(fun)))
IOW, it was designed like that because it's just syntactic sugar.
For proof, let's just decorate an existing function and not return a new one:
def dec1(f):
print(f"dec1: got {vars(f)}")
f.dec1 = True
return f
def dec2(f):
print(f"dec2: got {vars(f)}")
f.dec2 = True
return f
#dec1
#dec2
def foo():
pass
print(f"Fully decked out: {vars(foo)}")
prints out
dec2: got {}
dec1: got {'dec2': True}
Fully decked out: {'dec2': True, 'dec1': True}
TL;DR
g(f(x)) means applying f to x first, then applying g to the output.
Omit the parentheses, add # before and line break after each function name:
#g
#f
x
(Syntax only valid if x is the definition of a function/class.)
Abstract explanation
The reasoning behind this design decision becomes fairly obvious IMHO, if you remember what the decorator syntax - in its most abstract and general form - actually means. So I am going to try the abstract approach to explain this.
It is all about syntax
To be clear here, the distinguishing factor in the concept of the "decorator" is not the object underneath it (so to speak) nor the operation it performs. It is the special syntax and the restrictions for it. Thus, a decorator at its core is nothing more than feature of Python grammar.
The decorator syntax requires a target to be decorated. Initially (see PEP 318) the target could only be function definitions; later class definitions were also allowed to be decorated (see PEP 3129).
Minimal valid syntax
Syntactically, this is valid Python:
def f(): pass
#f
class Target: pass # or `def target: pass`
However, this will (perhaps unsuprisingly) cause a TypeError upon execution. As has been reiterated multiple times here and in other posts on this platform, the above is equivalent to this:
def f(): pass
class Target: pass
Target = f(Target)
Minimal working decorator
The TypeError stems from the fact that f lacks a positional argument. This is the obvious logical restriction imposed by what a decorator is supposed to do. Thus, to achieve not only syntactically valid code, but also have it run without errors, this is sufficient:
def f(x): pass
#f
class Target: pass
This is still not very useful, but it is enough for the most general form of a working decorator.
Decoration is just application of a function to the target and assigning the output to the target's name.
Chaining functions ⇒ Chaining decorators
We can ignore the target and what it is or does and focus only on the decorator. Since it merely stands for applying a function, the order of operations comes into play, as soon as we have more than one. What is the order of operation, when we chain functions?
def f(x): pass
def g(x): pass
class Target: pass
Target = g(f(Target))
Well, just like in the composition of purely mathematical functions, this implies that we apply f to Target first and then apply g to the result of f. Despite g appearing first (i.e. further left), it is not what is applied first.
Since stacking decorators is equivalent to nesting functions, it seems obvious to define the order of operation the same way. This time, we just skip the parentheses, add an # symbol in front of the function name and a line break after it.
def f(x): pass
def g(x): pass
#g
#f
class Target: pass
But, why though?
If after the explanation above (and reading the PEPs for historic background), the reasoning behind the order of operation is still not clear or still unintuitive, there is not really any good answer left, other than "because the devs thought it made sense, so get used to it".
PS
I thought I'd add a few things for additional context based on all the comments around your question.
Decoration vs. calling a decorated function
A source of confusion seems to be the distinction between what happens when applying the decorator versus calling the decorated function.
Notice that in my examples above I never actually called target itself (the class or function being decorated). Decoration is itself a function call. Adding #f above the target is calling the f and passing the target to it as the first positional argument.
A "decorated function" might not even be a function
The distinction is very important because nowhere does it say that a decorator actually needs to return a callable (function or class). f being just a function means it can return whatever it wants. This is again valid and working Python code:
def f(x): return 3.14
#f
def target(): return "foo"
try:
target()
except Exception as e:
print(repr(e))
print(target)
Output:
TypeError("'float' object is not callable")
3.14
Notice that the name target does not even refer to a function anymore. It just holds the 3.14 returned by the decorator. Thus, we cannot even call target. The entire function behind it is essentially lost immediately before it is even available to the global namespace. That is because f just completely ignores its first positional argument x.
Replacing a function
Expanding this further, if we want, we can have f return a function. Not doing that seems very strange, considering it is used to decorate a function. But it doesn't have to be related to the target at all. Again, this is fine:
def bar(): return "bar"
def f(x): return bar
#f
def target(): return "foo"
print(target())
print(target is bar)
Output:
bar
True
It comes down to convention
The way decorators are actually overwhelmingly used out in the wild, is in a way that still keeps a reference to the target being decorated around somewhere. In practice it can be as simple as this:
def f(x):
print(f"applied `f({x.__name__})`")
return
#f
def target(): return "foo"
Just running this piece of code outputs applied f(target). Again, notice that we don't call target here, we only called f. But now, the decorated function is still target, so we could add the call print(target()) at the bottom and that would output foo after the other output produced by f.
The fact that most decorators don't just throw away their target comes down to convention. You (as a developer) would not expect your function/class to simply be thrown away completely, when you use a decorator.
Decoration with wrapping
This is why real-life decorators typically either return the reference to the target at the end outright (like in the last example) or they return a different callable, but that callable itself calls the target, meaning a reference to the target is kept in that new callable's local namespace . These functions are what is usually referred to as wrappers:
def f(x):
print(f"applied `f({x.__name__})`")
def wrapper():
print(f"wrapper executing with {locals()=}")
return x()
return wrapper
#f
def target(): return "foo"
print(f"{target()=}")
print(f"{target.__name__=}")
Output:
applied `f(target)`
wrapper executing with locals()={'x': <function target at 0x7f1b2f78f250>}
target()='foo'
target.__name__='wrapper'
As you can see, what the decorator left us is wrapper, not what we originally defined as target. And the wrapper is what we call, when we write target().
Wrapping wrappers
This is the kind of behavior we typically expect, when we use decorators. And therefore it is not surprising that multiple decorators stacked together behave the way they do. The are called from the inside out (as explained above) and each adds its own wrapper around what it receives from the one applied before:
def f(x):
print(f"applied `f({x.__name__})`")
def wrapper_from_f():
print(f"wrapper_from_f executing with {locals()=}")
return x()
return wrapper_from_f
def g(x):
print(f"applied `g({x.__name__})`")
def wrapper_from_g():
print(f"wrapper_from_g executing with {locals()=}")
return x()
return wrapper_from_g
#g
#f
def target(): return "foo"
print(f"{target()=}")
print(f"{target.__name__=}")
Output:
applied `f(target)`
applied `g(wrapper_from_f)`
wrapper_from_g executing with locals()={'x': <function f.<locals>.wrapper_from_f at 0x7fbfc8d64f70>}
wrapper_from_f executing with locals()={'x': <function target at 0x7fbfc8d65630>}
target()='foo'
target.__name__='wrapper_from_g'
This shows very clearly the difference between the order in which the decorators are called and the order in which the wrapped/wrapping functions are called.
After the decoration is done, we are left with wrapper_from_g, which is referenced by our target name in global namespace. When we call it, wrapper_from_g executes and calls wrapper_from_f, which in turn calls the original target.
This simple Python code gets "False".
def foo():
def bar():
return 0
return bar
print(foo() == foo())
When I request
print(foo(),foo())
I get
<function foo.<locals>.bar at 0x03A0BC40> <function foo.<locals>.bar at 0x03C850B8>
So does Python store the result of the bar function every time in the new memory slot? I'd be happy if someone explain how it works behind the scene and possibly how this code can be a little bit modified to get "True" (which still seems logical to me!).
Each def statement defines a new function. The same name and body doesn't really matter. You're basically doing something like this:
def foo():
pass
old_foo = foo
def foo():
pass
assert old_foo == foo # will fail
bar is locally defined within foo the same way a local variable would be each time you call the foo().
As you can see by the defult __repr__, they have different memory addresses, and since bar does not implement __eq__, the two instances of bar will not be equal.
I have recently started using the mock framework in python. It seems that if I patch a function, the actual code is not called - which means that the database changes etc that this actual function does is not implemented.
I have been trying to go around it by calling the function before hand and storing the return value and passing it as arg in patch(), but
is there a better way to do it? Ideally, I would want a code that works as a silent observer and i can simply ask it if a certain observed function was called or not, how many times, and with what arguments
My current code
return_val = funct()
# C: Now call me again and assert that these intensive computation functions are not called but taken from cache
with patch('funct', return_value=return_val) as mock_task:
me_response = self.client.get(me_url, format='json') #should fetch from cache
assert not mock_task.called
To mock a method called, you should use the wraps keyword. Consider the following:
class Foo(object):
def do_thing(self, a):
print("A: %s" % a)
self._do_private_thing(a)
def _do_private_thing(self, a):
print("PRIVATE STUFF HAPPENING.")
print("A: %s" % a)
Then In your test you would have something like:
import mock
a = Foo()
with mock.patch.object(a, '_do_private_thing', wraps=a._do_private_thing) as private_mock:
a.do_thing("lol")
private_mock.assert_called_with("lol")
Hope this helps.
You could set the Mock#side_effect attribute to your original function.
orig = funct
funct = Mock(side_effect=orig)
I do find loganasherjones' answer more elegant.
Just adding another possibility for those who may need it.
I'd like to modify the arguments passed to a method in a module, as opposed to replacing its return value.
I've found a way around this, but it seems like something useful and has turned into a lesson in mocking.
module.py
from third_party import ThirdPartyClass
ThirdPartyClass.do_something('foo', 'bar')
ThirdPartyClass.do_something('foo', 'baz')
tests.py
#mock.patch('module.ThirdPartyClass.do_something')
def test(do_something):
# Instead of directly overriding its return value
# I'd like to modify the arguments passed to this function.
# change return value, no matter inputs
do_something.return_value = 'foo'
# change return value, based on inputs, but have no access to the original function
do_something.side_effect = lambda x, y: y, x
# how can I wrap do_something, so that I can modify its inputs and pass it back to the original function?
# much like a decorator?
I've tried something like the following, but not only is it repetitive and ugly, it doesn't work. After some PDB introspection.. I'm wondering if it's simply due to however this third party library works, as I do see the original functions being called successfully when I drop a pdb inside the side_effect.
Either that, or some auto mocking magic I'm just not following that I'd love to learn about.
def test():
from third_party import ThirdPartyClass
original_do_something = ThirdPartyClass.do_something
with mock.patch('module.ThirdPartyClass.do_something' as mocked_do_something:
def side_effect(arg1, arg2):
return original_do_something(arg1, 'overridden')
mocked_do_something.side_effect = side_effect
# execute module.py
Any guidance is appreciated!
You may want to use parameter wraps for the mock call. (Docs for reference.) This way the original function will be called, but it will have everything from Mock interface.
So for changing parameters called to original function you may want to try it like that:
org.py:
def func(x):
print(x)
main.py:
from unittest import mock
import org
of = org.func
def wrapped(a):
of('--{}--'.format(a))
with mock.patch('org.func', wraps=wrapped):
org.func('x')
org.func.assert_called_with('x')
result:
--x--
The trick is to pass the original underlying function that you still want to access as a parameter to the function.
Eg, for race condition testing, have tempfile.mktemp return an existing pathname:
def mock_mktemp(*, orig_mktemp=tempfile.mktemp, **kwargs):
"""Ensure mktemp returns an existing pathname."""
temp = orig_mktemp(**kwargs)
open(temp, 'w').close()
return temp
Above, orig_mktemp is evaluated when the function is declared, not when it is called, so all invocations will have access to the original method of tempfile.mktemp via orig_mktemp.
I used it as follows:
#unittest.mock.patch('tempfile.mktemp', side_effect=mock_mktemp)
def test_retry_on_existing_temp_path(self, mock_mktemp):
# Simulate race condition: creation of temp path after tempfile.mktemp
...
I have been working at learning Python over the last week and it has been going really well, however I have now been introduced to custom functions and I sort of hit a wall. While I understand the basics of it, such as:
def helloworld():
print("Hello World!")
helloworld()
I know this will print "Hello World!".
However, when it comes to getting information from one function to another, I find that confusing. ie: function1 and function2 have to work together to perform a task. Also, when to use the return command.
Lastly, when I have a list or a dictionary inside of a function. I'll make something up just as an example.
def my_function():
my_dict = {"Key1":Value1,
"Key2":Value2,
"Key3":Value3,
"Key4":Value4,}
How would I access the key/value and be able to change them from outside of the function? ie: If I had a program that let you input/output player stats or a character attributes in a video game.
I understand bits and pieces of this, it just confuses me when they have different functions calling on each other.
Also, since this was my first encounter with the custom functions. Is this really ambitious to pursue and this could be the reason for all of my confusion? Since this is the most complex program I have seen yet.
Functions in python can be both, a regular procedure and a function with a return value. Actually, every Python's function will return a value, which might be None.
If a return statement is not present, then your function will be executed completely and leave normally following the code flow, yielding None as a return value.
def foo():
pass
foo() == None
>>> True
If you have a return statement inside your function. The return value will be the return value of the expression following it. For example you may have return None and you'll be explicitly returning None. You can also have return without anything else and there you'll be implicitly returning None, or, you can have return 3 and you'll be returning value 3. This may grow in complexity.
def foo():
print('hello')
return
print('world')
foo()
>>>'hello'
def add(a,b):
return a + b
add(3,4)
>>>7
If you want a dictionary (or any object) you created inside a function, just return it:
def my_function():
my_dict = {"Key1":Value1,
"Key2":Value2,
"Key3":Value3,
"Key4":Value4,}
return my_dict
d = my_function()
d['Key1']
>>> Value1
Those are the basics of function calling. There's even more. There are functions that return functions (also treated as decorators. You can even return multiple values (not really, you'll be just returning a tuple) and a lot a fun stuff :)
def two_values():
return 3,4
a,b = two_values()
print(a)
>>>3
print(b)
>>>4
Hope this helps!
The primary way to pass information between functions is with arguments and return values. Functions can't see each other's variables. You might think that after
def my_function():
my_dict = {"Key1":Value1,
"Key2":Value2,
"Key3":Value3,
"Key4":Value4,}
my_function()
my_dict would have a value that other functions would be able to see, but it turns out that's a really brittle way to design a language. Every time you call my_function, my_dict would lose its old value, even if you were still using it. Also, you'd have to know all the names used by every function in the system when picking the names to use when writing a new function, and the whole thing would rapidly become unmanageable. Python doesn't work that way; I can't think of any languages that do.
Instead, if a function needs to make information available to its caller, return the thing its caller needs to see:
def my_function():
return {"Key1":"Value1",
"Key2":"Value2",
"Key3":"Value3",
"Key4":"Value4",}
print(my_function()['Key1']) # Prints Value1
Note that a function ends when its execution hits a return statement (even if it's in the middle of a loop); you can't execute one return now, one return later, keep going, and return two things when you hit the end of the function. If you want to do that, keep a list of things you want to return and return the list when you're done.
You send information into and out of functions with arguments and return values, respectively. This function, for example:
def square(number):
"""Return the square of a number."""
return number * number
... recieves information through the number argument, and sends information back with the return ... statement. You can use it like this:
>>> x = square(7)
>>> print(x)
49
As you can see, we passed the value 7 to the function, and it returned the value 49 (which we stored in the variable x).
Now, lets say we have another function:
def halve(number):
"""Return half of a number."""
return number / 2.0
We can send information between two functions in a couple of different ways.
Use a temporary variable:
>>> tmp = square(6)
>>> halve(tmp)
18.0
use the first function directly as an argument to the second:
>>> halve(square(8))
32.0
Which of those you use will depend partly on personal taste, and partly on how complicated the thing you're trying to do is.
Even though they have the same name, the number variables inside square() and halve() are completely separate from each other, and they're invisible outside those functions:
>>> number
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'number' is not defined
So, it's actually impossible to "see" the variable my_dict in your example function. What you would normally do is something like this:
def my_function(my_dict):
# do something with my_dict
return my_dict
... and define my_dict outside the function.
(It's actually a little bit more complicated than that - dict objects are mutable (which just means they can change), so often you don't actually need to return them. However, for the time being it's probably best to get used to returning everything, just to be safe).