Using the example
def foo(a):
def bar(b):
return a+b
return bar
d = {1:foo(1), 2:foo(2)}
It appears that pickle module will not work with a function not defined at the module scope, so pickling 'd' will not work. Is there another pickling mechanism available that I should consider?
I'm afraid that you can't pickle nested functions.
The pickle module serializes functions by name. That is, if you have a function myfunc in a module mymodule it simply saves the name mymodule.myfunc and looks it up again when unserializing. (This is an important security and compatibility issue, as it guarantees that the unserializing code uses its own definition for the function, rather than the original definition which might be compromised or obsolete.)
Alas, pickle can't do that with nested functions, because there's no way to directly address them by name. Your bar function, for instance, can't be accessed from outside of foo.
If you need a serializable object that works like a function, you can instead make a class with a __call__ method:
class foo(object):
def __init__(self, a):
self.a = a
def __call__(self, b): # the function formerly known as "bar"
return self.a + b
This works just like the nested functions in the question, and should pose no problem to pickle. Do be aware though, that you'll need to have the same class definition available when you unserialize a foo instance.
You can pickle nested functions if you use dill instead of pickle.
>>> import dill
>>>
>>> def foo(a):
... def bar(b):
... return a+b
... return bar
...
>>> d = {1:foo(1), 2:foo(2)}
>>>
>>> _d = dill.dumps(d)
>>> d_ = dill.loads(_d)
>>> d_
{1: <function bar at 0x108cfe848>, 2: <function bar at 0x108cfe8c0>}
>>> d[1](0) + d[2](10)
13
>>>
according to Blckknght's answersing. if nested function is the only extrac serialized type and will use it as decorator, you can just add functools.warps at top of inner function defination to lead other interpret find the correct name:
from functools import warps
def foo(func):
#wraps(func)
def bar(b):
return func(b)
return bar
#foo
def zzz(b):
return b
Related
I have a class A with some member functions that all do the same thing.
class A:
def a():
... boilerplate code ...
b = c = d = a
For debugging reasons, I would like to know the name of each member function at runtime. But since they all point to the same memory address, they will have the same __name__ attribute and I cannot figure out a way to distinguish between A.a and A.b just by looking at the object.
a = A.a
b = A.b
a.__name__ == b.__name__ # this is true
# how do I tell the difference between a and b?
Is there a way to achieve this without manually creating the functions b, c and d with the same boilerplate code?
No. Objects and names in Python live in separate spaces. There's only one function object there, and the function object doesn't know through what name it was conjured.
If you were a masochist, I suppose it would be possible to get a traceback and look at the line of code that called you, but that's just not practical.
You could do something like:
def reala(self,me=None):
pass
def a(self):
return reala('a')
def b(self):
return reala('b')
...
Consider:
class A(object):
def f(self): print("A")
class B(A):
def f(self): print("B")
b = B()
I can call A.f on b by doing:
A.f(b)
Is there an easy way to "wrap" b such that wrap(b).f() calls A.f for any f?
Here is my solution which copies the methods from the most upper base class:
import types, copy
def get_all_method_names(clazz):
return [func for func in dir(clazz) if callable(getattr(clazz, func))]
def wrap(obj):
obj = copy.copy(obj)
obj_clazz = obj.__class__
base_clazz = obj_clazz.__bases__[-1] # the one which directly inherits from object
base_methods = get_all_method_names(base_clazz) # list of all method names in base_clazz
for base_method_name in base_methods:
base_method = getattr(base_clazz, base_method_name) # get the method object
if isinstance(base_method, types.FunctionType): # skip dunder methods like __class__, __init__
setattr(obj, base_method_name, base_method) # copy it into our object
return obj
# class declaration from question here
wrapped_b = wrap(b)
wrapped_b.f(wrapped_b) # prints A, unfortunately we have to pass the self parameter explicitly
b.f() # prints B, proof that the original object is untouched
This feels dirty to me, but it also seems to work. I'm not sure I'd rely on this for anything important.
import copy
def upcast(obj, clazz):
if not isinstance(obj, clazz): # make sure we're actually "upcasting"
raise TypeError()
wrapped = copy.copy(obj)
wrapped.__class__ = clazz
return wrapped
This results in
>>> a = A()
>>> a.f()
A
>>> b = B()
>>> b.f()
B
>>> upcast(b, A).f()
A
What I've really done here is essentially monkey-patch a clone of b and lied to it and told it it's actually an A, so when it comes time to resolve which version of f to call, it'll call the one from A.
Object Slicing is not supported in python the way it is done in C++ (The link you are pointing to takes a cpp example).
In Python Object Slicing is a rather different thing which means to slice up any object which supports sequence protocol (implements getitem() and len() methods).
Example :
A = [1,2,3,4,5,6,7,8]
print(A[1:3])
But in C++ Object Slicing is just cutting off the properties added by a base class instance when assigned to a parent class variable.
I need to patch current datetime in tests. I am using this solution:
def _utcnow():
return datetime.datetime.utcnow()
def utcnow():
"""A proxy which can be patched in tests.
"""
# another level of indirection, because some modules import utcnow
return _utcnow()
Then in my tests I do something like:
with mock.patch('***.utils._utcnow', return_value=***):
...
But today an idea came to me, that I could make the implementation simpler by patching __call__ of function utcnow instead of having an additional _utcnow.
This does not work for me:
from ***.utils import utcnow
with mock.patch.object(utcnow, '__call__', return_value=***):
...
How to do this elegantly?
When you patch __call__ of a function, you are setting the __call__ attribute of that instance. Python actually calls the __call__ method defined on the class.
For example:
>>> class A(object):
... def __call__(self):
... print 'a'
...
>>> a = A()
>>> a()
a
>>> def b(): print 'b'
...
>>> b()
b
>>> a.__call__ = b
>>> a()
a
>>> a.__call__ = b.__call__
>>> a()
a
Assigning anything to a.__call__ is pointless.
However:
>>> A.__call__ = b.__call__
>>> a()
b
TLDR;
a() does not call a.__call__. It calls type(a).__call__(a).
Links
There is a good explanation of why that happens in answer to "Why type(x).__enter__(x) instead of x.__enter__() in Python standard contextlib?".
This behaviour is documented in Python documentation on Special method lookup.
[EDIT]
Maybe the most interesting part of this question is Why I cannot patch somefunction.__call__?
Because the function don't use __call__'s code but __call__ (a method-wrapper object) use function's code.
I don't find any well sourced documentation about that, but I can prove it (Python2.7):
>>> def f():
... return "f"
...
>>> def g():
... return "g"
...
>>> f
<function f at 0x7f1576381848>
>>> f.__call__
<method-wrapper '__call__' of function object at 0x7f1576381848>
>>> g
<function g at 0x7f15763817d0>
>>> g.__call__
<method-wrapper '__call__' of function object at 0x7f15763817d0>
Replace f's code by g's code:
>>> f.func_code = g.func_code
>>> f()
'g'
>>> f.__call__()
'g'
Of course f and f.__call__ references are not changed:
>>> f
<function f at 0x7f1576381848>
>>> f.__call__
<method-wrapper '__call__' of function object at 0x7f1576381848>
Recover original implementation and copy __call__ references instead:
>>> def f():
... return "f"
...
>>> f()
'f'
>>> f.__call__ = g.__call__
>>> f()
'f'
>>> f.__call__()
'g'
This don't have any effect on f function. Note: In Python 3 you should use __code__ instead of func_code.
I Hope that somebody can point me to the documentation that explain this behavior.
You have a way to work around that: in utils you can define
class Utcnow(object):
def __call__(self):
return datetime.datetime.utcnow()
utcnow = Utcnow()
And now your patch can work like a charm.
Follow the original answer that I consider even the best way to implement your tests.
I've my own gold rule: never patch protected methods. In this case the things are little bit smoother because protected method was introduced just for testing but I cannot see why.
The real problem here is that you cannot to patch datetime.datetime.utcnow directly (is C extension as you wrote in the comment above). What you can do is to patch datetime by wrap the standard behavior and override utcnow function:
>>> with mock.patch("datetime.datetime", mock.Mock(wraps=datetime.datetime, utcnow=mock.Mock(return_value=3))):
... print(datetime.datetime.utcnow())
...
3
Ok that is not really clear and neat but you can introduce your own function like
def mock_utcnow(return_value):
return mock.Mock(wraps=datetime.datetime,
utcnow=mock.Mock(return_value=return_value)):
and now
mock.patch("datetime.datetime", mock_utcnow(***))
do exactly what you need without any other layer and for every kind of import.
Another solution can be import datetime in utils and to patch ***.utils.datetime; that can give you some freedom to change datetime reference implementation without change your tests (in this case take care to change mock_utcnow() wraps argument too).
As commented on the question, since datetime.datetime is written in C, Mock can't replace attributes on the class (see Mocking datetime.today by Ned Batchelder). Instead you can use freezegun.
$ pip install freezegun
Here's an example:
import datetime
from freezegun import freeze_time
def my_now():
return datetime.datetime.utcnow()
#freeze_time('2000-01-01 12:00:01')
def test_freezegun():
assert my_now() == datetime.datetime(2000, 1, 1, 12, 00, 1)
As you mention, an alternative is to track each module importing datetime and patch them. This is in essence what freezegun does. It takes an object mocking datetime, iterates through sys.modules to find where datetime has been imported and replaces every instance. I guess it's arguable whether you can do this elegantly in one function.
I have a base decorator that takes arguments but that also is built upon by other decorators. I can't seem to figure where to put the functools.wraps in order to preserve the full signature of the decorated function.
import inspect
from functools import wraps
# Base decorator
def _process_arguments(func, *indices):
""" Apply the pre-processing function to each selected parameter """
#wraps(func)
def wrap(f):
#wraps(f)
def wrapped_f(*args):
params = inspect.getargspec(f)[0]
args_out = list()
for ind, arg in enumerate(args):
if ind in indices:
args_out.append(func(arg))
else:
args_out.append(arg)
return f(*args_out)
return wrapped_f
return wrap
# Function that will be used to process each parameter
def double(x):
return x * 2
# Decorator called by end user
def double_selected(*args):
return _process_arguments(double, *args)
# End-user's function
#double_selected(2, 0)
def say_hello(a1, a2, a3):
""" doc string for say_hello """
print('{} {} {}'.format(a1, a2, a3))
say_hello('say', 'hello', 'arguments')
The result of this code should be and is:
saysay hello argumentsarguments
However, running help on say_hello gives me:
say_hello(*args, **kwargs)
doc string for say_hello
Everything is preserved except the parameter names.
It seems like I just need to add another #wraps() somewhere, but where?
I experimented with this:
>>> from functools import wraps
>>> def x(): print(1)
...
>>> #wraps(x)
... def xyz(a,b,c): return x
>>> xyz.__name__
'x'
>>> help(xyz)
Help on function x in module __main__:
x(a, b, c)
AFAIK, this has nothing to do with wraps itself, but an issue related to help. Indeed, because help inspects your objects to provide the information, including __doc__ and other attributes, this is why you get this behavior, although your wrapped function has different argument list. Though, wraps doesn't update that automatically (the argument list) what it really updates is this tuple and the __dict__ which is technically the objects namespace:
WRAPPER_ASSIGNMENTS = ('__module__', '__name__', '__qualname__', '__doc__',
'__annotations__')
WRAPPER_UPDATES = ('__dict__',)
If you aren't sure about how wraps work, probably it'll help if your read the the source code from the standard library: functools.py.
It seems like I just need to add another #wraps() somewhere, but where?
No, you don't need to add another wraps in your code, help as I stated above works that way by inspecting your objects. The function's arguments are associated with code objects (__code__) because your function's arguments are stored/represented in that object, wraps has no way to update the argument of the wrapper to be like the wrapped function (continuing with the above example):
>>> xyz.__code__.co_varnames
>>> xyz.__code__.co_varnames = x.__code__.co_varnames
AttributeError: readonly attribute
If help displayed that function xyz has this argument list () instead of (a, b, c) then this is clearly wrong! And the same applies for wraps, to change the argument list of the wrapper to the wrapped, would be cumbersome! So this should not be a concern at all.
>>> #wraps(x, ("__code__",))
... def xyz(a,b,c): pass
...
>>> help(xyz)
Help on function xyz in module __main__:
xyz()
But xyz() returns x():
>>> xyz()
1
For other references take a look at this question or the Python Documentation
What does functools.wraps do?
direprobs was correct in that no amount of functools wraps would get me there. bravosierra99 pointed me to somewhat related examples. However, I couldn't find a single example of signature preservation on nested decorators in which the outer decorator takes arguments.
The comments on Bruce Eckel's post on decorators with arguments gave me the biggest hints in achieving my desired result.
The key was in removing the middle function from within my _process_arguments function and placing its parameter in the next, nested function. It kind of makes sense to me now...but it works:
import inspect
from decorator import decorator
# Base decorator
def _process_arguments(func, *indices):
""" Apply the pre-processing function to each selected parameter """
#decorator
def wrapped_f(f, *args):
params = inspect.getargspec(f)[0]
args_out = list()
for ind, arg in enumerate(args):
if ind in indices:
args_out.append(func(arg))
else:
args_out.append(arg)
return f(*args_out)
return wrapped_f
# Function that will be used to process each parameter
def double(x):
return x * 2
# Decorator called by end user
def double_selected(*args):
return _process_arguments(double, *args)
# End-user's function
#double_selected(2, 0)
def say_hello(a1, a2,a3):
""" doc string for say_hello """
print('{} {} {}'.format(a1, a2, a3))
say_hello('say', 'hello', 'arguments')
print(help(say_hello))
And the result:
saysay hello argumentsarguments
Help on function say_hello in module __main__:
say_hello(a1, a2, a3)
doc string for say_hello
I have code which contains the following two lines in it:-
instanceMethod = new.instancemethod(testFunc, None, TestCase)
setattr(TestCase, testName, instanceMethod)
How could it be re-written without using the "new" module? Im sure new style classes provide some kind of workaround for this, but I am not sure how.
There is a discussion that suggests that in python 3, this is not required. The same works in Python 2.6
http://mail.python.org/pipermail/python-list/2009-April/531898.html
See:
>>> class C: pass
...
>>> c=C()
>>> def f(self): pass
...
>>> c.f = f.__get__(c, C)
>>> c.f
<bound method C.f of <__main__.C instance at 0x10042efc8>>
>>> c.f
<unbound method C.f>
>>>
Reiterating the question for every one's benefit, including mine.
Is there a replacement in Python3 for new.instancemethod? That is, given an arbitrary instance (not its class) how can I add a new appropriately defined function as a method to it?
So following should suffice:
TestCase.testFunc = testFunc.__get__(None, TestCase)
You can replace "new.instancemethod" by "types.MethodType":
from types import MethodType as instancemethod
class Foo:
def __init__(self):
print 'I am ', id(self)
def bar(self):
print 'hi', id(self)
foo = Foo() # prints 'I am <instance id>'
mm = instancemethod(bar, foo) # automatically uses foo.__class__
mm() # prints 'I have been bound to <same instance id>'
foo.mm # traceback because no 'field' created in foo to hold ref to mm
foo.mm = mm # create ref to bound method in foo
foo.mm() # prints 'I have been bound to <same instance id>'
This will do the same:
>>> Testcase.testName = testFunc
Yeah, it's really that simple.
Your line
>>> instanceMethod = new.instancemethod(testFunc, None, TestCase)
Is in practice (although not in theory) a noop. :) You could just as well do
>>> instanceMethod = testFunc
In fact, in Python 3 I'm pretty sure it would be the same in theory as well, but the new module is gone so I can't test it in practice.
To confirm that it's not needed to use new.instancemthod() at all since Python v2.4, here's an example how to replace an instance method. It's also not needed to use descriptors (even though it works).
class Ham(object):
def spam(self):
pass
h = Ham()
def fake_spam():
h._spam = True
h.spam = fake_spam
h.spam()
# h._spam should be True now.
Handy for unit testing.