someone can tell me why this is incorrect as a singleton pattern:
class preSingleton(object):
def __call__(self):
return self
singleton = preSingleton()
# singleton is actually the singleton
a = singleton()
b = singleton()
print a==b
a.var_in_a = 100
b.var_in_b = 'hello'
print a.var_in_b
print b.var_in_a
Edit: The above code prints:
True
hello
100
thank you very much
Part Two
Maybe this is better?
class Singleton(object):
def __new__(cls):
return cls
a = Singleton()
b = Singleton()
print a == b
a.var_in_a = 100
b.var_in_b = 'hello'
print a.var_in_b
print b.var_in_a
Edit: The above code prints:
True
hello
100
Thanks again.
Singletons are actually really simple to make in Python. The trick is to have the module do your encapsulation for you and not make a class.
The module will only be initialized once
The module will not be initialized until the first time it is imported
Any attempts to re-import the module will return a pointer to the existing import
And if you want to pretend that the module is an instance of a class, you can do the following
import some_module
class SomeClass(object):
def __init__(self):
self.singleton = some_module
Because this is not a singleton. Singleton must be single, your object is not.
>>> class preSingleton(object):
... def __call__(self):
... return self
...
>>> singleton = preSingleton()
>>> singleton2 = preSingleton()
>>> singleton
<__main__.preSingleton object at 0x00C6D410>
>>> singleton2
<__main__.preSingleton object at 0x00C6D290>
This is actualy the Borg pattern. Multiple objects that share state.
That's not to say there's anything wrong with it, and for most if not all use cases it's functionaly equivalent to a singleton, but since you asked...
edit: Of course since they're Borg objects, each instance uses up more memory so if you're creating tons of them this will make a difference to resource usage.
Here's a sexy little singleton implemented as a decorator:
def singleton(cls):
"""Decorate a class with #singleton when There Can Be Only One."""
instance = cls()
instance.__call__ = lambda: instance
return instance
Use it like this:
#singleton
class MySingleton:
def spam(self):
print id(self)
What happens is that outside of the class definition, MySingleton will actually refer to the one and only instance of the class that exists, and you'll be left with no mechanism for creating any new instances. Calling MySingleton() will simply return the exact same instance. For example:
>>> MySingleton
<__main__.MySingleton instance at 0x7f474b9265a8>
>>> MySingleton()
<__main__.MySingleton instance at 0x7f474b9265a8>
>>> MySingleton() is MySingleton
True
>>> MySingleton.spam()
139944187291048
>>> MySingleton().spam()
139944187291048
I don't see the problem (if it walks like a duck and quacks like a duck...). Looks like a singleton to me.
It works differently from a Java singleton (for example) because Python uses the same syntax to call a function as to create a new instance of an object. So singleton() is actually calling the singleton object, which returns itself.
You can do this with your class:
>>> class preSingleton(object):
... def __call__(self):
... return self
...
>>> x = preSingleton()
>>> y = preSingleton()
>>> x == y
False
So, more than one instances of the class can be created and it violates the Singleton pattern.
Related
How can I get the class that defined a method in Python?
I'd want the following example to print "__main__.FooClass":
class FooClass:
def foo_method(self):
print "foo"
class BarClass(FooClass):
pass
bar = BarClass()
print get_class_that_defined_method(bar.foo_method)
import inspect
def get_class_that_defined_method(meth):
for cls in inspect.getmro(meth.im_class):
if meth.__name__ in cls.__dict__:
return cls
return None
I don't know why no one has ever brought this up or why the top answer has 50 upvotes when it is slow as hell, but you can also do the following:
def get_class_that_defined_method(meth):
return meth.im_class.__name__
For python 3 I believe this changed and you'll need to look into .__qualname__.
In Python 3, if you need the actual class object you can do:
import sys
f = Foo.my_function
vars(sys.modules[f.__module__])[f.__qualname__.split('.')[0]] # Gets Foo object
If the function could belong to a nested class you would need to iterate as follows:
f = Foo.Bar.my_function
vals = vars(sys.modules[f.__module__])
for attr in f.__qualname__.split('.')[:-1]:
vals = vals[attr]
# vals is now the class Foo.Bar
Thanks Sr2222 for pointing out I was missing the point...
Here's the corrected approach which is just like Alex's but does not require to import anything. I don't think it's an improvement though, unless there's a huge hierarchy of inherited classes as this approach stops as soon as the defining class is found, instead of returning the whole inheritance as getmro does. As said, this is a very unlikely scenario.
def get_class_that_defined_method(method):
method_name = method.__name__
if method.__self__:
classes = [method.__self__.__class__]
else:
#unbound method
classes = [method.im_class]
while classes:
c = classes.pop()
if method_name in c.__dict__:
return c
else:
classes = list(c.__bases__) + classes
return None
And the Example:
>>> class A(object):
... def test(self): pass
>>> class B(A): pass
>>> class C(B): pass
>>> class D(A):
... def test(self): print 1
>>> class E(D,C): pass
>>> get_class_that_defined_method(A().test)
<class '__main__.A'>
>>> get_class_that_defined_method(A.test)
<class '__main__.A'>
>>> get_class_that_defined_method(B.test)
<class '__main__.A'>
>>> get_class_that_defined_method(C.test)
<class '__main__.A'>
>>> get_class_that_defined_method(D.test)
<class '__main__.D'>
>>> get_class_that_defined_method(E().test)
<class '__main__.D'>
>>> get_class_that_defined_method(E.test)
<class '__main__.D'>
>>> E().test()
1
Alex solution returns the same results. As long as Alex approach can be used, I would use it instead of this one.
Python 3
Solved it in a very simple way:
str(bar.foo_method).split(" ", 3)[-2]
This gives
'FooClass.foo_method'
Split on the dot to get the class and the function name separately
I found __qualname__ is useful in Python3.
I test it like that:
class Cls(object):
def func(self):
print('1')
c = Cls()
print(c.func.__qualname__)
# output is: 'Cls.func'
def single_func():
print(2)
print(single_func.__module__)
# output: '__main__'
print(single_func.__qualname__)
# output: 'single_func'
After my test, I found another answer here.
I started doing something somewhat similar, basically the idea was checking whenever a method in a base class had been implemented or not in a sub class. Turned out the way I originally did it I could not detect when an intermediate class was actually implementing the method.
My workaround for it was quite simple actually; setting a method attribute and testing its presence later. Here's an simplification of the whole thing:
class A():
def method(self):
pass
method._orig = None # This attribute will be gone once the method is implemented
def run_method(self, *args, **kwargs):
if hasattr(self.method, '_orig'):
raise Exception('method not implemented')
self.method(*args, **kwargs)
class B(A):
pass
class C(B):
def method(self):
pass
class D(C):
pass
B().run_method() # ==> Raises Exception: method not implemented
C().run_method() # OK
D().run_method() # OK
UPDATE: Actually call method() from run_method() (isn't that the spirit?) and have it pass all arguments unmodified to the method.
P.S.: This answer does not directly answer the question. IMHO there are two reasons one would want to know which class defined a method; first is to point fingers at a class in debug code (such as in exception handling), and the second is to determine if the method has been re-implemented (where method is a stub meant to be implemented by the programmer). This answer solves that second case in a different way.
if you get this error:
'function' object has no attribute 'im_class'
try this:
import inspect
def get_class_that_defined_method(meth):
class_func_defided = meth.__globals__[meth.__qualname__.split('.')[0]]
#full_func_name = "%s.%s.%s"%(class_func_defided.__module__,class_func_defided.__name__,meth.__name__)
if inspect.isfunction(class_func_defided):
print("%s is not part of a class."%meth.__name__)
return None
return class_func_defided
sample test:
class ExampleClass:
#staticmethod
def ex_static_method():
print("hello from static method")
def ex_instance_method(self):
print("hello from instance method")
def ex_funct(self):
print("hello from simple function")
if __name__ == "__main__":
static_method_class = get_class_that_defined_method(ExampleClass.ex_static_method)
static_method_class.ex_static_method()
instance_method_class = get_class_that_defined_method(ExampleClass.ex_instance_method)
instance_method_class().ex_instance_method()
function_class = get_class_that_defined_method(ex_funct)
How can I get the class that defined a method in Python?
I'd want the following example to print "__main__.FooClass":
class FooClass:
def foo_method(self):
print "foo"
class BarClass(FooClass):
pass
bar = BarClass()
print get_class_that_defined_method(bar.foo_method)
import inspect
def get_class_that_defined_method(meth):
for cls in inspect.getmro(meth.im_class):
if meth.__name__ in cls.__dict__:
return cls
return None
I don't know why no one has ever brought this up or why the top answer has 50 upvotes when it is slow as hell, but you can also do the following:
def get_class_that_defined_method(meth):
return meth.im_class.__name__
For python 3 I believe this changed and you'll need to look into .__qualname__.
In Python 3, if you need the actual class object you can do:
import sys
f = Foo.my_function
vars(sys.modules[f.__module__])[f.__qualname__.split('.')[0]] # Gets Foo object
If the function could belong to a nested class you would need to iterate as follows:
f = Foo.Bar.my_function
vals = vars(sys.modules[f.__module__])
for attr in f.__qualname__.split('.')[:-1]:
vals = vals[attr]
# vals is now the class Foo.Bar
Thanks Sr2222 for pointing out I was missing the point...
Here's the corrected approach which is just like Alex's but does not require to import anything. I don't think it's an improvement though, unless there's a huge hierarchy of inherited classes as this approach stops as soon as the defining class is found, instead of returning the whole inheritance as getmro does. As said, this is a very unlikely scenario.
def get_class_that_defined_method(method):
method_name = method.__name__
if method.__self__:
classes = [method.__self__.__class__]
else:
#unbound method
classes = [method.im_class]
while classes:
c = classes.pop()
if method_name in c.__dict__:
return c
else:
classes = list(c.__bases__) + classes
return None
And the Example:
>>> class A(object):
... def test(self): pass
>>> class B(A): pass
>>> class C(B): pass
>>> class D(A):
... def test(self): print 1
>>> class E(D,C): pass
>>> get_class_that_defined_method(A().test)
<class '__main__.A'>
>>> get_class_that_defined_method(A.test)
<class '__main__.A'>
>>> get_class_that_defined_method(B.test)
<class '__main__.A'>
>>> get_class_that_defined_method(C.test)
<class '__main__.A'>
>>> get_class_that_defined_method(D.test)
<class '__main__.D'>
>>> get_class_that_defined_method(E().test)
<class '__main__.D'>
>>> get_class_that_defined_method(E.test)
<class '__main__.D'>
>>> E().test()
1
Alex solution returns the same results. As long as Alex approach can be used, I would use it instead of this one.
Python 3
Solved it in a very simple way:
str(bar.foo_method).split(" ", 3)[-2]
This gives
'FooClass.foo_method'
Split on the dot to get the class and the function name separately
I found __qualname__ is useful in Python3.
I test it like that:
class Cls(object):
def func(self):
print('1')
c = Cls()
print(c.func.__qualname__)
# output is: 'Cls.func'
def single_func():
print(2)
print(single_func.__module__)
# output: '__main__'
print(single_func.__qualname__)
# output: 'single_func'
After my test, I found another answer here.
I started doing something somewhat similar, basically the idea was checking whenever a method in a base class had been implemented or not in a sub class. Turned out the way I originally did it I could not detect when an intermediate class was actually implementing the method.
My workaround for it was quite simple actually; setting a method attribute and testing its presence later. Here's an simplification of the whole thing:
class A():
def method(self):
pass
method._orig = None # This attribute will be gone once the method is implemented
def run_method(self, *args, **kwargs):
if hasattr(self.method, '_orig'):
raise Exception('method not implemented')
self.method(*args, **kwargs)
class B(A):
pass
class C(B):
def method(self):
pass
class D(C):
pass
B().run_method() # ==> Raises Exception: method not implemented
C().run_method() # OK
D().run_method() # OK
UPDATE: Actually call method() from run_method() (isn't that the spirit?) and have it pass all arguments unmodified to the method.
P.S.: This answer does not directly answer the question. IMHO there are two reasons one would want to know which class defined a method; first is to point fingers at a class in debug code (such as in exception handling), and the second is to determine if the method has been re-implemented (where method is a stub meant to be implemented by the programmer). This answer solves that second case in a different way.
if you get this error:
'function' object has no attribute 'im_class'
try this:
import inspect
def get_class_that_defined_method(meth):
class_func_defided = meth.__globals__[meth.__qualname__.split('.')[0]]
#full_func_name = "%s.%s.%s"%(class_func_defided.__module__,class_func_defided.__name__,meth.__name__)
if inspect.isfunction(class_func_defided):
print("%s is not part of a class."%meth.__name__)
return None
return class_func_defided
sample test:
class ExampleClass:
#staticmethod
def ex_static_method():
print("hello from static method")
def ex_instance_method(self):
print("hello from instance method")
def ex_funct(self):
print("hello from simple function")
if __name__ == "__main__":
static_method_class = get_class_that_defined_method(ExampleClass.ex_static_method)
static_method_class.ex_static_method()
instance_method_class = get_class_that_defined_method(ExampleClass.ex_instance_method)
instance_method_class().ex_instance_method()
function_class = get_class_that_defined_method(ex_funct)
assume following class definition:
class A:
def f(self):
return 'this is f'
#staticmethod
def g():
return 'this is g'
a = A()
So f is a normal method and g is a static method.
Now, how can I check if the funcion objects a.f and a.g are static or not? Is there a "isstatic" funcion in Python?
I have to know this because I have lists containing many different function (method) objects, and to call them I have to know if they are expecting "self" as a parameter or not.
Lets experiment a bit:
>>> import types
>>> class A:
... def f(self):
... return 'this is f'
... #staticmethod
... def g():
... return 'this is g'
...
>>> a = A()
>>> a.f
<bound method A.f of <__main__.A instance at 0x800f21320>>
>>> a.g
<function g at 0x800eb28c0>
>>> isinstance(a.g, types.FunctionType)
True
>>> isinstance(a.f, types.FunctionType)
False
So it looks like you can use types.FunctionType to distinguish static methods.
Your approach seems a bit flawed to me, but you can check class attributes:
(in Python 2.7):
>>> type(A.f)
<type 'instancemethod'>
>>> type(A.g)
<type 'function'>
or instance attributes in Python 3.x
>>> a = A()
>>> type(a.f)
<type 'method'>
>>> type(a.g)
<type 'function'>
To supplement the answers here, in Python 3 the best way is like so:
import inspect
class Test:
#staticmethod
def test(): pass
isstatic = isinstance(inspect.getattr_static(Test, "test"), staticmethod)
We use getattr_static rather than getattr, since getattr will retrieve the bound method or function, not the staticmethod class object. You can do a similar check for classmethod types and property's (e.g. attributes defined using the #property decorator)
Note that even though it is a staticmethod, don't assume it was defined inside the class. The method source may have originated from another class. To get the true source, you can look at the underlying function's qualified name and module. For example:
class A:
#staticmethod:
def test(): pass
class B: pass
B.test = inspect.getattr_static(A, "test")
print("true source: ", B.test.__qualname__)
Technically, any method can be used as "static" methods, so long as they are called on the class itself, so just keep that in mind. For example, this will work perfectly fine:
class Test:
def test():
print("works!")
Test.test()
That example will not work with instances of Test, since the method will be bound to the instance and called as Test.test(self) instead.
Instance and class methods can be used as static methods as well in some cases, so long as the first arg is handled properly.
class Test:
def test(self):
print("works!")
Test.test(None)
Perhaps another rare case is a staticmethod that is also bound to a class or instance. For example:
class Test:
#classmethod
def test(cls): pass
Test.static_test = staticmethod(Test.test)
Though technically it is a staticmethod, it is really behaving like a classmethod. So in your introspection, you may consider checking the __self__ (recursively on __func__) to see if the method is bound to a class or instance.
I happens to have a module to solve this. And it's Python2/3 compatible solution. And it allows to test with method inherit from parent class.
Plus, this module can also test:
regular attribute
property style method
regular method
staticmethod
classmethod
For example:
class Base(object):
attribute = "attribute"
#property
def property_method(self):
return "property_method"
def regular_method(self):
return "regular_method"
#staticmethod
def static_method():
return "static_method"
#classmethod
def class_method(cls):
return "class_method"
class MyClass(Base):
pass
Here's the solution for staticmethod only. But I recommend to use the module posted here.
import inspect
def is_static_method(klass, attr, value=None):
"""Test if a value of a class is static method.
example::
class MyClass(object):
#staticmethod
def method():
...
:param klass: the class
:param attr: attribute name
:param value: attribute value
"""
if value is None:
value = getattr(klass, attr)
assert getattr(klass, attr) == value
for cls in inspect.getmro(klass):
if inspect.isroutine(value):
if attr in cls.__dict__:
bound_value = cls.__dict__[attr]
if isinstance(bound_value, staticmethod):
return True
return False
Why bother? You can just call g like you call f:
a = A()
a.f()
a.g()
I have a two part question.
>>> class One(object):
... pass
...
>>> class Two(object):
... pass
...
>>> def digest(constr):
... c = apply(constr)
... print c.__class__.__name__
... print constr.__class__.__name__
...
>>> digest(Two)
Two
type
How would one create object 'Two'? Neither constr() or c() work; and it seems that apply turns it into a type.
What happens when you pass a class rather and an instance into a method?
Classes are high level objects, so you can simply pass them like this:
def createMyClass ( myClass ):
obj = myClass()
return obj
class A ( object ):
pass
>>> x = createMyClass( A )
>>> type( x )
<class '__main__.A'>
How would one create object 'Two'?
Neither constr() or c() work; and it
seems that apply turns it into a
type.
The above comment was made in regards to this code:
>>> def digest(constr):
... c = apply(constr)
... print c.__class__.__name__
... print constr.__class__.__name__
apply (deprecated: see #pyfunc's answer) certainly does not turn the class Two into a type: It already is one.
>>> class Two(object): pass
...
>>> type(Two)
<type 'type'>
Classes are first class objects: they're instances of type. This makes sense if you look at the next example.
>>> two = Two()
>>> type(two)
<class '__main__.Two'>
You can see that a class very clearly functions as a type because it can be returned from type. Here's another example.
>>> Three = type('Three', (Two, ), {'foo': 'bar'})
>>> type(Three)
<type 'type'>
>>> three = Three()
>>> type(three)
<class '__main__.Three'>
You can see that type is a class that can be instantiated. Its constructor takes three arguments: the name of the class, a tuple of base classes and a dictionary containing the class attributes. It returns a new type aka class.
As to your final question,
What happens when you pass a class
rather and an instance into a method?
You're going to have to be more specific. Classes are just instances of type and so are first class objects. Asking what happens if I pass a class into a method is like asking what happens if I pass an integer into a method: It depends entirely on what the method is expecting.
Just another one example:
def InstanceFactory(classname):
cls = globals()[classname]
return cls()
class A(object):
def start(self):
print "a.start"
class B(object):
def start(self):
print "b.start"
InstanceFactory("A").start()
InstanceFactory("B").start()
If the class belongs to another module:
def InstanceFactory(modulename, classname):
if '.' in modulename:
raise ValueError, "can't handle dotted modules yet"
mod = __import__(modulename)
cls = getattr(mod, classname]
return cls()
I am confused though. Wasn't apply() deprecated since 2.3
http://www.python.org/dev/peps/pep-0290/
We don't need this any more.
apply(f, args, kwds) --> f(*args, **kwds)
Others have been moved / considered deprecated in modern usage:
buffer()
coerce()
and intern()
Simply use : Classname() to create an object.
How can I get the class that defined a method in Python?
I'd want the following example to print "__main__.FooClass":
class FooClass:
def foo_method(self):
print "foo"
class BarClass(FooClass):
pass
bar = BarClass()
print get_class_that_defined_method(bar.foo_method)
import inspect
def get_class_that_defined_method(meth):
for cls in inspect.getmro(meth.im_class):
if meth.__name__ in cls.__dict__:
return cls
return None
I don't know why no one has ever brought this up or why the top answer has 50 upvotes when it is slow as hell, but you can also do the following:
def get_class_that_defined_method(meth):
return meth.im_class.__name__
For python 3 I believe this changed and you'll need to look into .__qualname__.
In Python 3, if you need the actual class object you can do:
import sys
f = Foo.my_function
vars(sys.modules[f.__module__])[f.__qualname__.split('.')[0]] # Gets Foo object
If the function could belong to a nested class you would need to iterate as follows:
f = Foo.Bar.my_function
vals = vars(sys.modules[f.__module__])
for attr in f.__qualname__.split('.')[:-1]:
vals = vals[attr]
# vals is now the class Foo.Bar
Thanks Sr2222 for pointing out I was missing the point...
Here's the corrected approach which is just like Alex's but does not require to import anything. I don't think it's an improvement though, unless there's a huge hierarchy of inherited classes as this approach stops as soon as the defining class is found, instead of returning the whole inheritance as getmro does. As said, this is a very unlikely scenario.
def get_class_that_defined_method(method):
method_name = method.__name__
if method.__self__:
classes = [method.__self__.__class__]
else:
#unbound method
classes = [method.im_class]
while classes:
c = classes.pop()
if method_name in c.__dict__:
return c
else:
classes = list(c.__bases__) + classes
return None
And the Example:
>>> class A(object):
... def test(self): pass
>>> class B(A): pass
>>> class C(B): pass
>>> class D(A):
... def test(self): print 1
>>> class E(D,C): pass
>>> get_class_that_defined_method(A().test)
<class '__main__.A'>
>>> get_class_that_defined_method(A.test)
<class '__main__.A'>
>>> get_class_that_defined_method(B.test)
<class '__main__.A'>
>>> get_class_that_defined_method(C.test)
<class '__main__.A'>
>>> get_class_that_defined_method(D.test)
<class '__main__.D'>
>>> get_class_that_defined_method(E().test)
<class '__main__.D'>
>>> get_class_that_defined_method(E.test)
<class '__main__.D'>
>>> E().test()
1
Alex solution returns the same results. As long as Alex approach can be used, I would use it instead of this one.
Python 3
Solved it in a very simple way:
str(bar.foo_method).split(" ", 3)[-2]
This gives
'FooClass.foo_method'
Split on the dot to get the class and the function name separately
I found __qualname__ is useful in Python3.
I test it like that:
class Cls(object):
def func(self):
print('1')
c = Cls()
print(c.func.__qualname__)
# output is: 'Cls.func'
def single_func():
print(2)
print(single_func.__module__)
# output: '__main__'
print(single_func.__qualname__)
# output: 'single_func'
After my test, I found another answer here.
I started doing something somewhat similar, basically the idea was checking whenever a method in a base class had been implemented or not in a sub class. Turned out the way I originally did it I could not detect when an intermediate class was actually implementing the method.
My workaround for it was quite simple actually; setting a method attribute and testing its presence later. Here's an simplification of the whole thing:
class A():
def method(self):
pass
method._orig = None # This attribute will be gone once the method is implemented
def run_method(self, *args, **kwargs):
if hasattr(self.method, '_orig'):
raise Exception('method not implemented')
self.method(*args, **kwargs)
class B(A):
pass
class C(B):
def method(self):
pass
class D(C):
pass
B().run_method() # ==> Raises Exception: method not implemented
C().run_method() # OK
D().run_method() # OK
UPDATE: Actually call method() from run_method() (isn't that the spirit?) and have it pass all arguments unmodified to the method.
P.S.: This answer does not directly answer the question. IMHO there are two reasons one would want to know which class defined a method; first is to point fingers at a class in debug code (such as in exception handling), and the second is to determine if the method has been re-implemented (where method is a stub meant to be implemented by the programmer). This answer solves that second case in a different way.
if you get this error:
'function' object has no attribute 'im_class'
try this:
import inspect
def get_class_that_defined_method(meth):
class_func_defided = meth.__globals__[meth.__qualname__.split('.')[0]]
#full_func_name = "%s.%s.%s"%(class_func_defided.__module__,class_func_defided.__name__,meth.__name__)
if inspect.isfunction(class_func_defided):
print("%s is not part of a class."%meth.__name__)
return None
return class_func_defided
sample test:
class ExampleClass:
#staticmethod
def ex_static_method():
print("hello from static method")
def ex_instance_method(self):
print("hello from instance method")
def ex_funct(self):
print("hello from simple function")
if __name__ == "__main__":
static_method_class = get_class_that_defined_method(ExampleClass.ex_static_method)
static_method_class.ex_static_method()
instance_method_class = get_class_that_defined_method(ExampleClass.ex_instance_method)
instance_method_class().ex_instance_method()
function_class = get_class_that_defined_method(ex_funct)