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unable to use attribute from a python function in another python script

I want to use one of the attributes returned by a function in a python script (x) into a python script (y)
The communication between both scripts works well; I can get functions and attributes, but doesn't allow me to attributes returned by a function.
Here is how I worked:
x.py
def func():
b = 10
a = 15
return [a,b]
c = 20
y.py
from x import func
import x
print (x.c)
print (func.b)
I get the "c" value and the following error AttributeError: 'function' object has no attribute 'b'
I have tried also to print x.b, and I've got AttributeError: module 'WorkingLSTM' has no attribute 'b'
Thanks in advance

The way to call func is by using func(), which would give you [a,b].
example:
funcResult = func()
a = funcResult[0]
b = funcResult[1]
funcResults is the return value from func(), that is the list [a,b].

That's not allowed, you have to call the function to get the value from the functions returned list.
a, b = func()
print(b)
# or just...
print(func()[1])
PS: It's "not allowed" because it doesn't make sense in any way; when there is no function call, there is not variable b at all. You might take a look at classes, they can hold static variables.

you cannot access local variables of a function.
these variables exist only during the the time where func is executed and are destroyed afterwards.
You can of course call the function and look at the result, but the result is just a list with two values
rslt = func()
print("A = ", rslt[0])
print("B = ", rslt[1])

The variable was declared inside a function making it a local variable and as such it can"t be accessed outside the function.
The variable is declared outside of the function making it a global variable and is not in anyway tied to your function.
The concept of attributes relates to Classes and you are dealing with a function so you might have to treat it as a class object.
If you are concerned bout accessing the local variables, you might as well do this:
y.py
from x import *
d = func() # func returns a list which is now
# identified/referenced by variable d
# displays the elements in the list using index position
print(d[0])
print(d[1])

If you want to use attributes, you may create a callable class instead of function:
class Func:
def __init__(self):
self.b = 10
self.a = 15
def __call__():
return [self.a, self.b]
func = Func()

Python has the concept of the scope. Local variables have no effect outside the function.
If you want to use it, use class and self or make getter function(but it's not Pythonic).
x.py
class X:
def __init__(self):
self.b = 10
self.a = 15
self.c = 20
def func(self):
return [self.a, self.b]
y.py
from x import X
x = X()
print(x.c)
print(x.func()[1])

List of objects function not working

Sorry for the title, I hope it reflects correctly my problem :
In the following code, I was expecting the result to be result 0 1 2 but instead I have 2 2 2. The code inside my_function seems to be interpreted with the last instance of obj. What is wrong ?
class Example:
def __init__(self, x):
self.x = x
def get(self):
return self.x
a_list = []
for index in range(3):
obj = Example(index)
def my_function(x):
#some stuff with x like obj.another_function(x)
return obj.get()
a_list.append(my_function)
for c in a_list:
print(c())

When you define this
def my_function():
return obj.get()
Python will understand that my_function should run the get() method of an object called obj and return the value. It won't know the value of obj and what the get() method does until you attempt to call it.
So, you are actually defining three different functions that will eventually do the same thing. And, in the end, running the same code thrice.
But why is the return 2 2 2?
Because after the last iteration, the value of obj is Example(2)* because you redefine its value at every iteration, and the last one remains.
*
because of this line obj = Example(index)

Understanding a few things about how python works will help you understand what's happening here. Here obj is a closure, closures are evaluated at call time, not when the function is defined so if I do this:
x = "hello"
def printX():
print x
x = "goodbye"
printX() # goodbye
I get "goodbye" because printX is referencing a global variable in my module, which changes after I create printX.
What you want to do is create a function with a closure that references a specific object. The functional way to do this is to create a function that returns another function:
x = "hello"
def makePrintX(a):
def printX():
# We print a, the object passed to `makePrintX`
print a
return printX
# x is evaluated here when it is still "hello"
myPrintX = makePrintX(x)
x = "goodbye"
myPrintX() # "hello"
If you're having trouble understanding the above example I would recommend reading up on python's scoping rules. For your example, you could do something like this:
class Example:
def __init__(self, x):
self.x = x
def get(self):
return self.x
def makeObjFunction(obj):
def objFunction(x):
return obj.get()
return objFunction
a_list = []
for index in range(3):
obj = Example(index)
my_function = makeObjFunction(obj)
a_list.append(my_function)
for c in a_list:
print(c("some value"))

You are appending three my_functions to the a_list which are all closures over the same Example object. Try:
def my_function():
return obj
<__main__.Example object at 0x0054EDF0>
<__main__.Example object at 0x0054EDF0>
<__main__.Example object at 0x0054EDF0>
You can see they have the same id so calling get() on each should give the same answer.
If you just append the obj.get function (and drop the my_function) it'll work fine.
a_list.append(obj.get)
....
0
1
2
Edit: You've updated your question so to let you do more stuff in my_function(). It's still basically a scoping problem.
def my_func_factory(p_obj):
def my_function(x):
#some stuff with x like obj.another_function(x)
return p_obj.get()
return my_function
for index in range(3):
obj = Example(index)
a_list.append(my_func_factory(obj))
Since my_function can't see obj being reassigned, each instance doesn't pick up the change.

I think append() during the for just append the function address in a_list[]. After for iteration, the a_list is really given the number. Then it discovers the address of my_function, and they get the number in my_function, this is, 2. That's why you get [2,2,2].
Or maybe, in my_function, function give the method of "obj". But for iteration change the "obj" memory address each time, so the symbol "obj" always aim to the newest object Example. Due to my_function always get "obj", you get the same number from the last object.

Static variable in Python?

In C++ we have static keyword which in loops is something like this:
for(int x=0; x<10; x++)
{
for(int y=0; y<10; y++)
{
static int number_of_times = 0;
number_of_times++;
}
}
static here makes number_of_times initialized once. How can I do same thing in python 3.x?
EDIT: Since most of the people got confused I would like to point out that the code I gave is just example of static usage in C++. My real problem is that I want to initialize only ONE time variable in function since I dont want it to be global(blah!) or default parameter..

Assuming what you want is "a variable that is initialised only once on first function call", there's no such thing in Python syntax. But there are ways to get a similar result:
1 - Use a global. Note that in Python, 'global' really means 'global to the module', not 'global to the process':
_number_of_times = 0
def yourfunc(x, y):
global _number_of_times
for i in range(x):
for j in range(y):
_number_of_times += 1
2 - Wrap you code in a class and use a class attribute (ie: an attribute that is shared by all instances). :
class Foo(object):
_number_of_times = 0
#classmethod
def yourfunc(cls, x, y):
for i in range(x):
for j in range(y):
cls._number_of_times += 1
Note that I used a classmethod since this code snippet doesn't need anything from an instance
3 - Wrap you code in a class, use an instance attribute and provide a shortcut for the method:
class Foo(object):
def __init__(self):
self._number_of_times = 0
def yourfunc(self, x, y):
for i in range(x):
for j in range(y):
self._number_of_times += 1
yourfunc = Foo().yourfunc
4 - Write a callable class and provide a shortcut:
class Foo(object):
def __init__(self):
self._number_of_times = 0
def __call__(self, x, y):
for i in range(x):
for j in range(y):
self._number_of_times += 1
yourfunc = Foo()
4 bis - use a class attribute and a metaclass
class Callable(type):
def __call__(self, *args, **kw):
return self._call(*args, **kw)
class yourfunc(object):
__metaclass__ = Callable
_numer_of_times = 0
#classmethod
def _call(cls, x, y):
for i in range(x):
for j in range(y):
cls._number_of_time += 1
5 - Make a "creative" use of function's default arguments being instantiated only once on module import:
def yourfunc(x, y, _hack=[0]):
for i in range(x):
for j in range(y):
_hack[0] += 1
There are still some other possible solutions / hacks, but I think you get the big picture now.
EDIT: given the op's clarifications, ie "Lets say you have a recursive function with default parameter but if someone actually tries to give one more argument to your function it could be catastrophic", it looks like what the OP really wants is something like:
# private recursive function using a default param the caller shouldn't set
def _walk(tree, callback, level=0):
callback(tree, level)
for child in tree.children:
_walk(child, callback, level+1):
# public wrapper without the default param
def walk(tree, callback):
_walk(tree, callback)
Which, BTW, prove we really had Yet Another XY Problem...

You can create a closure with nonlocal to make them editable (python 3.x only). Here's an example of a recursive function to calculate the length of a list.
def recursive_len(l):
res = 0
def inner(l2):
nonlocal res
if l2:
res += 1
inner(l2[1:])
inner(l)
return res
Or, you can assign an attribute to the function itself. Using the trick from here:
def fn(self):
self.number_of_times += 1
fn.func_defaults = (fn,)
fn.number_of_times = 0
fn()
fn()
fn()
print (fn.number_of_times)

Python doesn't have static variables by design. For your example, and use within loop blocks etc. in general, you just use a variable in an outer scope; if that makes it too long-lived, it might be time to consider breaking up that function into smaller ones.
For a variable that continues to exist between calls to a function, that's just reimplementing the basic idea of an object and a method on that object, so you should make one of those instead.

The another function-based way of doing this in python is:
def f(arg, static_var=[0]):
static_var[0] += arg
As the static_var object is initialised at the function definition, and then reused for all the calls, it will act like a static variable. Note that you can't just use an int, as they are immutable.
>>> def f(arg, static_var=[0]):
... static_var[0] += arg
... print(static_var[0])
...
>>> f(1)
1
>>> f(2)
3
>>> f(3)
6

You can also use the global keyword:
def main(args):
for i in xrange(10):
print i
global tmp
tmp = i
But be careful... I most cases it will add more issues than it solves.

Use defaultdict:
from collections import defaultdict
static = defaultdict(lambda: 0)
def myfunc():
for x in range(10):
for y in range(10):
static['number_of_times'] += 1

Static member of a function in Python ? [duplicate]

This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Static class variables in Python
What is the Python equivalent of static variables inside a function?
How can I use static fields in Python ?
for example i want to count how many times the function has been called - how can i do this ?

If you wish to count how many times a method has been called, no matter which instance called it, you could use a class member like this:
class Foo(object):
calls=0 # <--- call is a class member
def baz(self):
Foo.calls+=1
foo=Foo()
bar=Foo()
for i in range(100):
foo.baz()
bar.baz()
print('Foo.baz was called {n} times'.format(n=foo.calls))
# Foo.baz was called 200 times
When you define calls this way:
class Foo(object):
calls=0
Python places the key-value pair ('calls', 0) in Foo.__dict__.
It can be accessed with Foo.calls.
Instances of Foo, such as foo=Foo(), can access it with foo.calls as well.
To assign new values to Foo.calls you must use Foo.calls = ....
Instances can not use foo.calls = ... because that causes Python to place a new and different key-value pair in foo.__dict__, where instance members are kept.

Here's a decorator adding counting to a function.
import functools
def count_calls(func):
#functools.wraps(func)
def decor(*args, **kwargs):
decor.count += 1
return func(*args, **kwargs)
decor.count = 0
return decor
Usage:
>>> #count_calls
... def foo():
... pass
...
>>> foo.count
0
>>> foo()
>>> foo.count
1

Here is some example counting the number of calls of all objects of the same class:
class Swallow():
i = 0 # will be used for counting calls of fly()
def fly(self):
Swallow.i += 1
And this is the proof:
>>> a = Swallow()
>>> b = Swallow()
>>> a.fly()
>>> a.i
1
>>> Swallow.i
1
>>> b.fly()
>>> b.i
2
>>> Swallow.i
2
so you can read it by giving the object name or class name.

Here's one simplistic way to do it:
def func():
if not hasattr(func, 'counter'):
func.counter = 0
func.counter += 1
counter = 0 # Not the same as `func.counter`
print(func.counter)
Or if you don't like the if being executed on every call, you can do:
def func():
func.counter += 1
print(func.counter)
func.counter = 0

Is it possible to make Python functions behave like instances?

I understand that functions can have attributes. So I can do the following:
def myfunc():
myfunc.attribute += 1
print(myfunc.attribute)
myfunc.attribute = 1
Is it possible by any means to make such a function behave as if it were an instance? For example, I'd like to be able to do something like this:
x = clever_wrapper(myfunc)
y = clever_wrapper(myfunc)
x.attribute = 5
y.attribute = 9
x() # I want this to print 6 (from the 5 plus increment)
y() # I want this to print 10 (from the 9 plus increment)
As it stands, there is only one "instance" of the function, so attribute only exists once. Modifying it by either x or y changes the same value. I'd like each of them to have their own attribute. Is that possible to do at all? If so, can you provide a simple, functional example?
It is important that I be able to access attribute from inside of the function but have the value of attribute be different depending on which "instance" of the function is called. Essentially, I'd like to use attribute as if it were another parameter to the function (so that it could change the behavior of the function) but not pass it in. (Suppose that the signature of the function were fixed so that I cannot change the parameter list.) But I need to be able to set the different values for attribute and then call the functions in sequence. I hope that makes sense.
The main answers seem to be saying to do something like this:
class wrapper(object):
def __init__(self, target):
self.target = target
def __call__(self, *args, **kwargs):
return self.target(*args, **kwargs)
def test(a):
return a + test.attribute
x = wrapper(test)
y = wrapper(test)
x.attribute = 2
y.attribute = 3
print(x.attribute)
print(y.attribute)
print(x(3))
print(y(7))
But that doesn't work. Maybe I've done it incorrectly, but it says that test does not have attribute. (I'm assuming that it's because wrapper actually has the attribute.)
The reason I need this is because I have a library that expects a function with a particular signature. It's possible to put those functions into a pipeline of sorts so that they're called in order. I'd like to pass it multiple versions of the same function but change their behavior based on an attribute's value. So I'd like to be able to add x and y to the pipeline, as opposed to having to implement a test1 function and a test2 function that both do almost exactly the same thing (except for the value of the attribute).

You can make a class with a __call__ method which would achieve a similar thing.
Edit for clarity: Instead of making myfunc a function, make it a callable class. It walks like a function and it quacks like a function, but it can have members like a class.

A nicer way:
def funfactory( attribute ):
def func( *args, **kwargs ):
# stuff
print( attribute )
# more stuff
return func
x = funfactory( 1 )
y = funfactory( 2 )
x( ) # 1
y( ) # 2
This works because the functions are closures, so they will grab all local variables in their scope; this causes a copy of attribute to be passed around with the function.

class Callable(object):
def __init__(self, x):
self.x = x
def __call__(self):
self.x += 1
print self.x
>> c1 = Callable(5)
>> c2 = Callable(20)
>> c1()
6
>> c1()
7
>> c2()
21
>> c2()
22

A generator might be an alternate solution here:
def incgen(init):
while True:
init += 1
print init
yield
x = incgen(5)
y = incgen(9)
x.next() # prints 6
y.next() # prints 10
y.next() # prints 11
x.next() # prints 7
You can't dig back in to the generator and manipulate the data though.

#!/usr/bin/env python
# encoding: utf-8
class Callable(object):
attribute = 0
def __call__(self, *args, **kwargs):
return self.attribute
def main():
c = Callable()
c.attribute += 1
print c()
if __name__ == '__main__':
main()