65

I am trying to access a local function variable outside the function in Python. So, for example,

bye = ''
def hi():
    global bye
    something
    something
    bye = 5
    sigh = 10

hi()
print bye

The above works fine as it should. Since I want to find out if I can access bye outside hi() without using global bye, I tried:

def hi():
    something
    something
    bye = 5 
    sigh = 10
    return

hi()
x = hi()
print x.bye 

The above gives AttributeError: 'NoneType' object has no attribute 'bye'.

Then, I tried:

def hi():
    something
    something
    bye = 5
    sigh = 10
    return bye 
hi()
x = hi()
print x.bye

This time it doesn't give even an error.

So, is there a way to access a local function variable (bye) outside its function (hi()) without using globals and without printing out variable sigh as well? (Question was edited to include sigh after @hcwhsa 's comment below.

martineau
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askance
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    `bye = hi();print bye` – Ashwini Chaudhary Oct 11 '13 at 19:45
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    What are you trying to achieve? What is the use case? – Rod Oct 11 '13 at 19:50
  • Thanks @hcwhsa ! This works. A follow-up question. Suppose the function defines more than 1 variable but I want to print just the `bye` variable. This solution, however, would print both. I'll edit my question accordingly. – askance Oct 11 '13 at 19:53
  • @Rod - My use case involves using the local variable outside the function. At the same time, the function has multiple variables and so, I have included the `sigh` variable. – askance Oct 11 '13 at 20:01
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    Are you really just interested in learning how to use `return` or something? – binki Jan 11 '17 at 17:44
  • @binki - Maybe the question implies it and if so , i still need to learn more about the return fn. could you pls clarify how the question connects with the properties of the rtn fn? Thx , – askance Jan 16 '17 at 14:14
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    If you are using the function to calculate a value and then using that value from outside of the function, then you might as well just directly return the calculated value. Based on the code you’ve shown, you don’t need static variables and you don’t need to access the value through the function itself. You’re just trying to figure out how to give the caller the value you calculated. Returning values is what [`return`](https://docs.python.org/3/reference/simple_stmts.html#return) does. See [this gist](https://gist.github.com/binki/cebd6e5305aa799b386a27c4e12a9873) and let me know. – binki Jan 16 '17 at 16:43

5 Answers5

127

You could do something along these lines (which worked in both Python v2.7.17 and v3.8.1 when I tested it/them):

def hi():
    # other code...
    hi.bye = 42  # Create function attribute.
    sigh = 10

hi()
print(hi.bye)  # -> 42

Functions are objects in Python and can have arbitrary attributes assigned to them.

If you're going to be doing this kind of thing often, you could implement something more generic by creating a function decorator that adds a this argument to each call to the decorated function.

This additional argument will give functions a way to reference themselves without needing to explicitly embed (hardcode) their name into the rest of the definition and is similar to the instance argument that class methods automatically receive as their first argument which is usually named self — I picked something different to avoid confusion, but like the self argument, it can be named whatever you wish.

Here's an example of that approach:

def add_this_arg(func):
    def wrapped(*args, **kwargs):
        return func(wrapped, *args, **kwargs)
    return wrapped

@add_this_arg
def hi(this, that):
    # other code...
    this.bye = 2 * that  # Create function attribute.
    sigh = 10

hi(21)
print(hi.bye)  # -> 42

Note

This doesn't work for class methods. Just use the instance argument, named self by convention, that's already passed to methods instead of the method's name. You can reference class-level attributes through type(self). See Function's attributes when in a class.

martineau
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    You're welcome, but I must add that this practice is fairly unusual. may be hard to debug, and makes the code hard to maintain -- because essentially it's the same thing as using global variables, and which have long [been considered harmful](http://c2.com/cgi/wiki?GlobalVariablesConsideredHarmful). – martineau Oct 12 '13 at 02:11
  • This do not work if you use it in this way `if hi(): print hi.bye`. Anyone know why? – m3nda Dec 06 '15 at 19:00
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    @erm3nda: As defined, the `hi()` function has no return value, so it effectively returns `None` which is considered a `False` value, so the `if` doesn't execute the conditional `print` statement. – martineau Dec 06 '15 at 19:27
  • @ReeshabhRanjan: Yes it does. The `print hi.bye` needs to be changed to `print(hi.bye)` because of the differences between Python 2 and 3—see [**Print Is A Function**](https://docs.python.org/3.0/whatsnew/3.0.html#print-is-a-function) in the [**_What’s New In Python 3.0_**](https://docs.python.org/3.0/whatsnew/3.0.html) document. Since the newer syntax would work in both versions in this particular case, I'll update my answer to avoid confusing folks... – martineau Sep 10 '17 at 15:19
  • @martineau I tried it but it didn't work. Will try again and update you. – Reeshabh Ranjan Sep 10 '17 at 17:35
  • @martineau I am extremely sorry, I forgot. Well, this doesn't work in Python 3.6 atleast: https://codepad.co/snippet/UPYhKngG – Reeshabh Ranjan Nov 08 '17 at 17:58
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    @ReeshabhRanjan: As I said before, it **does** work, at least for me anyway, using Python 3.6.3 (and there's really no reason why it wouldn't)—I just ran it again to make sure. If it doesn't work for you, then there must be something else wrong. Note that you have to call the function at least once before trying to access the attribute so it will be created. – martineau Nov 08 '17 at 18:21
  • @martineau great tip, thank you – dimButTries Jan 07 '22 at 09:39
10

The problem is you were calling print x.bye after you set x as a string. When you run x = hi() it runs hi() and sets the value of x to 5 (the value of bye; it does NOT set the value of x as a reference to the bye variable itself). EX: bye = 5; x = bye; bye = 4; print x; prints 5, not 4

Also, you don't have to run hi() twice, just run x = hi(), not hi();x=hi() (the way you had it it was running hi(), not doing anything with the resulting value of 5, and then rerunning the same hi() and saving the value of 5 to the x variable.

So full code should be

def hi():
    something
    something
    bye = 5
    return bye 
x = hi()
print x

If you wanted to return multiple variables, one option would be to use a list, or dictionary, depending on what you need.

ex:

def hi():
    something
    xyz = { 'bye': 7, 'foobar': 8}
    return xyz
x = hi()
print x['bye']

more on python dictionaries at http://docs.python.org/2/tutorial/datastructures.html#dictionaries

John
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    Thanks. This is helpful. Though I can't shake the feeling that there is a clean/documented trick to achieve it - to define multiple local variables and then call just one of them outside the function. – askance Oct 11 '13 at 21:42
3

I've experienced the same problem. One of the responds to your question led me to the following idea (which worked eventually). I use Python 3.7.

    # just an example 
    def func(): # define a function
       func.y = 4 # here y is a local variable, which I want to access; func.y defines 
                  # a method for my example function which will allow me to access 
                  # function's local variable y
       x = func.y + 8 # this is the main task for the function: what it should do
       return x

    func() # now I'm calling the function
    a = func.y # I put it's local variable into my new variable
    print(a) # and print my new variable

Then I launch this program in Windows PowerShell and get the answer 4. Conclusion: to be able to access a local function's variable one might add the name of the function and a dot before the name of the local variable (and then, of course, use this construction for calling the variable both in the function's body and outside of it). I hope this will help.

2

You could do something along this lines:

def static_example():
   if not hasattr(static_example, "static_var"):
       static_example.static_var = 0
   static_example.static_var += 1
   return static_example.static_var

print static_example()
print static_example()
print static_example()
0
 def hi():
     bye = 5
     return bye  

print hi()
Allan Mwesigwa
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