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How to import a function with a variable and make that variable global [duplicate]

Suppose I have a function like:
def foo():
x = 'hello world'
How do I get the function to return x, in such a way that I can use it as the input for another function or use the variable within the body of a program? I tried using return and then using the x variable in another function, but I get a NameError that way.
For the specific case of communicating information between methods in the same class, it is often best to store the information in self. See Passing variables between methods in Python? for details.

def foo():
x = 'hello world'
return x # return 'hello world' would do, too
foo()
print(x) # NameError - x is not defined outside the function
y = foo()
print(y) # this works
x = foo()
print(x) # this also works, and it's a completely different x than that inside
# foo()
z = bar(x) # of course, now you can use x as you want
z = bar(foo()) # but you don't have to

Effectively, there are two ways: directly and indirectly.
The direct way is to return a value from the function, as you tried, and let the calling code use that value. This is normally what you want. The natural, simple, direct, explicit way to get information back from a function is to return it. Broadly speaking, the purpose of a function is to compute a value, and return signifies "this is the value we computed; we are done here".
Directly using return
The main trick here is that return returns a value, not a variable. So return x does not enable the calling code to use x after calling the function, and does not modify any existing value that x had in the context of the call. (That's presumably why you got a NameError.)
After we use return in the function:
def example():
x = 'hello world'
return x
we need to write the calling code to use the return value:
result = example()
print(result)
The other key point here is that a call to a function is an expression, so we can use it the same way that we use, say, the result of an addition. Just as we may say result = 'hello ' + 'world', we may say result = foo(). After that, result is our own, local name for that string, and we can do whatever we want with it.
We can use the same name, x, if we want. Or we can use a different name. The calling code doesn't have to know anything about how the function is written, or what names it uses for things.1
We can use the value directly to call another function: for example, print(foo()).2 We can return the value directly: simply return 'hello world', without assigning to x. (Again: we are returning a value, not a variable.)
The function can only return once each time it is called. return terminates the function - again, we just determined the result of the calculation, so there is no reason to calculate any further. If we want to return multiple pieces of information, therefore, we will need to come up with a single object (in Python, "value" and "object" are effectively synonyms; this doesn't work out so well for some other languages.)
We can make a tuple right on the return line; or we can use a dictionary, a namedtuple (Python 2.6+), a types.simpleNamespace (Python 3.3+), a dataclass (Python 3.7+), or some other class (perhaps even one we write ourselves) to associate names with the values that are being returned; or we can accumulate values from a loop in a list; etc. etc. The possibilities are endless..
On the other hand, the function returns whether you like it or not (unless an exception is raised). If it reaches the end, it will implicitly return the special value None. You may or may not want to do it explicitly instead.
Indirect methods
Other than returning the result back to the caller directly, we can communicate it by modifying some existing object that the caller knows about. There are many ways to do that, but they're all variations on that same theme.
If you want the code to communicate information back this way, please just let it return None - don't also use the return value for something meaningful. That's how the built-in functionality works.
In order to modify that object, the called function also has to know about it, of course. That means, having a name for the object that can be looked up in a current scope. So, let's go through those in order:
Local scope: Modifying a passed-in argument
If one of our parameters is mutable, we can just mutate it, and rely on the caller to examine the change. This is usually not a great idea, because it can be hard to reason about the code. It looks like:
def called(mutable):
mutable.append('world')
def caller():
my_value = ['hello'] # a list with just that string
called(my_value)
# now it contains both strings
If the value is an instance of our own class, we could also assign to an attribute:
class Test:
def __init__(self, value):
self.value = value
def called(mutable):
mutable.value = 'world'
def caller():
test = Test('hello')
called(test)
# now test.value has changed
Assigning to an attribute does not work for built-in types, including object; and it might not work for some classes that explicitly prevent you from doing it.
Local scope: Modifying self, in a method
We already have an example of this above: setting self.value in the Test.__init__ code. This is a special case of modifying a passed-in argument; but it's part of how classes work in Python, and something we're expected to do. Normally, when we do this, the calling won't actually check for changes to self - it will just use the modified object in the next step of the logic. That's what makes it appropriate to write code this way: we're still presenting an interface, so the caller doesn't have to worry about the details.
class Example:
def __init__(self):
self._words = ['hello']
def add_word(self):
self._words.append('world')
def display(self):
print(*self.words)
x = Example()
x.add_word()
x.display()
In the example, calling add_word gave information back to the top-level code - but instead of looking for it, we just go ahead and call display.3
See also: Passing variables between methods in Python?
Enclosing scope
This is a rare special case when using nested functions. There isn't a lot to say here - it works the same way as with the global scope, just using the nonlocal keyword rather than global.4
Global scope: Modifying a global
Generally speaking, it is a bad idea to change anything in the global scope after setting it up in the first place. It makes code harder to reason about, because anything that uses that global (aside from whatever was responsible for the change) now has a "hidden" source of input.
If you still want to do it, the syntax is straightforward:
words = ['hello']
def add_global_word():
words.append('world')
add_global_word() # `words` is changed
Global scope: Assigning to a new or existing global
This is actually a special case of modifying a global. I don't mean that assignment is a kind of modification (it isn't). I mean that when you assign a global name, Python automatically updates a dict that represents the global namespace. You can get that dict with globals(), and you can modify that dict and it will actually impact what global variables exist. (I.e., the return from globals() is the dictionary itself, not a copy.)5
But please don't. That's even worse of an idea than the previous one. If you really need to get the result from your function by assigning to a global variable, use the global keyword to tell Python that the name should be looked up in the global scope:
words = ['hello']
def replace_global_words():
global words
words = ['hello', 'world']
replace_global_words() # `words` is a new list with both words
Global scope: Assigning to or modifying an attribute of the function itself
This is a rare special case, but now that you've seen the other examples, the theory should be clear. In Python, functions are mutable (i.e. you can set attributes on them); and if we define a function at top level, it's in the global namespace. So this is really just modifying a global:
def set_own_words():
set_own_words.words = ['hello', 'world']
set_own_words()
print(*set_own_words.words)
We shouldn't really use this to send information to the caller. It has all the usual problems with globals, and it's even harder to understand. But it can be useful to set a function's attributes from within the function, in order for the function to remember something in between calls. (It's similar to how methods remember things in between calls by modifying self.) The functools standard library does this, for example in the cache implementation.
Builtin scope
This doesn't work. The builtin namespace doesn't contain any mutable objects, and you can't assign new builtin names (they'll go into the global namespace instead).
Some approaches that don't work in Python
Just calculating something before the function ends
In some other programming languages, there is some kind of hidden variable that automatically picks up the result of the last calculation, every time something is calculated; and if you reach the end of a function without returning anything, it gets returned. That doesn't work in Python. If you reach the end without returning anything, your function returns None.
Assigning to the function's name
In some other programming languages, you are allowed (or expected) to assign to a variable with the same name as the function; and at the end of the function, that value is returned. That still doesn't work in Python. If you reach the end without returning anything, your function still returns None.
def broken():
broken = 1
broken()
print(broken + 1) # causes a `TypeError`
It might seem like you can at least use the value that way, if you use the global keyword:
def subtly_broken():
global subtly_broken
subtly_broken = 1
subtly_broken()
print(subtly_broken + 1) # 2
But this, of course, is just a special case of assigning to a global. And there's a big problem with it - the same name can't refer to two things at once. By doing this, the function replaced its own name. So it will fail next time:
def subtly_broken():
global subtly_broken
subtly_broken = 1
subtly_broken()
subtly_broken() # causes a `TypeError`
Assigning to a parameter
Sometimes people expect to be able to assign to one of the function's parameters, and have it affect a variable that was used for the corresponding argument. However, this does not work:
def broken(words):
words = ['hello', 'world']
data = ['hello']
broken(data) # `data` does not change
Just like how Python returns values, not variables, it also passes values, not variables. words is a local name; by definition the calling code doesn't know anything about that namespace.
One of the working methods that we saw is to modify the passed-in list. That works because if the list itself changes, then it changes - it doesn't matter what name is used for it, or what part of the code uses that name. However, assigning a new list to words does not cause the existing list to change. It just makes words start being a name for a different list.
For more information, see How do I pass a variable by reference?.
1 At least, not for getting the value back. If you want to use keyword arguments, you need to know what the keyword names are. But generally, the point of functions is that they're an abstraction; you only need to know about their interface, and you don't need to think about what they're doing internally.
2 In 2.x, print is a statement rather than a function, so this doesn't make an example of calling another function directly. However, print foo() still works with 2.x's print statement, and so does print(foo()) (in this case, the extra parentheses are just ordinary grouping parentheses). Aside from that, 2.7 (the last 2.x version) has been unsupported since the beginning of 2020 - which was nearly a 5 year extension of the normal schedule. But then, this question was originally asked in 2010.
3Again: if the purpose of a method is to update the object, don't also return a value. Some people like to return self so that you can "chain" method calls; but in Python this is considered poor style. If you want that kind of "fluent" interface, then instead of writing methods that update self, write methods that create a new, modified instance of the class.
4 Except, of course, that if we're modifying a value rather than assigning, we don't need either keyword.
5 There's also a locals() that gives you a dict of local variables. However, this cannot be used to make new local variables - the behaviour is undefined in 2.x, and in 3.x the dict is created on the fly and assigning to it has no effect. Some of Python's optimizations depend on the local variables for a function being known ahead of time.

>>> def foo():
return 'hello world'
>>> x = foo()
>>> x
'hello world'

You can use global statement and then achieve what you want without returning value from
the function. For example you can do something like below:
def foo():
global x
x = "hello world"
foo()
print x
The above code will print "hello world".
But please be warned that usage of "global" is not a good idea at all and it is better to avoid usage that is shown in my example.
Also check this related discussion on about usage of global statement in Python.

Changing globally defined list as locally

I have a question about the python variables. As you see below, I have defined with L a list
and with I an integer variable after that, i have added new number to list inside function as locally.I can use the changed list outside the function as globally; on the other hand same situation is not for the integer variable.
Can someone help me please ? What is the difference?
PS:I know that ,I use integer variable inside the function as local variable because of that, it doesn't change on global. but why not a list variable?
L=[]
I=5
def change_var(I,L):
I=10
L.append(12)
print('Inside function I',I)
print('Inside function L',L)
return I,L
change_var(I,L)
print('Outside function I',I)
print('Outside function L',L)
Result
>>>Inside function I 10
>>>Inside function L [12]
>>>Outside function I 5
>>>Outside function L [12]```

If a variable gets affected a value anywhere inside the body of a function (as in variable = ...), then it is considered to be local, unless you explicitely declare it global in this function.
You treat I and L very differently:
You assign a value to I in I = 10, so it is considered local, and so a different variable from the other, global I. The changes you make to it can't be reflected on the other, global I. If you wanted to act on the global I, you should have declared global I at the start of your function.
You don't assign anything to L, you just mutate the unique, already existing global list.

You would have to change the name of the variables you are using inside the function, since you are creating new variables for I and L when inside the function, if you don't. In my example, I have changed them to lowercase(i and l). Also, when you put a number/variable inside the function parenthesis, it sets the variables inside the function to that number/variable, rather than putting that number inside the variable. So, when you put change_var(I,L), it effectively running the function, but putting the lines i=I and l=L at the start, rather than replacing all is with Is. To fix this, you will have to make I global when inside the function, so it affects code outside of the function, and then make I equal i.
The full code would be
L=[]
I=5
def change_var(i,l):
i=10
l.append(12)
print('Inside function I',i)
print('Inside function L',l)
global I
I=i
return i,l
change_var(I,L)
print('Outside function I',I)
print('Outside function L',L)
The code should work, but I'm sorry if my explanation didn't make any sense - I'm still quite new to answering on Stack Overflow.

In Python, make a variable act like a function, call a function without parentheses, alias a function as a variable, etc?

I am working with slightly modifying someone else's code for my needs and want to replace what is currently a fixed variable with a function. But adding a () to each time the variable is referenced later to get the value is simply not feasible in this situation, or would not be worth the amount of work required. I need a way to define a variable as a function such that while it is referenced as a variable in all further code, it actually checks what the value should be each time it is queried as if I had added parentheses to each reference. I do not care how this is achieved, but it should not require any changes to the code that references the former-variable.
The function in this case is random.random(). I want to make a variable always have a random value within a certain range whenever checked, WITHOUT adding parentheses to each and every time the variable is referenced. Is there any way to do this? I know I'm not giving a whole lot of information about context here, but in this case the whole point is that no knowledge or adjustment of any of the other code should be required.

You could maybe get there with a class with a property.
import random as r
class AlwaysRandom:
#property
def random(self):
return r.random()
Then create an instance and assign it to the name random in your namespace:
random = AlwaysRandom()
Now, any time you assign random, you assign a return value from r.random():
>>> random.random
0.1993064343052221
>>> random.random
0.9121594527461093
>>> [random.random for _ in range(5)]
[0.0800719907184344, 0.14744257667766358, 0.5809572562744559, 0.337413501046831, 0.52033363367589]

assignment of variable after function definition in Python

I am new to Python so I am unsure about the difference in variable assignment before or after the function definition.
Specifically, the first example was adopted from Lutz's book.
def tester(start):
print("inside tester")
def nested(label):
print("inside nested")
print(label,nested.state)
nested.state += 1
print("done with nested")
nested.state = start
print("done with tester")
return nested
F = tester(0)
F('spam')
F.state
F.state
The objective of the code is to store the state information without using nonlocal.
I am unsure what nested.state means here. I am unsure because nested.state is used inside nested() function (i.e. nested.state +=1) and outside nested() function (i.e. nested.state = start).
I modified the code above to see whether Python accepts assigning variable after function declaration for nested() and to see whether there is any concept I am missing relating to function.variable call (i.e. nested.state call).
def tester(start):
def nested(label):
print(label, state)
state += 1 #replaced 'nested.state' with 'state' here
state = start #replaced 'nested.state' with 'state' here
return nested
F=tester(0)
F('spam')
F('ham')
Unfortunately, above code generates error local variable 'state' referenced before assignment. This tells me that I am missing some concept about function.variable (i.e. nested.state).
Can someone please help me understand three things:
I. why it is that the code with nested.state doesn't generate any error but state does?
II. what does nested.state mean? If nested.state is a mechanism to access function's variables, why is it that the call inside nested() function also uses nested.state and not state?
III. If nested.state is a mechanism to access variable inside function, then why is it that PyCharm fails to show state under dropdown when I type nested.?
I'd appreciate any help. I research SO, and couldn't find any explanation on such problems.

The reason the first code example worked is because it was assigning and referencing an attribute of the nested function object. The key concept to understand here, is that Python allows you to assign new, arbitrary attributes to objects - including functions:
>>> def func(a, b):
return a + b
>>> func(1, 2)
3
>>> func.attr = 5
>>> func.attr
5
The first code example takes advantage of this fact by using the nested function object to store the necessary state. This is the same concept as using any other object to store the state. It's particularly convenient, however, to use a function since it's readily available.
In the second example, a normal variable is used. Because of this, normal scoping rules apply which means simply that the state variable defined in tester is not the state variable being referenced in nested. Thus, an error is raised.

Actually, I think you're asking a question about scope in Python, ignoring your code, check this:
def scope_level_1():
variable = 'Scope_level_1'
def scope_level_2():
variable = 'Scope_level_2'
def core():
nonlocal variable
variable += '_TOUCHED_AND_MODIFIED_BY_CORE'
print(variable)
return core()
return scope_level_2()
scope_level_1()
# 'Scope_level_2_TOUCHED_AND_MODIFIED_BY_CORE'
Don't worry about the keyword nonlocal, treat it just as a declaring to make code more readable.

First, remember a += b is the same as a = a + b. So a must exist before getting to the +=.
Simply put, in the first example the function nested has an attribute called state (accessed by nested.state). It is an attribute, which means that once you tell nested that it has an attribute called state (you are doing this in line 9 when nested.state = start) it keep that attribute. So, in the first example nested.state exists when you get to the +=.
In the second example, you are declaring a variable called state in tester, and another variable called state in nested. The one in nested could be called potato for all that matters, because it is not the same variable. Therefore when you arrive to the +=, the variable state does not exist!

How to make a variable global without the global keyword? [duplicate]

Suppose I have a function like:
def foo():
x = 'hello world'
How do I get the function to return x, in such a way that I can use it as the input for another function or use the variable within the body of a program? I tried using return and then using the x variable in another function, but I get a NameError that way.
For the specific case of communicating information between methods in the same class, it is often best to store the information in self. See Passing variables between methods in Python? for details.

def foo():
x = 'hello world'
return x # return 'hello world' would do, too
foo()
print(x) # NameError - x is not defined outside the function
y = foo()
print(y) # this works
x = foo()
print(x) # this also works, and it's a completely different x than that inside
# foo()
z = bar(x) # of course, now you can use x as you want
z = bar(foo()) # but you don't have to

Effectively, there are two ways: directly and indirectly.
The direct way is to return a value from the function, as you tried, and let the calling code use that value. This is normally what you want. The natural, simple, direct, explicit way to get information back from a function is to return it. Broadly speaking, the purpose of a function is to compute a value, and return signifies "this is the value we computed; we are done here".
Directly using return
The main trick here is that return returns a value, not a variable. So return x does not enable the calling code to use x after calling the function, and does not modify any existing value that x had in the context of the call. (That's presumably why you got a NameError.)
After we use return in the function:
def example():
x = 'hello world'
return x
we need to write the calling code to use the return value:
result = example()
print(result)
The other key point here is that a call to a function is an expression, so we can use it the same way that we use, say, the result of an addition. Just as we may say result = 'hello ' + 'world', we may say result = foo(). After that, result is our own, local name for that string, and we can do whatever we want with it.
We can use the same name, x, if we want. Or we can use a different name. The calling code doesn't have to know anything about how the function is written, or what names it uses for things.1
We can use the value directly to call another function: for example, print(foo()).2 We can return the value directly: simply return 'hello world', without assigning to x. (Again: we are returning a value, not a variable.)
The function can only return once each time it is called. return terminates the function - again, we just determined the result of the calculation, so there is no reason to calculate any further. If we want to return multiple pieces of information, therefore, we will need to come up with a single object (in Python, "value" and "object" are effectively synonyms; this doesn't work out so well for some other languages.)
We can make a tuple right on the return line; or we can use a dictionary, a namedtuple (Python 2.6+), a types.simpleNamespace (Python 3.3+), a dataclass (Python 3.7+), or some other class (perhaps even one we write ourselves) to associate names with the values that are being returned; or we can accumulate values from a loop in a list; etc. etc. The possibilities are endless..
On the other hand, the function returns whether you like it or not (unless an exception is raised). If it reaches the end, it will implicitly return the special value None. You may or may not want to do it explicitly instead.
Indirect methods
Other than returning the result back to the caller directly, we can communicate it by modifying some existing object that the caller knows about. There are many ways to do that, but they're all variations on that same theme.
If you want the code to communicate information back this way, please just let it return None - don't also use the return value for something meaningful. That's how the built-in functionality works.
In order to modify that object, the called function also has to know about it, of course. That means, having a name for the object that can be looked up in a current scope. So, let's go through those in order:
Local scope: Modifying a passed-in argument
If one of our parameters is mutable, we can just mutate it, and rely on the caller to examine the change. This is usually not a great idea, because it can be hard to reason about the code. It looks like:
def called(mutable):
mutable.append('world')
def caller():
my_value = ['hello'] # a list with just that string
called(my_value)
# now it contains both strings
If the value is an instance of our own class, we could also assign to an attribute:
class Test:
def __init__(self, value):
self.value = value
def called(mutable):
mutable.value = 'world'
def caller():
test = Test('hello')
called(test)
# now test.value has changed
Assigning to an attribute does not work for built-in types, including object; and it might not work for some classes that explicitly prevent you from doing it.
Local scope: Modifying self, in a method
We already have an example of this above: setting self.value in the Test.__init__ code. This is a special case of modifying a passed-in argument; but it's part of how classes work in Python, and something we're expected to do. Normally, when we do this, the calling won't actually check for changes to self - it will just use the modified object in the next step of the logic. That's what makes it appropriate to write code this way: we're still presenting an interface, so the caller doesn't have to worry about the details.
class Example:
def __init__(self):
self._words = ['hello']
def add_word(self):
self._words.append('world')
def display(self):
print(*self.words)
x = Example()
x.add_word()
x.display()
In the example, calling add_word gave information back to the top-level code - but instead of looking for it, we just go ahead and call display.3
See also: Passing variables between methods in Python?
Enclosing scope
This is a rare special case when using nested functions. There isn't a lot to say here - it works the same way as with the global scope, just using the nonlocal keyword rather than global.4
Global scope: Modifying a global
Generally speaking, it is a bad idea to change anything in the global scope after setting it up in the first place. It makes code harder to reason about, because anything that uses that global (aside from whatever was responsible for the change) now has a "hidden" source of input.
If you still want to do it, the syntax is straightforward:
words = ['hello']
def add_global_word():
words.append('world')
add_global_word() # `words` is changed
Global scope: Assigning to a new or existing global
This is actually a special case of modifying a global. I don't mean that assignment is a kind of modification (it isn't). I mean that when you assign a global name, Python automatically updates a dict that represents the global namespace. You can get that dict with globals(), and you can modify that dict and it will actually impact what global variables exist. (I.e., the return from globals() is the dictionary itself, not a copy.)5
But please don't. That's even worse of an idea than the previous one. If you really need to get the result from your function by assigning to a global variable, use the global keyword to tell Python that the name should be looked up in the global scope:
words = ['hello']
def replace_global_words():
global words
words = ['hello', 'world']
replace_global_words() # `words` is a new list with both words
Global scope: Assigning to or modifying an attribute of the function itself
This is a rare special case, but now that you've seen the other examples, the theory should be clear. In Python, functions are mutable (i.e. you can set attributes on them); and if we define a function at top level, it's in the global namespace. So this is really just modifying a global:
def set_own_words():
set_own_words.words = ['hello', 'world']
set_own_words()
print(*set_own_words.words)
We shouldn't really use this to send information to the caller. It has all the usual problems with globals, and it's even harder to understand. But it can be useful to set a function's attributes from within the function, in order for the function to remember something in between calls. (It's similar to how methods remember things in between calls by modifying self.) The functools standard library does this, for example in the cache implementation.
Builtin scope
This doesn't work. The builtin namespace doesn't contain any mutable objects, and you can't assign new builtin names (they'll go into the global namespace instead).
Some approaches that don't work in Python
Just calculating something before the function ends
In some other programming languages, there is some kind of hidden variable that automatically picks up the result of the last calculation, every time something is calculated; and if you reach the end of a function without returning anything, it gets returned. That doesn't work in Python. If you reach the end without returning anything, your function returns None.
Assigning to the function's name
In some other programming languages, you are allowed (or expected) to assign to a variable with the same name as the function; and at the end of the function, that value is returned. That still doesn't work in Python. If you reach the end without returning anything, your function still returns None.
def broken():
broken = 1
broken()
print(broken + 1) # causes a `TypeError`
It might seem like you can at least use the value that way, if you use the global keyword:
def subtly_broken():
global subtly_broken
subtly_broken = 1
subtly_broken()
print(subtly_broken + 1) # 2
But this, of course, is just a special case of assigning to a global. And there's a big problem with it - the same name can't refer to two things at once. By doing this, the function replaced its own name. So it will fail next time:
def subtly_broken():
global subtly_broken
subtly_broken = 1
subtly_broken()
subtly_broken() # causes a `TypeError`
Assigning to a parameter
Sometimes people expect to be able to assign to one of the function's parameters, and have it affect a variable that was used for the corresponding argument. However, this does not work:
def broken(words):
words = ['hello', 'world']
data = ['hello']
broken(data) # `data` does not change
Just like how Python returns values, not variables, it also passes values, not variables. words is a local name; by definition the calling code doesn't know anything about that namespace.
One of the working methods that we saw is to modify the passed-in list. That works because if the list itself changes, then it changes - it doesn't matter what name is used for it, or what part of the code uses that name. However, assigning a new list to words does not cause the existing list to change. It just makes words start being a name for a different list.
For more information, see How do I pass a variable by reference?.
1 At least, not for getting the value back. If you want to use keyword arguments, you need to know what the keyword names are. But generally, the point of functions is that they're an abstraction; you only need to know about their interface, and you don't need to think about what they're doing internally.
2 In 2.x, print is a statement rather than a function, so this doesn't make an example of calling another function directly. However, print foo() still works with 2.x's print statement, and so does print(foo()) (in this case, the extra parentheses are just ordinary grouping parentheses). Aside from that, 2.7 (the last 2.x version) has been unsupported since the beginning of 2020 - which was nearly a 5 year extension of the normal schedule. But then, this question was originally asked in 2010.
3Again: if the purpose of a method is to update the object, don't also return a value. Some people like to return self so that you can "chain" method calls; but in Python this is considered poor style. If you want that kind of "fluent" interface, then instead of writing methods that update self, write methods that create a new, modified instance of the class.
4 Except, of course, that if we're modifying a value rather than assigning, we don't need either keyword.
5 There's also a locals() that gives you a dict of local variables. However, this cannot be used to make new local variables - the behaviour is undefined in 2.x, and in 3.x the dict is created on the fly and assigning to it has no effect. Some of Python's optimizations depend on the local variables for a function being known ahead of time.

>>> def foo():
return 'hello world'
>>> x = foo()
>>> x
'hello world'

You can use global statement and then achieve what you want without returning value from
the function. For example you can do something like below:
def foo():
global x
x = "hello world"
foo()
print x
The above code will print "hello world".
But please be warned that usage of "global" is not a good idea at all and it is better to avoid usage that is shown in my example.
Also check this related discussion on about usage of global statement in Python.