How do you dynamically set variables as global in Python 3 functions?
Something like this:
def func1(i):
global i
#Some operation on i
How would I get the global variable to set whatever pre-existing variable is passed into the function? Such that:
func1(foo)
Would preform the operation on the variable foo globally rather than creating a new global variable called i?
If I got it right, your problem is:
I've bound a name to some object at module level; now I want to write a function that changes the binding of that global name to another object passed as an argument to the function.
In Python global names can be referenced inside a function (provided that you don't bind another object to the same name), but to change their binding you must first declare those names as global. The reason is simple: by default all names bound inside a function have function scope.
GLOBAL_NAME = 12
ANOTHER_GLOBAL_NAME = 50
def func(value):
global GLOBAL_NAME
GLOBAL_NAME = value
print(GLOBAL_NAME, ANOTHER_GLOBAL_NAME)
When you call func(33), GLOBAL_NAME will be bound to the object 33. ANOTHER_GLOBAL_NAME is just a reference to a global name; since that name is not bound inside the function, Python look for that name in the global scope.
For further insights on Python scope enclosing you can refer to the Python documentation on the global statement, to PEP-0227 and to PEP-3104.
Related
For example if I write
a=1
def func():
return a
func()
It will return 1. Is this normal behavior? Are variables supposed to act in a global way in python?
If you create a variable with the same name inside a function, this variable will be local, and can only be used inside the function. The global variable with the same name will remain as it was, global and with the original value.
That is the way python language works !
Hey where doe Python store variables in a function and how do i access them
Lets say i have the following code:
import ctypes
def reveal():
me = "where"
print(id(me))
reveal() # memory_address
reveal() # same memory_address
ctypes.c_long.from_address(memory_address).value #1
from this i conclude that the variable has to be stored somewhere
i cant find it in globals() and it is not in the function __dict__
i dont know where to look for it or is it not accessible from outside of the function scope
No way. This variable is declared in the function's local scope and cannot be accessed outside of this function.
As you may know, the scope of a variable is statically determined in python (What are the rules for local and global variables in Python?).
For instance :
a = "global"
def function():
print(a)
a = "local"
function()
# UnboundLocalError: local variable 'a' referenced before assignment
The same rule applies to classes, but it seems to default to the global scope instead of raising an AttributeError:
a = "global"
class C():
print(a)
a = "local"
# 'global'
Moreover, in the case of a nested function, the behavior is the same (without using nonlocal or global) :
a = "global"
def outer_func():
a = "outer"
def inner_func():
print(a)
a = "local"
inner_func()
outer_func()
# UnboundLocalError: local variable 'a' referenced before assignment
But in the case of nested classes, it still defaults to the global scope, and not the outer scope (again without using global or nonlocal) :
a = "global"
def outer_func():
a = "outer"
class InnerClass:
print(a)
a = "local"
outer_func()
# 'global'
The weirdest part is that the nested class default to the outer scope when there is no declaration of a :
a = "global"
def outer_func():
a = "outer"
class InnerClass:
print(a)
outer_func()
# 'outer'
So my questions are :
Why the discrepancy between functions and classes (one raising an exception, the other defaulting to the global scope.
In nested classes, why the default scope has to become global instead of keeping using the outer one when using a variable defined afterward?
The answer is given in great detail in Section 9.2 of the official docs. The crux of the matter is
... On the other hand, the actual search for names is done dynamically, at run time — however, the language definition is evolving towards static name resolution, at “compile” time, so don’t rely on dynamic name resolution! (In fact, local variables are already determined statically.)
When you are in the class definition, which at the moment of its execution is the innermost scope, dynamic name resolution applies. You therefore see printouts of the global value of a.
If the name resolution were static, as in function definitions, the name a would be recognized as a local name even in the print statement. That is why you can't print a in a function before assigning to it.
The rules for class body scoping are alluded to in Section 4.2.2:
Class definition blocks and arguments to exec() and eval() are special in the context of name resolution. A class definition is an executable statement that may use and define names. These references follow the normal rules for name resolution with an exception that unbound local variables are looked up in the global namespace.
Let's parse that last sentence carefully, because it fully covers your last two examples. First off, what is an unbound local variable in this context? A class body creates a new namespace, just like entering a function. If a name is bound somewhere in a class body, it is a local variable. This is determined statically, as mentioned above. If you attempt to reference the name before it is first bound, you have an unbound local variable. Instead of raising an error, as a function call would do, python jumps straight to the global namespace to perform the lookup (and ignores builtins as well). In all other cases (not local variables), normal LEGB lookup order applies.
This is indeed a bit counter-intuitive, and I would argue that it pushes if not outright breaks the rule of least surprise.
Just a beginner question about local and global scope in python
X = 100
#is X a global variable?.We defined it outside the function scope
def foo():
print(X)
return X
#it prints 100 and even returns it
def foo():
X = X + 10
#local error
#UnboundLocalError: local variable 'X' referenced before assignment
def foo():
global X
# if X is a global variable why specify again?
X = X + 10
return X
To modify global copy of a variable you need the to use the global keyword, but you don't need global if you are only accessing that.
from the python website:
In Python, variables that are only referenced inside a function are implicitly global. If a variable is assigned a value anywhere within the function’s body, it’s assumed to be a local unless explicitly declared as global.
this means that you can access a global variable inside a function without a global keyword. if you want to change it though, you must use the global keyword beforehand.
global and nonlocal are very strange things when I was a beginner.
Just think about it: why do we need them in Python?
It is because we don't need var, let and such similar things to declare variables.
Think about Javascript, it is dynamic script language too and very alike to python, but it needs var or let or const to declare variables.
The most important thing of declaring variables is to determine scope.
So, in Python, our variables have implicit default scope: current scope where they are defined, and if we want to change scope of some variables, we need use global or nonlocal explicitly .
All names on the left side of = mean to define variables.
Before executing code of some certain scope, Python will pre-compute all local variables, which are those on the left side of =. This is why you got UnboundLocalError: local variable 'X' referenced before assignment in:
def foo():
X = X + 10
So, if we look up those names not in defined current scope, just
follow the rules of scope chain: up, up, up and until built_in.
Remember: scope of any name on the left side of = is default current scope, and you have to assign it(bind something to it) before referring it.
I came across some code which kind of puzzled me. Here's a minimal example which shows this:
# of course, the ... are not part of the actual code
some_var = {"key1":"value1" ... "keyN":"valueN"}
def some_func():
v = some_var["key1"]
The code works, but the fact that I can access some_var directly confuses me. The last time I had to write some Python code, I remember having to write some_func like this:
def some_func():
global some_var
v = some_var["key1"]
I am using Python 2.7.1 on a Windows 7 PC. Did something change in the 2.7 release that allows for this?
No, you just can't reassign some_var in a local scope. Consider the following:
some_var = {}
def some_func():
# some_var[5] = 6
some_var = {1:2}
some_var[3] = 4
some_func()
print (repr(some_var)) # {}
You'll see the assignment in some_func actually creates a local variable which shadows the global one. Therefore, uncommenting the line would result in an UnboundLocalError - you can't access variables before they're defined.
There's a difference between using (e.g. calling or using in an expression) a name from an outer scope and assigning it (and there's a difference between assigning a bare variable and assigning a member of an object pointed to by a variable - x.y = ... and x[...] = ... count as method calls!).
You only need to declare that a variable is from an outer scope if you're re-assigning it. In Python 2, you can only do that with global variables (via global var), in Python 3 you can do it for abritarily nested scopes (e.g. closures) using nonlocal var.
Using a nonlocal variable as in you example doesn't require global. It does, however, as soon as you assign the variable (with the aforementioned definition of assignment) anywhere within that scope - so add a line some_var = ... after the line where you're using it and you'll get an UnboundLocalError. Refer to the documentation for the nitty gritty details.
You only have to use global if you want to assign a new value to that variable.
Nested scope was introduced in Python 2.1 (and enabled by default in Python 2.2) (emphasis mine):
Put simply, when a given variable name is not assigned a value within a function (by an assignment, or the def, class, or import statements), references to the variable will be looked up in the local namespace of the enclosing scope. A more detailed explanation of the rules, and a dissection of the implementation, can be found in the PEP.
You only need to use global if you intend to assign to the variable, for reading the variable this is not necessary. This difference is not arbitrary, though it may seem like it at first glance.
When reading a value the interpreter can just look for a local variable named some_var, if it cannot find it then it looks for a global variable of that name. These are simple and straight forward semantics.
When assigning values to a variable the interpreter needs to know whether you intend to assign to a local variable some_var or a global variable. The interpreter assumes that some_var = 2 when called within a function is assigning to a local variable, this makes sense since this is the most common case. For the relatively rare times when you want to assign to a global variable from within a function then you use the global modifier global some_var = 2.
It depends on the usage of the variable in the function
Python variable scope error
Assigning a value to a name makes the name local, unless the name is explicitly declared global.
a = 12
def foo():
a = 42
print a # uses local
foo()
>>> 42
def foo():
global a
a = 42
foo()
print a
>>> 42
If a name is not assigned to, it is global.
a = 12
def foo():
print a # uses global
foo()
>>> 12
In short, you only have to explicitly declare a name global if you will be assigning to it. If you are just reading from it, you can use it at will. However, if you ever assign to the variable, it will be considered local in that function unless you declared it global.
b = 5
def foo():
print b
b = 7
foo()
>>> ???
Since b is assigned to in foo() and not declared global, Python decides at compile time that b is a local name. Therefore b is a local name throughout the whole function, including at the print statement before the assignment.
Therefore the print statement gives you an error, because the local name b has not been defined!