What does "nonlocal" do in Python 3?
Question:
What does nonlocal
do in Python 3.x?
To close debugging questions where OP needs nonlocal
and doesn’t realize it, please use Is it possible to modify variable in python that is in outer, but not global, scope? instead.
Although Python 2 is officially unsupported as of January 1, 2020, if for some reason you are forced to maintain a Python 2.x codebase and need an equivalent to nonlocal
, see nonlocal keyword in Python 2.x.
Answers:
In short, it lets you assign values to a variable in an outer (but non-global) scope. See PEP 3104 for all the gory details.
A google search for “python nonlocal” turned up the Proposal, PEP 3104, which fully describes the syntax and reasoning behind the statement. in short, it works in exactly the same way as the global
statement, except that it is used to refer to variables that are neither global nor local to the function.
Here’s a brief example of what you can do with this. The counter generator can be rewritten to use this so that it looks more like the idioms of languages with closures.
def make_counter():
count = 0
def counter():
nonlocal count
count += 1
return count
return counter
Obviously, you could write this as a generator, like:
def counter_generator():
count = 0
while True:
count += 1
yield count
But while this is perfectly idiomatic python, it seems that the first version would be a bit more obvious for beginners. Properly using generators, by calling the returned function, is a common point of confusion. The first version explicitly returns a function.
Compare this, without using nonlocal
:
x = 0
def outer():
x = 1
def inner():
x = 2
print("inner:", x)
inner()
print("outer:", x)
outer()
print("global:", x)
# inner: 2
# outer: 1
# global: 0
To this, using nonlocal
, where inner()
‘s x
is now also outer()
‘s x
:
x = 0
def outer():
x = 1
def inner():
nonlocal x
x = 2
print("inner:", x)
inner()
print("outer:", x)
outer()
print("global:", x)
# inner: 2
# outer: 2
# global: 0
If we were to use global
, it would bind x
to the properly "global" value:
x = 0
def outer():
x = 1
def inner():
global x
x = 2
print("inner:", x)
inner()
print("outer:", x)
outer()
print("global:", x)
# inner: 2
# outer: 1
# global: 2
It takes the one "closest" to the point of reference in the source code.
This is called "Lexical Scoping" and is standard for >40 years now.
Python’s class members are really in a dictionary called __dict__
and will never be reached by lexical scoping.
If you don’t specify nonlocal
but do x = 7
, it will create a new local variable "x".
If you do specify nonlocal
, it will find the "closest" "x" and assign to that.
If you specify nonlocal
and there is no "x", it will give you an error message.
The keyword global
has always seemed strange to me since it will happily ignore all the other "x" except for the outermost one.
My personal understanding of the “nonlocal” statement (and do excuse me as I am new to Python and Programming in general) is that the “nonlocal” is a way to use the Global functionality within iterated functions rather than the body of the code itself. A Global statement between functions if you will.
help(‘nonlocal’)
The nonlocal
statement
nonlocal_stmt ::= "nonlocal" identifier ("," identifier)*
The nonlocal
statement causes the listed identifiers to refer to
previously bound variables in the nearest enclosing scope. This is
important because the default behavior for binding is to search the
local namespace first. The statement allows encapsulated code to
rebind variables outside of the local scope besides the global
(module) scope.
Names listed in a nonlocal
statement, unlike to those listed in a
global
statement, must refer to pre-existing bindings in an
enclosing scope (the scope in which a new binding should be created
cannot be determined unambiguously).
Names listed in a nonlocal
statement must not collide with pre-
existing bindings in the local scope.
See also:
PEP 3104 – Access to Names in Outer Scopes
The specification for the nonlocal
statement.
Related help topics: global, NAMESPACES
Source: Python Language Reference
a = 0 #1. global variable with respect to every function in program
def f():
a = 0 #2. nonlocal with respect to function g
def g():
nonlocal a
a=a+1
print("The value of 'a' using nonlocal is ", a)
def h():
global a #3. using global variable
a=a+5
print("The value of a using global is ", a)
def i():
a = 0 #4. variable separated from all others
print("The value of 'a' inside a function is ", a)
g()
h()
i()
print("The value of 'a' global before any function", a)
f()
print("The value of 'a' global after using function f ", a)
with ‘nonlocal’ inner functions(ie;nested inner functions) can get read & ‘write‘ permission for that specific variable of the outer parent function. And nonlocal can be used only inside inner functions
eg:
a = 10
def Outer(msg):
a = 20
b = 30
def Inner():
c = 50
d = 60
print("MU LCL =",locals())
nonlocal a
a = 100
ans = a+c
print("Hello from Inner",ans)
print("value of a Inner : ",a)
Inner()
print("value of a Outer : ",a)
res = Outer("Hello World")
print(res)
print("value of a Global : ",a)
Quote from the Python 3 Reference:
The nonlocal statement causes the listed identifiers to refer to previously bound variables in the nearest enclosing scope excluding globals.
As said in the reference, in case of several nested functions only variable in the nearest enclosing function is modified:
def outer():
def inner():
def innermost():
nonlocal x
x = 3
x = 2
innermost()
if x == 3: print('Inner x has been modified')
x = 1
inner()
if x == 3: print('Outer x has been modified')
x = 0
outer()
if x == 3: print('Global x has been modified')
# Inner x has been modified
The “nearest” variable can be several levels away:
def outer():
def inner():
def innermost():
nonlocal x
x = 3
innermost()
x = 1
inner()
if x == 3: print('Outer x has been modified')
x = 0
outer()
if x == 3: print('Global x has been modified')
# Outer x has been modified
But it cannot be a global variable:
def outer():
def inner():
def innermost():
nonlocal x
x = 3
innermost()
inner()
x = 0
outer()
if x == 3: print('Global x has been modified')
# SyntaxError: no binding for nonlocal 'x' found
The documentation says below:
The nonlocal statement causes the listed identifiers to refer to
previously bound variables in the nearest enclosing scope excluding
globals. …
So for example, nonlocal foo
in inner()
can access the non-local variable foo = 10
in middle()
but not the non-local variable foo = 5
in outer()
or the global variable foo = 0
outside outer()
as shown below:
foo = 0 # <- ✖
def outer():
foo = 5 # <- ✖
def middle():
foo = 10 # <- 〇
def inner():
nonlocal foo # Here
foo += 1
print(foo) # 11
inner()
middle()
outer()
What does nonlocal
do in Python 3.x?
To close debugging questions where OP needs nonlocal
and doesn’t realize it, please use Is it possible to modify variable in python that is in outer, but not global, scope? instead.
Although Python 2 is officially unsupported as of January 1, 2020, if for some reason you are forced to maintain a Python 2.x codebase and need an equivalent to nonlocal
, see nonlocal keyword in Python 2.x.
In short, it lets you assign values to a variable in an outer (but non-global) scope. See PEP 3104 for all the gory details.
A google search for “python nonlocal” turned up the Proposal, PEP 3104, which fully describes the syntax and reasoning behind the statement. in short, it works in exactly the same way as the global
statement, except that it is used to refer to variables that are neither global nor local to the function.
Here’s a brief example of what you can do with this. The counter generator can be rewritten to use this so that it looks more like the idioms of languages with closures.
def make_counter():
count = 0
def counter():
nonlocal count
count += 1
return count
return counter
Obviously, you could write this as a generator, like:
def counter_generator():
count = 0
while True:
count += 1
yield count
But while this is perfectly idiomatic python, it seems that the first version would be a bit more obvious for beginners. Properly using generators, by calling the returned function, is a common point of confusion. The first version explicitly returns a function.
Compare this, without using nonlocal
:
x = 0
def outer():
x = 1
def inner():
x = 2
print("inner:", x)
inner()
print("outer:", x)
outer()
print("global:", x)
# inner: 2
# outer: 1
# global: 0
To this, using nonlocal
, where inner()
‘s x
is now also outer()
‘s x
:
x = 0
def outer():
x = 1
def inner():
nonlocal x
x = 2
print("inner:", x)
inner()
print("outer:", x)
outer()
print("global:", x)
# inner: 2
# outer: 2
# global: 0
If we were to use global
, it would bind x
to the properly "global" value:
x = 0
def outer():
x = 1
def inner():
global x
x = 2
print("inner:", x)
inner()
print("outer:", x)
outer()
print("global:", x)
# inner: 2
# outer: 1
# global: 2
It takes the one "closest" to the point of reference in the source code.
This is called "Lexical Scoping" and is standard for >40 years now.
Python’s class members are really in a dictionary called __dict__
and will never be reached by lexical scoping.
If you don’t specify nonlocal
but do x = 7
, it will create a new local variable "x".
If you do specify nonlocal
, it will find the "closest" "x" and assign to that.
If you specify nonlocal
and there is no "x", it will give you an error message.
The keyword global
has always seemed strange to me since it will happily ignore all the other "x" except for the outermost one.
My personal understanding of the “nonlocal” statement (and do excuse me as I am new to Python and Programming in general) is that the “nonlocal” is a way to use the Global functionality within iterated functions rather than the body of the code itself. A Global statement between functions if you will.
help(‘nonlocal’)
Thenonlocal
statement
nonlocal_stmt ::= "nonlocal" identifier ("," identifier)*
The
nonlocal
statement causes the listed identifiers to refer to
previously bound variables in the nearest enclosing scope. This is
important because the default behavior for binding is to search the
local namespace first. The statement allows encapsulated code to
rebind variables outside of the local scope besides the global
(module) scope.Names listed in a
nonlocal
statement, unlike to those listed in a
global
statement, must refer to pre-existing bindings in an
enclosing scope (the scope in which a new binding should be created
cannot be determined unambiguously).Names listed in a
nonlocal
statement must not collide with pre-
existing bindings in the local scope.See also:
PEP 3104 – Access to Names in Outer Scopes
The specification for thenonlocal
statement.Related help topics: global, NAMESPACES
Source: Python Language Reference
a = 0 #1. global variable with respect to every function in program
def f():
a = 0 #2. nonlocal with respect to function g
def g():
nonlocal a
a=a+1
print("The value of 'a' using nonlocal is ", a)
def h():
global a #3. using global variable
a=a+5
print("The value of a using global is ", a)
def i():
a = 0 #4. variable separated from all others
print("The value of 'a' inside a function is ", a)
g()
h()
i()
print("The value of 'a' global before any function", a)
f()
print("The value of 'a' global after using function f ", a)
with ‘nonlocal’ inner functions(ie;nested inner functions) can get read & ‘write‘ permission for that specific variable of the outer parent function. And nonlocal can be used only inside inner functions
eg:
a = 10
def Outer(msg):
a = 20
b = 30
def Inner():
c = 50
d = 60
print("MU LCL =",locals())
nonlocal a
a = 100
ans = a+c
print("Hello from Inner",ans)
print("value of a Inner : ",a)
Inner()
print("value of a Outer : ",a)
res = Outer("Hello World")
print(res)
print("value of a Global : ",a)
Quote from the Python 3 Reference:
The nonlocal statement causes the listed identifiers to refer to previously bound variables in the nearest enclosing scope excluding globals.
As said in the reference, in case of several nested functions only variable in the nearest enclosing function is modified:
def outer():
def inner():
def innermost():
nonlocal x
x = 3
x = 2
innermost()
if x == 3: print('Inner x has been modified')
x = 1
inner()
if x == 3: print('Outer x has been modified')
x = 0
outer()
if x == 3: print('Global x has been modified')
# Inner x has been modified
The “nearest” variable can be several levels away:
def outer():
def inner():
def innermost():
nonlocal x
x = 3
innermost()
x = 1
inner()
if x == 3: print('Outer x has been modified')
x = 0
outer()
if x == 3: print('Global x has been modified')
# Outer x has been modified
But it cannot be a global variable:
def outer():
def inner():
def innermost():
nonlocal x
x = 3
innermost()
inner()
x = 0
outer()
if x == 3: print('Global x has been modified')
# SyntaxError: no binding for nonlocal 'x' found
The documentation says below:
The nonlocal statement causes the listed identifiers to refer to
previously bound variables in the nearest enclosing scope excluding
globals. …
So for example, nonlocal foo
in inner()
can access the non-local variable foo = 10
in middle()
but not the non-local variable foo = 5
in outer()
or the global variable foo = 0
outside outer()
as shown below:
foo = 0 # <- ✖
def outer():
foo = 5 # <- ✖
def middle():
foo = 10 # <- 〇
def inner():
nonlocal foo # Here
foo += 1
print(foo) # 11
inner()
middle()
outer()