Asked  7 Months ago    Answers:  4   Viewed   31 times

What does the Python nonlocal statement do (in Python 3.0 and later)?

There's no documentation on the official Python website and help("nonlocal") does not work, either.

 Answers

60

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
Tuesday, June 1, 2021
 
tompave
answered 7 Months ago
50

You could use a loop:

conditions = (check_size, check_color, check_tone, check_flavor)
for condition in conditions:
    result = condition()
    if result:
        return result

This has the added advantage that you can now make the number of conditions variable.

You could use map() + filter() (the Python 3 versions, use the future_builtins versions in Python 2) to get the first such matching value:

try:
    # Python 2
    from future_builtins import map, filter
except ImportError:
    # Python 3
    pass

conditions = (check_size, check_color, check_tone, check_flavor)
return next(filter(None, map(lambda f: f(), conditions)), None)

but if this is more readable is debatable.

Another option is to use a generator expression:

conditions = (check_size, check_color, check_tone, check_flavor)
checks = (condition() for condition in conditions)
return next((check for check in checks if check), None)
Sunday, June 27, 2021
 
jedwards
answered 6 Months ago
14

The problem is that the variable cache is not in the scope of the function match. This is not a problem if you only want to read it as in your second example, but if you're assigning to it, python interprets it as a local variable. If you're using python 3 you can use the nonlocal keyword to solve this problem - for python 2 there's no simple workaround, unfortunately.

def f():
    v = 0

    def x():
        return v    #works because v is read from the outer scope

    def y():
        if v == 0:  #fails because the variable v is assigned to below
            v = 1

    #for python3:
    def z():
        nonlocal v  #tell python to search for v in the surrounding scope(s)
        if v == 0:
            v = 1   #works because you declared the variable as nonlocal

The problem is somewhat the same with global variables - you need to use global every time you assign to a global variable, but not for reading it.

A short explanation of the reasons behind that: The python interpreter compiles all functions into a special object of type function. During this compilation, it checks for all local variables the function creates (for garbage collection etc). These variable names are saved within the function object. As it is perfectly legal to "shadow" an outer scopes variable (create a variable with the same name), any variable that is assigned to and that is not explicitly declared as global (or nonlocal in python3) is assumed to be a local variable.

When the function is executed, the interpreter has to look up every variable reference it encounters. If the variable was found to be local during compilation, it is searched in the functions f_locals dictionary. If it has not been assigned to yet, this raises the exception you encountered. If the variable is not assigned to in the functions scope and thus is not part of its locals, it is looked up in the surrounding scopes - if it is not found there, this raises a similar exception.

Sunday, August 22, 2021
 
IcedAnt
answered 4 Months ago
25

In Python 3, I believe, you can use the nonlocal keyword to get permission to modify a variable in an enclosing non-global scope. In Python 2, you cannot reassign foo in an enclosing scope; instead, set foo equal to a mutable object like a list [] and then stick the value you want stored in the list:

def func1():
    foo = [None]
    def func2():
        foo[0] = 'Test'
Friday, October 8, 2021
 
Scott Kausler
answered 2 Months ago
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