Is it possible to create a nested function with state in Python?

Question

I'm trying to make a callback that updates if it's been more than a second since the last report. It looks something like this

def main_func()
    ...
    start_time = time.time()
    def callback():
        current_time = time.time()
        if current_time - start_time > 1.0:
            print 'working'
            start_time = time.time()
    work_function(args, callback)

Since start_time is local the global keyword won't work inside callback. Is there a Python paradigm that can handle adding state to local functions?

I think you need to better define what you mean by state? often you can .... but this does not seem to be an optimal solution based on the info you have provided ... — Joran Beasley, Jul 03 '14 at 17:22
Reference: (not duplicate) http://stackoverflow.com/q/233673/758446 — BlackVegetable, Jul 03 '14 at 17:24
Use nonlocal. http://stackoverflow.com/questions/1261875/python-nonlocal-statement — Joonazan, Jul 03 '14 at 17:25
This is a duplicate, but it's a very well-asked and useful one - it should help search engine results and won't mislead future readers. — Karl Knechtel, Sep 13 '22 at 15:46

score 3 · Accepted Answer · answered Jul 03 '14 at 18:00

As already suggested, you're probably better off using a class, and saving the state as a member.

Also as already suggested, you can use a global variable, or the nonlocal statement.

A third (and more original) alternative would be to set the state as an attribute of the function object:

def main_func()
    ...
    def callback():
        current_time = time.time()
        if current_time - callback.start_time > 1.0:
            print 'working'
            callback.start_time = time.time()
    callback.start_time = time.time()
    work_function(args, callback)

score 0 · Answer 2 · answered Jul 03 '14 at 18:05

You can make this work, though the details will depend on which version of Python you're using.

In Python 3, you can use a nonlocal statement to allow your callback function to assign a new value to the start_time variable in the outer function's scope.

def main_func()
    ...
    start_time = time.time()
    def callback():
        nonlocal start_time            # new line here
        current_time = time.time()
        if current_time - start_time > 1.0:
            print 'working'
            start_time = time.time()   # this assignment now works as intended
    work_function(args, callback)

In Python 2, nonlocal is not available, but you can wrap the value up in a mutable container (like a list) and modify the value without reassigning the variable:

def main_func()
    ...
    start_time = [time.time()]             # wrap time value in a list
    def callback():
        current_time = time.time()
        if current_time - start_time > 1.0:
            print 'working'
            start_time[0] = time.time()    # modify the list in place
    work_function(args, callback)

score 0 · Answer 3 · answered Jul 03 '14 at 18:27

Kinda, but it's durdly.

    def foo(a):
        def inner(b):
            if b > c[0]:
                c[0] > b
            else:
                print "too low"
        c = [a]
        return inner

>>> bar = foo(5)
>>> bar(5)
too low
>>> bar(6) # silence on success
>>> bar(5)
too low

As others have pointed out, it's probably best to just use an object (which you kind of do with this solution -- lists are objects after all)

This sort of approach is only required in 2.x, where `nonlocal` isn't available. — Karl Knechtel, Sep 13 '22 at 15:48

Is it possible to create a nested function with state in Python?

3 Answers3