How nonlocal keyword works?

Question

In the below code,

def makeAverage():
    series = []
    def average(newValue):
        series.append(newValue)
        total = sum(series)
        return total/len(series)
    return average

python interpreter does not expect series to be nonlocal in average().

But in the below code

def makeAverage():
    count = 0
    total = 0
    def average(newValue):
        nonlocal count, total
        count += 1
        total += newValue
        return total/count
    return average

Question:

Why python interpreter expects count & total declared nonlocal in average()?

The same way `makeAverage` would expect them to be `global` if you were doing `count += 1` in `makeAverage`. — Ashwini Chaudhary, Jul 17 '17 at 04:30

score 4 · Accepted Answer · answered Jul 17 '17 at 04:35

A variable is considered local to a function if you assign to it anywhere in that function and you don't mark it otherwise (with global or nonlocal). In your first example, there is no assignment to series inside average, so it is not considered local to average, so the version from the enclosing function is used. In the second example, there are assignments to total and count inside average, so they need to be marked nonlocal to access them from the enclosing function. (Otherwise you will get an UnboundLocalError because average tries to read their values before assigning to them for the first time.)

How nonlocal keyword works?

1 Answers1