Python Local Variable As Counter

Question

Using Python 3.7.

Is there any way to use a local variable as a counter? I tried and received error:

UnboundLocalError

I did find a recommendation to use a glbal variable which is working, but if possible I would prefer to use a local variable.

Thanks, -w

Working code using global variable for counter:

count = 0
def my_collatz(number):

    global count
    count +=1
    if int(number)%2 == 0:
        r = int((number)//2)
    else:
        r = int(((number * 3) + 1))
    print('Attempt : ' + str(count) + ',' + str(r))

    if r != 1:
        return my_collatz(int(r))

print('Please enter a number : ')
number=input()
my_collatz(int(number))

Answer 1

One possible solution is to move the count variable as parameter to the function and increment it each call:

def my_collatz(number, count=1):    
    if number % 2 == 0:
        r = number // 2
    else:
        r = (number * 3) + 1

    print('Attempt : ' + str(count) + ',' + str(r))

    if r != 1:
        return my_collatz(r, count + 1)

print('Please enter a number : ')
number=input()
my_collatz(int(number))

Prints:

Please enter a number : 
6
Attempt : 1,3
Attempt : 2,10
Attempt : 3,5
Attempt : 4,16
Attempt : 5,8
Attempt : 6,4
Attempt : 7,2
Attempt : 8,1

---

Other solution is to not use count at all, and instead make the function a generator (using yield). Then you can use enumerate() to obtain your number of steps:

def my_collatz(number):
    if number % 2 == 0:
        r = number // 2
    else:
        r = (number * 3) + 1

    yield r

    if r != 1:
        yield from my_collatz(r)

print('Please enter a number : ')
number=input()

for count, r in enumerate(my_collatz(int(number)), 1):
    print('Attempt : ' + str(count) + ',' + str(r))