Simple number generator [closed]

Question

How do I write a function that simply returns incrementing numbers on each call?

print counter() # 0
print counter() # 1
print counter() # 2
print counter() # 3
etc

For clarity, I'm not looking for a generator (although the answer would probably use it). The function should return an integer, not an iterable object. I'm also NOT looking for a solution involving global or otherwise shared variables (classes, function attribute etc).

See here http://www.valuedlessons.com/2008/01/monads-in-python-with-nice-syntax.html for some useful insights about the matter.

Answer 1

You don't need to implement that function -- it already exists:

counter = itertools.count().next

or in Python 3.x

counter = itertools.count().__next__

More generally, if you want a callable that has additional state, there are several solutions:

1. A class:

   class Counter(object):
       def __init__(self, start=0):
           self.count = 0
       def __call__(self):
           val = self.count
           self.count += 1
           return val
   

2. A mutable default argument:

   def counter(_count=[0]):
       val = _count[0]
       _count[0] += 1
       return val
   

3. A closure:

   def make_counter(start=0):
       count = [start]
       def counter():
           val = count[0]
           count[0] += 1
           return val
       return counter
   

   In Python 3.x, you wouldn't need the above hack using a list any more, since you could use a nonlocal declaration:

   def make_counter(start=0):
       count = start
       def counter():
           nonlocal count
           val = count
           count += 1
           return val
       return counter
   

4. Function arguments:

   def counter():
       val = counter.count
       counter.count += 1
       return val
   counter.count = 0
   

5. A generator:

   def count(start=0):
       while True:
           yield start
           start += 1
   counter = count().next   ## or .__next__ in Python 3.x
   

Of course all these solution have to store the current count somewhere.

Answer 2

In short,

from itertools import count
counter = lambda c=count(): next(c)