Recently reviewed some python for a friend of mine. He needed a counter while iterating items of a list.

stuff = ["some thing", "some thing else"]

i = 0
for thing in stuff:
  print "processing item {} of {}".format(i + 1, len(stuff))
  i += 1

Having an extra counter var always annoyed me.

Python comes with this nice for i in list syntax, so taking care of a counter separately seamed wrong or at least not elegant.

update

Big thanks to Rafa Rodríguez for pointing me to enumerate

stuff = ["some thing", "some thing else"]

for i, thing in enumerate(stuff):
  print "processing item {} of {}".format(i + 1, len(stuff))

You can skipp the rest :) .

Indeed I came up with the following refactoring:

stuff = ["some thing", "some thing else"]

for thing, i in zip(stuff, range(len(stuff))):
  print "processing item {} of {}".format(i + 1, len(stuff))

zip takes n lists and combines those elements into tuples, drawing from all lists, stopping if one of the lists is exhausted. This makes it very useful in combination with infinite lists.

This takes care of incrementing i for us :). However it's still cumbersome to have to specify the range. What we really want is a lazy sequence of integers that just counts up starting at a specified value. Python itertools to the rescue. The iter tools provide a count() function that returns this generator:

def count(start=0, step=1):
    # count(10) --> 10 11 12 13 14 ...
    # count(2.5, 0.5) -> 2.5 3.0 3.5 ...
    n = start
    while True:
        yield n
        n += step

So we can rewrite our code to:

from itertools import count

stuff = ["some thing", "some thing else"]

for thing, i in zip(stuff, count(1))):
  print "processing item {} of {}".format(i, len(stuff))

Note that by calling count(1) the sequence starts at 1, removing the need to call i + 1 in the print statement all together.

In the end I was very pleased to finally having found - what I belief to be - a beautiful solution to this problem in python.

Feel free to comment or contact me on twitter.