How to enumerate a range of numbers starting at 1

Question

As you already mentioned, this is straightforward to do in Python 2.6 or newer:

enumerate(range(2000, 2005), 1)

Python 2.5 and older do not support the start parameter so instead you could create two range objects and zip them:

r = xrange(2000, 2005)
r2 = xrange(1, len(r) + 1)
h = zip(r2, r)
print h

Result:

[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]

If you want to create a generator instead of a list then you can use izip instead.

Just to put this here for posterity sake, in 2.6 the "start" parameter was added to enumerate like so:

enumerate(sequence, start=1)

Python 3

Official Python documentation: enumerate(iterable, start=0)

You don't need to write your own generator as other answers here suggest. The built-in Python standard library already contains a function that does exactly what you want:

>>> seasons = ['Spring', 'Summer', 'Fall', 'Winter']
>>> list(enumerate(seasons))
[(0, 'Spring'), (1, 'Summer'), (2, 'Fall'), (3, 'Winter')]
>>> list(enumerate(seasons, start=1))
[(1, 'Spring'), (2, 'Summer'), (3, 'Fall'), (4, 'Winter')]

---

The built-in function is equivalent to this:

def enumerate(sequence, start=0):
  n = start
  for elem in sequence:
    yield n, elem
    n += 1

Easy, just define your own function that does what you want:

def enum(seq, start=0):
    for i, x in enumerate(seq):
        yield i+start, x