How to enumerate a range of numbers starting at 1

Question

I am using Python 2.5, I want an enumeration like so (starting at 1 instead of 0):

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

Answer 1

>>> list(enumerate(range(1999, 2005)))[1:]
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]

Answer 2

>>> h = enumerate(range(2000, 2005))
>>> [(tup[0]+1, tup[1]) for tup in h]
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]

Since this is somewhat verbose, I'd recommend writing your own function to generalize it:

def enumerate_at(xs, start):
    return ((tup[0]+start, tup[1]) for tup in enumerate(xs))

Answer 3

from itertools import count, izip

def enumerate(L, n=0):
    return izip( count(n), L)

# if 2.5 has no count
def count(n=0):
    while True:
        yield n
        n+=1

Now h = list(enumerate(xrange(2000, 2005), 1)) works.

Answer 4

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

Answer 5

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.

Answer 6

Python 3

Official documentation: enumerate(iterable, start=0)

So you would use it like this:

>>> 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')]