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)]

I know in Python 2.6 you can do: h = enumerate(range(2000, 2005), 1) to give the above result but in python2.5 you cannot...

Using python2.5:

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

Does anyone know a way to get that desired result in python 2.5?

Thanks,

Jeff

Answer 1

h = [(i + 1, x) for i, x in enumerate(xrange(2000, 2005))]

Answer 2

You could create two range objects and zip them:

r = xrange(2000, 2005)
h = zip(xrange(1, len(r) + 1), 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 3

>>> 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 4

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 5

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 6

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

Answer 7

Simplest way to do in Python 2.5 exactly what you ask about:

import itertools as it

... it.izip(it.count(1), xrange(2000, 2005)) ...

If you want a list, as you appear to, use zip in lieu of it.izip.

(BTW, as a general rule, the best way to make a list out of a generator or any other iterable X is not [x for x in X], but rather list(X)).

Answer 8

Ok, I feel a bit stupid here... what's the reason not to just do it with something like
[(a+1,b) for (a,b) in enumerate(r)] ? If you won't function, no problem either:

>>> r = range(2000, 2005)
>>> [(a+1,b) for (a,b) in enumerate(r)]
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]

>>> enumerate1 = lambda r:((a+1,b) for (a,b) in enumerate(r)) 

>>> list(enumerate1(range(2000,2005)))   # note - generator just like original enumerate()
[(1, 2000), (2, 2001), (3, 2002), (4, 2003), (5, 2004)]