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

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

enumerate(sequence, start=1)

Answer 2

enumerate is trivial, and so is re-implementing it to accept a start:

def enumerate(iterable, start = 0):
    n = start
    for i in iterable:
        yield n, i
        n += 1

Note that this doesn't break code using enumerate without start argument. Alternatively, this oneliner may be more elegant and possibly faster, but breaks other uses of enumerate:

enumerate = ((index+1, item) for index, item)

The latter was pure nonsense. @Duncan got the wrapper right.

Answer 3

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 4

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

Answer 5

I don't know how these posts could possibly be made more complicated then the following:

# Just pass the start argument to enumerate ...
for i,word in enumerate(allWords, 1):
    word2idx[word]=i
    idx2word[i]=word

Answer 6

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)]