Python enumerate start at 1

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 Python 2.5:

>>> 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?

Seanny123

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asked Jul 21, 2010 at 20:37

3

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.

answered Jul 21, 2010 at 20:41

Mark ByersMark Byers

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

enumerate(sequence, start=1)

answered Feb 6, 2013 at 18:28

dhacknerdhackner

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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

answered Jun 23, 2019 at 11:52

winklerrrwinklerrr

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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

answered Jul 21, 2010 at 20:44

DuncanDuncan

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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)).

answered Jul 21, 2010 at 21:48

Alex MartelliAlex Martelli

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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.

answered Jul 21, 2010 at 20:43

Jochen RitzelJochen Ritzel

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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.

answered Jul 21, 2010 at 20:49

1

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

answered Jul 21, 2010 at 21:06

JABJAB

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

answered Jul 21, 2010 at 20:39

Chris B.Chris B.

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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)]

answered Jul 22, 2010 at 3:55

Nas BanovNas Banov

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>>> 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))

answered Jul 21, 2010 at 20:41

Eli CourtwrightEli Courtwright

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1

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

answered May 27, 2019 at 17:00

bmcbmc

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