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 Python 2 5   gt  gt  gt  h   enumerate range 2000  2005    gt  gt  gt   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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

Just to put this here for posterity sake  in 2 6 the  start  parameter was added to enumerate like so   enumerate sequence  start 1

User · Answer

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

User · Answer

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   gt  gt  gt  seasons     Spring    Summer    Fall    Winter    gt  gt  gt  list enumerate seasons     0   Spring     1   Summer     2   Fall     3   Winter     gt  gt  gt  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

User · Answer

h     i   1  x  for i  x in enumerate xrange 2000  2005

User · Answer

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

User · Answer

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    gt  gt  gt  r   range 2000  2005   gt  gt  gt    a 1 b  for  a b  in enumerate r     1  2000    2  2001    3  2002    4  2003    5  2004     gt  gt  gt  enumerate1   lambda r   a 1 b  for  a b  in enumerate r      gt  gt  gt  list enumerate1 range 2000 2005        note - generator just like original enumerate     1  2000    2  2001    3  2002    4  2003    5  2004

User · Answer

gt  gt  gt  h   enumerate range 2000  2005    gt  gt  gt    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

User · Answer

gt  gt  gt  list enumerate range 1999  2005    1     1  2000    2  2001    3  2002    4  2003    5  2004

User · Answer

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

[python] How to enumerate a range of numbers starting at 1

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