Does python enumerate start at 1?

In Python, you can get the element and index [count] from iterable objects such as list and tuple in for loop with the built-in function enumerate[].

Built-in Functions - enumerate[] — Python 3.8.5 documentation

This article describes the following contents.

How to use enumerate[]
- Normal for loop
- for loop with enumerate[]
Start index at 1 with enumerate[]
Set step with enumerate[]

See the following articles for more information about for loop and how to use enumerate[] and zip[] together.

for loop in Python [with range, enumerate, zip, etc.]
Use enumerate[] and zip[] together in Python

How to use `enumerate[]`

Normal `for` loop

l = ['Alice', 'Bob', 'Charlie']

for name in l:
    print[name]
# Alice
# Bob
# Charlie

`for` loop with `enumerate[]`

By passing an iterable object to enumerate[], you can get index, element.

for i, name in enumerate[l]:
    print[i, name]
# 0 Alice
# 1 Bob
# 2 Charlie

Start index at 1 with `enumerate[]`

As in the example above, by default, the index of enumerate[] starts at 0.

If you want to start from another number, pass the number to the second argument of enumerate[].

Start at 1:

for i, name in enumerate[l, 1]:
    print[i, name]
# 1 Alice
# 2 Bob
# 3 Charlie

Start at the other number:

for i, name in enumerate[l, 42]:
    print[i, name]
# 42 Alice
# 43 Bob
# 44 Charlie

For example, this is useful when generating sequential number strings starting from 1. It is smarter to pass the starting number to the second argument of enumerate[] than to calculate i + 1.

for i, name in enumerate[l, 1]:
    print['{:03}_{}'.format[i, name]]
# 001_Alice
# 002_Bob
# 003_Charlie

Set step with `enumerate[]`

There is no argument like step to specify increment to enumerate[], but it can be done as follows.

step = 3
for i, name in enumerate[l]:
    print[i * step, name]
# 0 Alice
# 3 Bob
# 6 Charlie

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

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

>>> 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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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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Does Python enumerate start at 0?

As in the example above, by default, the index of enumerate[] starts at 0. If you want to start from another number, pass the number to the second argument of enumerate[] .

How do you enumerate beginning from 1 in Python?

Enumerate Start at 1 in Python In Python, enumerate[] takes an iterable and an optional start , as arguments. The value provided to start parameter acts as a starting point for the counter. To enumerate with a start of 1 , pass start=1 to enumerate[] function.

What does enumerate [] do in Python?

Python enumerate[] Function The enumerate[] function takes a collection [e.g. a tuple] and returns it as an enumerate object. The enumerate[] function adds a counter as the key of the enumerate object.

How do you enumerate a number in Python?

Using the enumerate[] Function The enumerate[] function takes in an iterable as an argument, such as a list, string, tuple, or dictionary. In addition, it can also take in an optional argument, start, which specifies the number we want the count to start at [the default is 0]. And that's it!

How to use enumerate[]

Normal for loop

for loop with enumerate[]

Start index at 1 with enumerate[]

Set step with enumerate[]