How to add indices to a list?

Question

For example, I have a list = ['a', 'b', 'c'] .
What I want is a list indexed(list) == [(1, 'a'), (2, 'b'), (3, 'c)] .

Is there a builtin or module for this?

Answer 1

You can enumerate it by doing the following,

for i,n in enumerate(list):
# where i is the index and n is the value of the list

Answer 2

You can use the built-in function enumerate to achieve that.

In [1]: a = ['a', 'b', 'c']

In [2]: b = [(idx, item) for idx,item in enumerate(a)]

In [3]: b
Out[3]: [(0, 'a'), (1, 'b'), (2, 'c')]

Note: the default indice would start with 0 , but you could try to add start=1 to config that, eg

In [4]: c = [(idx, item) for idx,item in enumerate(a, start=1)]

In [5]: c
Out[5]: [(1, 'a'), (2, 'b'), (3, 'c')]

Hope it helps.

Answer 3

You can do something like

indexed_list = [(i + 1, elem) for i, elem in enumerate(your_list)]

I'm assuming that you need the index to start from 1. Otherwise you can just do a list comprehension on the enumerate results directly without adding 1 to the index.

EDIT: Updated according to @pault 's suggestion, ie using the built-in argument

indexed_list = [indexed for indexed in enumerate(your_list, 1)]

Or simply

indexed_list = list(enumerate(your_list, 1))

Answer 4

Indices in Python begin at 0 not 1 . You can do what you want by using the built-in zip() function along with the count() generator function in the itertools module.

It's also necessary to explicitly convert zip() 's result to a list if that's what you want (which is why I changed the name of your variable to my_list to prevent it from "hiding" the built-in class with the same name — a good thing to always do):

from itertools import count

my_list = ['a', 'b', 'c']

indexed_my_list = list(zip(count(), my_list))
print(indexed_my_list)  # -> [(0, 'a'), (1, 'b'), (2, 'c')]

It's unclear why you would need to do this because you can use the built-in enumerate() function to get the indices whenever they're needed, as illustrated in a number of the other answers.

Answer 5

Code below is what you want:

>>>l = ['a', 'b', 'c']
>>>indl = [(i + 1, val) for i, val in enumerate(l)]
>>> indl
[(1, 'a'), (2, 'b'), (3, 'c')]

Edit: according to @pault's suggestion, code is changed as below:

>>> yourList = ['a', 'b', 'c']
>>> listInd = [(i, val) for i, val in enumerate(yourList, 1)]
>>> listInd
[(1, 'a'), (2, 'b'), (3, 'c')]
>>> 

Answer 6

You can use enumerate and it's start argument for this too:

l = ['a', 'b', 'c']
indexed_list = list(enumerate(l, 1))

As for a function called indexed you could make one

Note! Never replace a builtin keyword! list list is one of them

>>> def indexed(l, start=1):
    ...    return list(enumerate(l, start))
>>> l = ['a', 'b', 'c']
>>> indexed(l)
[(1, 'a'), (2, 'b'), (3, 'c)]

This defaults to a start value of 1.

Answer 7

使用列表理解和枚举

[(i,l) for i,l in enumerate(list)]