Accessing the index in 'for' loops

How do I access the index in a for loop?

xs = [8, 23, 45]

for x in xs:
    print("item #{} = {}".format(index, x))

Desired output:

item #1 = 8
item #2 = 23
item #3 = 45
Note that indexes in python start from 0, so the indexes for your example list are 0 to 4 not 1 to 5

M
Mateen Ulhaq

Use the built-in function enumerate():

for idx, x in enumerate(xs):
    print(idx, x)

It is non-pythonic to manually index via for i in range(len(xs)): x = xs[i] or manually manage an additional state variable.

Check out PEP 279 for more.


As Aaron points out below, use start=1 if you want to get 1-5 instead of 0-4.
Does enumerate not incur another overhead?
@TheRealChx101 according to my tests (Python 3.6.3) the difference is negligible and sometimes even in favour of enumerate.
@TheRealChx101: It's lower than the overhead of looping over a range and indexing each time, and lower than manually tracking and updating the index separately. enumerate with unpacking is heavily optimized (if the tuples are unpacked to names as in the provided example, it reuses the same tuple each loop to avoid even the cost of freelist lookup, it has an optimized code path for when the index fits in ssize_t that performs cheap in-register math, bypassing Python level math operations, and it avoids indexing the list at the Python level, which is more expensive than you'd think).
@user2585501. It does: for i in range(5) or for i in range(len(ints)) will do the universally common operation of iterating over an index. But if you want both the item and the index, enumerate is a very useful syntax. I use it all the time.
R
Russia Must Remove Putin

Using a for loop, how do I access the loop index, from 1 to 5 in this case?

Use enumerate to get the index with the element as you iterate:

for index, item in enumerate(items):
    print(index, item)

And note that Python's indexes start at zero, so you would get 0 to 4 with the above. If you want the count, 1 to 5, do this:

count = 0 # in case items is empty and you need it after the loop
for count, item in enumerate(items, start=1):
    print(count, item)

Unidiomatic control flow

What you are asking for is the Pythonic equivalent of the following, which is the algorithm most programmers of lower-level languages would use:

index = 0 # Python's indexing starts at zero for item in items: # Python's for loops are a "for each" loop print(index, item) index += 1

Or in languages that do not have a for-each loop:

index = 0 while index < len(items): print(index, items[index]) index += 1

or sometimes more commonly (but unidiomatically) found in Python:

for index in range(len(items)): print(index, items[index])

Use the Enumerate Function

Python's enumerate function reduces the visual clutter by hiding the accounting for the indexes, and encapsulating the iterable into another iterable (an enumerate object) that yields a two-item tuple of the index and the item that the original iterable would provide. That looks like this:

for index, item in enumerate(items, start=0):   # default is zero
    print(index, item)

This code sample is fairly well the canonical example of the difference between code that is idiomatic of Python and code that is not. Idiomatic code is sophisticated (but not complicated) Python, written in the way that it was intended to be used. Idiomatic code is expected by the designers of the language, which means that usually this code is not just more readable, but also more efficient.

Getting a count

Even if you don't need indexes as you go, but you need a count of the iterations (sometimes desirable) you can start with 1 and the final number will be your count.

count = 0 # in case items is empty
for count, item in enumerate(items, start=1):   # default is zero
    print(item)

print('there were {0} items printed'.format(count))

The count seems to be more what you intend to ask for (as opposed to index) when you said you wanted from 1 to 5.

Breaking it down - a step by step explanation

To break these examples down, say we have a list of items that we want to iterate over with an index:

items = ['a', 'b', 'c', 'd', 'e']

Now we pass this iterable to enumerate, creating an enumerate object:

enumerate_object = enumerate(items) # the enumerate object

We can pull the first item out of this iterable that we would get in a loop with the next function:

iteration = next(enumerate_object) # first iteration from enumerate
print(iteration)

And we see we get a tuple of 0, the first index, and 'a', the first item:

(0, 'a')

we can use what is referred to as "sequence unpacking" to extract the elements from this two-tuple:

index, item = iteration
#   0,  'a' = (0, 'a') # essentially this.

and when we inspect index, we find it refers to the first index, 0, and item refers to the first item, 'a'.

>>> print(index)
0
>>> print(item)
a

Conclusion

Python indexes start at zero

To get these indexes from an iterable as you iterate over it, use the enumerate function

Using enumerate in the idiomatic way (along with tuple unpacking) creates code that is more readable and maintainable:

So do this:

for index, item in enumerate(items, start=0):   # Python indexes start at zero
    print(index, item)

Does the "Getting a count" example work when items is empty?
@Bergi: It won't, but you can just add count = 0 before the loop to ensure it has a value (and it's the correct one when the loop never assigns to count, since by definition there were no items).
S
SuperStormer

It's pretty simple to start it from 1 other than 0:

for index, item in enumerate(iterable, start=1):
   print index, item  # Used to print in python<3.x
   print(index, item) # Migrate to print() after 3.x+
   

The question was about list indexes; since they start from 0 there is little point in starting from other number since the indexes would be wrong (yes, the OP said it wrong in the question as well). Otherwise, calling the variable that is tuple of index, item just index is very misleading, as you noted. Just use for index, item in enumerate(ints).
Better is to enclose index inside parenthesis pairs as (index), it will work on both the Python versions 2 and 3.
@AnttiHaapala The reason, I presume, is that the question's expected output starts at index 1 instead 0
@hygull: Turning index into (index) won't change a thing on either Py2 or Py3. I feel like maybe you're thinking of the change to print; the only way to make that work on both Py2 and Py3 is to add from __future__ import print_function to the top of your file to get consistent Py3-style print, and change the print to print(index, item). Or you read an earlier edit of the question when index was the original tuple, not unpacked to two names, but the parentheses still don't change anything if you fail to unpack.
m
mhhabib
for i in range(len(ints)):
   print(i, ints[i]) # print updated to print() in Python 3.x+ 

That should probably be xrange for pre-3.0.
Use enumerate instead
For Python 2.3 above, use enumerate built-in function since it is more Pythonic.
Enumerate is not always better - it depends on the requirements of the application. In my current situation the relationships between the object lengths is meaningful to my application. Although I started out using enumerate, I switched to this approach to avoid having to write logic to select which object to enumerate.
@adg I don't see how avoid enumerate saves any logic; you still have to select which object to index with i, no?
P
Peter Mortensen

As is the norm in Python, there are several ways to do this. In all examples assume: lst = [1, 2, 3, 4, 5]

Using enumerate (considered most idiomatic)

for index, element in enumerate(lst):
    # Do the things that need doing here

This is also the safest option in my opinion because the chance of going into infinite recursion has been eliminated. Both the item and its index are held in variables and there is no need to write any further code to access the item.

Creating a variable to hold the index (using for)

for index in range(len(lst)):   # or xrange
    # you will have to write extra code to get the element

Creating a variable to hold the index (using while)

index = 0
while index < len(lst):
    # You will have to write extra code to get the element
    index += 1  # escape infinite recursion

There is always another way

As explained before, there are other ways to do this that have not been explained here and they may even apply more in other situations. For example, using itertools.chain with for. It handles nested loops better than the other examples.


A
Andy Jazz

Here's how you can access the indices and array's elements using for-in loops.

1. Looping elements with counter and += operator.

items = [8, 23, 45, 12, 78]
counter = 0

for value in items:
    print(counter, value)
    counter += 1

Result:

#    0 8
#    1 23
#    2 45
#    3 12
#    4 78

2. Looping elements using enumerate() method.

items = [8, 23, 45, 12, 78]

for i in enumerate(items):
    print("index/value", i)

Result:

#    index/value (0, 8)
#    index/value (1, 23)
#    index/value (2, 45)
#    index/value (3, 12)
#    index/value (4, 78)

3. Using index and value separately.

items = [8, 23, 45, 12, 78]

for index, value in enumerate(items):
    print("index", index, "for value", value)

Result:

#    index 0 for value 8
#    index 1 for value 23
#    index 2 for value 45
#    index 3 for value 12
#    index 4 for value 78

4. You can change the index number to any increment.

items = [8, 23, 45, 12, 78]

for i, value in enumerate(items, start=1000):
    print(i, value)

Result:

#    1000 8
#    1001 23
#    1002 45
#    1003 12
#    1004 78

5. Automatic counter incrementation with range(len(...)).

items = [8, 23, 45, 12, 78]

for i in range(len(items)):
    print("Index:", i, "Value:", items[i])

Result:

#    ('Index:', 0, 'Value:', 8)
#    ('Index:', 1, 'Value:', 23)
#    ('Index:', 2, 'Value:', 45)
#    ('Index:', 3, 'Value:', 12)
#    ('Index:', 4, 'Value:', 78)

6. Using for-in loop inside function.

items = [8, 23, 45, 12, 78]

def enum(items, start=0):
    counter = start

    for value in items:
        print(counter, value)
        counter += 1
    
enum(items)

Result:

#    0 8
#    1 23
#    2 45
#    3 12
#    4 78

7. Of course, we can't forget about while loop.

items = [8, 23, 45, 12, 78]
counter = 0

while counter < len(items):
    print(counter, items[counter])
    counter += 1

Result:

#    0 8
#    1 23
#    2 45
#    3 12
#    4 78

8. yield statement returning a generator object.

def createGenerator():        
    items = [8, 23, 45, 12, 78]

    for (j, k) in enumerate(items):
        yield (j, k)
        

generator = createGenerator()

for i in generator:
    print(i)

Result:

#    (0, 8)
#    (1, 23)
#    (2, 45)
#    (3, 12)
#    (4, 78)

9. Inline expression with for-in loop and lambda.

items = [8, 23, 45, 12, 78]

xerox = lambda upperBound: [(i, items[i]) for i in range(0, upperBound)]
print(xerox(5))

Result:

#    [(0, 8), (1, 23), (2, 45), (3, 12), (4, 78)]

m
mhhabib

Old fashioned way:

for ix in range(len(ints)):
    print(ints[ix])

List comprehension:

[ (ix, ints[ix]) for ix in range(len(ints))]

>>> ints
[1, 2, 3, 4, 5]
>>> for ix in range(len(ints)): print ints[ix]
... 
1
2
3
4
5
>>> [ (ix, ints[ix]) for ix in range(len(ints))]
[(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)]
>>> lc = [ (ix, ints[ix]) for ix in range(len(ints))]
>>> for tup in lc:
...     print(tup)
... 
(0, 1)
(1, 2)
(2, 3)
(3, 4)
(4, 5)
>>> 

This is not wrong and is used in C/C++ and others. It's considered as non-pythonic, but can also be used in python. Like simple solutions that break it down to the source :+
Some python extremists would say, don't do this. But I said it only to indicate that there is more than one possible way
C
Community

Accessing indexes & Performance Benchmarking of approaches

The fastest way to access indexes of list within loop in Python 3.7 is to use the enumerate method for small, medium and huge lists.

Please see different approaches which can be used to iterate over list and access index value and their performance metrics (which I suppose would be useful for you) in code samples below:

# Using range
def range_loop(iterable):
    for i in range(len(iterable)):
        1 + iterable[i]

# Using enumerate
def enumerate_loop(iterable):
    for i, val in enumerate(iterable):
        1 + val

# Manual indexing
def manual_indexing_loop(iterable):
    index = 0
    for item in iterable:
        1 + item
        index += 1

See performance metrics for each method below:

from timeit import timeit

def measure(l, number=10000):
    print("Measure speed for list with %d items" % len(l))
    print("range: ", timeit(lambda :range_loop(l), number=number))
    print("enumerate: ", timeit(lambda :enumerate_loop(l), number=number))
    print("manual_indexing: ", timeit(lambda :manual_indexing_loop(l), number=number))

# Measure speed for list with 1000 items
measure(range(1000))
# range:  1.161622366
# enumerate:  0.5661940879999996
# manual_indexing:  0.610455682

# Measure speed for list with 100000 items
measure(range(10000))
# range:  11.794482958
# enumerate:  6.197628574000001
# manual_indexing:  6.935181098000001

# Measure speed for list with 10000000 items
measure(range(10000000), number=100)
# range:  121.416859069
# enumerate:  62.718909123
# manual_indexing:  69.59575057400002

As the result, using enumerate method is the fastest method for iteration when the index needed.

Adding some useful links below:

What is the difference between range and xrange functions in Python 2.X?

What is faster for loop using enumerate or for loop using xrange in Python?

range(len(list)) or enumerate(list)?


"readability counts" The speed difference in the small <1000 range is insignificant. It is 3% slower on an already small time metric.
How about updating the answer to Python 3?
@Georgy makes sense, on python 3.7 enumerate is total winner :)
L
Lars

You can use enumerate and embed expressions inside string literals to obtain the solution.

This is a simple way:

a=[4,5,6,8]
for b, val in enumerate(a):
    print('item #{} = {}'.format(b+1, val))

P
Peter Mortensen

First of all, the indexes will be from 0 to 4. Programming languages start counting from 0; don't forget that or you will come across an index-out-of-bounds exception. All you need in the for loop is a variable counting from 0 to 4 like so:

for x in range(0, 5):

Keep in mind that I wrote 0 to 5 because the loop stops one number before the maximum. :)

To get the value of an index, use

list[index]

L
Liam

You can do it with this code:

ints = [8, 23, 45, 12, 78]
index = 0

for value in (ints):
    index +=1
    print index, value

Use this code if you need to reset the index value at the end of the loop:

ints = [8, 23, 45, 12, 78]
index = 0

for value in (ints):
    index +=1
    print index, value
    if index >= len(ints)-1:
        index = 0

P
Peter Mortensen

According to this discussion: object's list index

Loop counter iteration

The current idiom for looping over the indices makes use of the built-in range function:

for i in range(len(sequence)):
    # Work with index i

Looping over both elements and indices can be achieved either by the old idiom or by using the new zip built-in function:

for i in range(len(sequence)):
    e = sequence[i]
    # Work with index i and element e

or

for i, e in zip(range(len(sequence)), sequence):
    # Work with index i and element e

via PEP 212 – Loop Counter Iteration.


This won't work for iterating through generators. Just use enumerate().
Nowadays, the current idiom is enumerate, not the range call.
and it's the same as the older answer: stackoverflow.com/a/522576/6451573
G
Georgy

In your question, you write "how do I access the loop index, from 1 to 5 in this case?"

However, the index for a list runs from zero. So, then we need to know if what you actually want is the index and item for each item in a list, or whether you really want numbers starting from 1. Fortunately, in Python, it is easy to do either or both.

First, to clarify, the enumerate function iteratively returns the index and corresponding item for each item in a list.

alist = [1, 2, 3, 4, 5]

for n, a in enumerate(alist):
    print("%d %d" % (n, a))

The output for the above is then,

0 1
1 2
2 3
3 4
4 5

Notice that the index runs from 0. This kind of indexing is common among modern programming languages including Python and C.

If you want your loop to span a part of the list, you can use the standard Python syntax for a part of the list. For example, to loop from the second item in a list up to but not including the last item, you could use

for n, a in enumerate(alist[1:-1]):
    print("%d %d" % (n, a))

Note that once again, the output index runs from 0,

0 2
1 3
2 4

That brings us to the start=n switch for enumerate(). This simply offsets the index, you can equivalently simply add a number to the index inside the loop.

for n, a in enumerate(alist, start=1):
    print("%d %d" % (n, a))

for which the output is

1 1
2 2
3 3
4 4
5 5

s
so2

If I were to iterate nums = [1, 2, 3, 4, 5] I would do

for i, num in enumerate(nums, start=1):
    print(i, num)

Or get the length as l = len(nums)

for i in range(l):
    print(i+1, nums[i])

G
Georgy

If there is no duplicate value in the list:

for i in ints:
    indx = ints.index(i)
    print(i, indx)

Note that the first option should not be used, since it only works correctly only when each item in the sequence is unique.
First option is O(n²), a terrible idea. If your list is 1000 elements long, it'll take literally a 1000 times longer than using enumerate. You should delete this answer.
A
Ashok Kumar Jayaraman

You can also try this:

data = ['itemA.ABC', 'itemB.defg', 'itemC.drug', 'itemD.ashok']
x = []
for (i, item) in enumerate(data):
      a = (i, str(item).split('.'))
      x.append(a)
for index, value in x:
     print(index, value)

The output is

0 ['itemA', 'ABC']
1 ['itemB', 'defg']
2 ['itemC', 'drug']
3 ['itemD', 'ashok']

P
Peter Mortensen

You can use the index method:

ints = [8, 23, 45, 12, 78]
inds = [ints.index(i) for i in ints]

It is highlighted in a comment that this method doesn’t work if there are duplicates in ints. The method below should work for any values in ints:

ints = [8, 8, 8, 23, 45, 12, 78]
inds = [tup[0] for tup in enumerate(ints)]

Or alternatively

ints = [8, 8, 8, 23, 45, 12, 78]
inds = [tup for tup in enumerate(ints)]

if you want to get both the index and the value in ints as a list of tuples.

It uses the method of enumerate in the selected answer to this question, but with list comprehension, making it faster with less code.


A
Amar Kumar

A simple answer using a while loop:

arr = [8, 23, 45, 12, 78]
i = 0
while i < len(arr):
    print("Item ", i + 1, " = ", arr[i])
    i += 1

Output:

Item  1  =  8
Item  2  =  23
Item  3  =  45
Item  4  =  12
Item  5  =  78

Please review Should we edit a question to transcribe code from an image to text? and Why not upload images of code/errors when asking a question? (e.g., "Images should only be used to illustrate problems that can't be made clear in any other way, such as to provide screenshots of a user interface.") and take the appropriate action (it covers answers as well). Thanks in advance.
S
Sam

You can simply use a variable such as count to count the number of elements in the list:

ints = [8, 23, 45, 12, 78]
count = 0
for i in ints:
    count = count + 1
    print('item #{} = {}'.format(count, i))

P
Peter Mortensen

To print a tuple of (index, value) in a list comprehension using a for loop:

ints = [8, 23, 45, 12, 78]
print [(i,ints[i]) for i in range(len(ints))]

Output:

[(0, 8), (1, 23), (2, 45), (3, 12), (4, 78)]

R
Ran Turner

You can use range(len(some_list)) and then lookup the index like this

xs = [8, 23, 45]
for i in range(len(xs)):
    print("item #{} = {}".format(i + 1, xs[i]))

Or use the Python’s built-in enumerate function which allows you to loop over a list and retrieve the index and the value of each item in the list

xs = [8, 23, 45]
for idx, val in enumerate(xs, start=1):
    print("item #{} = {}".format(idx, val))

k
kamalesh d

It can be achieved with the following code:

xs = [8, 23, 45]
for x, n in zip(xs, range(1, len(xs)+1)):
    print("item #{} = {}".format(n, x))

Here, range(1, len(xs)+1); If you expect the output to start from 1 instead of 0, you need to start the range from 1 and add 1 to the total length estimated since python starts indexing the number from 0 by default.

Final Output:
item #1 = 8
item #2 = 23
item #3 = 45

P
Peter Mortensen

A loop with a "counter" variable set as an initialiser that will be a parameter, in formatting the string, as the item number.

The for loop accesses the "listos" variable which is the list. As we access the list by "i", "i" is formatted as the item price (or whatever it is).

listos = [8, 23, 45, 12, 78]
counter = 1
for i in listos:
    print('Item #{} = {}'.format(counter, i))
    counter += 1

Output:

Item #1 = 8
Item #2 = 23
Item #3 = 45
Item #4 = 12
Item #5 = 78

@calculuswhiz the while loop is an important code snippet. the initialiser "counter" is used for item number. About Indentation: The guy must be enough aware about programming that indentation matters. so after you do your special attribute...{copy paste} you can still edit the indentation. And the code runs very nicely and smoothly
@LunaticXXD10 In your original post, which you can see in the edit history, counter += 1 was indented at the same level as the for loop. In that case, it would have updated per iteration of the for loop. Here, it updates per iteration of the while loop. My question on the while loop remains: Why is it necessary to use a while loop when counter can be updated in the for loop? This is plainly visible in Rahul's answer.
Re "...must be enough aware about programming that indentation matters": Well, computers are quite literally minded.
P
Peter Mortensen

This serves the purpose well enough:

list1 = [10, 'sumit', 43.21, 'kumar', '43', 'test', 3]
for x in list1:
    print('index:', list1.index(x), 'value:', x)

This will break down if there are repeated elements in the list as index() will search for the first occurrence of x, not mentioning the O( n^2 ) time required to look up each element.
totally agreed that it won't work for duplicate elements in the list. afterall I'm also learning python.
The accepted suggested edit by user Raj kumar resulted in the error "NameError: name 'list1' is not defined"