Accessing the index in 'for' loops

Question:

How do I access the index while iterating over a sequence with 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
Asked By: Joan Venge

||

Answers:

for i in range(len(ints)):
   print(i, ints[i]) # print updated to print() in Python 3.x+ 
Answered By: David Hanak

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)
>>> 
Answered By: Charlie Martin

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.

Answered By: Mike Hordecki

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.

Answered By: thinker007

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+
   
Answered By: A.J.

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)

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]
Answered By: rassa45

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
Answered By: Liam

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:

Answered By: Andriy Ivaneyko

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

  1. 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.

  1. 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
  1. 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
  1. 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.

Answered By: Charitoo

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']

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)
Answered By: Sumit Kumar

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)]
Answered By: lola

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.

Answered By: PyRsquared

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])
Answered By: Ankur Kothari

If there is no duplicate value in the list:

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

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
Answered By: DrM

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)]
Answered By: Andy Jazz

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
Answered By: Amar Kumar

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))
Answered By: Rahul

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))
Answered By: Kofi

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
Answered By: LunaticXXD10

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
Answered By: kamalesh d

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))
Answered By: Ran Turner
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