Accessing the index in 'for' loops

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.

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

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)]

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)
>>> 

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))

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])

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

If there is no duplicate value in the list:

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

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']

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 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

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)]

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))

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))

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

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

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)