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)

You can use this, if you already have the list / array but don't know the size yet.

for idx,data in enumerate(content_list):
    print(idx) # will print the index of the loop starting from 0
    print(data) # will print the actual data,
    print(content_list[idx]) # will manually access the list using the index. both this and data will print the same thing

If you want to do a for loop with a set range, do this,

for i in range(1,12+1): #need to do plus 1 because python is upper bound exclusive
    print(i)

or a while loop like this.

i = 1
while i <= 12: #loop will end when i becomes 12.
    print(i)
    i += 1 #increments by 1

If each item is a separate variable and not a list, try this,

for i in range(1,12+1): 
    if eval(f'response{i}.status_code != 200'): #this is an eval function with an f string, the curly braces will dynamically assign the variable on each loop
        print("error in response" + str(i)) #you cant concat a string and int together unless you convert the int to a string like this.

for i in data: loops directly over data, thus i will be each element in data, not its index. If you need the index, use enumerate():

output = []
for i, elem in enumerate(data):
  x = data[i] - data[i+1]
  output.append(data)

However, that won't work, due to an out-of-bounds error that will occur at the very end. Instead, you can use built-in numpy functionality:

output = data[:-1] - data[1:]
print(output)

Output:

array([-5.5, -5.5, -5.5, -5.5, -5.5, -5.5, -5.5, -5.5, -5.5])

Or possible solutions:

np.full(9, -5.5)

np.ones(9) * -5.5