Just another option:

from collections import deque
def reverse(iterable):
    d = deque()
    d.extendleft(iterable)
    return ''.join(d)

What is the best way of implementing a reverse function for strings?

My own experience with this question is academic. However, if you're a pro looking for the quick answer, use a slice that steps by -1:

>>> 'a string'[::-1]
'gnirts a'

or more readably (but slower due to the method name lookups and the fact that join forms a list when given an iterator), str.join:

>>> ''.join(reversed('a string'))
'gnirts a'

or for readability and reusability, put the slice in a function

def reversed_string(a_string):
    return a_string[::-1]

and then:

>>> reversed_string('a_string')
'gnirts_a'

Longer explanation

If you're interested in the academic exposition, please keep reading.

There is no built-in reverse function in Python's str object.

Here is a couple of things about Python's strings you should know:

1. In Python, strings are immutable. Changing a string does not modify the string. It creates a new one.

2. Strings are sliceable. Slicing a string gives you a new string from one point in the string, backwards or forwards, to another point, by given increments. They take slice notation or a slice object in a subscript:

   string[subscript]
   

The subscript creates a slice by including a colon within the braces:

    string[start:stop:step]

To create a slice outside of the braces, you'll need to create a slice object:

    slice_obj = slice(start, stop, step)
    string[slice_obj]

A readable approach:

While ''.join(reversed('foo')) is readable, it requires calling a string method, str.join, on another called function, which can be rather relatively slow. Let's put this in a function - we'll come back to it:

def reverse_string_readable_answer(string):
    return ''.join(reversed(string))

Most performant approach:

Much faster is using a reverse slice:

'foo'[::-1]

But how can we make this more readable and understandable to someone less familiar with slices or the intent of the original author? Let's create a slice object outside of the subscript notation, give it a descriptive name, and pass it to the subscript notation.

start = stop = None
step = -1
reverse_slice = slice(start, stop, step)
'foo'[reverse_slice]

Implement as Function

To actually implement this as a function, I think it is semantically clear enough to simply use a descriptive name:

def reversed_string(a_string):
    return a_string[::-1]

And usage is simply:

reversed_string('foo')

What your teacher probably wants:

If you have an instructor, they probably want you to start with an empty string, and build up a new string from the old one. You can do this with pure syntax and literals using a while loop:

def reverse_a_string_slowly(a_string):
    new_string = ''
    index = len(a_string)
    while index:
        index -= 1                    # index = index - 1
        new_string += a_string[index] # new_string = new_string + character
    return new_string

This is theoretically bad because, remember, strings are immutable - so every time where it looks like you're appending a character onto your new_string, it's theoretically creating a new string every time! However, CPython knows how to optimize this in certain cases, of which this trivial case is one.

Best Practice

Theoretically better is to collect your substrings in a list, and join them later:

def reverse_a_string_more_slowly(a_string):
    new_strings = []
    index = len(a_string)
    while index:
        index -= 1                       
        new_strings.append(a_string[index])
    return ''.join(new_strings)

However, as we will see in the timings below for CPython, this actually takes longer, because CPython can optimize the string concatenation.

Timings

Here are the timings:

>>> a_string = 'amanaplanacanalpanama' * 10
>>> min(timeit.repeat(lambda: reverse_string_readable_answer(a_string)))
10.38789987564087
>>> min(timeit.repeat(lambda: reversed_string(a_string)))
0.6622700691223145
>>> min(timeit.repeat(lambda: reverse_a_string_slowly(a_string)))
25.756799936294556
>>> min(timeit.repeat(lambda: reverse_a_string_more_slowly(a_string)))
38.73570013046265

CPython optimizes string concatenation, whereas other implementations may not:

... do not rely on CPython's efficient implementation of in-place string concatenation for statements in the form a += b or a = a + b . This optimization is fragile even in CPython (it only works for some types) and isn't present at all in implementations that don't use refcounting. In performance sensitive parts of the library, the ''.join() form should be used instead. This will ensure that concatenation occurs in linear time across various implementations.

Not wanting to repeat what AJNeufeld has already said, I was curious how long it would take your code to reverse the full text of Moby Dick. While running this experiment, I noticed some funny behavior with spaces. For example, when your code is given " " (3 spaces), it returns " " (two spaces).

You might consider trying some quick tests such as these:

assert reverse_string("Happy new year! 2022") == "2022 year! new Happy"
assert reverse_string(" ") == " "
assert reverse_string("   ") == "   "
assert reverse_string(" a") == "a "
assert reverse_string("a ") == " a"
assert reverse_string(" a ") == " a "
assert reverse_string(" a  bc ") == " bc  a "
assert reverse_string("nospacesinthisone") == "nospacesinthisone"

On my computer, by the way, your approach handles Moby Dick in about 14 seconds! When I initially tried to refactor your code, I achieved a similar result:

Refactor Attempt #1: 13-14 seconds

def reverse_string(string: str) -> str:
    """Reverse the words of a string.

    >>> reverse_words('Happy new year! 2022')
    '2022 year! new Happy'
    """
    new_string, word = "", ""
    for char in string:
        if char == " ":
            new_string = char + word + new_string
            word = ""
        else:
            word += char
    n_missing_chars = len(string) - len(new_string)
    if n_missing_chars != 0:
        new_string = string[-n_missing_chars:] + new_string
    return new_string

Eventually, as you'll see below (if you try it), I managed to reduce the time to 0.1 seconds. It seems driven by replacing new = c + w + new with new += w + c. That this change would make things faster doesn't feel surprising, but I don't truly understand what is happening differently behind the scenes.

Refactor Attempt #2: 0.1 seconds

def reverse_string(string: str) -> str: 
    """Reverse the words of a string.

    >>> reverse_words('Happy new year! 2022')
    '2022 year! new Happy'
    """
    new_string, word = "", ""
    for char in reversed(string):
        if char == " ":
            new_string += word[::-1] + char
            word = ""
        else:
            word += char
    n_missing_chars = len(string) - len(new_string)
    if n_missing_chars != 0:
        new_string += string[:n_missing_chars]
    return new_string

While still adhering to your constraints, I wonder if there is a way to achieve similar or better performance than this without iterating through the string backwards.

Code Snippet: Trying Moby Dick yourself

If you want to reproduce the Moby Dick test yourself, you can use this snippet. You will need to save the full text of Moby Dick as a text file in the same directory as your Python script.

import time


with open("moby_dick.txt", "r") as f:
    string = f.read()

start = time.time()
reversed_moby_dick = reverse_string(string)
end = time.time()
print(f"Execution time: {end - start: .1f} seconds")

As a final comment, were it not for your rules about "built-in functions of string", I likely would have tried using str.join at some point.

This has been a very fun exercise. Thank you for posting it!

def rev(s):
    s1=""
    s2=""
    for i in range(len(s)):
        if s[i]!=" " and i!=len(s)-1:
            s1+=s[i]
        elif s[i]==" ":
            s2+=s1[::-1]+" "
            s1=""
        else:
            s1+=s[i]
            s2+=s1[::-1]
    return s2
if__name__="__main__"
s=input("Enter a sentense : ")
print("Sentence with individual reverse word : ",rev(s))