The following does the job by considering a different view of the original array. This means no additional arrays (and thus memory) are required. And no for loops :)

import numpy as np

A = np.array(['2', '3', '5', '7', '11', '13', '17', '19', '23', '29', '31', '37',
       '41', '43', '47', '53', '59', '61', '67', '71', '73', '79', '83',
       '89', '97'], dtype='<U2')

B = A.view(np.uint32)           # Interpret as individual numbers
B = B.reshape((A.shape[0],2))   # Group numbers in pairs
B[B[:,1] != 0, :] = B[B[:,1] != 0, ::-1] # Flip pairs if more than 1 digit

print(A) # ['2' '3' '5' '7' '11' '31' '71' '91' '32' '92' '13' '73' '14' '34' '74' '35' '95' '16' '76' '17' '37' '97' '38' '98' '79']

For short arrays, like the one you gave, @hpaulj's answer is faster. But for larger arrays, this method is faster. As an example, for an array of 100,000 elements, this method is approximately 5x faster compared to @hpaulj's fastest method ([s[::-1] for s in arr.tolist()]).

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.