Python strings are immutable, you change them by making a copy.
The easiest way to do what you want is probably:

text = "Z" + text[1:]

The text[1:] returns the string in text from position 1 to the end, positions count from 0 so ‘1’ is the second character.

edit:
You can use the same string slicing technique for any part of the string

text = text[:1] + "Z" + text[2:]

Or if the letter only appears once you can use the search and replace technique suggested
below

new = text[:1] + 'Z' + text[2:]

Like other people have said, generally Python strings are supposed to be immutable.

However, if you are using CPython, the implementation at python.org, it is possible to use ctypes to modify the string structure in memory.

Here is an example where I use the technique to clear a string.

Mark data as sensitive in python

I mention this for the sake of completeness, and this should be your last resort as it is hackish.

This code is not mine. I couldn’t recall the site form where, I took it. Interestingly, you can use this to replace one character or more with one or more charectors.
Though this reply is very late, novices like me (anytime) might find it useful.

Change Text function.

mytext = 'Hello Zorld'
# change all Z(s) to "W"
while "Z" in mytext:
      # replace "Z" to "W"
      mytext = mytext.replace('Z', 'W')
print(mytext)

Starting with python 2.6 and python 3 you can use bytearrays which are mutable (can be changed element-wise unlike strings):

s = "abcdefg"
b_s = bytearray(s)
b_s[1] = "Z"
s = str(b_s)
print s
aZcdefg

edit: Changed str to s

edit2: As Two-Bit Alchemist mentioned in the comments, this code does not work with unicode.

Fastest method?

There are three ways. For the speed seekers I recommend ‘Method 2’

Method 1

Given by this answer

text = 'abcdefg'
new = list(text)
new[6] = 'W'
''.join(new)

Which is pretty slow compared to ‘Method 2’

timeit.timeit("text = 'abcdefg'; s = list(text); s[6] = 'W'; ''.join(s)", number=1000000)
1.0411581993103027

Method 2 (FAST METHOD)

Given by this answer

text = 'abcdefg'
text = text[:1] + 'Z' + text[2:]

Which is much faster:

timeit.timeit("text = 'abcdefg'; text = text[:1] + 'Z' + text[2:]", number=1000000)
0.34651994705200195

Method 3:

Byte array:

timeit.timeit("text = 'abcdefg'; s = bytearray(text); s[1] = 'Z'; str(s)", number=1000000)
1.0387420654296875

Actually, with strings, you can do something like this:

oldStr = 'Hello World!'    
newStr = ''

for i in oldStr:  
    if 'a' < i < 'z':    
        newStr += chr(ord(i)-32)     
    else:      
        newStr += i
print(newStr)

'HELLO WORLD!'

Basically, I’m “adding”+”strings” together into a new string :).

if your world is 100% ascii/utf-8(a lot of use cases fit in that box):

b = bytearray(s, 'utf-8')
# process - e.g., lowercasing: 
#    b[0] = b[i+1] - 32
s = str(b, 'utf-8')

python 3.7.3

I would like to add another way of changing a character in a string.

>>> text = '~~~~~~~~~~~'
>>> text = text[:1] + (text[1:].replace(text[0], '+', 1))
'~+~~~~~~~~~'

How faster it is when compared to turning the string into list and replacing the ith value then joining again?.

List approach

>>> timeit.timeit("text = '~~~~~~~~~~~'; s = list(text); s[1] = '+'; ''.join(s)", number=1000000)
0.8268570480013295

My solution

>>> timeit.timeit("text = '~~~~~~~~~~~'; text=text[:1] + (text[1:].replace(text[0], '+', 1))", number=1000000)
0.588400217000526

Strings are immutable in Python, which means you cannot change the existing string.
But if you want to change any character in it, you could create a new string out it as follows,

def replace(s, position, character):
    return s[:position] + character + s[position+1:]

replace(‘King’, 1, ‘o’)
// result: Kong

Note: If you give the position value greater than the length of the string, it will append the character at the end.

replace(‘Dog’, 10, ‘s’)
// result: Dogs

A solution combining find and replace methods in a single line if statement could be:

```python
my_var = "stackoverflaw"
my_new_var = my_var.replace('a', 'o', 1) if my_var.find('s') != -1 else my_var
print(f"my_var = {my_var}")           # my_var = stackoverflaw
print(f"my_new_var = {my_new_var}")   # my_new_var = stackoverflow
```

try this :

old_string = "mba"
string_list = list(old_string)
string_list[2] = "e"
//Replace 3rd element

new_string = "".join(string_list)

print(new_string)

I like f-strings:

text = f'{text[:1]}Z{text[2:]}'

In my machine this method is 10% faster than the "fast method" of using + to concatenate strings:

>>> timeit.timeit("text = 'abcdefg'; text = text[:1] + 'Z' + text[2:]", number=1000000)
1.1691178000000093
>>> timeit.timeit("text = 'abcdefg'; text = f'{text[:1]}Z{text[2:]}'", number =1000000)
0.9047831999999971
>>>

To replace a character in a string

You can use either of the method:

Method 1

In general,

string = f'{string[:index]}{replacing_character}{string[index+1:]}'

Here

text = f'{text[:1]}Z{text[2:]}'

Method 2

In general,

string = string[:index] + replacing_character + string[index+1:]

Here,

text = text[:1] + 'Z' + text[2:]