Changing one character in a string

Question

What is the easiest way in Python to replace a character in a string?

For example:

text = \"abcdefg\";
text[1] = \"Z\";
           ^

Answer 1

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'
mytext = mytext.replace('Z', 'W')
print mytext,

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

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