Replacing instances of a character in a string

Question

Strings in python are immutable, so you cannot treat them as a list and assign to indices.

Use .replace() instead:

line = line.replace(';', ':')

If you need to replace only certain semicolons, you'll need to be more specific. You could use slicing to isolate the section of the string to replace in:

line = line[:10].replace(';', ':') + line[10:]

That'll replace all semi-colons in the first 10 characters of the string.

You can do the below, to replace any char with a respective char at a given index, if you wish not to use .replace()

word = 'python'
index = 4
char = 'i'

word = word[:index] + char + word[index + 1:]
print word

o/p: pythin

Turn the string into a list; then you can change the characters individually. Then you can put it back together with .join:

s = 'a;b;c;d'
slist = list(s)
for i, c in enumerate(slist):
    if slist[i] == ';' and 0 <= i <= 3: # only replaces semicolons in the first part of the text
        slist[i] = ':'
s = ''.join(slist)
print s # prints a:b:c;d

If you want to replace a single semicolon:

for i in range(0,len(line)):
 if (line[i]==";"):
     line = line[:i] + ":" + line[i+1:]

Havent tested it though.

You cannot simply assign value to a character in the string. Use this method to replace value of a particular character:

name = "India"
result=name .replace("d",'*')

Output: In*ia

Also, if you want to replace say * for all the occurrences of the first character except the first character, eg. string = babble output = ba**le

Code:

name = "babble"
front= name [0:1]
fromSecondCharacter = name [1:]
back=fromSecondCharacter.replace(front,'*')
return front+back

This should cover a slightly more general case, but you should be able to customize it for your purpose

def selectiveReplace(myStr):
    answer = []
    for index,char in enumerate(myStr):
        if char == ';':
            if index%2 == 1: # replace ';' in even indices with ":"
                answer.append(":")
            else:
                answer.append("!") # replace ';' in odd indices with "!"
        else:
            answer.append(char)
    return ''.join(answer)