Python: String replace index

Question

I mean, i want to replace str[9:11] for another string. If I do str.replace(str[9:11], "###") It doesn't work, because the sequence [9:11] can be more than one time. If str is "cdabcjkewabcef" i would get "cd###jkew###ef" but I only want to replace the second.

• Luke: Using 'string.replace' converts every occurrence of the given text to the text you input. You are not wanting to do this, you just want to replace the text based on its position (index), not based on its contents.

Answer 1

you can do

s="cdabcjkewabcef"
snew="".join((s[:9],"###",s[12:]))

which should be faster than joining like snew=s[:9]+"###"+s[12:] on large strings

Answer 2

You can achieve this by doing:

yourString = "Hello"
yourIndexToReplace = 1 #e letter
newLetter = 'x'
yourStringNew="".join((yourString[:yourIndexToReplace],newLetter,yourString[yourIndexToReplace+1:]))

Answer 3

You can use join() with sub-strings.

s = 'cdabcjkewabcef'
sequence = '###'
indicies = (9,11)
print sequence.join([s[:indicies[0]-1], s[indicies[1]:]])
>>> 'cdabcjke###cef'

Answer 4

Given txt and s - the string you want to replace:

txt.replace(s, "***", 1).replace(s, "###").replace("***", s)

Another way:

txt[::-1].replace(s[::-1], "###", 1)[::-1]

Answer 5

str = "cdabcjkewabcef"
print((str[::-1].replace('cba','###',1))[::-1])

• Perry: This would be a better answer if you explained how the code you provided answers the question.

Answer 6

Here is a sample code:

word = "astalavista"
index = 0
newword = ""
addon = "xyz"
while index < 8:
 newword = newword + word[index]
 index += 1
 ind = index
i = 0
while i < len(addon):
 newword = newword + addon[i]
 i += 1
while ind < len(word):
 newword = newword + word[ind]
 ind += 1
print newword