Python: String replace index

7
votes

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.

6
pythonstringpython-3.x
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. - Luke

6 Answers

14
votes

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

3
votes

You can achieve this by doing:

yourString = "Hello"
yourIndexToReplace = 1 #e letter
newLetter = 'x'
yourStringNew="".join((yourString[:yourIndexToReplace],newLetter,yourString[yourIndexToReplace+1:]))
2
votes

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'
1
votes

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]
1
votes
str = "cdabcjkewabcef"
print((str[::-1].replace('cba','###',1))[::-1])
0
votes

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