Python - replace multiple characters without .replace()

Question

The task is to transform any string into any string without built-in .replace(). I failed because I forgot that technically space is also a string character. Firstly I transformed this string into the list, but now I see I did it unnecessarily. However, it still doesn't work.

I can replace "cat" into "dog"
I can replace "c" into "dog"

I can't replace "a cat" into "a dog".

I tried with lambda or zip, but I don't really know how to do it. Do you have any clue?

string = "Alice has a cat, a cat has Alice."
old = "a cat"
new = "a dog"

def rplstr(string,old,new):
    """ docstring"""

    result = ''
    for i in string:
        if i == old:
            i = new
        result += i
    return result

print rplstr(string, old, new)

Hmm... given strings are immutable, really the assignment doesn't make sense since you could just return new... — Claudiu
– Claudiu, Commented Feb 8, 2016 at 18:11

Martin Evans · Accepted Answer · 2016-02-08 18:39:48Z

5

This solution avoids string concatenation which can be less efficient. It creates a list of segments to join together at the end:

string = "Alice has a cat, a cat has Alice."
old = "a cat"
new = "a dog"

def rplstr(string, old, new):
    """ docstring"""

    output = []
    index = 0

    while True:
        next = string.find(old, index)

        if next == -1:
            output.append(string[index:])
            return ''.join(output)
        else:
            output.append(string[index:next])
            output.append(new)
            index = next + len(old)

print rplstr(string, old, new)

Giving:

Alice has a dog, a dog has Alice.

answered Feb 8, 2016 at 18:39

Martin Evans

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Comments

Brendan Abel · Accepted Answer · 2016-02-08 19:20:23Z

3

You can step through the string, one character at a time, and test to see if it matches with the first character of your old string. If it matches, keep a reference to the index, then continue stepping through the characters, now trying to match against the 2nd char of old. Keep going until you match the entire old string. If the full match succeeds, use the index of the first character match and the length of the old string to create a new string with the new string inserted into it.

def replstr(orig, old, new):
    i = 0
    output = ''
    temp = ''
    for c in orig:
        if c == old[i]:
            i += 1
            temp += c
        else:
            i = 0
            if temp:
                output += temp
                temp = ''
            output += c
        if len(temp) == len(old):
            output += new
            temp = ''
            i = 0
    else:
        if temp:
            output += temp

edited Feb 8, 2016 at 19:20

answered Feb 8, 2016 at 18:13

Brendan Abel

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2 Comments

Tom Wojcik Over a year ago

>>>Alice hasa a dogt, a dog hasa aAalaiacaea.a That's what I have, but probably it's just the matter of a few small things in the code. A cat was changed into a dog. I'll read your code few times. Thank you.

Brendan Abel Over a year ago

@TomWojcik Ah, yes, I missed a line. Updated.

pp_ · Accepted Answer · 2016-02-08 18:21:03Z

2

You can do it with slices:

def rplstr(string, old, new):
    for i in xrange(len(string)):
        if old == string[i:i+len(old)]:
            string = string[:i] + new + string[i+len(old):]
    return string

answered Feb 8, 2016 at 18:21

pp_

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2 Comments

Brendan Abel Over a year ago

I believe if the new string is longer than the old string, then letters on the end of the string won't get tested. Or if the new string contains the old string.

Tom Wojcik Over a year ago

You, sir, know how to code. Thank you. Works perfect and it's relatively simple.

Prashanth · Accepted Answer · 2016-02-08 20:16:55Z

2

You can do it in a simple and tiny way by using regular expressions.

import re

my_string = "Alice has a cat, a cat has Alice."
new_string = re.sub(r'a cat', 'a dog', my_string)
print new_string

answered Feb 8, 2016 at 20:16

Prashanth

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1 Comment

Tom Wojcik Over a year ago

Indeed, it does work but my tutor said that we cannot use any built-in methods, nor regular expressions.

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