Implement python replace() function without using regexp

Question

I'm trying to rewrite the equivalent of the python replace() function without using regexp. Using this code, i've managed to get it to work with single chars, but not with more than one character:

def Replacer(self, find_char, replace_char):
    s = []
    for char in self.base_string:
        if char == find_char:
            char = replace_char
        #print char
        s.append(char)
    s = ''.join(s)

my_string.Replacer('a','E')

Anybody have any pointers how to make this work with more than one character? example:

my_string.Replacer('kl', 'lll')

technically, if you are doing a search like this you will be creating a regular expression. Whether you avoid using python's convenient regex syntax is up to you however — deontologician
– deontologician, Commented Nov 15, 2011 at 16:35
@habitue: Actually, you'll probably do a lot worse. Modern regex engines have all kinds of neat tricks, that simple algorithms are not going to benefit from. — Tyler Eaves
– Tyler Eaves, Commented Nov 15, 2011 at 16:38
Rewriting this method gives me in-depth understanding of string operations. Just by look @CedricJulien code I've learned a lot and I appreciate that he took the effort to share the implementation — jwesonga
– jwesonga, Commented Nov 15, 2011 at 17:19

MikeyB · Accepted Answer · 2011-11-15 19:42:37Z

4

How clever are you trying to be?

def Replacer(self, find, replace):
    return(replace.join(self.split(find)))

>>> Replacer('adding to dingoes gives diamonds','di','omg')
'adomgng to omgngoes gives omgamonds'

answered Nov 15, 2011 at 19:42

MikeyB

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Andrew Clark · Accepted Answer · 2017-04-04 20:13:16Z

3

Here is a method that should be pretty efficient:

def replacer(self, old, new):
    return ''.join(self._replacer(old, new))

def _replacer(self, old, new):
    oldlen = len(old)
    i = 0
    idx = self.base_string.find(old)
    while idx != -1:
        yield self.base_string[i:idx]
        yield new
        i = idx + oldlen
        idx = self.base_string.find(old, i)
    yield self.base_string[i:]

edited Apr 4, 2017 at 20:13

answered Nov 15, 2011 at 17:46

Andrew Clark

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Comments

Cédric Julien · Accepted Answer · 2011-11-15 16:42:28Z

0

Let's try with some slices (but you really should consider using the builtin method of python) :

class ReplacableString:
    def __init__(self, base_string):
        self.base_string =base_string

    def replacer(self, to_replace, replacer):
        for i in xrange(len(self.base_string)):
            if to_replace == self.base_string[i:i+len(to_replace)]:
                self.base_string = self.base_string[:i] + replacer + self.base_string[i+len(to_replace):]

    def __str__(self):
        return str(self.base_string)


test_str = ReplacableString("This is eth string")
test_str.replacer("eth", "the")
print test_str

>>> This is the string

answered Nov 15, 2011 at 16:42

Cédric Julien

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2 Comments

jwesonga Over a year ago

Thanks Cedric, I noticed that if u call replacer twice with different params, it replaces the characters based on the last value of self.base_string. Anyway to call replacer and have it change the value of the original self.base_string?

Cédric Julien Over a year ago

@jwesonga : if you don't want to change the original string, juste create a new_string = self.base_string[:] at the beginning of the replacer method, and then, use it everywhere instead of self.base_string, and at the end, return new_string

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