Replace a pattern in python

Question

How to replace the pattern in the string with

     decoded_str=" Name(++info++)Age(++info++)Adress of the emp(++info++)"

 The first pattern "(++info++)" needs to replaced with (++info a++)
 The second pattern "(++info++)" needs to replaced with (++info b++)
 The third pattern "(++info++)" needs to replaced with (++info c++)
 If there many more then it should be replaced accordingly

There are some good O(N) answers here, why accept one of the O(N^2) ones? — John La Rooy
– John La Rooy, Commented Aug 20, 2010 at 6:51
@gnibbler Would you mind tagging the O(N) ones? I'm not quite used to looking at real examples of that sort of thing. Does N here refer to the count of '++info++' elements? — aaronasterling
– aaronasterling, Commented Aug 20, 2010 at 7:13
@aaron, The O(N^2) answers are the ones that are making copies of the entire string for each (++info++). In this case it is mostly the ones using the 1 argument for replace() — John La Rooy
– John La Rooy, Commented Aug 20, 2010 at 7:28
I could be wrong, but I sense that these strings will be fairly short. — carl
– carl, Commented Aug 20, 2010 at 7:38
@gnibbler. Thanks. I see that now. I think my mistake was not knowing the string manipulation facilities as well as I should. Thanks again. — aaronasterling
– aaronasterling, Commented Aug 20, 2010 at 7:55

carl · Accepted Answer · 2010-08-20 07:38:38Z

4

This should be simple enough:

for character in range(ord('a'), ord('z')):
    if "(++info++)" not in decoded_str:
        break
    decoded_str = decoded_str.replace("(++info++)", "(++info {0}++)".format(chr(character)), 1)

print decoded_str

It has the added benefit of stopping at 'z'. If you want to wrap around:

import itertools

for character in itertools.cycle(range(ord('a'), ord('z'))):
    if "(++info++)" not in decoded_str:
        break
    decoded_str = decoded_str.replace("(++info++)", "(++info {0}++)".format(chr(character)), 1)

print decoded_str

And just for fun, a one-liner, and O(n):

dstr = "".join(x + "(++info {0}++)".format(chr(y)) for x, y in zip(dstr.split("(++info++)"), range(ord('a'), ord('z'))))[:-len("(++info a++)")]

edited Aug 20, 2010 at 7:38

answered Aug 20, 2010 at 6:26

carl

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Comments

killown · Accepted Answer · 2010-08-20 23:35:36Z

3

import string

decoded_str = " Name(++info++)Age(++info++)Adress of the emp(++info++)"
s = decoded_str.replace('++info++', '++info %s++')
s % tuple(i for i in string.ascii_lowercase[:s.count('%s')])

edited Aug 20, 2010 at 23:35

answered Aug 20, 2010 at 7:16

killown

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Comments

Andrew Hare · Accepted Answer · 2010-08-20 06:23:40Z

2

Here is a rather ugly yet pragmatic solution:

import string

decoded_str = " Name(++info++)Age(++info++)Adress of the emp(++info++)"
letters = list(string.lowercase)
token = "(++info++)"
rep_token = "(++info %s++)"

i = 0
while (token in decoded_str):
    decoded_str = decoded_str.replace(token, rep_token % letters[i], 1)
    i += 1

print decoded_str

answered Aug 20, 2010 at 6:23

Andrew Hare

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Comments

kennytm · Accepted Answer · 2010-08-20 06:30:07Z

1

>>> import re
>>> rx = re.compile(r'\(\+\+info\+\+\)')
>>> s = "Name(++info++)Age(++info++)Adress of the emp(++info++)"
>>> atoz = iter("abcdefghijklmnopqrstuvwxyz")
>>> rx.sub(lambda m: '(++info ' + next(atoz) + '++)', s)
'Name(++info a++)Age(++info b++)Adress of the emp(++info c++)'

answered Aug 20, 2010 at 6:30

kennytm

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Comments

phimuemue · Accepted Answer · 2010-08-20 06:32:44Z

1

Here's a quick hack to do it:

string=" Name(++info++)Age(++info++)Adress of the emp(++info++)"

def doit(s):
    import string
    allTheLetters = list(string.lowercase)
    i=0
    s2 = s.replace("++info++","++info "+allTheLetters[i]+"++",1)
    while (s2!=s):
        s=s2
        i=i+1
        s2 = s.replace("++info++","++info "+allTheLetters[i]+"++",1)
    return s

Note that performance is probably not very great.

edited Aug 20, 2010 at 6:32

answered Aug 20, 2010 at 6:21

phimuemue

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John La Rooy · Accepted Answer · 2010-08-20 06:37:36Z

1

import re, string

decoded_str=" Name(++info++)Age(++info++)Adress of the emp(++info++)"

sub_func=('(++info %s++)'%c for c in '.'+string.ascii_lowercase).send
sub_func(None)
print re.sub('\(\+\+info\+\+\)', sub_func, decoded_str)

answered Aug 20, 2010 at 6:37

John La Rooy

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John La Rooy · Accepted Answer · 2010-08-20 08:45:17Z

1

from itertools import izip
import string
decoded_str=" Name(++info++)Age(++info++)Adress of the emp(++info++)"
parts = iter(decoded_str.split("(++info++)"))
first_part = next(parts)
tags = iter(string.ascii_lowercase)
encoded_str=first_part+"".join("(++info %s++)%s"%x for x in izip(tags, parts))
print encoded_str

answered Aug 20, 2010 at 8:45

John La Rooy

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1 Comment

aaronasterling Over a year ago

Thanks for the example. It thought me one new thing and reminded me of about five I've forgotten.

ebt · Accepted Answer · 2010-08-20 06:36:13Z

0

decoded_str=" Name(++info++)Age(++info++)Adress of the emp(++info++)"

import re
for i, f in enumerate(re.findall(r"\(\+\+info\+\+\)",decoded_str)):
    decoded_str = re.sub(r"\(\+\+info\+\+\)","(++info %s++)"%chr(97+i),decoded_str,1)
print decoded_str

answered Aug 20, 2010 at 6:36

ebt

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