append special chars with "\"

Duncan Booth duncan at NOSPAMrcp.co.uk
Tue Sep 30 10:26:03 CEST 2003


tertius <tcronj at ananzi.co.za> wrote in 
news:3f793544$0$64721 at hades.is.co.za:

> Is there a better way to append certain chars in a string with a 
> backslash that the example below?
> 
> chr = "#$%^&_{}"          # special chars to look out for
> str = "123 45^ & 00 0_"   # string to convert
> n = ""                    # init new string
> for i in str:
>      if i in chr:          # if special character in str
>          n+='\\'           # append it with a backslash
>      n+=i
> print "old:",str
> print "new:",n
> 
Not using builtins as variable names would be better for a start. 'chr' is 
a builtin function, 'str' is a builtin type.

Building a string by appending characters is never a good idea, if you try 
this code with longer strings you will rapidly find it grinds to a halt. 
The Pythonic way to build up a string from substrings is either to use 
str.join on a list of strings, or to use StringIO.

Regular expressions are rarely the best solution to a problem, but in this 
case it seems they could be:

>>> import re
>>> myStr = "123 45^ & 00 0_"
>>> print re.sub("([#$%^&_{}])", r"\\\1", myStr)
123 45\^ \& 00 0\_

The only real catch is that if your list of special characters is variable 
and sometimes contains ']' you have a bit of work to build the regular 
expression.

A solution without regular expressions (so it works for any string of 
special characters) would be:

>>> specialChars = "#$%^&_{}"
>>> myStr = "123 45^ & 00 0_"
>>> specials = dict([(c, '\\'+c) for c in specialChars])
>>> print str.join('', [ specials.get(c, c) for c in myStr ])
123 45\^ \& 00 0\_

-- 
Duncan Booth                                             duncan at rcp.co.uk
int month(char *p){return(124864/((p[0]+p[1]-p[2]&0x1f)+1)%12)["\5\x8\3"
"\6\7\xb\1\x9\xa\2\0\4"];} // Who said my code was obscure?




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