Need help converting "sassy" to "$a55y" using a regular expression? - regex

Question

Any s at the beginning of the word should be converted to a $.
Any s inside the word should be converted to a 5.

Answer 1

To match an s at the start of the word, use \b to match word boundaries and \w to match alphanumerics:
/\bs\w/
(as #Matthew points out, the \w is really superfluous:)
/\bs/
Once you've replaced all s at the start of a word, then the only remaining ones are inside the word (I'm assuming that you also want to replace s at the end of a word with 5) so you can simply use
/s/
For completeness, here's how to put it all together (I'm going to assume JavaScript):
function pimpMyEsses(str)
{
return str.replace(/\bs/gi, '$').replace(/s/gi, '5');
}
console.log(pimpMyEsses('slither quantum Sassy. arcades'));
// > "$lither quantum $a55y. arcade5"

Answer 2

Depending on the language it may be possible to capture the substitutions with a single regular expression and replace them procedurally. Here's a PHP example:
<?php
$word = 'sassy';
preg_match_all('/\b(s)|([^s]+)|(s)/', $word, $matches, PREG_SET_ORDER);
/* captures:
* $matches = array(
* array('s','s'),
* array('a','','a'),
* array('s','','','s'),
* array('s','','','s'),
* array('y','','y')
* )
*/
$newword = '';
foreach ($matches as $m){
if ($m[1]) $newword .= '$'; # leading s --> $
elseif ($m[2]) $newword .= $m[2]; # not an s --> as-is
else $newword .= '5'; # any other s --> 5
}
echo $newword;
Because I've used \b to match a word-boundary before the "leading s", the string 'sassy socks' becomes '$a55y $ock5'
If you want only the s at the start of "sassy" to become a $, change the regular expression to:
'/^(s)|([^s]+)|(s)/'

Answer 3

You can do:
/^(s)/ to select only the first "s";
/(?:[^s])(?:(s)[^s]*)+ to select all other "s". Note that the first character will be skipped (which is independent of);
Explain:ignore first character;Repeat one or more: get a "s" and ignore others character that not "s";
Next step: you need to determinate what language you will use.