Sure, here's a start:

  $n =~ s/(^| )$a( |$)/$1$b$2/i;
[download]

You can avoid the $1..$2 in the substitution if you can avoid capturing the spaces, which you can do by matching with lookahead/lookbehind:

  $n =~ s/(^|(?<= ))$a((?= )|$))/$b/i;
[download]

Or inverse the sense of the lookarounds: "not after a non-space" rather than "after a space, or after nothing":

  $n =~ s/(?<![^ ])$a(?![%a])/$b/i;
[download]

Note that the first version is adequate for most purposes, but may trip you up if you want to do a global substitution for all copies of $a:

  "FOO foo" =~ s/(^| )foo( |$)/${1}bar$2/ig
[download]

The other two version do not suffer from that problem, since the lookarounds don't affect where a future match will start.

Hugo

Thankfully for you, there's an easy way to specify this: \b

my $x = 'a';
my $y = 'b';

while (<DATA>) {
    chomp;
    s/\b$x\b/$y/g;
    print "$_\n";
}

__DATA__
a
a 
 a
 a
[download]

But you can't be sure that $x starts and ends with an alphanumeric character, in which case \b$x\b wouldn't match when $x is surrounded by spaces.

---

Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart

$n =~ s/^( ?)$a( ?)$/$1$b$2/i;
[download]

You and sh1tn both made a fairly common mistake on this type of problem -- of the possibilities, you made one item required, and one item optional. So, instead of matching words in the middle of a longer string of words, it'll now only match single word strings, optionally padded with spaces.

TedPride had the method that I'd typically use, but it'll actually match much more than what water originally asked for. (it'll match if $a is near punctuation, which may result in substituting out part of a hyphenated compound word)

To match exactly what was asked for, I'd have used the same logic as hv's first response. (I've never been comfortable with look(ahead|behind) assertions). If I needed to do multiple replacements, I'd get around his mentioned problem this method by using \G :

$n =~ s/(^|\G| )$a($| )/$1$b$2/ig;

Not the prettiest, but this should work:

  $n =~ s/(?:^|([ ]))$a(?:([ ])|$)/($1 ? $1 : "") . $b . ($2 ? $2 : ""
+)/ie;
[download]

$n =~ s/^(\s*)$a(\s*)$/$1$b$2/i;
[download]