I think there should be a way to do this simply, but I can't find it in the docs.
Syslog records contain literal dates that do not have leading zeros, but rather leading spaces. So Feb 1st is 'Feb..1' (2 spaces).
I want to remove one of those spaces (so I can 'split()' the line on spaces consistantly). Simple: 
$a="test string x  x\nFeb  1 09:12:33  (";
if ($a =~ /^\w{3}\s\s\d/m)  { # /^Feb  1/
  $a =~ s/^\w{3}(\s\s)\d/ /;
  print "<$a>\n";
}else{
  print "no match\n";
}
#<x  x
Feb  1 09:12:33  (>

[download]

But it seems the substtitution isn't as smart as a regex. The program runs, but doesn't modify the string. It obviously recognizes the parens for grouping purposes, but only as insertions into the substituting text. But it can do: 
my $var = 'testing';
$_ = 'In this string we are $var the "e" modifier.';

s/(\$\w+)/$1/ee;

print;

>In this string we are testing the "e" modifier.

[download]
It seem intuitive that a grouped substring would be the target of a substitution, if it exists for no other purpose. Or how about: 
  $a=~ s/$a =~ s/^\w{3}(\s\s)\d/$1/ /; #3 partr substitute like sed, $
+1 with a space?

[download]
So, is there a simple way to do substitution within a context?

It is always better to have seen your target for yourself, rather than depend upon someone else's description.

In reply to contextual substitution with s///? by Wiggins

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