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copied!<p>I am no great expert in perl, so there is quite possibly a quicker way to this, but it seems like the match operator "<code>//</code>" in list context is what you need. When you assign the result of a match operation to a list, the match operator takes on list context and returns a list with each of the parenthesis delimited sub-expressions. If you specify global matches with the "<code>g</code>" flag, it will return a list of all the matches of each sub-expression. Example:</p> <pre><code># print a list of each match for "x" in "xxx" @aList = ("xxx" =~ /(x)/g); print(join(".", @aList)); </code></pre> <p>Will print out</p> <pre><code>x.x.x </code></pre> <p>I'm assuming you have a regular expression for each of those 5 types <code>h</code>, <code>D</code>, <code>s</code>, <code>E</code>, and <code>x</code>. You didn't say whether each of these parts is a single character or multiple, so I'm going to assume they can be multiple characters. If so, your solution might be something like this:</p> <pre><code>$h = ""; # Insert regex to match "h" $D = ""; # Insert regex to match "D" $s = ""; # Insert regex to match "s" $E = ""; # Insert regex to match "E" $x = ""; # Insert regex to match "x" $sequenceRE = "($h){3}($D){3}($s){2}($E){2}($x)($D)" if ($line =~ /$sequenceRE/) { $hPart = $1; $sPart = $3; $xPart = $5; @hValues = ($hPart =~ /($h)/g); @sValues = ($sPart =~ /($s)/g); @xValues = ($xPart =~ /($x)/g); } </code></pre> <p>I'm sure there is something I've missed, and there are some subtleties of perl that I have overlooked, but this should get you most of the way there. For more information, read up on perl's <a href="http://www.perl.com/doc/manual/html/pod/perlop.html#Regexp_Quote_Like_Operators" rel="nofollow noreferrer">match operator</a>, and <a href="http://www.perl.com/doc/manual/html/pod/perlre.html" rel="nofollow noreferrer">regular expressions</a>.</p>

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