No I cannot, sorry.
I've yet to find a language that gives me as much power despite the terseness as Perl. If I understand correcly, awk does not have an inbuilt array sort routine, but I may be mistaken.

The only other non-Perl and non-Ruby method I could think of leverages the shell's multiple-key sort command, the algorithm for which would be something like -

(1) Read a line and form the start-end pairs separated by a delimiter.
(2) Redirect the pairs, one per line, to a temporary file.
(3) Sort the temporary file on the multiple numeric keys.
(4) Read them back and print to stdout.
(5) Repeat (1) through (4) till eof.

tyler_durden

---------- Post updated at 05:15 PM ---------- Previous update was at 04:16 PM ----------

Well ok, here's an implementation of algorithm posted earlier, in a Bash shell script -

Code:

$
$
$ cat f4
No Start End
A 22,222,33,22,1233,3232,44 555,333,222,55,1235,3235,66
B 33,333,22,44 66,340,44,55
C 66,55,555 75,58,560
$
$
$ cat f4.sh
#!/usr/bin/bash
LNUM=1
while read NO START END; do
  if [ "$LNUM" -eq "1" ]; then
    echo "$NO $START $END"
  else
    echo $START | tr -s "," "\n" >f1.tmp
    echo $END | tr -s "," "\n" >f2.tmp
    paste -d~ f1.tmp f2.tmp | sort -t~ -nk1,1 -nk2,2 >f3.tmp
    IFS="~"
    INDX=1
    while read NUM1 NUM2; do
      if [ "$INDX" -eq "1" ]; then
        LIST1=$NUM1
        LIST2=$NUM2
      else
        LIST1="$LIST1,$NUM1"
        LIST2="$LIST2,$NUM2"
      fi
      INDX=`expr $INDX + 1`
    done <f3.tmp
    echo "$NO $LIST1 $LIST2"
    unset IFS
  fi
  LNUM=`expr $LNUM + 1`
done <f4
rm f1.tmp f2.tmp f3.tmp
$
$
$
$ . f4.sh
No Start End
A 22,22,33,44,222,1233,3232 55,555,222,66,333,1235,3235
B 22,33,44,333 44,66,55,340
C 55,66,555 58,75,560
$
$
$

I believe it's not good enough if the data is huge. Use of arrays instead of temp files should make it faster. It's definitely slower than the Perl script.

tyler_durden