[ILUG] sed question

[Brian Foster]

  | Date: Thu, 19 Feb 2004 19:48:40 +0000
  | From: Padraig Brady <padraig.brady at corvil.com>
  | 
  | OK I've a (very big) file like:
  | 
  | 0
  | 0
  | 1
  | 1
  | 2
  | 2
  | 3
  | 3
  | :
  | 
  | I would like to find and print the "2" lines,
  | AND THEN STOP PRCOCESSING the humungous file
  | [ using sed(1) ].

 if the above really is what the file is like,
 I'd consider something like:

    sed -n -e '/3/q' -e '/2/p'  |  ... processing `2' lines ...

 the sed(1) prints all (and only) `2' lines, and
 stops when it finds a (_known_) line which first
 occurs (soon) after the `2' lines; in this case,
 a `3' line.

 otherwise:

    sed -n -e '
    /2/{
       p
       x
       b
    }
    s/^.*$/anything which does not match the RE/
    x
    /2/q'

 where “the RE‟ is /2/.  the key here is to use
 the hold space, which starts out empty.  as long
 as the hold space does not match /2/, then we
 either have not seen, or are still processing,
 the `2'-lines; as soon as it does match /2/,
 then we are done with all the `2'-lines.

 the above, of course, can be written more tersely,
 something like (NOT tested!):

    sed -n -e '/2/{p;x;b;}' -e 's/^.*$/no/' -e x -e /2/q
 or
    sed -n -e '/2/{p;x;b;};s/^.*$/no/;x;/2/q'

 you can probably replace the `b' with `d'; and/or
 omit the `s/^.*$/no/' altogether; and I am sure
 other tweaks are possible .....

 it is equally trivial in awk(1) --- which _may_
 not be as slow as you think! --- but sed(1) is
 probably faster if a tiny amount more obscure.

cheers!
	-blf-
--
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