On Wed, 2013-05-15 at 16:02 +0100, Dave Mitchell wrote:
> On Tue, May 14, 2013 at 12:31:42PM -0430, Patrick O'Callaghan wrote:
> > If not, then "cat *.list|sort -u" is enough.
>
> Or even just
>
> sort -u *.list
Indeed. Kernighan has a comment somewhere on the psychology of why
people prefer
On Tue, May 14, 2013 at 12:31:42PM -0430, Patrick O'Callaghan wrote:
> If not, then "cat *.list|sort -u" is enough.
Or even just
sort -u *.list
--
Spock (or Data) is fired from his high-ranking position for not being able
to understand the most basic nuances of about one in three sentences
On Tue, 2013-05-14 at 09:56 +0100, Frank Murphy wrote:
> Having looked at "man join" wasn't sure of it's use here.
'join' is more like a database table join (think of two tables being
joined horizontally).
> Unknown number of files, constant is extension .list
> (For testing purposes only using t
On Tue, 14 May 2013 11:09:19 +0200
Joachim Backes wrote:
>
> Hi Frank,
>
> give
> cat *.list >> output.joined | sort -u | uniq
> --all-repeated
>
> a try. If the output is empty ===> no dupes!
>
> Kind regards
>
Thanks Joachim,
Indeed it's empty.
--
Regards,
Frank - I check for new
On 05/14/2013 10:56 AM, Frank Murphy wrote:
Having looked at "man join" wasn't sure of it's use here.
Unknown number of files, constant is extension .list
(For testing purposes only using two)
cat *.list >> output.joined | sort -u
How can I test if the output.joined,
is indeed the combined two
