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
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 lists with dupes removed.
--
Regards,
Fran
