On Tue, 8 Oct 2002 dman(_at_)nomotek(_dot_)com wrote:
[...]
though it's now only one recipe. Okay, how about assigning to
each of the words different incremental spreads? Is there
an anticipated maximum number of times the words will appear?
Say, under 10 times each? If under ten, you could assign increments
of 1 to word1, 10 to word2, and 100 to word3. A final score of
564 means word1 appeared five times, word2 six times, and word3 four.
Or increment word1 by 1, word2 by 1000, and word3 by 100000.
Nice trick! Remind my another scenario when you do not need to
count the appearances:
:0 B
* 1^0 word1
* 2^0 word2
* 4^0 word3
* 8^0 word4
...
In this case you can use the $= to figure out the missing words.
Of course we have a limit here.
Thanks,
Udi
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