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Re: Union of sets (lists
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Re: Union of sets (lists


  • Subject: Re: Union of sets (lists
  • From: Michelle Steiner <email@hidden>
  • Date: Sun, 12 Jan 2003 09:34:57 -0700

On Sunday, January 12, 2003, at 08:18 AM, Kai wrote:

This may have something to do with the original situation, which the example
wouldn't necessarily indicate. However, if it's not critical which list is
added to which, it may be worth considering reversing the switch - so that
the repeat loop is based on the shorter list (thus speeding up the process)
- perhaps something like this:

*nod* For some reason, I had it in my head that I had to go through the long list to match against the short list in order to ensure that all items are covered.

This still assumes that the original lists contain no duplicates within
themselves - otherwise Deivy's point would also need to be considered:

If I recall set theory correctly (and it's been some 40+ years since I studied it), sets contain only unique items, so there wouldn't be any duplicates.

I was trying to solve a problem posted on the applscript newsgroup; I had only the information provided by the original poster. There was another problem too--intersecting sets--the result is a set containing items that appear on both sets, eliminating all that appear in only one set. That was a trivial modification of a working union script, though.

--Michelle

"There's some good in the world, Mr. Frodo, and it's worth fighting for."
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  • Follow-Ups:
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      • From: Gary Lists <email@hidden>
References: 
 >Re: Union of sets (lists (From: Kai <email@hidden>)

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