On Jun 22, 2011, at 10:58 PM, Pete wrote:
> I may be wrong, but I don't think intersect solves the problem. If I
> understand the dictionary correctly, this will remove all the t1 keys that
> do NOT match t2 keys so it does the opposite of what is needed.
Thanks, Pete. I misunderstood the OP.
He wants the symmetric difference.
<http://en.wikipedia.org/wiki/Symmetric_difference>
Here's a way:
function symmetricDifferenceOfLists t1, t2
split t1 by cr and tab
split t2 by cr and tab
return the keys of symmeticDifferenceOfArrays( t1, t2 )
end symmetricDifferenceOfLists
function symmeticDifferenceOfArrays p1, p2
local tResult
put differenceOfArrays( p1, p2 ) into tResult
union tResult with differenceOfArrays( p2, p1 )
return tResult
end symmeticDifferenceOfArrays
function differenceOfArrays t1, p1
repeat for each key tKey in p1
delete variable t1[ tKey ]
end repeat
return t1
end differenceOfArrays
In comparison, Jacque's code is looking lean.
-- Dick
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