I have two lists: excluded and values. I would like to efficiently determine whether excluded and values are disjoint. How does one go about doing this (preferably using FreeQ)?
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In version 10.0 you have DisjointQ (and conversely IntersectingQ) to test this. 10.2 adds the Contains* family of function with ContainsNone being equivalent to DisjointQ. For earlier versions you could build this yourself:
ClearAll[disjointQ]
disjointQ[a_List, b_List] := Intersection[a, b] === {}
disjointQ[{1, 2, 3}, {6, 4, 5}]
(* True *)
disjointQ[{1, 2, 3}, {1, 4, 5}]
(* False *)
answered Jul 6 '15 at 14:34
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I'd expect this might be faster than intersection on larger lists:
With[{j = Join[DeleteDuplicates@#1, DeleteDuplicates@#2]}, DeleteDuplicates@j == j] &[l1, l2]
Addendum - a little testing, does seem to have advantage when both lists large, otherwise a bit of a wash between this and using intersection... perhaps others can test on non-loungbook environments - I get wildly varying results depending on how I produce the random test lists :-|
Addendum 2: Per comments, differences were from packed/unpacked lists, and in my limited tests the above is faster for unpacked... carry on...
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$\begingroup$ Could you give an example of the wildly varying results? I presume it's not simply packed vs. unpacked? $\endgroup$ – Mr.Wizard♦ Jul 6 '15 at 17:04
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$\begingroup$ @Mr.Wizard: {l1,l2}=Random[...,{2,x}] vs l1=Random[...,x];l2=Random[...,x], for example - I'd imagine is cache related on the hampsterbox... $\endgroup$ – ciao Jul 6 '15 at 17:15
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$\begingroup$ And you are running 10.1.0? (Referring to Why does list assignment with a packed array result in unpacked values?) $\endgroup$ – Mr.Wizard♦ Jul 6 '15 at 17:19
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$\begingroup$ @Mr.Wizard: Nope, 10.x still on my voodoo list. And herp-a-derp on my part - could have sworn I checked packing when I got differing results and both cases were unpacked, appears I inadvertently tested same set, so that explains it. $\endgroup$ – ciao Jul 6 '15 at 17:22
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While highly inefficient you asked about FreeQ and you could do this:
f0 = FreeQ[#, Alternatives @@ #2] &;
More practically here is a condensed version of rasher/ciao's method:
f1 = DuplicateFreeQ[Join @@ DeleteDuplicates /@ {##}] &
answered Jul 6 '15 at 17:07
