Symmetric difference of a sample of lists

Question

Consider the following:

list1={a,b,c,d};
list2={c,d,e,f};
list3={g,d,c,h};

Now I would like to get the symmetric difference. Hence, the result must be {a,b,e,f,g,h} (whether the result is sorted or not is irrelevant)

BTW: Using Complement[list1,list2,list3] returns {a,b}, since Complement only "gives the elements in e_all which are not in any of the e_i." (Mathematica Documentation Center)

What you ask for is not the symmetric difference. The symmetric difference of the three sets you posted is {a, b, c, d, e, f, g, h}... — Henrik Schumacher, Apr 14 '18 at 21:16

score 8 · Accepted Answer · 2012-04-19T14:48:51.100

8

I think you are looking for this:

Complement[Join[list1, list2, list3], 
 Intersection[list1, list2, list3]]

or as suggested in the comments:

Complement[Union[list1, list2, list3], 
 Intersection[list1, list2, list3]]

(*
=> {a, b, e, f, g, h}
*)

Note, that the h is present here but not the d. I assume this is a typo in the question.

edited Apr 19 '12 at 14:48

answered Apr 19 '12 at 14:28

1

So that's the list of elements that are missing from at least one set. Then h should be in the expected result too. – Szabolcs Apr 19 '12 at 14:30
@Szabolcs, that's a valid point. I have updated the post. – Apr 19 '12 at 14:36
1

@Szabolcs, I assume that the d<=>h is a typo. We will see what OP has to say. – Apr 19 '12 at 14:40
Totaly what I was looking for. Thx! – John Apr 19 '12 at 14:45
1

@John, most welcome. May I suggest that you fix the question to reflect this? – Apr 19 '12 at 14:46
From mathematical point of view I would rather use Union[list1, list2, list3] instead of Join[list1, list2, list3]. – Artes Apr 19 '12 at 14:46
@Artes, added. Thanks. – Apr 19 '12 at 14:49
@Artes: May I ask why? – John Apr 19 '12 at 14:52
1

@John With the Join you get duplicates, while with Union you don't. Union is implementation of sum of sets, Join is just a concatenation of lists. – Artes Apr 19 '12 at 14:56

score 2 · Answer 2 · answered Dec 13 '23 at 10:40

x = {a, b, c, d};
y = {c, d, e, f};
z = {g, d, c, h}

Nowadays we get the wanted result more easily with:

Flatten @ UniqueElements[{x, y, z}]

{a, b, e, f, g, h}

As Henrik commented the question title is misleading:

SymmetricDifference[x, y, z]

{a, b, c, d, e, f, g, h}

UniqueElements and SymmetricDifference came with V 13.1

E. Chan-López · Answer 3 · 2023-12-13T18:13:58.337

1

Another way using Counts and Select:

x = {a, b, c, d};
y = {c, d, e, f};
z = {g, d, c, h};
Keys@Select[Counts[Catenate@{x, y, z}], # == 1 &]
({a, b, e, f, g, h})

edited Dec 13 '23 at 18:13

answered Dec 13 '23 at 18:02

E. Chan-López

23,117
3
21
44

Symmetric difference of a sample of lists

3 Answers3

Linked