Pattern matching deletion of list items

Question

Possible Duplicate:
How can I remove B -> A from a list if A -> B is in the list?

I'm working toward building a graph and use some code to construct my rules which results in a form as follows:

g = {1 -> 2, 1 -> 3, 2 -> 1, 2 -> 3, 2 -> 4, 3 -> 2, 3 -> 4, 3 -> 1, 4 -> 3, 4 -> 2}

I would like to delete all reverse duplicate entries form the list, i.e. both 1->2 and 2->1 appear in the list. I would like to get rid of one of them and do this for all reverse duplicates in the list g.

All: please try a bit harder to find duplicate questions; thank you. — Mr.Wizard, Jun 27 '12 at 20:46
@Tom That case is already covered by my own answer to the previous question. If you feel that this closure was in error feel free to vote to reopen. IMHO the fact that identical answers were posted indicates it is a true duplicate. (ps. please use the @name notification when responding; I didn't see your message until now.) — Mr.Wizard, Jun 28 '12 at 07:10

score 5 · Answer 1 · answered Jun 27 '12 at 16:45

5

Union[Sort /@ g] seems to be what you need...

answered Jun 27 '12 at 16:45

J. M.'s missing motivation

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score 4 · Accepted Answer · answered Jun 27 '12 at 16:49

4

This satisfies your needs

DeleteDuplicates @ (Sort /@ g)

{1 -> 2, 1 -> 3, 2 -> 3, 2 -> 4, 3 -> 4}

answered Jun 27 '12 at 16:49

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Kind of a waste to f @ (stuff) when you can just f[stuff], no? – J. M.'s missing motivation Jun 27 '12 at 16:52
Well, no unnecessary signs : Sort /@ g // DeleteDuplicates – Artes Jun 27 '12 at 16:57
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@J.M. I also use this kind of @(...) syntax because of reasons given here. By removing the head you don't have to jump to the end to remove ]. – István Zachar Jun 28 '12 at 08:53
@István: still bloody ugly for me. To each his own, I suppose. – J. M.'s missing motivation Jun 28 '12 at 14:21

Pattern matching deletion of list items

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