Select elements in a list when the firsts two items are the same (no matter order) and its third element is lower

Question

I'm trying to solve a heuristic search in Mathematica 9. I get a list of succesor nodes called open with two nodes and a heuristic value, e.g.,

open={{node1,node2,H1},...,{node2,node1,Hi},...,{noden,noden,Hn}}

I need to select only when the two first elements are the same (no matter order) but where the heuristic is lower. Following the code that I used for sort the elements by the first item

Sort[open, #1[[1]] < #2[[1]] &]

I try to select cases when the same first and second numbers and the heuristic value where lower:

Select[open, #1[[1]]==#2[[1]] &&  #1[[2]] == #2[[2]] && #1[[3]]< #2[[3]] &]

But, returns {}...

My list

open = {{1, 2, 3}, {2, 1, 5}, {1, 2, 2}, {3, 4, 5}, {4, 3, 2}};

Between {1, 2, 3}, {1, 2, 2}, and {2, 1, 5} then choose {1, 2, 2} that has the lower heuristic. And, between {3, 4, 5} and {4, 3, 2} then {4, 3, 2}...

{{1, 2, 2},{4, 3, 2}}

Thanks,

Answer 1

You can try something like this:

First[Sort[#, #1[[3]] < #2[[3]] &]]&/@Gather[open, Complement[Most[#1], Most[#2]] === {}&]

{{1, 2, 2}, {4, 3, 2}}

this is a combination of two functions: First you gather the same edges and then you sort them according to their weight and take the first element (which has the lowest value.)

Answer 2

A combination of GatherBy[] and SortBy[] does the trick:

First[SortBy[#, Last]] & /@
GatherBy[{{1, 2, 3}, {2, 1, 5}, {1, 2, 2}, {3, 4, 5}, {4, 3, 2}}, Sort[Take[#, 2]] &]
   {{1, 2, 2}, {4, 3, 2}}

Answer 3

In versions 10+, you can use MinimalBy combined with GroupBy:

open = {{1, 2, 3}, {2, 1, 5}, {1, 2, 2}, {3, 4, 5}, {4, 3, 2}};

Values @ GroupBy[open, Sort[#[[;; 2]]] &, First @ MinimalBy @ Last]

{{1, 2, 2}, {4, 3, 2}}

Or, MinimalBy combined with Merge

Values @ Merge[First @ MinimalBy @ Last][Sort[#[[;;2]]] -> # &/@ open]

{{1, 2, 2}, {4, 3, 2}}

Or, combined with GatherBy:

First @ MinimalBy[Last] /@ GatherBy[open, Sort[#[[;; 2]]] &]

{{1, 2, 2}, {4, 3, 2}}

For older versions, a combination of SplitBy and SortBy also works

First /@ SplitBy[SortBy[open, {Sort[#[[;;2]]]&, Last[#]&}], Sort[#[[;;2]]]&]

{{1, 2, 2}, {4, 3, 2}}