DeleteCases depending on all elements

Question

I have a listed of Tupels list = {{10, 5}, {20, 4}, {20, 3}, {10, 6}}; If there are two elements a and b for which a[[1]] <= b[[1]] && a[[2]] <= b[[2]] then I would like to delete the element a. If I implement this via for loops, For[i = Length@list, i > 0, i--, For[j = Length@list, j > 0, j--, If[i != j && list[[i, 1]] <= list[[j, 1]] && list[[i, 2]] <= list[[j, 2]], list = Delete[list, i]; Break[];]; ]; ]; then this works but seems to be quite slow. As I need to implement this for many elements, I thought there could be a better/more performant way to solve this directly with Mathematica functions, like DeleteCases. Do you have an idea? Thanks a lot!

kglr · Accepted Answer

list = {{10, 5}, {20, 4}, {20, 3}, {10, 6}};

1. You can use the function Internal`ListMin as follows:

- Internal`ListMin[-list]

{{20, 4}, {10, 6}}

2. You can also use DeleteDuplicates:

DeleteDuplicates[ReverseSort@list, And @@ Thread[GreaterEqual[##]] &]

{{20, 4}, {10, 6}}

3. ... and SequenceReplace:

SequenceReplace[Sort @ list, 
  {a__} /; (And @@ Thread[LessEqual[a]]) :> Last[{a}]]

{{10, 6}, {20, 4}}

Pillsy · Answer

You can sort the list into ascending order using plain old Sort, and then Split it into sublists where the second element is ascending. The Last element of each such sublist will be the one you want to keep:

Last /@ Split[Sort[list], #1[[2]] <= #2[[2]] &]
(* {{10, 6}, {20, 4}} *)

If you need the operation to keep the remaining elements in their original order, you can sort again based on PositionIndex.

SortBy[
 Last /@ Split[Sort[list], #1[[2]] <= #2[[2]] &],
 PositionIndex[list]]
(* {{20, 4}, {10, 6}} *)

DeleteCases depending on all elements

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