arra 3 2 1 0 1 0 3 2 2 3 0 1 0 0 0 0 arrb 2 3 0 1 3 2 1 0 1 0 3 2 0 0 0 0 arra -:&(/:~) arrb 1
That is, (/:~ arra) -: /:~ arrb 1 On Thursday, July 10, 2014, Jon Hough <[email protected]> wrote: > The following two 4x4 arrays are rearrangements of each other's rows. > > > 3 2 1 0 > > > > 1 0 3 2 > > > > 2 3 0 1 > > > > 0 0 0 0 > > > > > > > > > 2 3 0 1 > > > > 3 2 1 0 > > > > 1 0 3 2 > > > > 0 0 0 0 > > > I would like to know a way to acknowledge two arrays as being > rearrangements of each other. Eventually my goal is to compare long lists > of such arrays and nub out duplicates - duplicates being rearrangements. > > > The only way I can think to do this is to cycle through all permutations > of the row of one of the arrays and test for equality with the other array, > using A. . Of course, there are 24 permutations to test for 4x4 arrays, but > obviously for bigger arrays things get worse. > > > Is there a faster way to check two arrays are (ignoring row permutations) > equivalent? > > > > > > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > -- Sent from Gmail Mobile ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
