Re: Ranks for an array of triplets

*To*: mathgroup at smc.vnet.net*Subject*: [mg87535] Re: Ranks for an array of triplets*From*: Ray Koopman <koopman at sfu.ca>*Date*: Sat, 12 Apr 2008 07:00:36 -0400 (EDT)*References*: <ftncq8$8a9$1@smc.vnet.net>

On Apr 11, 2:59 am, Claus <clausena... at gmail.com> wrote: > I can create an array with x,y,z triplets. x,y are on a regularly spaced > raster, z is a RandomReal. > For sorting the array according to the z values, I found two options, > the second of which is significantly faster. > However, my goal is not to sort, but to calculate the rank of the z > value within the triplet. > For example: > original array Ar: {0,0,9.8},{0,1,2.3},{1,1,12.6} > convert to: {0,0,2},{0,1,1},{1,1,3} > $B"*(B 2.3 is the smallest z-value, hence it gets assigned rank 1 > > In my case I can reach this converted array only with extra steps: > - separating the z-values from Ar, > - calculating the (standardized) "RanksOfAr", > - "gluing" the triplets back together. > > Is there a way to to this in one step? > > Thanks, > Claus > > Here is the mathematica code: > ----------------------------- > > Make up an array with (x, y) being coordinates and z being a value at > that (x, y) location > > Ar = Partition[Flatten[Table[{i, j, k}, > {i, 1, 10} > , {j, 1, 10} > , {k, {RandomReal[]}} > ]], 3]; > > Time two versions of Sorting the array Ar according to z > > Timing[SortBottomAATriples = Sort[Ar, #1[[3]] < #2[[3]] &]]; > > sll[ll_, elem_] := ll[[Ordering[ll[[All, elem]]]]] > Timing[OrdBotAATrip = sll[Ar, 3]]; > > Create the Ranks of z at the original position from Ar > > those ranks are scaled between [0, 1] > > RanksOfAr = Ordering[SortBottomAATriples]/Length[SortBottomAATriples]; > > Procedure to put it all back together > > x = Ar[[All, 1]]; > y = Ar[[All, 2]]; > FinalAr = Transpose[{x, y, RanksOfAr}] // N; Here's a smaller example of a method that should be faster: In[1]:= Ar = Flatten[Array[{##,Random[]}&,{nrows = 5, ncols = 4}],1] Out[1]= {{1,1,0.418682},{1,2,0.692552},{1,3,0.263399},{1,4,0.860788}, {2,1,0.379193},{2,2,0.339294},{2,3,0.293482},{2,4,0.842751}, {3,1,0.522699},{3,2,0.684685},{3,3,0.382372},{3,4,0.0906687}, {4,1,0.192081},{4,2,0.176807},{4,3,0.464003},{4,4,0.0851869}, {5,1,0.869599},{5,2,0.567554},{5,3,0.970354},{5,4,0.0748791}} In[2]:= N@MapThread[ReplacePart[#1,#2,3]&, {Ar,Ordering@Ordering@Ar[[All,3]]/(nrows*ncols)}] Out[2]= {{1.,1.,0.55},{1.,2.,0.8},{1.,3.,0.3},{1.,4.,0.9}, {2.,1.,0.45},{2.,2.,0.4},{2.,3.,0.35},{2.,4.,0.85}, {3.,1.,0.65},{3.,2.,0.75},{3.,3.,0.5},{3.,4.,0.15}, {4.,1.,0.25},{4.,2.,0.2},{4.,3.,0.6},{4.,4.,0.1}, {5.,1.,0.95},{5.,2.,0.7},{5.,3.,1.},{5.,4.,0.05}}