Mathematica fast 2D binning algorithm

问题:

I am having some trouble developing a suitably fast binning algorithm in Mathematica. I have a large (~100k elements) data set of the form T={{x1,y1,z1},{x2,y2,z2},....} and I want to bin it into a 2D array of around 100x100 bins, with the bin value being given by the sum of the Z values that fall into each bin.

Currently I am iterating through each element of the table, using Select to pick out which bin it is supposed to be in based on lists of bin boundaries, and adding the z value to a list of values occupying that bin. At the end I map Total onto the list of bins, summing their contents (I do this because I sometimes want to do other things, like maximize).

I have tried using Gather and other such functions to do this but the above method was ridiculously faster, though perhaps I am using Gather poorly. Anyway It still takes a few minutes to do the sorting by my method and I feel like Mathematica can do better. Does anyone have a nice efficient algorithm handy?

回答1:

Here is a method based on Szabolcs's post that is about about an order of magnitude faster.

data = RandomReal[5, {500000, 3}];
(*500k values*)
zvalues = data[[All, 3]];

epsilon = 1*^-10;(*prevent 101 index*)
(*rescale and round (x,y) coordinates to index pairs in the 1..100 range*)
indexes = 1 + Floor[(1 - epsilon) 100 Rescale[data[[All, {1, 2}]]]];

res2 = Module[{gb = GatherBy[Transpose[{indexes, zvalues}], First]}, 
    SparseArray[
     gb[[All, 1, 1]] -> 
      Total[gb[[All, All, 2]], {2}]]]; // AbsoluteTiming

Gives about {2.012217, Null}

AbsoluteTiming[
 System`SetSystemOptions[ 
  "SparseArrayOptions" -> {"TreatRepeatedEntries" -> 1}];
 res3 = SparseArray[indexes -> zvalues];
 System`SetSystemOptions[ 
  "SparseArrayOptions" -> {"TreatRepeatedEntries" -> 0}];
 ]

Gives about {0.195228, Null}

res3 == res2
True

"TreatRepeatedEntries" -> 1 adds duplicate positions up.

回答2:

I intend to do a rewrite of the code below because of Szabolcs' readability concerns. Until then, know that if your bins are regular, and you can use Round, Floor, or Ceiling (with a second argument) in place of Nearest, the code below will be much faster. On my system, it tests faster than the GatherBy solution also posted.

---

Assuming I understand your requirements, I propose:

data = RandomReal[100, {75, 3}];

bins = {0, 20, 40, 60, 80, 100};

Reap[
  Sow[{#3, #2}, bins ~Nearest~ #] & @@@ data,
  bins,
  Reap[Sow[#, bins ~Nearest~ #2] & @@@ #2, bins, Tr@#2 &][[2]] &
][[2]] ~Flatten~ 1 ~Total~ {3} // MatrixForm

---

Refactored:

f[bins_] := Reap[Sow[{##2}, bins ~Nearest~ #]& @@@ #, bins, #2][[2]] &

bin2D[data_, X_, Y_] := f[X][data, f[Y][#2, #2~Total~2 &] &] ~Flatten~ 1 ~Total~ {3}

Use:

bin2D[data, xbins, ybins]

回答3:

Here's my approach:

data = RandomReal[5, {500000, 3}]; (* 500k values *)

zvalues = data[[All, 3]];

epsilon = 1*^-10; (* prevent 101 index *)

(* rescale and round (x,y) coordinates to index pairs in the 1..100 range *)    
indexes = 1 + Floor[(1 - epsilon) 100 Rescale[data[[All, {1, 2}]]]];

(* approach 1: create bin-matrix first, then fill up elements by adding  zvalues *)
res1 = Module[
    {result = ConstantArray[0, {100, 100}]},
    Do[
      AddTo[result[[##]], zvalues[[i]]] & @@ indexes[[i]], 
      {i, Length[indexes]}
    ];
    result
    ]; // Timing

(* approach 2: gather zvalues by indexes, add them up, convert them to a matrix *)
res2 = Module[{gb = GatherBy[Transpose[{indexes, zvalues}], First]},
    SparseArray[gb[[All, 1, 1]] -> (Total /@ gb[[All, All, 2]])]
    ]; // Timing

res1 == res2

These two approaches (res1 & res2) can handle 100k and 200k elements per second, respectively, on this machine. Is this sufficiently fast, or do you need to run this whole program in a loop?

回答4:

Here's my approach using the function SelectEquivalents defined in What is in your Mathematica tool bag? which is perfect for a problem like this one.

data = RandomReal[100, {75, 3}];
bins = Range[0, 100, 20];
binMiddles = (Most@bins + Rest@bins)/2;
nearest = Nearest[binMiddles];

SelectEquivalents[
   data
   ,
   TagElement -> ({First@nearest[#[[1]]], First@nearest[#[[2]]]} &)
   ,
   TransformElement -> (#[[3]] &)
   ,
   TransformResults -> (Total[#2] &)
   ,
   TagPattern -> Flatten[Outer[List, binMiddles, binMiddles], 1]
   , 
   FinalFunction -> (Partition[Flatten[# /. {} -> 0], Length[binMiddles]] &)
]

If you would want to group according to more than two dimensions you could use in FinalFunction this function to give to the list result the desired dimension (I don't remember where I found it).

InverseFlatten[l_,dimensions_]:= Fold[Partition[#, #2] &, l, Most[Reverse[dimensions]]];