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Overlapping binning of differences of two lists

• To: mathgroup at smc.vnet.net
• Subject: [mg92607] Overlapping binning of differences of two lists
• From: Art <grenander at gmail.com>
• Date: Wed, 8 Oct 2008 06:24:28 -0400 (EDT)

Given two sorted vectors a and b of different lengths, what is the
best way to count the number of elements in the set of all differences
between elements of a and b that fall in overlapping bins of [-bsize -
i, bsize - i) for i in Range[-n, n], where bsize >= 1.

Below are 2 implementations I've tried which are two slow and memory
intensive. I haven't quite been able to do it using BinCounts,
Partition, and ListCorrelate.

Was wondering if there is a faster way.

(* Generate random a, b  *)
T = 500; bsize = 10; n = 20;
r := Rest@FoldList[Plus, 0, RandomReal[ExponentialDistribution[0.01],
{T}]]
a = r; b = r;

bindiff1[a_, b_, bsize_, n_] :=
  With[{d = Flatten@Outer[Subtract, a, b]},
  	Table[Count[d, _?(-bsize <= # - i < bsize &)], {i, -n, n}]]

bindiff2[a_, b_, bsize_, n_] :=
 Module[{os, i, j, s, tmp,
   d = Sort@Flatten@Outer[Subtract, a, b],
   c = ConstantArray[0, {2 n + 1}]},
  For[os = 0; j = 1; i = -n,  i <= n,  i++; j++,
   s = Flatten@Position[Drop[d, os], _?(# >= -bsize + i &), 1, 1];
   If[s == {}, Break[],
    os += s[[1]] - 1;
    tmp = Flatten@Position[Drop[d, os], _?(# > bsize + i &), 1, 1];
    c[[j]] = If[tmp == {}, Length[d] - os, First@tmp - 1]]];
  Return[c]]

First@Timing@bindiff[a,b, bsize, n] is about 36 seconds.

First@Timing@bindiff2[a, b, bsize, n] is about 3 seconds but still too
slow and d uses up too much memory.

Thanks!

• Follow-Ups:
  • Re: Re: Overlapping binning of differences of
    • From: danl@wolfram.com
  • Re: Overlapping binning of differences of two lists
    • From: Daniel Lichtblau <danl@wolfram.com>