Re: Overlapping binning of differences of two lists

*To*: mathgroup at smc.vnet.net*Subject*: [mg92717] Re: [mg92607] Overlapping binning of differences of two lists*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Sat, 11 Oct 2008 06:44:53 -0400 (EDT)*References*: <200810081024.GAA00469@smc.vnet.net>

Art wrote: > 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! Something similar to your bindiff2 can be sent through Compile. bindiff3=Compile[{{l1,_Real,1}, {l2,_Real,1}, {bsize,_Integer}, {n,_Integer}}, Module[{os, diffs, bins, j, k=-n, m}, diffs = Sort[Flatten[Outer[Plus, l1, -l2]]]; bins = ConstantArray[0, 2*n + 1]; For[j=1, j<=Length[diffs], j++, If [diffs[[j]]<-bsize+k,Continue[]]; While[diffs[[j]]>bsize+k && k<=n, k++]; m=k; If[k>n, Return[bins]]; While[-bsize+m<diffs[[j]]<bsize+m &&m<=n, bins[[m+n+1]]++; m++]; ]; bins ]]; With this I get around an order of magnitude improvement in speed on your example. Daniel Lichtblau Wolfram Research

**References**:**Overlapping binning of differences of two lists***From:*Art <grenander@gmail.com>