Re: Overlapping binning of differences of two lists

*To*: mathgroup at smc.vnet.net*Subject*: [mg92652] Re: Overlapping binning of differences of two lists*From*: Szabolcs Horvat <szhorvat at gmail.com>*Date*: Thu, 9 Oct 2008 06:38:23 -0400 (EDT)*Organization*: University of Bergen*References*: <gci2hv$t8$1@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. > The first thing that came to my mind was BinCounts and Partition. What was the trouble you ran into when using them? bindiff3[a_, b_, bsize_, n_] := With[{diffs = Flatten@Outer[Subtract, a, b]}, Total /@ Partition[BinCounts[diffs, {-bsize - n, bsize + n, 1}], 2 bsize, 1] ] In[7]:= Timing[r1 = bindiff1[a, b, bsize, n];] Timing[r2 = bindiff2[a, b, bsize, n];] Timing[r3 = bindiff3[a, b, bsize, n];] Out[7]= {43.7454, Null} Out[8]= {2.32665, Null} Out[9]= {0.479927, Null} In[10]:= r1 === r2 === r3 Out[10]= True (I haven't measured memory use, but I am guessing that the critical operation is Outer[], so the memory requirements of the three functions are likely to be similar.)