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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!
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