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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
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