[Date Index]
[Thread Index]
[Author Index]
Re: Overlapping binning of differences of two lists
*To*: mathgroup at smc.vnet.net
*Subject*: [mg92729] Re: Overlapping binning of differences of two lists
*From*: m.r at inbox.ru
*Date*: Sat, 11 Oct 2008 06:47:04 -0400 (EDT)
*References*: <gci2hv$t8$1@smc.vnet.net> <gckn7m$nom$1@smc.vnet.net>
Szabolcs Horvat wrote:
> 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.)
This can be sped up some more:
bindiff4[a_, b_, bsize_, n_] :=
ListConvolve[ConstantArray[1, 2 bsize],
BinCounts[Subtract @@ Transpose@Tuples@{a, b},
{-bsize - n, bsize + n}]]
In[6]:= Timing[r3 = bindiff3[a, b, bsize, n];]
Out[6]= {0.312, Null}
In[7]:= Timing[r4 = bindiff4[a, b, bsize, n];]
Out[7]= {0.11, Null}
The advantages are that it doesn't unpack and that Subtract is called
only once rather than repeatedly for each pair. Using
Total[Partition[..., 2 bsize, 1], {2}] gives a comparable timing.
Maxim Rytin
m.r at inbox.ru
Prev by Date:
**Re: Math Formulas**
Next by Date:
**Re: Unevaluated subtleties**
Previous by thread:
**Re: Overlapping binning of differences of two lists**
Next by thread:
**Re: Get Graphics Coordinates accuracy**
| |