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Re: Overlapping binning of differences of two lists
*To*: mathgroup at smc.vnet.net
*Subject*: [mg92654] Re: Overlapping binning of differences of two lists
*From*: Januk <ggroup at sarj.ca>
*Date*: Thu, 9 Oct 2008 06:38:45 -0400 (EDT)
*References*: <gci2hv$t8$1@smc.vnet.net>
Hi Art,
A very speedy way would be to take advantage of the fact that your
list d can be sorted. You can then use a binary search algorithm to
find the starting and ending positions of the particular bin you're
looking at. The difference will be the number of elements in that
range.
Here is a binary search algorithm taken almost verbatim from this list
(sorry, I can't find the reference):
FindPosSorted[lst_List, e_, p_: ((#1 <= #2) &)] := Module[
{lower = 1, upper = Length[lst] + 1},
While[lower != upper,
With[{mid = Floor[(lower + upper)/2]},
If[p[e, lst[[mid]]],
upper = mid,
lower = mid + 1
];
];
];
If[e == lst[[lower]], lower++];
Return[lower]
];
You can then use something like:
Timing[
tst = Module[{d2},
Table[
d2 = d - i;
FindPosSorted[d2, bsize] - FindPosSorted[d2, -bsize],
{i, -n, n}
]
];
]
Compare this to your first concept:
First@Timing[
tst2 = bindiff1[a, b, bsize, n]
]
On my system, there is an approximately 100x improvement in speed
using the binary search method.
I hope that helps.
Januk
On Oct 8, 6:37 am, Art <grenan... at gmail.com> 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!
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