Re: Re: Re: finding periodicity in a set
- To: mathgroup at smc.vnet.net
- Subject: [mg39321] Re: [mg39304] Re: [mg39283] Re: finding periodicity in a set
- From: "Mihajlo Vanevic" <mvane at eunet.yu>
- Date: Tue, 11 Feb 2003 04:41:58 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Here is something similar to Dr Bob's idea for small sample size... (but more accurate, I think) << Statistics`DescriptiveStatistics` Clear[mihajlo2]; mihajlo2[xlist_, dr_:0.1, error_:0.01] := Module[ {xxlist, gaps, ratios, rationalRatios, lcm, intRatios, xvals, linefit, a, b}, xxlist = Mean /@ Split[Sort[xlist], #2 - #1 < error &]; gaps = ListConvolve[{1., -1.}, xxlist]; ratios = gaps/Min[gaps]; rationalRatios = Rationalize[ratios, dr]; lcm = LCM @@ Denominator[rationalRatios]; intRatios = rationalRatios*lcm; xvals = FoldList[Plus, lcm, intRatios]; linefit = Fit[Transpose[{xvals, xxlist}], {1, x}, x]; {a, b} = CoefficientList[linefit, x]; {Mod[a, b], b, Max[xvals]} ]; test: xlist := 0.202 + 1.618 Table[0.001 Random[] + Random[Integer, {3, 17}], {20}]; TableForm[Table[{{mihajlo2[xlist]}}, {50}]] and xlist := 0.202 + 1.618 Table[0.001 Random[] + Random[Integer, {3, 17}], {4}]; TableForm[Table[{{mihajlo2[xlist]}}, {50}]] Mihajlo ************************************************************** * At 2003-02-10, 01:07:00 * Dr Bob, drbob at bigfoot.com wrote: ************************************************************** >Mihajlo's method doesn't work well with small sample sizes, so I'd suggest >this: > ><< Statistics`DescriptiveStatistics` >xlist = 0.202 + 1.618 Table[0.001 Random[] + Random[Integer, {3, 17}], >{20}]; >error = 0.5; >xxlist = Mean /@ Split[Sort[xlist], #2 - #1 < error &]; >Fit[xxlist, {1, x, x^2}, x]; >{a, b, c} = CoefficientList[, x]; >{Mod[a, b], b} >gaps = Subtract @@@ Transpose@{Rest@xxlist, Drop[xxlist, -1]}; >ratios = gaps/Min@gaps; >gcd[ratios_List] := Module[{n = 2, k}, > While[n < 10 && (k = n/GCD @@ Round[n*ratios]) â?¥ n, n++]; > k > ] >b2 = Min[gaps]/gcd@ratios; >{a2 = Mean@Mod[xxlist, b2], b2} > >{1.18984, 1.39711} (* Mihajlo's method with Mean in place of First *) >{0.207753, 1.6175} (* better *) > >With as few as four samples, this often returns the right period, while the >other method rarely gets it right even with 20 samples. > >You shouldn't make the error estimate too small --- a fourth of the period >is about right. (But you don't know the period in advance.) A more >complete method would adjust the error estimate after estimating the >period, to see if the answer changes. > >Bobby
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