Re: Time series minima and maxima

• To: mathgroup at smc.vnet.net
• Subject: [mg115942] Re: Time series minima and maxima
• From: Ray Koopman <koopman at sfu.ca>
• Date: Wed, 26 Jan 2011 05:30:38 -0500 (EST)
• References: <ihm4nf\$kgv\$1@smc.vnet.net>

```On Jan 25, 1:21 am, Jagra <jagra24... at mypacks.net> wrote:
> Working on some code to find minima and maxima of time series data I
> came across this earlier discussion on the forum:
>
>
> In it, Ray Koopman posted a neat way to do it, but I have some
> questions about what I need to do next.
>
> I took Ray's code and wrapped it in a function:
>
> findMinimaMaxima[data_, window_] := With[{k = window},
> minmaxdata[[k + Flatten[Position[Partition[data[[All, 2]], 2 k + 1, 1],
>         x_List /; x[[k + 1]] < Min[Delete[x, k + 1]] || [[k + 1]] >
> Max[Delete[x, k + 1]]]]]]]

There's a typo: after the || you should have x[[k+1]] > ... .
Also, shouldn't minmaxdata[[...]] be data[[...]]

>
> The findMinimaMaxima[] function has a window parameter, which I hoped
> would enable the function to look further left and right and thereby
> identify fewer minima and maxima, but more major ones (relative to
> the length of the window).
> I grab some time series data using FinancialData, strip off the
> associated date fields and replaced them with integer positions to
> make things easier to debug:
>
> data = FinancialData["SPY", {"May 1, 2006", "Jan. 21, 2011"}][[All,
> 2]];data = Transpose[{Range[Length@data], data}];
>
> Now.
>
> minimamaxima = findMinimaMaxima[data, 50];
> ListLinePlot[{data, minimamaxima}, Joined -> True, ImageSize -> 500]
>
> When you look at the plot, you'll see that with a 50 interval window
> (and probably almost any other lengths) and data from "SPY" some
> problems appear.
>
> (Note: You can run a couple of other instruments to see things that
> move up and down different ways at different times.  Try "GLD" or
> "TLT")
>
> Two minor problems.  The function doesn't identify the first point
> as either a minima or maxima and it doesn't identify what one would
> expect to be the last minima or maxima.

The window contains 2k+1 points. The first k points and last k points
are never considered as possible extrema. Also, the algorithm will
miss an extremum if there is more than one point with that value in
the window.

>
> More important, one would want the function to return a list which
> alternates between minima and maxima, but the chart shows that a
> couple of times, it returns 2 minima in a row or 2 maxima in a row.
> In some case it might return even more.  So, clearly expanding the
> window left and right more than the original value of 2 the Ray used
> introduces some problems.
>
> I first thought to add some code to get rid of the extra point after
> the calculation, but everything I thought of seemed clumsy and drifted
> away from the clarity of Ray's  original solution.
>
> I haven't downloaded Mathematica 8 yet.  Any chance it has a way to
> do this sort of thing?  This relates to data compression too I think,
> in getting rid of the extraneous data.
>
> Thanks in advance for any suggestions or directions pursue.

```

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