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AI-GEOSTATS: Ripley's K

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  • Mathias Tobler
    I am using Ripley s K to analize distribution patterns of trees in a forest plot. Here some basic information. The plot is 1000 x 1000 m. There are 7765 trees
    Message 1 of 2 , Nov 27, 2002
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      I am using Ripley's K to analize distribution patterns of trees in a forest
      plot. Here some basic information. The plot is 1000 x 1000 m. There are
      7765 trees that were measured throughout the plot. This is the first time I
      am unsing Ripley's K and I run into several questions / problems.

      I am using Crime Stat 2.0 and SPP (by Peter Haase) to calculate the
      statistcs but get very different results from the two packages. Both
      packages are supposed to output SQR(K(t) / Pi) - t for the data set and the
      confidence intervalls. The attached image showes the output from both
      packages. Both were run with 19 iterations and Rectangular Bordercorrection
      was used in Crime Stat. The third graph showes the Coefficient of
      Dispersion CD=SD^2/M for diffferent quadrant sizes.

      Does anyone have an idea what the problem could be? Why do the confidence
      intervals show a curved shape and are not around 0? Is there any other free
      and easy to use software to calculate Ripley's K (I know that there is R
      but I don't have the time right now to get familiar with it).

      Thanks for your help,

      Mathias


      [Non-text portions of this message have been removed]
    • Mathias Tobler
      I am using Ripley s K to analize distribution patterns of trees in a forest plot. Here some basic information. The plot is 1000 x 1000 m. There are 7765 trees
      Message 2 of 2 , Nov 27, 2002
      • 0 Attachment
        I am using Ripley's K to analize distribution patterns of trees in a forest
        plot. Here some basic information. The plot is 1000 x 1000 m. There are
        7765 trees that were measured throughout the plot. This is the first time I
        am unsing Ripley's K and I run into several questions / problems.

        I am using Crime Stat 2.0 and SPP (by Peter Haase) to calculate the
        statistcs but get very different results from the two packages. Both
        packages are supposed to output SQR(K(t) / Pi) - t for the data set and the
        confidence intervalls. The following link shows the output from both packages.

        http://www.botanypages.org/Mathias/ripleysk.jpg

        Both were run with 19 iterations and Rectangular Bordercorrection was used
        in Crime Stat. The third graph shows the Coefficient of Dispersion
        CD=SD^2/M for diffferent quadrant sizes.

        Does anyone have an idea what the problem could be? Why do the confidence
        intervals show a curved shape and are not around 0? Is there any other free
        and easy to use software to calculate Ripley's K (I know that there is R
        but I don't have the time right now to get familiar with it).

        Thanks for your help,

        Mathias


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