Accumulation of points within a shape

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11-13-2013 12:28 AM
RichardDixon
New Contributor
Dear All,

I hope I've come to the right location for this query - if not please point me towards the right forum! I've got a map lat/long points in ArcMap and also a simple rectangular shape. I am looking to calculate the location of the shape that covers the most points - i.e. move the shape around until it finds the most points within this box. Does anyone know if this is "do-able" in ArcMap?

Many thanks
Richard
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2 Replies
DanLee
by Esri Regular Contributor
Esri Regular Contributor
You didn't describe if your rectangle has a fixed ratio of the shorter side over the longer side or if it can be rotated. Here are some ideas that might work for you or give you some leads:

1. Run Buffer tool on the points with an "estimated" buffer distance. You can simply estimate by measuring distances between points on your map where more points are located; take about half of the average distance; or you can se Generate Near Table to find distances from points to other points, followed by other tools to derive a distance.
2. Dissolve the buffers, uncheck "Create multipart".
3. Run Spatial Join to find which buffers CONTAIN the points, specify JOIN_ONE_TO_ONE so you get the Join_Count field in the output.

From here, you can select the buffer(s) with high Join_Count and
- either use Feature To Point tool to get the CENTRIOD point. Put the center of your rectangle at that point. Hopefully it will cover max number of points or pretty close to that.
- or if the shape and orientation of your rectangle is flexible, you can use Minimum Bounding Geometry on the selected buffers, using one of the rectangle geometry types (by width or by area) to see if the resulting rectangles are suitable for your purposes.
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RichardDixon
New Contributor
Dan

Thanks for your thoughts. Yes my rectange would be flexible and I'm simply looking for the rectange that has the most points within it. What you've described is pretty new to me so I will go away and give it a go!

Cheers
Richard
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