[R-sig-Geo] Polygon width

[Barry Rowlingson]

Loop over the row indexes of an sf-class object `pctest`, applying a
function given the row index which returns the distance:

 > ppa =
sapply(1:nrow(pctest),function(i){max(st_distance(st_cast(st_geometry(pctest[i,]),"POINT")))})
 > ppa
  [1] 172.55598 360.77081 107.53889 137.17785  51.66645 132.82052 113.00875
  [8] 161.02432 141.13909  88.67002

If you want to make that a new column, do `pctest$maxdist = ppa`

And test on a simple example - make sure a unit square returns
approximately sqrt(2)!

Barry




On Wed, Apr 25, 2018 at 5:24 PM, Paulo Flores Ribeiro <
paulo.flores.mail at gmail.com> wrote:

> Thank you, Barry. I am using planar coordinates (meters) and Euclidean
> distances. The shape or size of the polygons is not important (for
> illustration purposes, imagine that polygons are farm boundaries and that
> each farm is a single polygon). I know how to extract the maximum
> distance in a single polygon by calculating the distance matrix and
> selecting the maximum value. My "difficulty" is exactly in the coding of
> the loop process (or in the construction of the "function" to be used in
> the apply approach), so that I can apply it "automatically" to 25,000
> polygons. I am using the rgadl package, but I can switch to sf.
> Thanks for any help.
> Cheers,
> PauloFR
>
> Às 12:27 de 25-04-2018, Barry Rowlingson escreveu:
>
> Do you want great-circle distance or is your space small enough that you
> can use planar coordinates?
>
> Are your polygons all single rings or are there islands and/or holes? Does
> that matter?
>
> The straightforward way would be to coerce the polygons to points, compute
> the distance matrix, then take the maximum. Depending on if you are reading
> your shapefile into sp or sf classes, the functions would be a bit
> different. You should try and implement the straightforward way, test it
> for correctness, and then worry about the "best way" if the straightforward
> way isn't what you need. Its often the case that "best" ways need fancy
> data structures or complex algorithms with more opportunity for bugs. Start
> simple, work up.
>
> For example, using sf, here's the max distance between any points in the
> first feature of `pcs`
>
> > max(st_distance(st_cast(st_geometry(pcs[1,]),"POINT")))
> 172.556 m
>
> loop from 1 to N or otherwise apply over the features, and you're done.
>
> Barry
>
>
>
> On Wed, Apr 25, 2018 at 11:26 AM, Paulo Flores Ribeiro <
> paulo.flores.mail at gmail.com> wrote:
>
>> Hello,
>>
>> I have a shapefile with ca. 25000 polygons. Each polygon has an average
>> of 40 vertices (nodes). I would like to extract, for each polygon, the
>> distance separating the two most distant vertices (aka "polygon diagonal"
>> or "maximum polygon width"). It is not important whether the polygon is
>> convex or concave, so the lines connecting the vertices can be inside or
>> outside the polygon. The desired result would be a two-column array, with a
>> number of rows equal to the number of polygons, and where the first column
>> is the id of the polygons, and the second the "maximum width" of the
>> corresponding polygon.
>>
>> What would be the best way to do this, considering that the calculation
>> will probably require frequent updates (e.g. due to changes in the shape of
>> the polygons)?
>>
>> Thanks in advance,
>>
>> PauloFR
>>
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>>
>
>
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