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#1 2018-10-15 11:12:58

nevw
Member
From: SE Queensland, Australia
Registered: 2013-04-11
Posts: 390

JOSM: Count of polygons in a relation

I am using Josm 'Merge Selection' to merge some protected area boundaries and this entails using 'Replace Geometry' and using shared ways at some sections of the polygons. At the final gathering of all the ways and polygons, I like to check if I have collected all the polygons in the relation.
Josm gives the 'number of members' in the relation, which is the number of ways, but I have found it is best to also add up the number of closed polygons in the relation as a final check to ensure I have not missed any.
Normally the datasets that are obtained from shapefiles have just polygons therefore the number of polygons is readily available. I do all the splitting up of the ways (and sharing of ways) in the merged dataset.

Am I missing something obvious...are the number of polygons in a relation readily seen in addition to the number of members?

It would be nice to have a polygon count in the relations window or the relation edit window.
I use the relations edit window to count the polygons but I tend to get a bit cross-eyed after counting over 20 polygons in one relation.

Last edited by nevw (2018-10-15 11:25:06)

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#2 2018-10-15 19:52:33

Klumbumbus
Member
Registered: 2014-01-12
Posts: 545
Website

Re: JOSM: Count of polygons in a relation

You could right click the relation in the relations panel and click select members. This gives you the number of ways in the selection panel. Then you could perform a search for closed ways within the selection and look at the number of ways in the selection panel again. For quicker access you could place that specific search as button in the tool bar.
See https://josm.openstreetmap.de/wiki/Help … aintoolbar and/or https://josm.openstreetmap.de/wiki/Help … rchresults

Last edited by Klumbumbus (2018-10-15 19:53:07)

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#3 2018-10-15 22:44:59

nevw
Member
From: SE Queensland, Australia
Registered: 2013-04-11
Posts: 390

Re: JOSM: Count of polygons in a relation

Thanks Klumbumbus, that's a much better way to do it.

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