Hi Martin,
Thanks for the hints,
Now the question is : do we prefer speed or robustness in case of weird
polygons
Or should we add options to our shapefile reader :
In this case, weird polygons are polygons where some shells are
described as if they were holes...
> There's a few fixes required:
> 1. if only one shell is present, do not run the PIP code, but simply
> assign the holes to the shell. (This fix has been made in the current
> GeoTools PolygonHandler code, and could be copied from there)
Currently, we check if holes are inside a shell with a PIP test and in
case a hole don't
find any shell, it is considered itself as a shell.
Not sure how often it happens and how arcgis would read such a polygon.
Also I don't remember what happens if we let holes which happen to be
out of the shell in a Polygon.
Would need to create such a shapefile to test, as I don't think I have one.
> 2. this can be extended to checking for only one *candidate* shell (a
> candidate shell is one whose envelope contains the hole envelope). This
> code is not yet developed.
Same case as above. If a single candidate shell is found, do we still
make a PIP test
to be sure the hole is included ? Currently, we do.
> 3. the code uses ArrayList.indexOf. This is inefficient, since it uses
> equals(), which does a full equality comparison. Instead, an iterator
> and == should be used instead (code below)
Thanks, will test that as soon as I could create a monstruous polygon.
> The performance improvement from just #1 was dramatic - reading a 1M
> feature shapefile went from over 1000 s to 32 s.
Wow, your polygon should be really complex !
Will do some test and I'll report my results on this list.
Michaël
>
> ----------------------------
>
> PolygonHandler change:
>
> ((ArrayList) holesForShells.get(findIndex(shells,
> minShell))).add(testHole);
>
>
> /**
> * Finds a object in a list. Should be much faster than indexof
> *
> * @param list
> * @param o
> * @return
> */
> private static int findIndex(ArrayList list, Object o)
> {
> int n = list.size();
> for (int i = 0; i< n; i++) {
> if (list.get(i) == o)
> return i;
> }
> return -1;
> }
>
>
>
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