Fix invalid polygon in Shapely

Question

Shapely defines a Polygon as invalid if any of its segments intersect, including segments that are colinear. Many software packages will create a region or area with a "cutout" as shown here which has colinear segments:

>>> pp = Polygon([(0,0), (0,3), (3,3), (3,0), (2,0), 
                  (2,2), (1,2), (1,1), (2,1), (2,0), (0,0)])
>>> pp.is_valid
WARNING:shapely.geos:Self-intersection at or near point 2 0
False

Naturally, the cutout can be implemented natively in Shapely, or this same geometry can be implemented as two valid polygons, but if I only have the list of points shown above, is there an easy to "fix" this (create valid geometry from this list of points)?

score 58 · Accepted Answer · edited Jun 03 '19 at 14:35

58

I found a solution that works for the specific case given:

>>> pp2 = pp.buffer(0)
>>> pp2.is_valid
True
>>> pp2.exterior.coords[:]
[(0.0, 0.0), (0.0, 3.0), (3.0, 3.0), (3.0, 0.0), (2.0, 0.0), (0.0, 0.0)]
>>> pp2.interiors[0].coords[:]
[(2.0, 1.0), (2.0, 2.0), (1.0, 2.0), (1.0, 1.0), (2.0, 1.0)]

edited Jun 03 '19 at 14:35

Georgy

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answered Jan 01 '14 at 22:08

jpcgt

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I find that this buffer( 0 ) method works in general for fixing polygons with coincident lines among sub-polygons. This is very useful. Do you know of a place where this trick is officially recommended/sanctioned? – M Katz Mar 23 '14 at 22:20
2

This method did not fix my issue: I had some self-intersecting multipolygons. – Rutger Hofste Sep 15 '17 at 16:35
29

You need to be a little careful with the `buffer(0)` technique. We've had bowtie cases where it destroyed the big part of the polygon and left just a small bowtied corner. YMMV. – Aidan Kane Dec 06 '17 at 11:31
For me it made a valid polygon but it no longer had any .exterior.coords so it still did not work for a .within(polygon) analysis. – Casivio Feb 12 '21 at 20:29
1

Does anyone knows if a solution exists for this now? As much as i undertand @AidanKane didn't recommended the `buffer(0)`. – José Rojas Jul 03 '21 at 07:08
@JoséRojas It's.....complicated. The solution we ended up using is to break invalid polygons into a map of valid polygons using shapely.ops.polygonize. We then iterated of each one and decided if it was inside or outside using winding rules. We then union the included ones together. – Aidan Kane Jul 08 '21 at 23:54
@JoséRojas See my new answer, looks like shapely have added a function to do this! https://shapely.readthedocs.io/en/latest/manual.html#validation.make_valid – Aidan Kane Jul 08 '21 at 23:58
buffer(0) also behaves different on different os (see my answer)! – LegNaiB Oct 03 '21 at 16:56

score 11 · Answer 2 · edited Oct 12 '22 at 23:43

11

Shapely implemented a solution for this. Through pip you can use any shapely version >= 1.8a3 and import this way:

from shapely.validation import make_valid

The current version of shapely available via pip and conda includes the make_valid function. If you need to install an older version of shapely, you can use the shapely implementation as shown below:

def make_valid(ob):

    from shapely.geometry.base import geom_factory
    from shapely.geos import lgeos
    
    if ob.is_valid:
        return ob

    return geom_factory(lgeos.GEOSMakeValid(ob._geom))

edited Oct 12 '22 at 23:43

Michael Delgado

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answered Sep 22 '21 at 12:10

Rafael Matias

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This doesn't work, the following exception is raised: AttributeError: 'LGEOS360' object has no attribute 'GEOSMakeValid' Any objections, how to solve this? I would like to use the make_valid function but on a stable release (and no alpha release) ... – LegNaiB Oct 01 '21 at 15:18
make_valid is better than a buffer(0) which can change the geometry completely. The the doc's [second example](https://shapely.readthedocs.io/en/latest/manual.html#id32) in particular is a good example to test both methods. – tgrandje Mar 07 '23 at 13:27

score 4 · Answer 3 · answered Jul 08 '21 at 23:56

4

Untested, but it appears that Shapely have added a function to support this now.

https://shapely.readthedocs.io/en/latest/manual.html#validation.make_valid

answered Jul 08 '21 at 23:56

Aidan Kane

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score 3 · Answer 4 · answered Nov 09 '21 at 02:01

3

simplest solution

new_polygon = pp.buffer(0)
new_polygon.is_valid

maybe it would be work.

answered Nov 09 '21 at 02:01

minsu

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5

score 2 · Answer 5 · answered Oct 03 '21 at 16:56

I have used the .buffer(0) method a lot, but it gave different results on Windows and on Linux. Therefore if you encounter such a problem, remember that:

Warning: The .buffer(0) function of shapely may behave different on different operating systems, i.e. Windows and Linux. I had examples, where on Linux an empty Polygon is returned while on Windows the correct non-empty Polygon is returned!

This problem took me days to solve, that's why I want to add that as an answer (my suggested edit to the accepted answer was sadly rejected).

score 1 · Answer 6 · answered Feb 23 '22 at 03:34

This is my first attempt at a fix geometry function. I had to handle the special case where a polygons are converted to multipolygons, where make_valid doesn't work b/c len(make_valid(feature)) !=1.

from shapely.validation import make_valid

def fix_geom(in_feature):

    # avoid changing original geodf
    in_feature = in_feature.copy(deep=True)    
        
    # drop any missing geometries
    in_feature = in_feature[~(in_feature.is_empty)]
    
    # Repair broken geometries
    for index, row in in_feature.iterrows(): # Looping over all polygons
        if row['geometry'].is_valid:
            next
        else:
            fix = make_valid(row['geometry'])

            try:
                in_feature.loc[[index],'geometry'] =  fix # issue with Poly > Multipolygon
            except ValueError:
                in_feature.loc[[index],'geometry'] =  in_feature.loc[[index], 'geometry'].buffer(0)
    return in_feature

Fix invalid polygon in Shapely

6 Answers6

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