blocks|key|1273241|text|你需要使用determine+which+points+lie+inside。删除这些点后，可以将一组“外部”点插入到另一组点中。您的插入点(例如，右侧图片中箭头的位置)是您必须从输入集中删除点的位置。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1273242|entityMap|0|LINK|mutability|MUTABLE|url|http://web.archive.org/web/20120323102807/http://local.wasp.uwa.edu.au/~pbourke/geometry/insidepoly/^0|5|X|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

You need to <a href="http://web.archive.org/web/20120323102807/http://local.wasp.uwa.edu.au/~pbourke/geometry/insidepoly/" rel="nofollow noreferrer">determine which points lie inside</a>. After removing these points, you can insert one set of "outside" points into the other. Your insertion points (e.g. where you have the arrow in the picture on the right) are where you had to remove points from the input sets.

blocks|key|1273275|text|问得好！我以前从未尝试过这一点，但现在我将尝试一下。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1273276|首先:你需要知道这两个形状重叠的地方。为此，您可以查看多边形A中的每条边，并查看它与多边形B中的边相交的位置。在此示例中，应该有两个交点。|1273277|然后:制作联合形状。您可以获取A和B中的所有顶点以及交点，然后排除包含在最终图形中的顶点。要找到这些点，看起来你可以找到A在B中的任何顶点，以及B在A中的任何顶点。|1273278|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

Good question! I've never attempted this before, but I'll take a crack at it now.

First: You need to know where these two shapes overlap. To do this, you could look at every edge in Polygon A and see where it intersects and edge in Polygon B. In this example, there should be two points of intersection.

Then: Make the union shape. You can take all of the vertices in A and B, and also the points of intersection, and then exclude the vertices that contained by the final shape. To find these points, it looks like you could just find any vertex of A that is inside B, and any vertext of B that is inside A.

blocks|key|1058618|text|试试gpc。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1058619|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cs.man.ac.uk/~toby/alan/software/^0|2|3|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Try <a href="http://www.cs.man.ac.uk/~toby/alan/software/" rel="nofollow noreferrer">gpc</a>.

blocks|key|1058793|text|我也遇到过同样的问题，我用下面的方法解决了这个问题|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1058794|用于Angus+Johnson的Clipper库(版本)的C%2B%2B翻译的Cython包装器6.4.2)+https://github.com/fonttools/pyclipper|offset|length|1058795|pc+=+pyclipper.Pyclipper()
def+get_poly_union(polygons):
++++pc.AddPaths(polygons,+pyclipper.PT_SUBJECT,+True)
++++solution+=+pc.Execute(pyclipper.CT_UNION,+pyclipper.PFT_NONZERO,+pyclipper.PFT_NONZERO)
++++return+solution[0]

print_image+=+image.copy()
solution+=+get_poly_union(polygons_array)+
#polygons_array=[polygon,polygon,polygon,+...,polygon]+and+polygon=[point,point,point...,point]

cv2.drawContours(print_image,+[np.asarray(solution)],+-1,+(0,+255,+0),+2)

plt.imshow(print_image)|code-block|syntax|javascript|1058796|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/fonttools/pyclipper^0|0|1F|12|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@$D|U|E|V|1|W]]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

I have faced the same problem and I solved the problem by using the following way

Cython wrapper for the C++ translation of the Angus Johnson's Clipper library (ver. 6.4.2)
<a href="https://github.com/fonttools/pyclipper" rel="nofollow noreferrer">https://github.com/fonttools/pyclipper</a>

<pre><code>pc = pyclipper.Pyclipper()
def get_poly_union(polygons):
 pc.AddPaths(polygons, pyclipper.PT_SUBJECT, True)
 solution = pc.Execute(pyclipper.CT_UNION, pyclipper.PFT_NONZERO, pyclipper.PFT_NONZERO)
 return solution[0]

print_image = image.copy()
solution = get_poly_union(polygons_array) 
#polygons_array=[polygon,polygon,polygon, ...,polygon] and polygon=[point,point,point...,point]

cv2.drawContours(print_image, [np.asarray(solution)], -1, (0, 255, 0), 2)

plt.imshow(print_image)
</code></pre>

blocks|key|1058746|text|这是一个非常老的问题，但是Boost的Union_函数对我很有效。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1058747|请看下面的代码片段：|1058748|#include+<iostream>
#include+<vector>

#include+<boost/geometry.hpp>
#include+<boost/geometry/geometries/point_xy.hpp>
#include+<boost/geometry/geometries/polygon.hpp>

#include+<boost/foreach.hpp>


int+main()
{
++++typedef+boost::geometry::model::polygon<boost::geometry::model::d2::point_xy<double>+>+polygon;

++++polygon+green,+blue;

++++boost::geometry::read_wkt(
++++++++"POLYGON((0+0,+0+10,+10+10,+10+0,+0+0))",+green);

++++boost::geometry::read_wkt(
++++++++"POLYGON((5+5,+5+15,+15+15,+15+5,+5+5))",+blue);

++++std::vector<polygon>+output;
++++boost::geometry::union_(green,+blue,+output);

++++int+i+=+0;
++++std::cout+<<+"green+%7C%7C+blue:"+<<+std::endl;
++++BOOST_FOREACH(polygon+const&+p,+output)
++++{
++++++++std::cout+<<+i%2B%2B+<<+":+"+<<+boost::geometry::area(p)+<<+std::endl;

++++++++for+(int+i+=+0;+i+<+p.outer().size();+i%2B%2B)
++++++++{
++++++++++++std::cout+<<+p.outer().at(i).x()+<<+"+"+<<+p.outer().at(i).y()+<<+std::endl;
++++++++}
++++}



++++return+0;
}|code-block|syntax|javascript|1058749|entityMap|0|LINK|mutability|MUTABLE|url|https://www.boost.org/doc/libs/1_58_0/libs/geometry/doc/html/geometry/reference/algorithms/union_.html^0|J|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

This is very old question but <a href="https://www.boost.org/doc/libs/1_58_0/libs/geometry/doc/html/geometry/reference/algorithms/union_.html" rel="nofollow noreferrer">Union_</a> function from Boost worked for me.

See this snippet below:

<pre><code>#include &lt;iostream&gt;
#include &lt;vector&gt;

#include &lt;boost/geometry.hpp&gt;
#include &lt;boost/geometry/geometries/point_xy.hpp&gt;
#include &lt;boost/geometry/geometries/polygon.hpp&gt;

#include &lt;boost/foreach.hpp&gt;


int main()
{
 typedef boost::geometry::model::polygon&lt;boost::geometry::model::d2::point_xy&lt;double&gt; &gt; polygon;

 polygon green, blue;

 boost::geometry::read_wkt(
 "POLYGON((0 0, 0 10, 10 10, 10 0, 0 0))", green);

 boost::geometry::read_wkt(
 "POLYGON((5 5, 5 15, 15 15, 15 5, 5 5))", blue);

 std::vector&lt;polygon&gt; output;
 boost::geometry::union_(green, blue, output);

 int i = 0;
 std::cout &lt;&lt; "green || blue:" &lt;&lt; std::endl;
 BOOST_FOREACH(polygon const&amp; p, output)
 {
 std::cout &lt;&lt; i++ &lt;&lt; ": " &lt;&lt; boost::geometry::area(p) &lt;&lt; std::endl;

 for (int i = 0; i &lt; p.outer().size(); i++)
 {
 std::cout &lt;&lt; p.outer().at(i).x() &lt;&lt; " " &lt;&lt; p.outer().at(i).y() &lt;&lt; std::endl;
 }
 }



 return 0;
}
</code></pre>

blocks|key|1058609|text|在对国家进行分组时，我希望没有重叠--你可以采用一个相当简单的算法来寻找共享顶点--一个简单的视图是迭代一个多边形上的点，看看它是否在你的任何其他多边形上，然后共享下一个或上一个点，看看是否有匹配。然后，只需移除共享顶点即可创建并集|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1058610|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

When grouping countries, I'd hope there be no overlap -- you could take a fairly naive algorithm that looks for shared vertices - a simple view would be to iterate through the points on one polygon, see if it's on any of your other polygons, and shares the same next or previous point to see if there is a match. Then just remove the shared vertex to create your union

blocks|key|1273439|text|我今天需要解决同样的问题，并使用这个库找到了解决方案：http://www.cs.man.ac.uk/~toby/alan/software/。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1273440|它有很多语言实现，包括Java、Obj-C、C#、Lua、the+list+here等等。|1273441|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cs.man.ac.uk/~toby/alan/software/|1|http://www.cs.man.ac.uk/~toby/alan/software/#ports^0|R|18|0|0|T|D|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]|N|$5|I|J|K|C|$L|O]]]]

I needed to solve this same problem today and found the solution with this lib: <a href="http://www.cs.man.ac.uk/~toby/alan/software/" rel="nofollow">http://www.cs.man.ac.uk/~toby/alan/software/</a>.

It have a lot of language implementations <a href="http://www.cs.man.ac.uk/~toby/alan/software/#ports" rel="nofollow">the list here</a> including Java, Obj-C, C#, Lua, python and more.

Given two polygons:

<pre><code>POLYGON((1 0, 1 8, 6 4, 1 0))
POLYGON((4 1, 3 5, 4 9, 9 5, 4 1),(4 5, 5 7, 6 7, 4 4, 4 5))
</code></pre>

How can I calculate the union (combined polygon)?

<a href="https://i.stack.imgur.com/N5P2R.png" rel="noreferrer"><img src="https://i.stack.imgur.com/N5P2R.png" alt="enter image description here"></a>

<a href="http://blogs.msdn.com/davidlean/archive/2008/10/25/sql-2008-spatial-samples-part-n-2-on-n-geometric-set-theory-methods.aspx" rel="noreferrer">Dave's example</a> uses SQL server to produce the union, but I need to accomplish the same in code. I'm looking for a mathematical formula or code example in any language that exposes the actual math. I am attempting to produce maps that combine countries dynamically into regions. I asked a related question here: <a href="https://stackoverflow.com/questions/2653812/grouping-geographical-shapes">Grouping geographical shapes</a>

How do I combine complex polygons?

Java

给定两个多边形：POLYGON((1 0, 1 8, 6 4, 1 0))POLYGON((4 1, 3 5, 4 9, 9 5, 4 1),(4 5, 5 7, 6 7, 4 4, 4 5))如何计算并集(组合多边形)？​​使用SQL server来生成联合，但我需要在代码中完成相同的操作。我正在寻找任何语言中的数学公式或代码示例，以揭示实际的数学知识。我正在尝试制作将国家动态组合成区域的地图。

问如何组合复杂的多边形？
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问如何组合复杂的多边形？
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