Re: [jgrapht-users] How to remove all edges of a certain vertex iteratively?

[Joris K. <de...@gm...>]

Java doesn't allow you to iterate over a list using a for each statement
and modify the list at the same time. That leads to an
ConcurrentModificationException. To resolve this issue, you either need to
create a separate list of vertices you want to delete and invoke
removeVertex on each of these vertices, or you should use an iterator and
invoke iterator.remove();

Finally, removeVertex also removes all edges incident to that vertex, so
you don't need a separate call to remove the edges of that vertex.

br,

Joris




On Fri, Aug 29, 2014 at 1:28 PM, Zhigang Wu <ep...@sc...> wrote:

> Dear jgrapht users,
>
>
>
> In my problem, I want to traverse the graph to find a certain vertex, then
> delete the vertex accompanied with all the edges connected to it. Then I
> will traverse the remained graph again, find another vertex by some special
> rules, deleted the vertex and all the connected edges…
>
>
>
> However, when I translate the procedure into JAVA codes, I always receive
> the same error message:
>
>
>
> java.util.ConcurrentModificationException
>
>          at java.util.ArrayList$Itr.checkForComodification(Unknown Source)
>
>          at java.util.ArrayList$Itr.next(Unknown Source)
>
>          at java.util.Collections$UnmodifiableCollection$1.next(Unknown
> Source)
>
>          at
> org.jgrapht.graph.AbstractGraph.removeAllEdges(AbstractGraph.java:89)
>
>          ……
>
>
>
> The following is my JAVA code related to this:
>
>
>
>                             *for* (Node myNode : myNodeList) {
>
>                                    // Some codes to find the certain
> node, the name is myNode
>
>                                    // Some operations
>
>                                    myGraph.removeAllEdges(myGraph
> .edgesOf(myNode));
>
>                                    myGraph.removeVertex(myNode);
>
>                             }
>
>
>
> This problem has bothered me for quite a long time, many thanks in advance!
>
>
>
> Best Regards
>
>
>
> Zhigang Wu
>
>
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