Usage Of ForEdges Iterator In Networkit (python)

I carefully read the docs, but it still is unclear to me how to use G.forEdges(), described as an 'experimental edge iterator interface'. Let's say that I want to decrease the den

Solution 1:

NetworKit iterators accept a callback function so if you want to iterate over edges (or nodes) you have to define a function and then pass it to the iterator as a parameter. You can find more information here. For example a simple function that just prints all edges is:

# Callback function.
# To iterate over edges it must accept 4 parameters
def myFunction(u, v, weight, edgeId):
    print("Edge from {} to {} has weight {} and id {}".format(u, v, weight, edgeId))

# Using iterator with callback function
G.forEdges(myFunction)

Now if you want to keep removing edges whose weight is inside your weightlist until the graph splits into two connected components you also have to update the connected components of the graph since ConnectedComponents will not do that for you automatically (this may be also one of the reasons why the iteration takes forever). To do this efficiently, you can use the DynConnectedComponents class (see my example below). In this case, I think that the edge iterator will not help you much so I would suggest you to keep using the for loop.

from networkit import *

# Efficiently updates connected components after edge updates
cc = components.DynConnectedComponents(G).run()

# Removes edges with weight equals to w until components split
def removeEdges(w):
    for (u, v) in G.edges():
        if G.weight(u, v) == weight:
            G.removeEdge(u, v)
            # Updating connected components
            event = dynamic.GraphEvent(dynamic.GraphEvent.EDGE_REMOVAL, u, v, weight)
            cc.update(event)
            if cc.numberOfComponents() > 1:
                # Components did split
                return True
    # Components did not split
    return False

if cc.numberOfComponents() == 1:
    for weight in weights:
        if removeEdges(weight):
            break

This should speed up a bit your original code. However, it is still sequential code so even if you run it on a multi-core machine it will use only one core.