Hi Alexandre, Thx for your reply! Indeed, your and my following optimizations are not gt-specific. I’m now here: # set layer to be drawn layer = 0 # create graph view v_filter = edge_list.groupby('layer')[['i', 'j']].apply(lambda x: set(x.values.flatten())).to_dict() # create dictionary with vertices in a layer vp_filter = g.new_vp('bool') vp_filter.a[list(v_filter[layer])] = True ep_filter = g.new_ep('bool') ep_filter.a[g.ep['layer'].a == layer] = True # replaces my or Alexandre's loop g_filter = GraphView(g, vfilt=vp_filter, efilt=ep_filter) # draw filtered graph graph_draw(g_filter, edge_color=g.ep.layer) This is already rather short and nice. Still, I'm wondering if there’s a way that is more gt-esque. Haiko Von: graph-tool [mailto:graph-tool-bounces@skewed.de] Im Auftrag von Alexandre Hannud Abdo Gesendet: Donnerstag, 29. August 2019 15:03 An: Main discussion list for the graph-tool project Betreff: Re: [graph-tool] Graph views via property maps Ni! Hi Haiko, GraphView can simultaneously take `vfilt` and `efilt` parameters, if that is what you wanna know. So, g_filter = GraphView(g, vfilt=vp_filter, efilt=ep_filter) should work just fine. Regarding your code snippet, the 'elegance' issues are kinda unrelated to graph_tool. For example, in the second part, you iterate the dataframe instead of simply iterating the graph edges and getting the edge layer from the internal property map. You also use an if clause to assign the boolean value of the if clause, instead of directly assigning it. So, this is equivalent to your `for` loop: for e in g.edges(): ep_filter[e] = g.edge_properties['layer'][e] == layer All the best, .~´ On Thu, Aug 29, 2019 at 10:49 AM Lietz, Haiko <Haiko.Lietz@gesis.org<mailto:Haiko.Lietz@gesis.org>> wrote: Hi all, gt allows creating graph views where vertices and/or edges are filtered out. The principle is described in the documentation: "Vertices or edges which are to be filtered should be marked with a PropertyMap with value type bool, and then set with set_vertex_filter() or set_edge_filter() methods. By default, vertex or edges with value '1' are kept in the graphs, and those with value '0' are filtered out." (https://graph-tool.skewed.de/static/doc/quickstart.html#graph-filtering) Given a layered graph, where layers are coded via an edge property, I want to draw the graphs layer by layer. This is an example graph: from graph_tool.all import * import numpy as np import pandas as pd # data edge_list = pd.DataFrame(np.array([[0, 1, 0], [1, 2, 1], [2, 0, 2]]), columns=['i', 'j', 'layer']) # create graph g = Graph(directed=False) ep_layer = g.new_edge_property('int') g.add_edge_list(edge_list.values, eprops=[ep_layer]) g.edge_properties['layer'] = ep_layer # draw graph with edge colors showing the layers graph_draw(g, edge_color=g.ep.layer) It is possible to draw a single layer (in this example layer 0) in this complicated way: # set layer to be drawn layer = 0 # create graph view ep_filter = g.new_edge_property('bool') for i in range(0, len(edge_list)): e = g.edge(edge_list['i'][i], edge_list['j'][i]) if edge_list['layer'][i] == layer: ep_filter[e] = True else: ep_filter[e] = False g_filter = GraphView(g, efilt=ep_filter) # draw filtered graph graph_draw(g_filter, edge_color=g.ep.layer) If I proceed this way I will also have to create a vertex property map to filter unused vertices and do these steps for all layers I want to draw. But isn’t there a more elegant way – preferably handling vertices and edges in the same step? Many thanks and best wishes Haiko _______________________________________________ graph-tool mailing list graph-tool@skewed.de<mailto:graph-tool@skewed.de> https://lists.skewed.de/mailman/listinfo/graph-tool