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getting key error for accessing the graph's weights, networkx

I am using networkx to create an algorithm to calculate the modularity for the different communities. Now I am getting this key problem when I was doing G[complst[i]][complst[j]]['weight'], whereas I printed out complst[i] and compost[j] and find these values are correct. Anyone can help? I tried many ways to debug it such as saving them in seperate variables but they don't help.
import networkx as nx
import copy
#load the graph made in previous task
G = nx.read_gexf("graph.gexf")
#set a global max modualrity value
maxmod = 0
#deep copy of the coriginal graph, since when removing edges, the graph will change
ori = copy.deepcopy(G)
#create an array for saving the edges to remove
arr = []
#see if all edges are broken, if not, keep looping, otherwise stop
while(G.number_of_edges()!=0):
#find the edge_betweeness for each edge
betweeness = nx.edge_betweenness_centrality(G,weight='weight',normalized=False)
print('------------------******************--------------------')
#sort the result in descending order and save all edges with the maximum betweenness to 'arr'
sortbet = {k: v for k, v in sorted(betweeness.items(), key=lambda item: item[1],reverse=True)}
#covert the dict to list for processing
betlst = list(sortbet)
for i in range(len(betlst)):
if betlst[i] == betlst[0]:
arr.append(betlst[i])
#remove all edges with maximum betweeness from the graph
G.remove_edges_from(arr)
#find the leftover component, and convert the result to list for further modualrity processing
lst = list(nx.connected_components(G))
#!!!!!!!!testing and debugging the value, now the value is printed correctly
print(G['pk_sullivan']['ChrisWarcraft']['weight'])
#create a variable cnt to represent modularity in this graph
cnt = 0
#iterate the lst, which is each component(each component is saved as python set)
for n in range(len(lst)):
#convert each component from set to list for processing
complst = list(lst[n])
#if this component is a singleton, the modualrity for this component 0, so add 0 the current cnt
if len(complst)==1:
cnt += 0
else:
# calulate the modularity for this component by using combinations of edges
for i in range(0,len(complst)):
if i+1 <=len(complst)-1:
for j in range(i+1,len(complst)):
#!!!!!!!!! there is a bunch of my testing and find the value are printed all fine until "print(G[a][b]['weight'])""
print(i)
print(j)
print(complst)
a = complst[i]
print(type(a))
b = complst[j]
print(type(b))
print(G[a][b]['weight'])
#calculate the modualrity by using equation M = 1/2m*(weight(a,b)-degree(a)*degree(b)/2m)
cnt += 1/(2*ori.number_of_edges())*(G[a][b]['weight']-ori.degree(a)*ori.degree(b)/(2*ori.number_of_edges()))
#find the maximum modualrity and save this split of graph, end!
if cnt>=maxmod:
maxmod = cnt
newgraph = copy.deepcopy(G)
print('maxmod is',maxmod)
here is the error, welcome to run the code and hope my code illustration can help!

It looks like you're trying to find the weight of all combinations of nodes within each connected component. But the problem, is that you're assuming that all nodes in a connected component are first degree connected, i.e. are connected through a single edge, which is wrong.
In your code, you have:
...
for i in range(0,len(complst)):
if i+1 <=len(complst)-1:
for j in range(i+1,len(complst)):
...
And then you try to find the weight of the edge that connects these two nodes. But every edge in a connected component is not connected to the rest. A connected component just means that all nodes are reachable from all others.
So you should be iterating over the edges in the subgraph generated by the connected component, or something along these lines.

Prohibitively slow execution of function compute_resilience in Python

The idea is to compute resilience of the network presented as an undirected graph in form
{node: (set of its neighbors) for each node in the graph}.
The function removes nodes from the graph in random order one by one and calculates the size of the largest remaining connected component.
The helper function bfs_visited() returns the set of nodes that are still connected to the given node.
How can I improve the implementation of the algorithm in Python 2? Preferably without changing the breadth-first algorithm in the helper function
def bfs_visited(graph, node):
"""undirected graph {Vertex: {neighbors}}
Returns the set of all nodes visited by the algrorithm"""
queue = deque()
queue.append(node)
visited = set([node])
while queue:
current_node = queue.popleft()
for neighbor in graph[current_node]:
if neighbor not in visited:
visited.add(neighbor)
queue.append(neighbor)
return visited
def cc_visited(graph):
""" undirected graph {Vertex: {neighbors}}
Returns a list of sets of connected components"""
remaining_nodes = set(graph.keys())
connected_components = []
for node in remaining_nodes:
visited = bfs_visited(graph, node)
if visited not in connected_components:
connected_components.append(visited)
remaining_nodes = remaining_nodes - visited
#print(node, remaining_nodes)
return connected_components
def largest_cc_size(ugraph):
"""returns the size (an integer) of the largest connected component in
the ugraph."""
if not ugraph:
return 0
res = [(len(ccc), ccc) for ccc in cc_visited(ugraph)]
res.sort()
return res[-1][0]
def compute_resilience(ugraph, attack_order):
"""
input: a graph {V: N}
returns a list whose k+1th entry is the size of the largest cc after
the removal of the first k nodes
"""
res = [len(ugraph)]
for node in attack_order:
neighbors = ugraph[node]
for neighbor in neighbors:
ugraph[neighbor].remove(node)
ugraph.pop(node)
res.append(largest_cc_size(ugraph))
return res

I received this tremendously great answer from Gareth Rees, which covers the question completely.
Review
The docstring for bfs_visited should explain the node argument.
The docstring for compute_resilience should explain that the ugraph argument gets modified. Alternatively, the function could take a copy of the graph so that the original is not modified.
In bfs_visited the lines:
queue = deque()
queue.append(node)
can be simplified to:
queue = deque([node])
The function largest_cc_size builds a list of pairs:
res = [(len(ccc), ccc) for ccc in cc_visited(ugraph)]
res.sort()
return res[-1][0]
But you can see that it only ever uses the first element of each pair (the size of the component). So you could simplify it by not building the pairs:
res = [len(ccc) for ccc in cc_visited(ugraph)]
res.sort()
return res[-1]
Since only the size of the largest component is needed, there is no need to build the whole list. Instead you could use max to find the largest:
if ugraph:
return max(map(len, cc_visited(ugraph)))
else:
return 0
If you are using Python 3.4 or later, this can be further simplified using the default argument to max:
return max(map(len, cc_visited(ugraph)), default=0)
This is now so simple that it probably doesn't need its own function.
This line:
remaining_nodes = set(graph.keys())
can be written more simply:
remaining_nodes = set(graph)
There is a loop over the set remaining_nodes where on each loop iteration you update remaining_nodes:
for node in remaining_nodes:
visited = bfs_visited(graph, node)
if visited not in connected_components:
connected_components.append(visited)
remaining_nodes = remaining_nodes - visited
It looks as if the intention of the code to avoid iterating over the nodes in visited by removing them from remaining_nodes, but this doesn't work! The problem is that the for statement:
for node in remaining_nodes:
only evaluates the expression remaining_nodes once, at the start of the loop. So when the code creates a new set and assigns it to remaining_nodes:
remaining_nodes = remaining_nodes - visited
this has no effect on the nodes being iterated over.
You might imagine trying to fix this by using the difference_update method to adjust the set being iterated over:
remaining_nodes.difference_update(visited)
but this would be a bad idea because then you would be iterating over a set and modifying it within the loop, which is not safe. Instead, you need to write the loop as follows:
while remaining_nodes:
node = remaining_nodes.pop()
visited = bfs_visited(graph, node)
if visited not in connected_components:
connected_components.append(visited)
remaining_nodes.difference_update(visited)
Using while and pop is the standard idiom in Python for consuming a data structure while modifying it — you do something similar in bfs_visited.
There is now no need for the test:
if visited not in connected_components:
since each component is produced exactly once.
In compute_resilience the first line is:
res = [len(ugraph)]
but this only works if the graph is a single connected component to start with. To handle the general case, the first line should be:
res = [largest_cc_size(ugraph)]
For each node in attack order, compute_resilience calls:
res.append(largest_cc_size(ugraph))
But this doesn't take advantage of the work that was previously done. When we remove node from the graph, all connected components remain the same, except for the connected component containing node. So we can potentially save some work if we only do a breadth-first search over that component, and not over the whole graph. (Whether this actually saves any work depends on how resilient the graph is. For highly resilient graphs it won't make much difference.)
In order to do this we'll need to redesign the data structures so that we can efficiently find the component containing a node, and efficiently remove that component from the collection of components.
This answer is already quite long, so I won't explain in detail how to redesign the data structures, I'll just present the revised code and let you figure it out for yourself.
def connected_components(graph, nodes):
"""Given an undirected graph represented as a mapping from nodes to
the set of their neighbours, and a set of nodes, find the
connected components in the graph containing those nodes.
Returns:
- mapping from nodes to the canonical node of the connected
component they belong to
- mapping from canonical nodes to connected components
"""
canonical = {}
components = {}
while nodes:
node = nodes.pop()
component = bfs_visited(graph, node)
components[node] = component
nodes.difference_update(component)
for n in component:
canonical[n] = node
return canonical, components
def resilience(graph, attack_order):
"""Given an undirected graph represented as a mapping from nodes to
an iterable of their neighbours, and an iterable of nodes, generate
integers such that the the k-th result is the size of the largest
connected component after the removal of the first k-1 nodes.
"""
# Take a copy of the graph so that we can destructively modify it.
graph = {node: set(neighbours) for node, neighbours in graph.items()}
canonical, components = connected_components(graph, set(graph))
largest = lambda: max(map(len, components.values()), default=0)
yield largest()
for node in attack_order:
# Find connected component containing node.
component = components.pop(canonical.pop(node))
# Remove node from graph.
for neighbor in graph[node]:
graph[neighbor].remove(node)
graph.pop(node)
component.remove(node)
# Component may have been split by removal of node, so search
# it for new connected components and update data structures
# accordingly.
canon, comp = connected_components(graph, component)
canonical.update(canon)
components.update(comp)
yield largest()
In the revised code, the max operation has to iterate over all the remaining connected components in order to find the largest one. It would be possible to improve the efficiency of this step by storing the connected components in a priority queue so that the largest one can be found in time that's logarithmic in the number of components.
I doubt that this part of the algorithm is a bottleneck in practice, so it's probably not worth the extra code, but if you need to do this, then there are some Priority Queue Implementation Notes in the Python documentation.
Performance comparison
Here's a useful function for making test cases:
from itertools import combinations
from random import random
def random_graph(n, p):
"""Return a random undirected graph with n nodes and each edge chosen
independently with probability p.
"""
assert 0 <= p <= 1
graph = {i: set() for i in range(n)}
for i, j in combinations(range(n), 2):
if random() <= p:
graph[i].add(j)
graph[j].add(i)
return graph
Now, a quick performance comparison between the revised and original code. Note that we have to run the revised code first, because the original code destructively modifies the graph, as noted in §1.2 above.
>>> from timeit import timeit
>>> G = random_graph(300, 0.2)
>>> timeit(lambda:list(resilience(G, list(G))), number=1) # revised
0.28782312001567334
>>> timeit(lambda:compute_resilience(G, list(G)), number=1) # original
59.46968446299434
So the revised code is about 200 times faster on this test case.

Variant of Dijkstra - no repeat groups

I'm trying to write an optimization process based on Dijkstra's algorithm to find the optimal path, but with a slight variation to disallow choosing items from the same group/family when finding the optimal path.
Brute force traversal of all edges to find the solution would be np-hard, which is why am attempting to (hopefully) use Dijkstra's algorithm, but I'm struggling to add in the no-repeat groups logic.
Think of it like a traveling salesman problem, but I want to travel from New Your to Los Angels, and have an interesting route (by never visiting 2 similar cities from same group) and minimize my fuel costs. There are approx 15 days and 40 cities, but for defining my program, I've pared it down to 4 cities and 3 days.
Valid paths don't have to visit every group, they just can't visit 2 cities in the same group. {XL,L,S} is a valid solution, but {XL,L,XL} is not valid because it visits the XL group twice. All Valid solutions will be the same length (15 days or edges) but can use any combination of groups (w/out duplicating groups) and need not use them all (since 15 days, but 40 different city groups).
Here's a picture I put together to illustrate a valid & invalid route: (FYI - groups are horizontal rows in the matrix)
**Day 1**
G1->G2 # $10
G3->G4 # $30
etc...
**Day 2**
G1->G3 # $50
G2->G4 # $10
etc...
**Day 3**
G1->G4 # $30
G2->G3 # $50
etc...
The optimal path would be G1->G2->G3, however a standard Dijkstra solution returns G1-
I found & tweaked this example code online, and name my nodes with the following syntax so I can quickly check what day & group they belong to: D[day#][Group#] by slicing the 3rd character.
## Based on code found here: https://raw.githubusercontent.com/nvictus/priority-queue-dictionary/0eea25fa0b0981558aa780ec5b74649af83f441a/examples/dijkstra.py
import pqdict
def dijkstra(graph, source, target=None):
"""
Computes the shortests paths from a source vertex to every other vertex in
a graph
"""
# The entire main loop is O( (m+n) log n ), where n is the number of
# vertices and m is the number of edges. If the graph is connected
# (i.e. the graph is in one piece), m normally dominates over n, making the
# algorithm O(m log n) overall.
dist = {}
pred = {}
predGroups = {}
# Store distance scores in a priority queue dictionary
pq = pqdict.PQDict()
for node in graph:
if node == source:
pq[node] = 0
else:
pq[node] = float('inf')
# Remove the head node of the "frontier" edge from pqdict: O(log n).
for node, min_dist in pq.iteritems():
# Each node in the graph gets processed just once.
# Overall this is O(n log n).
dist[node] = min_dist
if node == target:
break
# Updating the score of any edge's node is O(log n) using pqdict.
# There is _at most_ one score update for each _edge_ in the graph.
# Overall this is O(m log n).
for neighbor in graph[node]:
if neighbor in pq:
new_score = dist[node] + graph[node][neighbor]
#This is my attempt at tracking if we've already used a node in this group/family
#The group designator is stored as the 4th char in the node name for quick access
try:
groupToAdd = node[2]
alreadyVisited = predGroups.get( groupToAdd, False )
except:
alreadyVisited = False
groupToAdd = 'S'
#Solves OK with this line
if new_score < pq[neighbor]:
#Erros out with this line version
#if new_score < pq[neighbor] and not( alreadyVisited ):
pq[neighbor] = new_score
pred[neighbor] = node
#Store this node in the "visited" list to prevent future duplication
predGroups[groupToAdd] = groupToAdd
print predGroups
#print node[2]
return dist, pred
def shortest_path(graph, source, target):
dist, pred = dijkstra(graph, source, target)
end = target
path = [end]
while end != source:
end = pred[end]
path.append(end)
path.reverse()
return path
if __name__=='__main__':
# A simple edge-labeled graph using a dict of dicts
graph = {'START': {'D11':1,'D12':50,'D13':3,'D14':50},
'D11': {'D21':5},
'D12': {'D22':1},
'D13': {'D23':50},
'D14': {'D24':50},
'D21': {'D31':3},
'D22': {'D32':5},
'D23': {'D33':50},
'D24': {'D34':50},
'D31': {'END':3},
'D32': {'END':5},
'D33': {'END':50},
'D34': {'END':50},
'END': {'END':0}}
dist, path = dijkstra(graph, source='START')
print dist
print path
print shortest_path(graph, 'START', 'END')

Convert graph to have outdegree 1 (except extra zero weight edges)

I am reading graphs such as http://www.dis.uniroma1.it/challenge9/data/rome/rome99.gr from http://www.dis.uniroma1.it/challenge9/download.shtml in python. For example, using this code.
#!/usr/bin/python
from igraph import *
fname = "rome99.gr"
g = Graph.Read_DIMACS(fname, directed=True )
(I need to change the line "p sp 3353 8870" " to "p max 3353 8870" to get this to work using igraph.)
I would like to convert the graph to one where all nodes have outdegree 1 (except for extra zero weight edges we are allowed to add) but still preserve all shortest paths. That is a path between two nodes in the original graph should be a shortest path in the new graph if and only if it is a shortest path in the converted graph. I will explain this a little more after an example.
One way to do this I was thinking is to replace each node v by a little linear subgraph with v.outdegree(mode=OUT) nodes. In the subgraph the nodes are connected in sequence by zero weight edges. We then connect nodes in the subgraph to the first node in other little subgraphs we have created.
I don't mind using igraph or networkx for this task but I am stuck with the syntax of how to do it.
For example, if we start with graph G:
I would like to convert it to graph H:
As the second graph has more nodes than the first we need to define what we mean by its having the same shortest paths as the first graph. I only consider paths between either nodes labelled with simple letters of with nodes labelled X1. In other words, in this example a path can't start or end in A2 or B2. We also merge all versions of a node when considering a path. So a path A1->A2->D in H is regarded as the same as A->D in G.
This is how far I have got. First I add the zero weight edges to the new graph
h = Graph(g.ecount(), directed=True)
#Connect the nodes with zero weight edges
gtoh = [0]*g.vcount()
i=0
for v in g.vs:
gtoh[v.index] = i
if (v.degree(mode=OUT) > 1):
for j in xrange(v.degree(mode=OUT)-1):
h.add_edge(i,i+1, weight = 0)
i = i+1
i = i + 1
Then I add the main edges
#Now connect the nodes to the relevant "head" nodes.
for v in g.vs:
h_v_index = gtoh[v.index]
i = 0
for neighbour in g.neighbors(v, mode=OUT):
h.add_edge(gtoh[v.index]+i,gtoh[neighbour], weight = g.es[g.get_eid(v.index, neighbour)]["weight"])
i = i +1
Is there a nicer/better way of doing this? I feel there must be.

The following code should work in igraph and Python 2.x; basically it does what you proposed: it creates a "linear subgraph" for every single node in the graph, and connects exactly one outgoing edge to each node in the linear subgraph corresponding to the old node.
#!/usr/bin/env python
from igraph import Graph
from itertools import izip
def pairs(l):
"""Given a list l, returns an iterable that yields pairs of the form
(l[i], l[i+1]) for all possible consecutive pairs of items in l"""
return izip(l, l[1:])
def convert(g):
# Get the old vertex names from g
if "name" in g.vertex_attributes():
old_names = map(str, g.vs["name"])
else:
old_names = map(str, xrange(g.vcount))
# Get the outdegree vector of the old graph
outdegs = g.outdegree()
# Create a mapping from old node IDs to the ID of the first node in
# the linear subgraph corresponding to the old node in the new graph
new_node_id = 0
old_to_new = []
new_names = []
for old_node_id in xrange(g.vcount()):
old_to_new.append(new_node_id)
new_node_id += outdegs[old_node_id]
old_name = old_names[old_node_id]
if outdegs[old_node_id] <= 1:
new_names.append(old_name)
else:
for i in xrange(1, outdegs[old_node_id]+1):
new_names.append(old_name + "." + str(i))
# Add a sentinel element to old_to_new just to make our job easier
old_to_new.append(new_node_id)
# Create the edge list of the new graph and the weights of the new
# edges
new_edgelist = []
new_weights = []
# 1) Create the linear subgraphs
for new_node_id, next_new_node_id in pairs(old_to_new):
for source, target in pairs(range(new_node_id, next_new_node_id)):
new_edgelist.append((source, target))
new_weights.append(0)
# 2) Create the new edges based on the old ones
for old_node_id in xrange(g.vcount()):
new_node_id = old_to_new[old_node_id]
for edge_id in g.incident(old_node_id, mode="out"):
neighbor = g.es[edge_id].target
new_edgelist.append((new_node_id, old_to_new[neighbor]))
new_node_id += 1
print g.es[edge_id].source, g.es[edge_id].target, g.es[edge_id]["weight"]
new_weights.append(g.es[edge_id]["weight"])
# Return the graph
vertex_attrs = {"name": new_names}
edge_attrs = {"weight": new_weights}
return Graph(new_edgelist, directed=True, vertex_attrs=vertex_attrs, \
edge_attrs=edge_attrs)

My A Star implementation won't return the list of steps to get to a destination

I'll try to be brief here. I'm trying to implement A Star on Python, but obviously I'm doing something wrong, because when I test it, it doesn't return the list of steps to get to the destination.
Basically, the context is: I have a map, represented as a graph, formed by nodes. I have a Player class, a Node class, and a Graph class. That doens't matter much, but might be necessary. The player has to get to the nearest node with a Coin in it, which is also a Class.
My implementation is based on the Wikipedia pseudocode, but for some reason it won't work. I'm almost completely sure that my mistake is on A* Star, but i can't find it. Here, i'll put the two functions that i made regarding A Star. Hope it's not too messy, i'm just starting with programming and i like commenting a lot.
I would really appreciate any help to find the problem :)
Note: I'm not an english speaker, so i'm sorry for my mistakes. Wish, in a few years, i'll be able to comunicate better.
def A_Star(player,graph,array_of_available_coins):
# Define the initial position and the last position, where the coin is
initial_position=player.position # Player is a class. Position is of type Node
final_position=closest_cpin(player,graph,array_of_available_coins)
# Define the open_set, closed_set, and work with a Heap.
open_set=[initial_position] # Open_set will be initialized with the current position of the player
closed_set=[]
heapq.heapify(open_set) # Converts the open_set into a Python Heap (or Priority Queue)
came_from={} # It's a dictionary where each key is the a node, and the value is the previous node in the path
# Modify G and H, and therefore F, of the initial position. G of the inicial position is 0.
#And H of the initial position is the pitagoric distance.
initial_position.modify_g_and_h(0,initial_position.distance(final_position))
while open_set!=[]:
square=heapq.heappop(open_set) # Gets the least value of the open_set
if square.is_wall(): # If it's a Wall, the player can't move over it.
continue
if square==final_position:
movements=[] # Creates a empty array to save the movements
rebuild_path(came_from,square,movements) # Calls the function to rebuild the path
player.add_movements_array(movements) # Copies the movements into the movements array of the player
return
# In this point, the square is not a wall and it's not the final_position
closed_set.append(square) # Add the square into the closed_set
neighbours=graph.see_neighbours(square) # Checks all the neighbours of the current square
for neigh in neighbours:
if neigh.is_wall()==True:
continue
if neigh in closed_set:
continue
# Calculates the new G, H and F values
g_aux=square.see_g()+square.get_cost(neigh) # Current g + the cost to get from current to neighbour
h_aux=neigh.distance(final_position) # Pitagoric distance between the neighbour and the last position
f_aux=g_aux+h_aux # F=G+H
if neigh not in open_set:
heapq.heappush(open_set,neigh) # Adds the neigh into the open_set
is_better=True
elif f_aux<neigh.see_f():
is_better=True
else:
is_better=False
if is_better==True:
came_from[neigh]=square # The actual neigh came from the actual square
neigh.modify_g_and_h(g_aux,h_aux) #Modifies the g and h values of the actual neighbour
return None
def rebuild_path(came_from,square,array_of_movements):
array_of_movements.insert(0,square) # Adds, in the first position of the array, the square it gets by parameter
if not square in came_from: # if there is no key "square" in the came_from dictionary, then it's the first position
array_of_movements.remove(array_of_movements[0]) # Gets the first element of the array out (because i need it to be that way later)
return array_of_movements
rebuild_path(came_from,came_from[square],array_of_movements)
return
The thing is, i have to implement the algorithm, because it's part of an Excercise (much larger, with Pygame and everything), and this is the only thing that's making me nervous. If i use a library, it'll count as if i didn't do it, so i'll have to deliver it again :(

I would recommend networkx
import networkx
this can do this kind of stuff:
#!/usr/bin/env python
# encoding: utf-8
"""
Example of creating a block model using the blockmodel function in NX. Data used is the Hartford, CT drug users network:
#article{,
title = {Social Networks of Drug Users in {High-Risk} Sites: Finding the Connections},
volume = {6},
shorttitle = {Social Networks of Drug Users in {High-Risk} Sites},
url = {http://dx.doi.org/10.1023/A:1015457400897},
doi = {10.1023/A:1015457400897},
number = {2},
journal = {{AIDS} and Behavior},
author = {Margaret R. Weeks and Scott Clair and Stephen P. Borgatti and Kim Radda and Jean J. Schensul},
month = jun,
year = {2002},
pages = {193--206}
}
"""
__author__ = """\n""".join(['Drew Conway <drew.conway#nyu.edu>',
'Aric Hagberg <hagberg#lanl.gov>'])
from collections import defaultdict
import networkx as nx
import numpy
from scipy.cluster import hierarchy
from scipy.spatial import distance
import matplotlib.pyplot as plt
def create_hc(G):
"""Creates hierarchical cluster of graph G from distance matrix"""
path_length=nx.all_pairs_shortest_path_length(G)
distances=numpy.zeros((len(G),len(G)))
for u,p in path_length.items():
for v,d in p.items():
distances[u][v]=d
# Create hierarchical cluster
Y=distance.squareform(distances)
Z=hierarchy.complete(Y) # Creates HC using farthest point linkage
# This partition selection is arbitrary, for illustrive purposes
membership=list(hierarchy.fcluster(Z,t=1.15))
# Create collection of lists for blockmodel
partition=defaultdict(list)
for n,p in zip(list(range(len(G))),membership):
partition[p].append(n)
return list(partition.values())
if __name__ == '__main__':
G=nx.read_edgelist("hartford_drug.edgelist")
# Extract largest connected component into graph H
H=nx.connected_component_subgraphs(G)[0]
# Makes life easier to have consecutively labeled integer nodes
H=nx.convert_node_labels_to_integers(H)
# Create parititions with hierarchical clustering
partitions=create_hc(H)
# Build blockmodel graph
BM=nx.blockmodel(H,partitions)
# Draw original graph
pos=nx.spring_layout(H,iterations=100)
fig=plt.figure(1,figsize=(6,10))
ax=fig.add_subplot(211)
nx.draw(H,pos,with_labels=False,node_size=10)
plt.xlim(0,1)
plt.ylim(0,1)
# Draw block model with weighted edges and nodes sized by number of internal nodes
node_size=[BM.node[x]['nnodes']*10 for x in BM.nodes()]
edge_width=[(2*d['weight']) for (u,v,d) in BM.edges(data=True)]
# Set positions to mean of positions of internal nodes from original graph
posBM={}
for n in BM:
xy=numpy.array([pos[u] for u in BM.node[n]['graph']])
posBM[n]=xy.mean(axis=0)
ax=fig.add_subplot(212)
nx.draw(BM,posBM,node_size=node_size,width=edge_width,with_labels=False)
plt.xlim(0,1)
plt.ylim(0,1)
plt.axis('off')
plt.savefig('hartford_drug_block_model.png')