更优雅的方法从元组列表生成邻接矩阵

Question

Suppose we start with a "friendship" graph represented by a list of tuples,

friendships = [(0, 1), (0, 2), (1, 2), (1, 3), (2,
3), (3, 4),(4, 5), (5, 6), (5, 7), (6, 8), (7, 8), (8, 9)]

where element 0 is a friend of 1 (and hence 1 is a friend of 0).

I want to construct the adjacency matrix from scratch in a way that will always work for this type of tuple representation.

I have the following (repulsive) Python code:

def make_matrix(num_rows,num_cols,entry_fn):
    return [[entry_fn(i,j)
             for j in range(num_cols)]
            for i in range(num_rows)]
def adjacency(connections):
    new=connections+[(x[1],x[0]) for x in connections]
    elements=list(set([x[0] for x in connections]+ [x[1] for x in connections]))
    def test(i,j):
        if (elements[i],elements[j]) in new:
            return 1
        else: return 0
    return make_matrix(len(elements),len(elements),test)

I know that it is inefficient and very ugly. Is there a smarter way to solve this problem? The output for the example list that I gave above should be

[[0, 1, 1, 0, 0, 0, 0, 0, 0, 0],
 [1, 0, 1, 1, 0, 0, 0, 0, 0, 0],
 [1, 1, 0, 1, 0, 0, 0, 0, 0, 0],
 [0, 1, 1, 0, 1, 0, 0, 0, 0, 0],
 [0, 0, 0, 1, 0, 1, 0, 0, 0, 0],
 [0, 0, 0, 0, 1, 0, 1, 1, 0, 0],
 [0, 0, 0, 0, 0, 1, 0, 0, 1, 0],
 [0, 0, 0, 0, 0, 1, 0, 0, 1, 0],
 [0, 0, 0, 0, 0, 0, 1, 1, 0, 1],
 [0, 0, 0, 0, 0, 0, 0, 0, 1, 0]]

update: as per one of the answers, I have the following possible solution, although I don't know if it's any better

def adj(connections):
    ##step 1
    temp=(set(elem[0] for elem in connections).union(
        set(elem[1] for elem in connections)))
    n=max(temp)+1
    ans=[]
    ##step 2
    for i,_ in enumerate(temp):
        ans.append([])
        for j,_ in enumerate(temp):
            ans[i].append(0)
    ##step 3
    for pair in connections:
        ans[pair[0]][pair[1]]=1
        ans[pair[1]][pair[0]]=1
    return ans

Answer 1

My algorithm would be something like this:

1. Find the maximum vertex id. Call this n .
2. Create an n+1 by n+1 array of zeros. Call this M .
3. For each pair x, y in the input list, set M[x][y] = 1

To come up with this solution, I first thought of step 3. With the given input, this seems the most straightforward way to populate the adjacency matrix. However, it requires a 2D array of a fixed size. So the problem is how do I figure out n for step 2. From there it doesn't take much thought to figure out step 1 is what is needed.

The details are left as an exercise for the reader.

Answer 2

I am not much of a Python programmer, but something like...

def to_adjacency_matrix(pairs):
  size = 1 + max(map(max, pairs))
  a = [[0 for j in range(size)] for i in range(size)]
  for i, j in pairs:
    a[i][j] = a[j][i] = 1
  return a

Answer 3

If you want short and easy-readable solution (and you are planning to work with graph later) I recommend you to use networkx library. Your problem can be solved with pair of lines:

import networkx as nx
G = nx.Graph()
G.add_edges_from(friendships)
nx.to_numpy_matrix(G)

which will return numpy matrix:

matrix([[0., 1., 1., 0., 0., 0., 0., 0., 0., 0.],
        [1., 0., 1., 1., 0., 0., 0., 0., 0., 0.],
        [1., 1., 0., 1., 0., 0., 0., 0., 0., 0.],
        [0., 1., 1., 0., 1., 0., 0., 0., 0., 0.],
        [0., 0., 0., 1., 0., 1., 0., 0., 0., 0.],
        [0., 0., 0., 0., 1., 0., 1., 1., 0., 0.],
        [0., 0., 0., 0., 0., 1., 0., 0., 1., 0.],
        [0., 0., 0., 0., 0., 1., 0., 0., 1., 0.],
        [0., 0., 0., 0., 0., 0., 1., 1., 0., 1.],
        [0., 0., 0., 0., 0., 0., 0., 0., 1., 0.]])

or

nx.adjacency_matrix(G)

which will return scipy sparse matrix:

<10x10 sparse matrix of type '<class 'numpy.int64'>'
    with 24 stored elements in Compressed Sparse Row format>

Answer 4

I would do:

import numpy as np

def adjacency(connections):
    n_people = max(sum(connections, ())) + 1
    mat = np.zeros((n_people, n_people), dtype='int')
    for friend1, friend2 in connections:
        mat[friend1, friend2] = 1
        mat[friend2, friend1] = 1
    return mat

and if by "from scratch" you mean that you don't want to use numpy:

def adjacency(connections):
    n_people = max(sum(connections, ())) + 1
    mat = [[0 for _ in range(n_people)] for _ in range(n_people)]
    for friend1, friend2 in connections:
        mat[friend1][friend2] = 1
        mat[friend2][friend1] = 1
    return mat