从列表生成邻接矩阵，其中邻接意味着相等的元素

Question

I have a list like this:

lst = [0, 1, 0, 5, 0, 1]

I want to generate an adjacency matrix:

out = 
array([[ 1.,  0.,  1.,  0.,  1.,  0.],
       [ 0.,  1.,  0.,  0.,  0.,  1.],
       [ 1.,  0.,  1.,  0.,  1.,  0.],
       [ 0.,  0.,  0.,  1.,  0.,  0.],
       [ 1.,  0.,  1.,  0.,  1.,  0.],
       [ 0.,  1.,  0.,  0.,  0.,  1.]])

where out[i,j] = 1 if lst[i]==lst[j]

Here is my code with two for loops:

lst = np.array(lst)
label_lst = list(set(lst))
out = np.eye(lst.size, dtype=np.float32)
for label in label_lst:
  idx = np.where(lst == label)[0]
  for pair in itertools.combinations(idx,2):
    out[pair[0],pair[1]] = 1
    out[pair[1],pair[0]] = 1

But I feel there should be a way to improve this. Any suggestion?

Answer 1

Use broadcasted comparison -

np.equal.outer(lst, lst).astype(int) # or convert to float

Sample run -

In [787]: lst = [0, 1, 0, 5, 0, 1]

In [788]: np.equal.outer(lst, lst).astype(int)
Out[788]: 
array([[1, 0, 1, 0, 1, 0],
       [0, 1, 0, 0, 0, 1],
       [1, 0, 1, 0, 1, 0],
       [0, 0, 0, 1, 0, 0],
       [1, 0, 1, 0, 1, 0],
       [0, 1, 0, 0, 0, 1]])

Or convert to array and then manually extend to 2D and compare -

In [793]: a = np.asarray(lst)

In [794]: (a[:,None]==a).astype(int)
Out[794]: 
array([[1, 0, 1, 0, 1, 0],
       [0, 1, 0, 0, 0, 1],
       [1, 0, 1, 0, 1, 0],
       [0, 0, 0, 1, 0, 0],
       [1, 0, 1, 0, 1, 0],
       [0, 1, 0, 0, 0, 1]])

Answer 2

While the suggestion from @Divakar is very good I will leave this here as a solution without numpy.

lst = [0, 1, 0, 5, 0, 1]
print([[1 if x==y else 0 for x in lst ] for y in lst])

Also for large lists, the accepted solution is much faster.