从子阵列中的numpy 2D数组中提取交叉数组的索引

Question

I have two 2D numpy square arrays, A and B. B is an array extracted from A where a certain number of columns and rows (with the same indices) have been stripped. Both of them are symmetric. For instance, A and B could be:

A = np.array([[1,2,3,4,5],
              [2,7,8,9,10],
              [3,8,13,14,15],
              [4,9,14,19,20],
              [5,10,15,20,25]])
B = np.array([[1,3,5],
              [3,13,15],
              [5,15,25]])

such that the missing indices are [1,3] and intersecting indices are [0,2,4].

Is there a "smart" way to extract the indices in A corresponding to the rows/columns present in B that involves advanced indexing and such? All I could come up with was:

        import numpy as np
        index = np.array([],dtype=int)
        n,m = len(A),len(B)
        for j in range(n):
            k = 0
            while set(np.intersect1d(B[j],A[k])) != set(B[j]) and k<m:
                k+=1
            np.append(index,k)

which I'm aware is slow and resource-consuming when dealing with large arrays.

Thank you!

Edit: I did find a smarter way. I extract the diagonal from both arrays and perform the aforementioned loop on it with a simple equality check:

        index = []
        a = np.diag(A)
        b = np.diag(B)
        for j in range(len(b)):
            k = 0
            while a[j+k] != b[j] and k<n:
                k+=1
            index.append(k+j)

Although it still doesn't use advanced indexing and still iterates over a potentially long list, this partial solution looks cleaner and I'm going to stick with it for the time being.

Answer 1

Consider the easy case when all the values are distinct:

A = np.arange(25).reshape(5,5)
ans = [1,3,4]
B = A[np.ix_(ans, ans)]

In [287]: A
Out[287]: 
array([[ 0,  1,  2,  3,  4],
       [ 5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14],
       [15, 16, 17, 18, 19],
       [20, 21, 22, 23, 24]])

In [288]: B
Out[288]: 
array([[ 6,  8,  9],
       [16, 18, 19],
       [21, 23, 24]])

If we test the first row of B with each row of A, we will eventually come to the comparison of [6, 8, 9] with [5, 6, 7, 8, 9] from which we can glean the candidate solution of indices [1, 3, 4] .

We can generate a set of all possible candidate solutions by pairing the first row of B with each row of A.

If there is only one candidate, then we are done, since we are given that B is a submatrix of A and therefore there is always a solution.

If there is more than one candidate, then we can do the same thing with the second row of B, and take the intersection of the candidate solutions -- After all, a solution must be a solution for each and every row of B.

Thus we can loop through the rows of B and short-circuit once we find there is only one candidate. Again, we are assuming that B is always a submatrix of A.

The find_idx function below implements the idea described above:

import itertools as IT
import numpy as np

def find_idx_1d(rowA, rowB):
    result = []
    if np.in1d(rowB, rowA).all():
        result = [tuple(sorted(idx)) 
                  for idx in IT.product(*[np.where(rowA==b)[0] for b in rowB])]
    return result

def find_idx(A, B):
    candidates = set([idx for row in A for idx in find_idx_1d(row, B[0])])
    for Bi in B[1:]:
        if len(candidates) == 1:
            # stop when there is a unique candidate
            return candidates.pop()
        new = [idx for row in A for idx in find_idx_1d(row, Bi)]  
        candidates = candidates.intersection(new)
    if candidates:
        return candidates.pop()
    raise ValueError('no solution found')

---

Correctness : The two solutions you've proposed may not always return the correct result, particularly when there are repeated values. For example,

def is_solution(A, B, idx):
    return np.allclose(A[np.ix_(idx, idx)], B)

def find_idx_orig(A, B):
    index = []
    for j in range(len(B)):
        k = 0
        while k<len(A) and set(np.intersect1d(B[j],A[k])) != set(B[j]):
            k+=1
        index.append(k)
    return index

def find_idx_diag(A, B):
    index = []
    a = np.diag(A)
    b = np.diag(B)
    for j in range(len(b)):
        k = 0
        while a[j+k] != b[j] and k<len(A):
            k+=1
        index.append(k+j)
    return index

def counterexample():
    """
    Show find_idx_diag, find_idx_orig may not return the correct result
    """
    A = np.array([[1,2,0],
                  [2,1,0],
                  [0,0,1]])
    ans = [0,1]
    B = A[np.ix_(ans, ans)]
    assert not is_solution(A, B, find_idx_orig(A, B))
    assert is_solution(A, B, find_idx(A, B))

    A = np.array([[1,2,0],
                  [2,1,0],
                  [0,0,1]])
    ans = [1,2]
    B = A[np.ix_(ans, ans)]

    assert not is_solution(A, B, find_idx_diag(A, B))
    assert is_solution(A, B, find_idx(A, B))

counterexample()

---

Benchmark : Ignoring at our peril the issue of correctness, out of curiosity let's compare these functions on the basis of speed.

def make_AB(n, m):
    A = symmetrize(np.random.random((n, n)))
    ans = np.sort(np.random.choice(n, m, replace=False))
    B = A[np.ix_(ans, ans)]
    return A, B

def symmetrize(a):
    "http://stackoverflow.com/a/2573982/190597 (EOL)"
    return a + a.T - np.diag(a.diagonal())

if __name__ == '__main__':
    counterexample()
    A, B = make_AB(500, 450)
    assert is_solution(A, B, find_idx(A, B))

In [283]: %timeit find_idx(A, B)
10 loops, best of 3: 74 ms per loop

In [284]: %timeit find_idx_orig(A, B)
1 loops, best of 3: 14.5 s per loop

In [285]: %timeit find_idx_diag(A, B)
100 loops, best of 3: 2.93 ms per loop

So find_idx is much faster than find_idx_orig , but not as fast as find_idx_diag .