I have two numpy arrays:
A.shape = (nA,x,y)
and
B.shape = (nB,x,y).
I want to find all subarrays such that
A(i,:,:) == B(j,:,:).
I know I can write a double for loop and use
np.array_equal(A(i,:,:),B(j,:,:)
However, is there a more efficient method?
You should only need to loop through one of the arrays, since you wouldn't find any additional unique subarrays after that, and you can do this with a simple list comprehension.
subarrays = [x for x in A if x in B]
If you only want the indices instead of storing the whole subarray, you can do:
indices = [x[0] for x in enumerate(A) if x[1] in B]
Utilizing Steven Rouk's solution, here is a method to get the indices for the subarrays that are equal:
indicesForMatches = [(i,j) for i,subArrayOfA in enumerate(A) for j,subArrayOfB in enumerate(B) if np.array_equal(subArrayOfA,subArrayOfB)]
You can use NumPy broadcasting for a vectorized solution, like so -
mask = ((A[:,None,:,:] == B).all(2)).all(2)
A_idx,B_idx = np.where(mask)
You can use reshaping to avoid double .all() usages and get the mask, like so -
mask = (A.reshape(A.shape[0],1,-1) == B.reshape(B.shape[0],-1)).all(-1)
Sample run -
In [41]: # Setup input arrays and force some indices to be same between A and B
...: nA = 4
...: nB = 5
...: x = 3
...: y = 2
...:
...: A = np.random.randint(0,9,(nA,x,y))
...: B = np.random.randint(0,9,(nB,x,y))
...:
...: A[2,:,:] = B[1,:,:]
...: A[3,:,:] = B[4,:,:]
...:
In [42]: mask = ((A[:,None,:,:] == B).all(2)).all(2)
...: A_idx,B_idx = np.where(mask)
...:
In [43]: A_idx, B_idx
Out[43]: (array([2, 3]), array([1, 4]))
In [44]: mask = (A.reshape(A.shape[0],1,-1) == B.reshape(B.shape[0],-1)).all(-1)
...: A_idx,B_idx = np.where(mask)
...:
In [45]: A_idx, B_idx
Out[45]: (array([2, 3]), array([1, 4]))
