Find indices of common values in two arrays

np.unique and np.searchsorted could be used together to solve it -

def unq_searchsorted(A,B):    # Get unique elements of A and B and the indices based on the uniqueness    unqA,idx1 = np.unique(A,return_inverse=True)    unqB,idx2 = np.unique(B,return_inverse=True)    # Create mask equivalent to np.in1d(A,B) and np.in1d(B,A) for unique elements    mask1 = (np.searchsorted(unqB,unqA,'right') - np.searchsorted(unqB,unqA,'left'))==1    mask2 = (np.searchsorted(unqA,unqB,'right') - np.searchsorted(unqA,unqB,'left'))==1    # Map back to all non-unique indices to get equivalent of np.in1d(A,B),     # np.in1d(B,A) results for non-unique elements    return mask1[idx1],mask2[idx2]

Runtime tests and verify results -

In [233]: def org_app(A,B):     ...:     return np.in1d(A,B), np.in1d(B,A)     ...: In [234]: A = np.random.randint(0,10000,(10000))     ...: B = np.random.randint(0,10000,(10000))     ...: In [235]: np.allclose(org_app(A,B)[0],unq_searchsorted(A,B)[0])Out[235]: TrueIn [236]: np.allclose(org_app(A,B)[1],unq_searchsorted(A,B)[1])Out[236]: TrueIn [237]: %timeit org_app(A,B)100 loops, best of 3: 7.69 ms per loopIn [238]: %timeit unq_searchsorted(A,B)100 loops, best of 3: 5.56 ms per loop

If the two input arrays are already sorted and unique, the performance boost would be substantial. Thus, the solution function would simplify to -

def unq_searchsorted_v1(A,B):    out1 = (np.searchsorted(B,A,'right') - np.searchsorted(B,A,'left'))==1    out2 = (np.searchsorted(A,B,'right') - np.searchsorted(A,B,'left'))==1      return out1,out2

Subsequent runtime tests -

In [275]: A = np.random.randint(0,100000,(20000))     ...: B = np.random.randint(0,100000,(20000))     ...: A = np.unique(A)     ...: B = np.unique(B)     ...: In [276]: np.allclose(org_app(A,B)[0],unq_searchsorted_v1(A,B)[0])Out[276]: TrueIn [277]: np.allclose(org_app(A,B)[1],unq_searchsorted_v1(A,B)[1])Out[277]: TrueIn [278]: %timeit org_app(A,B)100 loops, best of 3: 8.83 ms per loopIn [279]: %timeit unq_searchsorted_v1(A,B)100 loops, best of 3: 4.94 ms per loop

A simple multiprocessing implementation will get you a little more speed:

import timeimport numpy as npfrom multiprocessing import Process, Queuea = np.random.randint(0, 20, 1000000)b = np.random.randint(0, 20, 1000000)def original(a, b, q):    q.put( np.in1d(a, b) )if __name__ == '__main__':    t0 = time.time()    q = Queue()    q2 = Queue()    p = Process(target=original, args=(a, b, q,))    p2 = Process(target=original, args=(b, a, q2))    p.start()    p2.start()    res = q.get()    res2 = q2.get()    print time.time() - t0>>> 0.21398806572 

Divakar's unq_searchsorted(A,B) method took 0.271834135056 seconds on my machine.