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Find common value indices in two arrays

I am using Python 2.7. I have two arrays: A and B. To find the indices of the elements from A that are present in B, I can do

A_inds = np.in1d(A,B)

I also want to get the indices of the elements from B that are present in A, i.e. the indices in B of the same overlapping elements that I found using the code above.

I am currently running the same line again:

 B_inds = np.in1d(B,A)

but this additional calculation seems that it should not be superfluous. Is there a more computationally efficient way to get both A_inds and B_inds ?

I am open to using list or array methods.

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2 answers

np.unique and np.searchsorted can 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]

Checking runtime and checking 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]: True In [236]: np.allclose(org_app(A,B)[1],unq_searchsorted(A,B)[1]) Out[236]: True In [237]: %timeit org_app(A,B) 100 loops, best of 3: 7.69 ms per loop In [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 , a performance improvement will be significant. Thus, the solution function will 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]: True In [277]: np.allclose(org_app(A,B)[1],unq_searchsorted_v1(A,B)[1]) Out[277]: True In [278]: %timeit org_app(A,B) 100 loops, best of 3: 8.83 ms per loop In [279]: %timeit unq_searchsorted_v1(A,B) 100 loops, best of 3: 4.94 ms per loop
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A simple multiprocessor implementation will give you a bit more speed:

 import time import numpy as np from multiprocessing import Process, Queue a = 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

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

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