CODEWITHSUNDEEP
pythonarraysperformancenumpyindices
berkelem
berkelem

Reputation: 2115

Find indices of common values in two arrays

I'm using Python 2.7. I have two arrays, A and B. To find the indices of the elements in 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 in B that are present in A, i.e. the indices in B of the same overlapping elements I found using the above code.

Currently I am running the same line again as follows:

B_inds = np.in1d(B,A)

but this extra calculation seems like it should be unnecessary. Is there a more computationally efficient way of obtaining both A_inds and B_inds?

I am open to using either list or array methods.

Upvotes: 5

Views: 5527

Answers (2)

kezzos
kezzos

Reputation: 3221

A simple multiprocessing implementation will get you a little 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

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

Upvotes: 1

Divakar
Divakar

Reputation: 221514

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]: 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, 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]: 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

Upvotes: 3

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